Information for Students in MATH 329 2010 01

McGILL UNIVERSITYFACULTY OF SCIENCE

DEPARTMENT OFMATHEMATICS AND STATISTICS

MATH 329 2010 01THEORY OF INTEREST

Information for Students(Winter Term, 2009/2010)

Pages 1 - 9 of these notes may be considered theCourse Outline for this course.

W. G. Brown

April 28, 2010

Information for Students in MATH 329 2010 01

Contents1 General Information 1

1.1 Instructor and Times . . . . . . 11.2 Course Description . . . . . . . 1

1.2.1 Calendar Description . . 11.2.2 Syllabus (in terms of sec-

tions of the text-book) . 11.2.3 “Verbal” arguments . . . 3

1.3 Evaluation of Your Progress . . 31.3.1 Term Mark . . . . . . . 31.3.2 Assignments. . . . . . . 41.3.3 Class Test . . . . . . . . 41.3.4 Final Examination . . . 41.3.5 Supplemental Assessments 41.3.6 Machine Scoring . . . . 51.3.7 Language of Written Sub-

missions . . . . . . . . 51.3.8 Plagiarism . . . . . . . 5

1.4 Published Materials . . . . . . . 51.4.1 Required Text-Book . . 51.4.2 Website . . . . . . . . . 51.4.3 Reference Books . . . . 6

1.5 Other information . . . . . . . . 61.5.1 Prerequisites . . . . . . 61.5.2 Calculators . . . . . . . 61.5.3 Self-Supervision . . . . 61.5.4 Escape Routes . . . . . 71.5.5 Showing your work; good

mathematical form; simpli-fying answers . . . . . . 7

2 Timetable 8

3 Assignments, Tests, and Solutions in MATH141 2010 01 1003.1 First Problem Assignment . . . 1003.2 Second Problem Assignment . . 1023.3 Third Problem Assignment . . . 1043.4 Draft Solutions, First Problem As-

signment . . . . . . . . . . . . . 1073.5 Draft Solutions, Second Problem

Assignment . . . . . . . . . . . 112

3.6 Draft Solutions, Third Problem As-signment . . . . . . . . . . . . . 118

3.7 Fourth Problem Assignment . . 1253.8 Class Tests . . . . . . . . . . . 127

3.8.1 Class Test, Version 1 . . 1273.8.2 Class Test, Version 2 . . 1333.8.3 Class Test, Version 3 . . 1393.8.4 Class Test, Version 4 . . 145

3.9 Draft Solutions, 2010 Class Test 1513.9.1 Problems on Nominal/Effective

Interest/Discount Rates . 1513.9.2 Problems on Force of In-

terest . . . . . . . . . . 1533.9.3 Problems on Increasing/Decreasing

Annuities . . . . . . . . 1563.9.4 Problems on Balloon and

Drop Payments . . . . . 1583.10 Draft Solutions, 2010 Class Test,

arranged by version . . . . . . . 1623.10.1 Version 1 . . . . . . . . 1623.10.2 Version 2 . . . . . . . . 1653.10.3 Version 3 . . . . . . . . 1683.10.4 Version 4 . . . . . . . . 172

3.11 Draft Solutions, Fourth Problem As-signment . . . . . . . . . . . . . 176

4 About the following references 501

5 References 501

6 Supplementary Lecture Notes 10016.1 These notes . . . . . . . . . . . 10016.2 Supplementary Notes for the Lec-

ture of January 04th, 2010 . . . 10016.2.1 §1.1 INTRODUCTION 10016.2.2 §1.2 THE ACCUMULA-

TION AND AMOUNT FUNC-TIONS . . . . . . . . . 1001

6.3 Supplementary Notes for the Lec-ture of January 06th, 2010 . . . 10036.3.1 §1.2 THE ACCUMULA-

TION AND AMOUNT FUNC-TIONS (conclusion) . . 1003


6.4 Supplementary Notes for the Lec-ture of January 08th, 2010 . . . 10096.4.1 §1.3 THE EFFECTIVE RATE

OF INTEREST . . . . . 10096.4.2 §1.4 SIMPLE INTEREST 10116.4.3 §1.5 COMPOUND INTER-

EST . . . . . . . . . . . 10126.5 Supplementary Notes for the Lec-

ture of January 11th, 2010 . . . 10146.5.1 §1.5 COMPOUND INTER-

EST (conclusion) . . . . 10146.5.2 §1.6 PRESENT VALUE 1016

6.6 Supplementary Notes for the Lec-ture of January 13th, 2010 . . . 10196.6.1 §1.7 THE EFFECTIVE RATE

OF DISCOUNT . . . . 10196.6.2 §1.8 NOMINAL RATES OF

INTEREST AND DISCOUNT 10236.7 Supplementary Notes for the Lec-

ture of January 15th, 2010 . . . 10246.7.1 §1.8 NOMINAL RATES OF

INTEREST AND DISCOUNT(conclusion) . . . . . . . 1024

6.8 Supplementary Notes for the Lec-ture of January 18th, 2010 . . . 10296.8.1 §1.9 FORCES OF INTER-

EST AND DISCOUNT . 10296.9 Supplementary Notes for the Lec-

ture of January 20th, 2010 . . . 10336.9.1 §1.9 FORCES OF INTER-

EST AND DISCOUNT (con-clusion, for now) . . . . 1033

6.9.2 §1.10 VARYING INTER-EST . . . . . . . . . . . 1034

6.9.3 §1.11 SUMMARY OF RE-SULTS . . . . . . . . . 1035

6.9.4 §2.1 INTRODUCTION 10366.9.5 §2.2 THE BASIC PROB-

LEM . . . . . . . . . . 10366.9.6 §2.3 EQUATIONS OF VALUE 1037

6.10 Supplementary Notes for the Lec-ture of January 22nd, 2010 . . . 1040

6.10.1 §2.3 EQUATIONS OF VALUE(conclusion) . . . . . . . 1040

6.10.2 §2.4 UNKNOWN TIME 10416.10.3 §2.5 UNKNOWN RATE OF

INTEREST . . . . . . . 10446.11 Supplementary Notes for the Lec-

ture of January 25th, 2010 . . . 10476.11.1 §2.6 DETERMINING TIME

PERIODS . . . . . . . . 10476.11.2 §2.7 PRACTICAL EXAM-

PLES . . . . . . . . . . 10486.11.3 §2.8 MISCELLANEOUS PROB-

LEMS . . . . . . . . . . 10516.11.4 §3.1 INTRODUCTION 10536.11.5 §3.2 ANNUITY-IMMEDIATE 1054

6.12 Supplementary Notes for the Lec-ture of January 27th, 2010 . . . 10556.12.1 §3.2 ANNUITY-IMMEDIATE

(conclusion) . . . . . . . 10556.13 Supplementary Notes for the Lec-

ture of January 29th, 2010 . . . 10576.13.1 §3.3 ANNUITY-DUE . 10576.13.2 §3.4 ANNUITY VALUES

ON ANY DATE . . . . 10586.14 Supplementary Notes for the Lec-

ture of February 01st, 2010 . . . 10616.14.1 §3.4 ANNUITY VALUES

ON ANY DATE (conclu-sion) . . . . . . . . . . 1061

6.14.2 §3.5 PERPETUITIES . 10646.15 Supplementary Notes for the Lec-

ture of February 03rd, 2010 . . . 10686.15.1 §3.6 UNKNOWN TIME 1068

6.16 Supplementary Notes for the Lec-ture of February 05th, 2010 . . . 10736.16.1 §3.7 UNKNOWN RATE OF

INTEREST . . . . . . . 10736.16.2 §3.8 VARYING INTEREST 10766.16.3 §3.9 ANNUITIES NOT IN-

VOLVING COMPOUNDINTEREST . . . . . . . 1077


6.16.4 §3.10 MISCELLANEOUSPROBLEMS . . . . . . 1078

6.16.5 APPENDIX 3 . . . . . . 10796.17 Supplementary Notes for the Lec-

ture of February 07th, 2010 . . . 10806.17.1 §4.1 INTRODUCTION 10806.17.2 §4.2 ANNUITIES PAYABLE

AT A DIFFERENT FRE-QUENCY THAN INTER-EST IS CONVERTIBLE 1080

6.18 Supplementary Notes for the Lec-ture of February 10th, 2010 . . . 10846.18.1 §4.3 ANNUITIES PAYABLE

LESS FREQUENTLY THANINTEREST IS CONVERT-IBLE . . . . . . . . . . 1084


LESS FREQUENTLY THANINTEREST IS CONVERT-IBLE . . . . . . . . . . 1088


LESS FREQUENTLY THANINTEREST IS CONVERT-IBLE (conclusion) . . . 1092

6.21 Supplementary Notes for the Lec-ture of February 12th, 2010 . . . 10946.21.1 §4.4 FURTHER ANALY-

SIS OF ANNUITIES PAYABLEMORE FREQUENTLY THANINTEREST IS CONVERT-IBLE . . . . . . . . . . 1094

6.22 Supplementary Notes for the Lec-ture of February 15th, 2010 . . . 10986.22.1 §4.5 CONTINUOUS AN-

NUITIES . . . . . . . . 10986.23 Supplementary Notes for the Lec-

ture of February 17th, 2010 . . . 1102

6.23.1 §4.6 PAYMENTS VARY-ING IN ARITHMETIC PRO-GRESSION . . . . . . . 1102

6.24 Supplementary Notes for the Lec-ture of February 19th, 2010 . . . 11046.24.1 §4.6 PAYMENTS VARY-

ING IN ARITHMETIC PRO-GRESSION (continued) 1104

6.25 Supplementary Notes for the Lec-ture of March 01st, 2010 . . . . 11076.25.1 §4.6 PAYMENTS VARY-

ING IN ARITHMETIC PRO-GRESSION (conclusion) 1107

6.26 Supplementary Notes for the Lec-ture of March 03rd, 2010 . . . . 11096.26.1 §4.7 PAYMENTS VARY-

ING IN GEOMETRIC PRO-GRESSION . . . . . . . 1109

6.26.2 §4.8 MORE GENERAL VARY-ING ANNUITIES . . . 1111

6.26.3 §4.10 MISCELLANEOUSPROBLEMS . . . . . . 1111

6.27 Supplementary Notes for the Lec-ture of March 05th, 2010 . . . . 11146.27.1 §5.1 INTRODUCTION 11146.27.2 §5.2 FINDING THE OUT-

STANDING LOAN BAL-ANCE . . . . . . . . . . 1114

6.28 Supplementary Notes for the Lec-ture of March 08th, 2010 . . . . 1117

6.29 Supplementary Notes for the Lec-ture of March 10th, 2010 . . . . 11186.29.1 §5.2 FINDING THE OUT-

STANDING LOAN BAL-ANCE (conclusion) . . . 1118

6.29.2 §5.3 AMORTIZATION SCHED-ULES . . . . . . . . . . 1119

6.30 Supplementary Notes for the Lec-ture of March 12th, 2010 . . . . 11276.30.1 §5.4 SINKING FUNDS 1127

6.31 Supplementary Notes for the Lec-ture of March 15th, 2010 . . . . 1131


6.31.1 §5.4 SINKING FUNDS (con-tinued) . . . . . . . . . 1131

6.32 Supplementary Notes for the Lec-ture of March 17th, 2010 . . . . 11336.32.1 §5.4 SINKING FUNDS (con-

clusion) . . . . . . . . . 11336.32.2 §5.5 DIFFERING PAYMENT

PERIODS AND INTERESTCONVERSION PERIODS 1133

6.32.3 §5.6 VARYING SERIES OFPAYMENTS . . . . . . 1134

6.32.4 §5.7 AMORTIZATION WITHCONTINUOUS PAYMENTS 1134

6.32.5 §5.8 STEP-RATE AMOUNTSOF PRINCIPAL . . . . 1134

6.32.6 §6.1 INTRODUCTION 11346.32.7 §6.2 TYPES OF SECURI-

TIES . . . . . . . . . . 11346.33 Supplementary Notes for the Lec-

ture of March 19th, 2010 . . . . 11366.33.1 §6.2 TYPES OF SECURI-

TIES (conclusion) . . . 11366.33.2 §6.3 PRICE OF A BOND 1137

6.34 Supplementary Notes for the Lec-ture of March 22nd, 2010 . . . . 11406.34.1 §6.3 PRICE OF A BOND

(conclusion) . . . . . . . 11406.35 Supplementary Notes for the Lec-

ture of March 24th, 2010 . . . . 11426.35.1 §6.4 PREMIUM AND DIS-

COUNT . . . . . . . . . 11426.35.2 §6.5 VALUATION BETWEEN

COUPON PAYMENT DATES(Omit) . . . . . . . . . . 1145

6.35.3 §6.6 DETERMINATION OFYIELD RATES (omit) . 1145

6.36 Supplementary Notes for the Lec-ture of March 26th, 2010 . . . . 11466.36.1 §6.7 CALLABLE AND PUTABLE

BONDS . . . . . . . . . 11466.37 Supplementary Notes for the Lec-

ture of March 29th, 2010 . . . . 1150

6.37.1 §7.1 INTRODUCTION (omit) 11506.37.2 §7.2 DISCOUNTED CASH

FLOW ANALYSIS (omit) 11506.37.3 §7.3 UNIQUENESS OF THE

YIELD RATE (part) . . 11506.37.4 §7.4 REINVESTMENT RATES 11506.37.5 §7.5 INTEREST MEASURE-

MENT OF A FUND (Omit) 11546.37.6 §7.6 TIME-WEIGHTED RATES

OF INTEREST (Omit) . 11546.37.7 §7.7 PORTFOLIO METH-

ODS AND INVESTMENTYEAR METHODS (Omit) 1154

6.37.8 §7.8 SHORT SALES (Omit) 11546.37.9 §7.9 CAPITAL BUDGET-

ING - BASIC TECHNIQUES(Omit) . . . . . . . . . . 1154

6.37.10 §7.10 CAPITAL BUDGET-ING - OTHER TECHNIQUES(Omit) . . . . . . . . . . 1154

6.37.11 Appendix 7 (Omit) . . . 11546.38 Supplementary Notes for the Lec-

ture of March 31st, 2010 . . . . 11556.39 Supplementary Notes for the Lec-

ture of April 07st, 2010 . . . . . 1156

7 Problem Assignments, Tests, and Ex-aminations from Previous Years 30017.1 2002/2003 . . . . . . . . . . . . 3001

7.1.1 First 2002/2003 Problem As-signment, with Solutions 3001

7.1.2 Second 2002/2003 ProblemAssignment, with Solutions 3006

7.1.3 Third 2002/2003 ProblemAssignment, with Solutions 3010

7.1.4 Fourth 2002/2003 ProblemAssignment, with Solutions 3017

7.1.5 Fifth 2002/2003 Problem As-signment, with Solutions 3026

7.1.6 2002/2003 Class Tests, withSolutions . . . . . . . . 3034

7.1.7 Final Examination, 2002/2003 3039


7.2 2003/2004 . . . . . . . . . . . . 30427.2.1 First 2003/2004 Problem As-

signment, with Solutions 30427.2.2 Second 2003/2004 Problem

Assignment, with Solutions 30457.2.3 Third 2003/2004 Problem

Assignment, with Solutions 30537.2.4 Fourth 2003/2004 Problem

Assignment, with Solutions 30647.2.5 Fifth 2003/2004 Problem As-

signment, with Solutions 30687.2.6 2003/2004 Class Test, Ver-

sion 1 . . . . . . . . . . 30747.2.7 2003/2004 Class Test, Ver-



sion 4 . . . . . . . . . . 30787.2.10 Solutions to Problems on

the 2003/2004 Class Tests 30797.2.11 Final Examination, 2003/2004 30887.2.12 Supplemental/Deferred Ex-

amination, 2003/2004 . 30907.3 2004/2005 . . . . . . . . . . . . 3093

7.3.1 First Problem Assignment,with Solutions . . . . . 3093

7.3.2 Second Problem Assignment,with Solutions . . . . . 3098

7.3.3 Third Problem Assignment,with Solutions . . . . . 3107

7.3.4 Fourth Problem Assignment,with Solutions . . . . . 3115

7.3.5 Fifth Problem Assignment,with Solutions . . . . . 3123

7.3.6 Solutions to Problems onthe Class Test . . . . . . 3130

7.3.7 Final Examination 2004/2005 31417.3.8 Supplemental/Deferred Ex-

amination 2004/2005 . . 31457.4 2008/2009 . . . . . . . . . . . . 3148

7.4.1 First Problem Assignment,with Solutions . . . . . 3148

7.4.2 Second Problem Assignment,with Solutions . . . . . 3152

7.4.3 Third Problem Assignment,with Solutions . . . . . 3155

7.4.4 Fourth Problem Assignment,with Solutions . . . . . 3161

7.4.5 Fifth Problem Assignment,with Solutions . . . . . 3167

7.4.6 Draft Solutions to Problemson the 2009 Class Tests . 3171

7.4.7 Final Examination 2008/2009 31817.4.8 Supplemental/Deferred Ex-

amination 2008/2009 . . 3184

Information for Students in MATH 329 2010 01 1

1 General InformationDistribution Date: Monday, January 04th, 2010

subject to correction

(All information is subject to change, either by announcements at lectures, on WebCT, or inprint. An updated version will be accessible via a link from WebCT. The Course Outline forMATH 329 2010 01 can be considered to be pages 1 through 9 of these notes.)

1.1 Instructor and Times

CRN: 610INSTRUCTOR: Prof. W. G. BrownOFFICE: BURN 1224OFFICE HRS. W 15:45→16:45 h. (as of 17 Feb., 2010);(subject to further F 10:00→11:00 h.;change) and by appointment

TELEPHONE: 398–3836E-MAIL: [email protected]: BURN 1B23 (until 08 January, 2010)

LEA 26 (beginning 11 January, 2010)CLASS HOURS: MWF 14:35–15:25 h.

Table 1: Instructor and Times

1.2 Course Description1.2.1 Calendar Description

THEORY OF INTEREST. (3 credits) (Prerequisite: MATH 141.) Simple and compoundinterest, annuities certain, amortization schedules, bonds, depreciation.

1.2.2 Syllabus (in terms of sections of the text-book)

The central part of the course consists of many of the topics in the first eight chapters of thetextbook [5]1; section numbers, where shown, refer to that book. In the list below I show thechapters and appendices of the textbook. Following each is a description as of the date of thisrevision, of the sections to be excluded. This list could be updated during the semester, asit becomes apparent that certain sections are not appropriate to the level of the course or thelecture time available.

1[n] refers to item n in the bibliography, page 501.

UPDATED TO April 28, 2010


Chapter 1. The Measurement of Interest §§1.1–1.11. Appendix 1.

Chapter 2. Solution of Problems in Interest §§2.1–2.7. Omit method of equated time, p.55 and Appendix 2.

Chapter 3. Basic Annuities §§3.1–3.6. In §3.7 students should know how to derive and usethe approximation formulæ from Appendix 3, but need not memorize the formulæ. Omit §3.8,with the exception of the definitions of the Portfolio and Yield Methods. Only one examplefrom §3.9 will be considered. Appendix 3 is included in connection with §3.7.

Chapter 4. More General Annuities §§4.1-4.7. Appendix 4. Omit §4.9.

Chapter 5. Amortization schedules and sinking funds §§5.1–5.4.

Chapter 6. Bonds and other securities §§6.1–6.4, and 6.7. Omit §§6.5, 6.6, 6.8–6.11 andAppendix 6.

Chapter 7. Yield rates §7.3 (part), §7.4. Omit §§7.1–7.2, 7.5–7.10, Appendix 7.

Chapter 8. Practical applications Omit this chapter.

Chapter 9. More advanced financial analysis Omit this chapter.

Chapter 10. The term structure of interest rates Omit this chapter.

Chapter 11. Duration, convexity and immunization Omit this chapter.

Chapter 12. Stochastic approaches to interest Omit this chapter.

Chapter 13. Options and other derivatives Omit this chapter.

Appendix A. Table numbering the days of the year Omit.

Appendix B. Illustrative mortgage loan amortization schedule This is an example of aschedule, but much longer than any that you might be expected to be able to generate manually.The type of schedule that you could be asked to generate, in connection with chapters 5 and 6could show more columns than this example.



Appendix C. Basic mathematical review Mostly definitions and a few formulæ that shouldbe known to all mathematics undergraduates. Sections E, F are from Calculus 3, and may bediscussed in the course.

Appendix D. Statistical background Omit.

Appendix E. Iteration methods The only method which will be discussed extensively willbe successive bisection [5, §B].

1.2.3 “Verbal” arguments

An essential feature of investment and insurance mathematics is the need to be able to un-derstand and to formulate “verbal” arguments; that is, explanations of the truth of an identitypresented verbally i.e., in words, rather then as an algebraic proof. In a verbal argument weseek more than mathematically correctness: we wish to see an explanation that could be pre-sented to a layman who is not competent in the mathematical bases of this subject, but is stillpossessed of reason, and needs to be assured that he is not being exploited. When the skill hasbeen mastered it can be used to verify the correctness of statements proved mathematically.Verbal arguments require some care with the underlying language; students who have diffi-culty with expression in English are reminded that McGill students have the right to submitany written materials in either English or French.2

1.3 Evaluation of Your ProgressIn the event of extraordinary circumstances beyond the University’s control, the content and/orevaluation scheme in this course is subject to change.

1.3.1 Term Mark

The Term Mark will be computed one-third from the assignment grades, and two-thirds fromthe class test. The Term Mark will count for 30 of the 100 marks in the final grade, but onlyif it exceeds 30% of the final examination percentage; otherwise the final examination will beused exclusively in the computation of the final grade.

2For a lexicon of actuarial terms in English/French, see The Canadian Institute of Actuaries English-Frenchlexicon [1], at

http://www.actuaries.ca/members/lexicon/index.html


1.3.2 Assignments.

A total of about 5 assignments will together be worth 10 of the 30 marks assigned to TermWork.

1.3.3 Class Test

A class test will be held on Monday, March 8th, 2010; the date was changed from the pre-viously announced tentative date of Wednesday, March 03rd, after consultation with studentsat the lectures — a procedure that had been announced in the original course outline, whichstated that “this date could be changed after discussion with the class at any scheduled lecturedate. Students who don’t come to class should ensure that they are aware of any changes inthe date of the test. There will be no ‘make-up’ test for persons who miss the test.”

1.3.4 Final Examination

Written examinations form an important part of the tradition of actuarial mathematics. Thefinal examination in MATH 329 2010 01 will count for either 70% or 100% of the numericalgrade from which the submitted final letter grade will be computed. Where a student’s FinalExamination percentage is superior to her Term Mark percentage, the Final Examination gradewill replace the Term Mark grade in the calculations.

A 3-hour-long final examination will be scheduled during the regular examination periodfor the winter term (April 15th, 2010 through April 30th, 2010). You are advised not to makeany travel arrangements that would prevent you from being present on campus at any timeduring this period. Students who have religious or other constraints that could affect theirability to write examinations at particular times should watch for the Preliminary ExaminationTimetable, as their rights to apply for special consideration at their faculty may have expiredby the time the Final Examination Timetable is published.

1.3.5 Supplemental Assessments

Supplemental Examination. For eligible students who obtain a Final Grade of F or D inthe course there will be a supplemental examination. (For information about SupplementalExaminations, see the McGill Calendar, [8].)

There is No Additional Work Option. “Will students with marks of D, F, or J have theoption of doing additional work to upgrade their mark?” No. (“Additional Work” refers to anoption available in certain Arts and Science courses, but not available in this course.)



1.3.6 Machine Scoring

“Will the final examination be machine scored?” While there could be Multiple Choice ques-tions on the Class Test and/or the Final Examination, such questions will not be machinescored.

1.3.7 Language of Written Submissions

In accord with McGill University’s Charter of Students’ Rights, students in this course havethe right to submit in English or in French any written work that is to be graded.

1.3.8 Plagiarism

While students are not discouraged from discussing assignment problems with their colleagues,the work that you submit — whether through homework, the class test, or on tutorial quizzesor the final examination should be your own. The Handbook on Student Rights and Responsi-bilities states in ¶15(a)3 that

“No student shall, with intent to deceive, represent the work of another personas his or her own in any academic writing, essay, thesis, research report, projector assignment submitted in a course or program of study or represent as his orher own an entire essay or work of another, whether the material so representedconstitutes a part or the entirety of the work submitted.”

You are also referred to the following URL:

http://www.mcgill.ca/integrity/studentguide/

1.4 Published Materials1.4.1 Required Text-Book

The textbook for the course this semester is [5] Stephen G. Kellison, The Theory of Interest,Third Edition. McGraw-Hill/Irwin, Boston, etc. (2009), ISBN 978-0-07-338244-9.

1.4.2 Website

These notes, and other materials distributed to students in this course, will be accessible viamyCourses (=WebCT) at the following URL:

http://www.math.mcgill.ca/brown/math329b.html

3http://upload.mcgill.ca/secretariat/greenbookenglish.pdf


The notes will be in “pdf” (.pdf) form, and can be read using the Adobe Acrobat reader, whichmany users have on their computers. This free software may be downloaded from the followingURL:

http://www.adobe.com/prodindex/acrobat/readstep.html4

Where revisions are made to distributed printed materials — for example these informationsheets — it is expected that the last version will be posted on the Web.

The notes will also be available via a link from the WebCT URL:

http://mycourses.mcgill.ca

but not all features of WebCT will be implemented.

1.4.3 Reference Books

The textbooks used in the course in previous years may be used as references.

1.5 Other information1.5.1 Prerequisites

It is your responsibility as a student to verify that you have the prescribed calculus prerequi-sites. It would be foolish to attempt to take the course without them.

1.5.2 Calculators

It is intended that the use of non-programmable, non-graphing calculators only will be permit-ted in homework, tests, or the final examination in this course. Students may be required toconvince examiners and invigilators that all memories have been cleared. The use of calcula-tors that are either graphing, programmable will not be permitted during test or examinations,in order to “level the playing field”. It is not intended that students should be permitted to usefinancial calculators.

1.5.3 Self-Supervision

This is not a high-school course, and McGill is not a high school. The monitoring of yourprogress before the final examination is largely your own responsibility. While the instructoris available to help you, he cannot do so unless and until you identify the need for help. Whilethe significance of the homework assignments and class test in the computation of your grade isminimal, these are important learning experiences, and can assist you in gauging your progress

4At the time of this writing the current version is Version 9.


in the course. This is not a course that can be crammed for: you must work steadily throughthe term if you wish to develop the facilities needed for a strong performance on the finalexamination.

Working Problems on Your Own. You are advised to work large numbers of problemsfrom your textbook. The skills you acquire in solving textbook problems could have muchmore influence on your final grade than either the homework or the class test.

1.5.4 Escape Routes

At any time, even after the last date for dropping the course, students who are experiencingmedical or personal difficulties should not hesitate to consult their advisors or the StudentAffairs office of their faculty. Don’t allow yourself to be overwhelmed by such problems; theUniversity has resource persons who may be able to help you.

1.5.5 Showing your work; good mathematical form; simplifying answers

When, in a test or examination problem, you are explicitly instructed to show all your work,failure to do so could result in a substantial loss of marks — possibly even all of the marks;this is the default. The guiding principle should be that you want to be able to communicateyour precise reasoning to others and to yourself. You are always expected to “simplify” anyalgebraic or numerical expressions that arise in your solutions or calculations. Verbal proofsare expected to be “convincing”: it will not be sufficient to simply describe mathematicalexpressions verbally.


2 TimetableDistribution Date: (original version) Monday, January 4th, 2010

this revision as of April 28, 2010(Subject to correction and change.)

Section numbers refer to the text-book.5

MONDAY WEDNESDAY FRIDAYJANUARY

04 §§1.1-1.2 06 §§1.1-1.2 08 §§1.3-1.511 §§1.5-1.6 13 §1.7-1.8 15 §§1.8

Course changes must be completed on MINERVA by Tuesday, Jan. 19, 201018 §1.9 20 §§1.9, 2.1-2.3 22 §§2.4-2.5

Deadline for withdrawal with fee refund = Jan. 24, 2010Verification Period: January 25 – 29, 2010

25 §§2.6-2.8, 3.1-3.2 1© 27 §§3.2 29 §§3.3-3.4FEBRUARY

01 §3.4, §3.5 03 §3.6 05 §3.7-3.1008 §4.1-4.2 2© 10 §4.3 12 §4.3

Deadline for web withdrawal (with W) from course via MINERVA = Feb. 14, 201015 §4.4 17 §4.5 19 §4.6

Study Break: February 21 – 27, 2010No lectures, no regular office hours!

22 NO LECTURE 24 NO LECTURE 26 NO LECTURE

5

Notation: n© = Assignment #n due todayR© = Read OnlyX = reserved for eXpansion or review


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3 Assignments, Tests, and Solutions in MATH 141 2010 01

3.1 First Problem AssignmentDistribution Date: 04 January, 2010

Solutions are to be submitted by Monday, January 25th, 2010

These problems are to be solved with full solutions, modelled either on the solu-tions to problems in the textbook, presented in class, or in the notes on the Web forthis or previous years. The essence is that the reader should be able to reconstructevery step of the proof from what you have written: getting the right answer isnever enough. You are not being graded for elegance, but simply for the proof be-ing logical, without serious gaps. While the data given may sometimes not justifya large number of decimal places, you should show your intermediate calculationsin sufficient detail that the grader can determine that your calculations are correct.The default number of decimal places given by your calculator may not always besufficient for this.

1. Assuming compound interest or discount, find the accumulated value of 5,000 at the endof 6 years

(a) if the nominal annual rate of discount is 8%, convertible quarterly;

(b) if the nominal annual rate of interest is 8%, convertible quarterly;

(c) if the nominal annual rate of interest is 100%, convertible annually;

(d) if the nominal annual rate of discount is 99%, convertible annually.

2. If, at compound interest or discount, 5,000 accumulates at the end of 6 years to 7,000,determine

(a) the nominal annual rate of interest, convertible every 3 months;

(b) the nominal annual rate of discount, convertible every every third of a year

(c) the nominal annual rate of interest, convertible every 3 years.

3. (a) At a given quarterly rate of compound interest an account with initial value of 1,000will accumulate to 2,500 in 20 years. What was the value of the account after 10years?

(b) At a given effective quarterly rate k of compound interest an account with initialvalue of 1,000 will accumulate to 2,500 in 20 years. What would the value of theaccount be after 20 years if the initial deposit of 1,000 had grown under an effectivequarterly rate k of compound discount?


4. A Credit Union makes money available to its members at the current rate charged bybanks in the area. However, this Credit Union views its mandate to be primarily pro-viding short-term financing. Accordingly, it wishes to discourage loans for longer than1 year. It proposes to do this by increasing the interest rate gradually after that time,implemented by a gradual increase in the force of interest.

(a) Determine the force of interest which the Credit Union will charge for the first year,if the interest rate throughout that year is to be an effective monthly rate of 0.5%.

(b) As a loan period extends beyond one year, the Credit Union will increase the forceof interest linearly (i.e., in such a way that d

dtδt is a constant k when t > 1), so thatthe amount owing at time t = 2 on a loan of 1 at time t = 0 is equivalent to whatwould have been owing had the effective monthly interest rate been 0.6% over theentire 2-year period of that loan. Determine the value of k, and the force of interestat any time t > 0. (Hint: You are assuming that δt is a function which is continuousfor t > 0, such that d

dtδt = k, a constant, when t > 1.)

(c) Determine the equivalent constant effective annual interest rate that a borrower willpay over the full period of her loan if she fully repays a loan taken out at time t = 0at the following time:

i. t = 1 years;ii. t = 1.5 years;

iii. t = 2 years.


3.2 Second Problem AssignmentDistribution Date: 04 January, 2010 (changed slightly on 21 January, 2010)

Solutions are to be submitted by Monday, February 08th, 2010

1. Short, direct calculations: Show all your work in each case.

(a) Determine whether it is better to pay 10,200 cash, or 900 per month for one year ata nominal annual rate of 10%, compounded monthly

i. if the monthly payments are at the beginning of the month, beginning imme-diately; and

ii. if the monthly payments are at the end of the month, first payment one monthfrom now.

(b) A home has just been bought, and the purchaser has signed a mortgage agreeing topay 1,600 at the end of each half-month for 15 years, where i(24) = 6.5%.Determine

i. the amount of the mortgage loan; andii. the cash price of the home, assuming that the down payment was 60,000.

2. On his granddaughter’s 10th birthday her grandfather deposits a total of 100,000 into afund that accumulates at a rate of 6%, compounded quarterly.

(a) The fund is to pay the child equal amounts on each of her 20th and her 21st birth-days. Determine the amounts of the equal payments.

(b) Suppose that the fund is to pay twice as much on the 21st birthday as on the 20thbirthday. Determine the amount of each of the payments.

(c) Suppose that the payment on the 20th birthday is to be 90,000, and that the balanceis to be paid on the 21st birthday. Determine the amount of the 21st-birthdaypayment.

(d) When the granddaughter reaches age 15 14 a baby brother is born. The grandfather

revises his plans. He has no more capital available, but he wishes that the fundshould now pay each of the children a payment on the children’s respective 20thand 21st birthdays; all four payments are to be equal. Determine the amount ofeach of those four equal payments.

3. (a) 70,000 is placed in an account which pays annual interest of 4% compoundedmonthly. If a withdrawal is made between interest payment dates, it earns no inter-est from the last payment date. Determine the payment date when the account willcontain at least 95,000.



(b) A loan today of 15000 is to be repaid by one payment of 10000 n years from now,and a payment of 20000 2n years from now; n need not be an integer. If futurepayments are discounted at an effective annual rate of 5%, determine n.

(c) A deposits 9000 in a “special interest” account, which, on interest-compoundingdates, pays 1% interest effective per quarter when the balance is less than 10000,and 1 1

2% effective per quarter when the balance is at least 10000. Determine theamount in the account after 10 years.

4. The force of interest δt is known to be of the form δt = k√

t, where t is time, in years.

(a) Determine the value of k such that money will triple in 9 years.

(b) With this force of interest how long will it take for money invested at time t = 0 todouble?

(c) How long will it take for money invested at time t = 1 to double?

(d) What would be the equivalent effective constant annual rate of interest which wouldcause money to double in the same time as with the current variable force of inter-est?

5. Joan is borrowing 10,000 today from the bank.

(a) If she repays the loan over a 5-year period, by regular monthly payments on thefirst of the month, beginning immediately, what will be the amount of the monthlypayment at a nominal discount rate of 6%, compounded monthly?

(b) Joan finds the monthly payments inconvenient, as she receives her salary chequeevery half-month. Determine the semi-monthly payment she would have to pay, inadvance, which would be equivalent to the preceding monthly payment.

(c) What effective annual interest rate will Joan be paying?

6. If i(3) = 1.4%, find the present value of payments of 100 every 4 months, starting 4months from now, and continuing through 5 years from the present, together with pay-ments of 200 every 8 months thereafter, and continuing for 6 years.


3.3 Third Problem AssignmentDistribution Date: 25 January, 2010

Solutions are to be submitted by Monday, March 07th, 2010 Friday, March 05th, at thelecture, not in the Departmental mail box. This date was changed in consultation with the

class on 01 March, since the previously published dates were inconsistent.

These problems were to be solved with full solutions, modelled either on the solu-tions to problems in the textbook, presented in class, or in the notes on the Web forthis or previous years. The essence is that the reader should be able to reconstructevery step of the proof from what you have written: getting the right answer isnever enough. You are not being graded for elegance, but simply for the proof be-ing logical, without serious gaps. While the data given may sometimes not justifya large number of decimal places, you should show your intermediate calculationsin sufficient detail that the grader can determine that your calculations are correct.The default number of decimal places given by your calculator may not always besufficient for this.

1. Short annuity computations: For each of the following sequences of payments, writedown

• a formula in terms of annuity symbols; or

• a “simplified” formula in terms of the annuity and discount rates specified in theproblem.

Then calculate the value of the annuity.

(a) The value today of a 120-payment monthly annuity-due of 100, first payment im-mediately, at an effective annual rate of discount of d = 6%.

(b) The value as of the first payment, exactly one-half year from now, of a 30-paymentsemi-annual annuity of 10,000, where the (annual) force of interest is 6%.

(c) The value today of a perpetuity paying 40,000 at the end of each year forever, firstpayment 12 years from now, where interest is compounded semi-annually at a rateequivalent to i(

12 ) = 8%, i.e., a nominal annual rate of 8% compounded biyearly6.

2. (a) A loan of 1,000,000 is to be repaid by quarterly payments of 60,000, except for thelast payment, which could be less, at an interest rate of i(4) = 8%. Determine thedate and amount of that last payment.

6every 2 years


(b) A loan of 1,000,000 is to be repaid by a maximum number of quarterly paymentsof exactly 60,000, except for the last payment, which could be more, at an interestrate of i(4) = 8%. Again determine the date and amount of the last payment.

(c) A loan of 1,000,000 is to be repaid by quarterly payments of 60,000 at an interestrate of i(4) = 8% until the current value outstanding on the loan just after a paymentis not more than 500,000. From that payment onward the borrower is required topay 80,000 quarterly until the last payment, which may be less; the interest ratecharged beginning just after the last payment of 60,000 will be i(4) = 7%. De-termine the time of the first payment of 80,000, and the time and amount of thelast payment. (The intention is that you solve this problem systematically, not bytrial and error. This is a complicated problem: the purpose is to show you howyour methods are sufficiently powerful to solve any reasonably composed prob-lem. Don’t panic — it’s probably too complicated to be considered an examinationproblem.)

3. George buys a home for 300,000, paying a down payment of 120,000, and assuming amortgage to pay off the balance. The mortgage rate is initially a nominal annual rateof 5.25% compounded semi-annually for the first 5 years, but the initial payments arecalculated on the basis of the mortgage being paid off after 20 years at this rate. How-ever, the rate will be adjusted every five years: if George is not willing to accept therate offered after one of these 5-year intervals, he is obliged to pay the mortgage off

immediately. Should the rate be adjusted, a new payment will be calculated, so as tomaintain the plan to pay off the mortgage in 20 years. George has the right to pay, onany payment day, more than the required payment, thereby reducing the principal, butnot the payments which will still be recalculated every five years.

(a) Determine the semi-annual payment George must pay for the first 5 years.

(b) George finds the semi-annual payment onerous, as his salary is paid every half-month. Determine the semi-monthly payment he would have to make to be equiva-lent to the contracted semi-annual payment (for the first five years of the mortgage).

(c) George finds that he has more disposable income at the middle of the month thanat the end of the month, and would prefer that the odd-numbered semi-monthlypayments should be twice as large as those at the ends of the months. Calculate thetwo sizes of payments (for the first five years of the mortgage).

(d) George finds that he has more disposable income at the middle of the month thanat the end of the month, and would prefer that the odd-numbered semi-monthlypayments should be 1000 larger than those at the ends of the months. Calculate thetwo sizes of payments (for the first five years of the mortgage). The 1000 is to beconsidered part of his regular payments, not a voluntary, additional payment.


(e) Suppose that the interest rates for the remaining periods (all nominal annual ratescompounded semi-annually) are: 5.75% for the second 5-year period, 6% for thethird 5-year period, and 10% for the fourth 5-year period. Calculate the semi-annual payment due during each of these periods. All payments are at the end ofthe half-year. In each case the payment you compute is to be based on amortizationfor the number of years remaining from 20, even though it is likely that the paymentwill have to be adjusted for the next 5-year period.


3.4 Draft Solutions, First Problem AssignmentDistribution Date: 29 January, 2010

Solutions were to be submitted by Monday, January 25th, 2010(subject to correction — these solutions have not been fully checked!)


1. Assuming compound interest or discount, find the accumulated value of 5,000 at the endof 6 years

(a) if the nominal annual rate of discount is 8%, convertible quarterly;

(b) if the nominal annual rate of interest is 8%, convertible quarterly;

(c) if the nominal annual rate of interest is 100%, convertible annually;

(d) if the nominal annual rate of discount is 99%, convertible annually.

Solution:

(a) 5000(1 − 0.08

4

)−6×4

= 5000(0.98)−24 = 8, 119.78.

(b) 5000(1 +

0.084

)6×4

= 5000(1.02)24 = 8, 042.19.

(c) 5000(1 +

11

)6

= 5000(64) = 320, 000.

(d) 5000(1 − 0.99

1

)−6

= 5000(0.01)−6 = 5, 000, 000, 000, 000, 000.

2. If, at compound interest or discount, 5,000 accumulates at the end of 6 years to 7,000,determine

(a) the nominal annual rate of interest, convertible every 3 months;


(b) the nominal annual rate of discount, convertible every every third of a year

(c) the nominal annual rate of interest, convertible every 3 years.

Solution: In all of these problems the growth factor for 6 years is70005000

= 1.4.

(a)

(1 +

i(4)

4

)6×4

= 1.4

⇒ 1 +i(4)

4= (1.4)

124

⇒ i(4) = 4(1.4

124 − 1

)= 0.056473652 = 5.64735452% .

(b)

(1 − d(3)

3

)−6×3

= 1.4

⇒ 1 − d(3)

3= (1.4)−

118

⇒ d(3) = 3(1 − 1.4−

118)

= 0.0555578199 = 5.55578199% .

(c)

1 +i(

13 )13

6× 1

3

= 1.4

⇒ 1 + 3i(13 ) = (1.4)

12

⇒ i(13 ) =

1.412 − 13

= 0.06107198567 = 6.107198567% .

3. (a) At a given quarterly rate of compound interest an account with initial value of 1,000will accumulate to 2,500 in 20 years. What was the value of the account after 10years?

(b) At a given effective quarterly rate k of compound interest an account with initialvalue of 1,000 will accumulate to 2,500 in 20 years. What would the value of theaccount be after 20 years if the initial deposit of 1,000 had grown under an effectivequarterly rate k of compound discount?

Solution:


(a) Let the effective quarterly rate of compound interest be k. Then 1000 (1 + k)20×4 =

2500⇒ (1 + k)10×4 =

(25001000

) 12

⇒ 1000 (1 + k)10×4 = 1000√

2.5 = 1581.138830 ≈1583.14.

(b) (1 + k)20×4 = 2.5⇒ k = 2.5180−1. Hence, at an effective quarterly rate of compound

discount of k, 1,000 would grow in 20 years to

1000(1 − k)−80 = 1000(2 − 2.5

180)−20×4

= 2526.682793 = 2, 526.68.

4. A Credit Union makes money available to its members at the current rate charged bybanks in the area. However, this Credit Union views its mandate to be primarily pro-viding short-term financing. Accordingly, it wishes to discourage loans for longer than1 year. It proposes to do this by increasing the interest rate gradually after that time,implemented by a gradual increase in the force of interest.

(a) Determine the force of interest which the Credit Union will charge for the first year,if the interest rate throughout that year is to be an effective monthly rate of 0.5%.

(b) As a loan period extends beyond one year, the Credit Union will increase the forceof interest linearly (i.e., in such a way that d

dtδt is a constant k when t > 1), so thatthe amount owing at time t = 2 on a loan of 1 at time t = 0 is equivalent to whatwould have been owing had the effective monthly interest rate been 0.6% over theentire 2-year period of that loan. Determine the value of k, and the force of interestat any time t > 0. (Hint: You are assuming that δt is a function which is continuousfor t > 0, such that d

dtδt = k, a constant, when t > 1.)

(c) Determine the equivalent constant effective annual interest rate that a borrower willpay over the full period of her loan if she fully repays a loan taken out at time t = 0at the following time

i. t = 1 years;ii. t = 1.5 years;

iii. t = 2 years.

Solution:

(a) For t ≤ 1, the annual growth factor is (1.005)12, (i.e., a(t) = (1.005)12t for 0 ≤ t ≤1), so the force of interest is δt =

ddt

ln(1.005)12t =ddt

(12t ln 1.005) = 12 ln 1.005 =

0.05985049813 = 5.985049813%.


(b) We are assuming thatddtδt = k, a constant to be determined. It follows that, for

1 ≤ t, δt = kt + C, where C is a constant of integration. More precisely,

δt =

{0.05985049813 when 0 < t ≤ 1kt + C when t ≥ 1

Since δt is continuous at t = 1,

k · 1 + C = 0.05985049813 . (1)

When t = 2,

(1.006)24 = a(2) = e∫ 2

0 δr dr

= e∫ 1

0 δr dr+∫ 2

1 δr dr

= e∫ 1

0 0.05985049813 dr+∫ 2

1 (kr+C) dr

= e0.05985049813 · e∫ 2

1 (kr+C) dr

= (1.005)12 · e[k· r2

2 +Cr]2

1

= (1.005)12 · ek· 32 +C ,

so

3k2

+ C = ln(1.006)24

(1.005)12 = 24 ln 1.006 − 12 ln 1.005 = 0.08371922217 . (2)

Solving equations (1), (2) yields k = 0.04773744808 and C = 0.01211305005, so

δt = 0.04773744808t + 0.01211305005 for t ≥ 1 . (3)

We can verify that δ1 = 0.04773744808+0.01211305005 = 0.05985049813. (Notethat it would have been incorrect to argue that, around the time t = 2, a(t) =

(1.006)2·12t, the force of interest would be δ2 =ddt

ln(1.006)24t =ddt

(24t ln 1.006) =

24 ln 1.006 = 0.071784860140 = 7.178486014%. The error is that we are differ-entiating a function which is not an exponential to a constant base.)

(c) Only one of these three problems requires the use of equation (3).

i. For a loan held until time t = 1, the effective annual interest rate is (1.005)12 −1 = 0.061677812 = 6.1677812%. This could also have been computed using

the force of interest: a(1)− 1 = e

1∫0δt dt− 1 = e0.05985049813 − 1 = 6.1677812%.


ii. Let it denote the equivalent constant effective annual interest rate for a loanheld until time t, where 1 ≤ t ≤ 2. Then

(1 + i)t = a(t) = e∫ t

0 δt dt

= a(1)e∫ t

1 δt dt

= a(1) · e∫ t

1 (0.04773744808r+0.01211305005) dr

= a(1) · e 0.047737448082 r2+0.01211305005r]t

1

= a(1) · e0.02386872404t2+0.01211305005t−0.03598177409

= 1.024155862e0.02386872404t2+0.01211305005t .

It follows that

(1 + i1.5)1.5 = a(1.5)= 1.024155862e0.02386872404(1.5)2+0.01211305005(1.5)

= 1.024155862e0.07187420417 = 1.100476126 ,

so 1 + i1.5 = 1.1004761261

1.5 = 1.065909706, and i1.5 = 0.065909706 =

6.5909706%.iii.

(1 + i2)2 = a(2) = 1.024155862e0.02386872404(2)2+0.01211305005(2)

= 1.024155862e0.1197009963 = 1.154387292 ,

so 1+i2 = 1.15438729212 = 1.074424168, and i2 = 0.074424168 = 7.4424168%.

This could also have been computed as (1.006)12 − 1.


3.5 Draft Solutions, Second Problem AssignmentDistribution Date: Monday, February 15th, 2010

Corrected on 01 March, 2001; subject to further corrections.Solutions were to be submitted by Monday, February 08th, 2010

(subject to correction)

1. Short, direct calculations: Show all your work in each case.

(a) Determine whether it is better to pay 10,200 cash, or 900 per month for one year ata nominal annual rate of 10%, compounded monthly

i. if the monthly payments are at the beginning of the month, beginning imme-diately; and

ii. if the monthly payments are at the end of the month, first payment one monthfrom now.

(b) A home has just been bought, and the purchaser has signed a mortgage agreeing topay 1,600 at the end of each half-month for 15 years, where i(24) = 6.5%.Determine

i. the amount of the mortgage loan; andii. the cash price of the home, assuming that the down payment was 60,000.

Solution:

(a) i. The present value of the monthly payments is 900a12 1012 % = 900 ·

(1 +

0.112

)·

1 −(1 + 0.1

12

)−12

0.112

= 10, 322.36600 .

ii. The present value of the monthly payments is 900a12 1012 % = 900·

1 −(1 + 0.1

12

)−12

0.112

=

10, 237.05719.Thus both of the payment schemes cost more than the immediate cash payment.

(b) i. The face value of the mortgage is the present value of the mortgage payments,

1600 · a15×2×12 .06524

= 1600 ·1 −

(1 + 0.065

24

)−360

0.06524

= 367, 641.6941.

ii. The cash price of the home will be the sum of the down payment and the valueof the mortgage, i.e., 60, 000 + 367, 641.6941 = 427, 641.69.

2. On his granddaughter’s 10th birthday her grandfather deposits a total of 100,000 into afund that accumulates at a rate of 6%, compounded quarterly.


(a) The fund is to pay the child equal amounts on each of her 20th and her 21st birth-days. Determine the amounts of the equal payments.

(b) Suppose that the fund is to pay twice as much on the 21st birthday as on the 20thbirthday. Determine the amount of each of the payments.

(c) Suppose that the payment on the 20th birthday is to be 90,000, and that the balanceis to be paid on the 21st birthday. Determine the amount of the 21st-birthdaypayment.

(d) When the granddaughter reaches age 15 14 a baby brother is born. The grandfather

revises his plans. He has no more capital available, but he wishes that the fundshould now pay each of the children a payment on the children’s respective 20thand 21st birthdays; all four payments are to be equal. Determine the amount ofeach of those four equal payments.

Solution:

(a) Denote the amount of the equal payments by X. An equation of value as of thetime the fund is established is

100000 = X((1.015)−4×(20−10) + (1.015)−4×(21−10)

)

⇒ X =100000

1.015−40 + 1.015−44 = 93400.94421 .

(b) Denote the amounts of the payment at age 20 by X, so that the payment at age 21will be 2X. An equation of value as of the time the fund is established is

100000 = X(1 · (1.015)−4×(20−10) + 2 · (1.015)−4×(21−10)

)

⇒ X =100000

1.015−40 + 2 · 1.015−44 = 62891.35502 .

Thus the payments at ages 20 and 21 will respectively be 62,891.36 and 125,782.71.

(c) Denote the amount of the payment on the 21st birthday by X. An equation of valueas of the time the fund is established is

100000 = 90000 · (1.015)−4×(20−10) + X · (1.015)−4×(21−10)

⇒ X =100000 − 90000(1.015−40)

1.015−44 = 100000(1.015)44 − 90000(1.0154)

⇒ X = 97010.58231 .

(The second equation shows an equation of value on the child’s 21st birthday.)


(d) The grandson is born at the end of 4 ×(15 1

4 − 10)

= 21 interest periods. Thepayments to the granddaughter occur at the ends of the 40th and 44th periods; andthe payments to the grandson are at the ends of the 101st and 105th periods. Therelative times are shown in the following table (which could have been representedby lines in a time diagram):

Quarter-year 0 21 40 44 101 105Granddaughter’s age 10 151

4 20 21 35 14 36 1

4Grandson’s age 0 43

4 534 20 21

Let the amounts of the four equal payments be denoted by X. An equation of valueas of the date of establishment of the fund is

100000 = X(1.015−40 + 1.015−44

) (1 + 1.015−21

)

which implies that

X =100000(

1.015−40 + 1.015−44) (1 + 1.015−61) = 66, 560.57488 .

3. (a) 70,000 is placed in an account which pays annual interest of 4% compoundedmonthly. If a withdrawal is made between interest payment dates, it earns no inter-est from the last payment date. Determine the payment date when the account willcontain at least 95,000.

(b) A loan today of 15000 is to be repaid by one payment of 10000 n years from now,and a payment of 20000 2n years from now; n need not be an integer. If futurepayments are discounted at an effective annual rate of 5%, determine n.

(c) A deposits 9000 in a “special interest” account, which, on interest-compoundingdates, pays 1% interest effective per quarter when the balance is less than 10000,and 1 1

2% effective per quarter when the balance is at least 10000. Determine theamount in the account after 10 years.

Solution:

(a) Let n denote the number of months required for the balance to be not less than

95,000. An inequality of value as of the date of deposit is 70000(1 +

1300

)n

≥

95000, which is equivalent to(1 +

1300

)n

≥ 9500070000

. We can solve this inequality

by taking logarithms of both sides; since the logarithm function is increasing, it

preserves the inequality, giving n ≥ ln 95 − ln 70ln 1.00333333333

= 91.76711031. Thus


the first payment date when the account exceeds 95,000 is d91.76711031e = 92months7 after the original deposit, i.e., 7 2

3 years after the deposit.

(b) An equation of value as of today is 15000 = 10000(1−d)n + 20000(1−d)2n, which

is a quadratic equation in (1−d)n; solving this, we obtain (1−d)n =−1 ± √13

4. Of

the two solutions, the one with the − is extraneous, as it yields a negative value for

the exponential. Hence (1 − d)n =−1 +

√13

4= 0.6513878188. Taking (natural)

logarithms yields

n =ln 0.6513878188

ln 0.95= 8.356844523 years.

(c) Let n be the number of quarter-years that need to pass for the balance to exceed10000. Then n is the smallest integer solution to the inequality 9000(1.01)n ≥10000 ⇒ (1.01)n ≥ 10000

9000⇒ n ≥ ln 10 − ln 9

ln 1.01= 10.58864445, from which it

follows that n = 11 quarter-years.8 The balance does not exceed 10,000 until afterthe d10.58864445eth, i.e., 11th payment. It is only after this 11th payment that theaccount qualifies for the higher interest rate, which is applicable first at the 12thpayment. After 40 quarter-years the account will have grown to 9000 · (1.01)11 ·(1.015)40−11 = 15, 462.96762.

4. The force of interest δt is known to be of the form δt = k√

t, where t is time, in year.

(a) Determine the value of k such that money will triple in 9 years.

(b) With this force of interest how long will it take for money invested at time t = 0 todouble?

(c) With this force of interest how long will it take for money invested at time t = 1 todouble?

(d) What would be the equivalent effective constant annual rate of interest which wouldcause money to double in the same time as with the current variable force of inter-est?

Solution:

(a)

3 = a(9) = e∫ 9

0 k√

t dt = ek· 23 ·

[t

32]9

0 = ek(18−0) ⇒ k =ln 318

= 0.06103401606 .

7dxe is the smallest integer n such that n ≥ x.8The previously posted draft solution was corrected at this point.


(b) Let n denote the length of time for money invested at time t = 0 to double. Then

2 = a(n) = e∫ n

0 k√

t dt = ek[t

32]n

0 = ek· 23 ·n32 ⇒ ln 2 =

23· k · n 3

2 ,

hence n = 9(ln 2ln 3

) 23

= 6.620587276 years.

(c) Let n denote the length of time for money invested at time t = 1 to double. Then

2 =a(n + 1)

a(1)= e

∫ n+11 k

√t dt = e

k· 23[t

32]n+1

1 = ek· 23 ·((n+1)

32 −1

)⇒ ln 2 =

23·k ·

((n + 1)

32 − 1

),

hence n = 5.877212176 years.

(d) With n as before, we solve (1+ i)n = 2, to obtain i = 21

6.620587276 −1 = 0.110372700 =

11.03727%.

5. Joan is borrowing 10,000 today from the bank.

(a) If she repays the loan over a 5-year period, by regular monthly payments on thefirst of the month, beginning immediately, what will be the amount of the monthlypayment at a nominal discount rate of 6%, compounded monthly?

(b) Joan finds the monthly payments inconvenient, as she receives her salary chequeevery half-month. Determine the semi-monthly payment she would have to pay, inadvance, which would be equivalent to the preceding monthly payment.

(c) What effective annual interest rate will Joan be paying?

Solution:

(a) Denote the monthly payment in advance by M. An equation of value today isMa5×12 d= 6

12 % = 10000 which implies that

M =10000

a5×12 d= 612 %

=10000d1 − v60 =

10000d1 − (1 − d)60 =

10000(0.005)1 − (0.995)60 = 192.5009023 .

Note that the symbol I have used, with a subscript that reads d = 612 , is non-

standard.9

9Mathematicians are accustomed to making their own notation as they need it; but actuaries establish precisenotational conventions and play by the agreed rules.


(b) Denote the equivalent semi-monthly payment by N. The effective semi-monthlydiscount rate which is equivalent to an effective monthly discount rate of 1

2% is1 − √0.995 = 0.0025031328, or 0.25031328%. We can use this rate to solve

the problem as previously, M =10000

a5×24 d=0.25031328%

=10000d1 − v120 =

10000d1 − (1 − d)120 =

10000(0.0025031328)1 − (0.995)60 = 96.37106451. Alternatively, we can solve the equation

N(1 + (1 − d)) = M ⇒ N =192.5009023

2 − 0.0025031328= 96.37106595. We could also

have used, instead of 1 + (1−d), a2 d=0.25031328%. (The slight difference in the valuesis due to rounding errors.)

(c) The effective monthly discount rate is d = 0.005%, so the effective monthly interest

rate isd

1 − d=

0.0050.995

= 0.005025125628; the effective annual interest rate is

(1 +

0.0050.995

)− 1 = 0.061996372 = 6.1996372% .

6. If i(3) = 1.4%, find the present value of payments of 100 every 4 months, starting 4months from now, and continuing through 5 years from the present, together with pay-ments of 200 every 8 months thereafter, and continuing for 6 years.

Solution: The effective interest rate per one-third year is 1.43 %. The present value of the

payments of 100 is

100a 513

1.43 %

=1000.014

3

·1 −

(1 +

.0143

)−15 = 1445.450441 .

For the payments of 200 the effective interest rate per payment interval of 8 monthsis

(1 + 0.014

3

)2 − 1 = 0.009355112 = 0.9355112%; as of the date of the last pay-

ment of 100, the value of the payments of 200 is 200a 623

0.9355112%=

2000.009355112

·(1 − (1.009355112)−9

)= 1718.612932 . The value of these payments 5 years earlier is

thus(1.014)−5×3 · 1718.612932 = 1602.685007 ,

so the total value now of all the payments is 1445.450441+1602.685007 = 3, 048.135448.


3.6 Draft Solutions, Third Problem AssignmentDistribution Date: 05 March, 2010 (after the deadline for submission of solutions)

Solutions were to be submitted by the class of Friday, March 05th, 2010(subject to correction — these solutions may not have been fully checked!)


1. Short annuity computations: For each of the following sequences of payments, writedown

• a formula in terms of annuity symbols; or

• a “simplified” formula in terms of the annuity and discount rates specified in theproblem.

Then calculate the value of the annuity.

(a) The value today of a 120-payment monthly annuity-due of 100, first payment im-mediately, at an effective annual rate of discount of d = 6%.

(b) The value as of the first payment, exactly one-half year from now, of a 30-paymentsemi-annual annuity of 10,000, where the (annual) force of interest is 6%.

(c) The value today of a perpetuity paying 40,000 at the end of each year forever, firstpayment 12 years from now, where interest is compounded semi-annually at a rateequivalent to i(

12 ) = 8%, i.e., a nominal annual rate of 8% compounded biyearly10.

Solution:

(a) The annual rate of interest corresponding to d = 6% is i =d

1 − d=

694

. The

equivalent effective monthly interest rate is(1 +

694

) 112

− 1 = 0.0051696, or j =

10every 2 years


0.51696%. Since the first payment is immediate, the annuity has been advanced 1month, so its present value is

100(1.0051696)+1a120 0.51696% = 100·1.0051696· 1 − (1.0051696)−120

0.0051696= 8, 971.10 .

(b) Let i denote the effective semi-annual rate of interest. Then 0.06 = ln((1 + i)2

)=

2 ln(1 + i), so 1 + i = eδ2 = e0.03 = 1.030454534, and i = 3.0454534%. The value

of all the payments as of today is

10000a30 3.0454534% = 10000 · 1 − (1.030454534)−30

0.030454534= 194, 857.80 .

But we were asked for the value as of the first payment: this will be

(1.03045434)(194, 857.80) = 200, 792.0657 .

(c) This perpetuity can be viewed as either a perpetuity-immediate deferred 11 years ora perpetuity-immediate which begins now (with first payment in 1 year) diminishedby the present value of an 11-payment annuity-immediate which also begins now.If we denote the effective annual interest rate by i, then i(

12 ) = 8%, so 1+i =

√1.16,

so i = 7.7032961%. The present value of this annuity is

40000(a∞ i − a11 i

)=

400000.077032961

·(1 −

(1 − 1.16−

112))

= 229, 543.22 .

2. (a) A loan of 1,000,000 is to be repaid by quarterly payments of 60,000, except for thelast payment, which could be less, at an interest rate of i(4) = 8%. Determine thedate and amount of that last payment.

(b) A loan of 1,000,000 is to be repaid by a maximum number of quarterly paymentsof exactly 60,000, except for the last payment, which could be more, at an interestrate of i(4) = 8%. Again determine the date and amount of the last payment.

(c) A loan of 1,000,000 is to be repaid by quarterly payments of 60,000 at an interestrate of i(4) = 8% until the current value outstanding on the loan just after a paymentis not more than 500,000. From that payment onward the borrower is required topay 80,000 quarterly until the last payment, which may be less; the interest ratecharged beginning just after the last payment of 60,000 will be i(4) = 7%. De-termine the time of the first payment of 80,000, and the time and amount of thelast payment. (The intention is that you solve this problem systematically, not bytrial and error. This is a complicated problem: the purpose is to show you howyour methods are sufficiently powerful to solve any reasonably composed prob-lem. Don’t panic — it’s probably too complicated to be considered an examinationproblem.)

Solution:


(a) (I am solving this problem on the assumption that the first payment will be in 3

months.) The effective quarterly interest rate isi(4)

4= 2%. Let the total number

of payments be n. Then the condition on the last payment is that n should be thesmallest integer which satisfies the inequality

60000an 2% ≥ 1000000

⇒ 1 − (1.02)−n

0.02≥ 1000000

60000

⇒ vn ≤ 23

⇒ n ≥− ln

23

ln 1.02= 20.47531885 .

Thus the last payment is #21. Its amount will be

(1.02)21 ·(1, 000, 000 − 60, 000 · a20 2%

)

= (1.02)21 ·(1, 000, 000 − 60, 000

0.02·(1 − (1.02)−20

))= 28, 667.31247

i.e., 28,667.31.

(b) In this case the condition on n is that it be the largest integer which satisfies theinequality

60000an 2% ≤ 1000000

⇒ 1 − (1.02)−n

0.02≤ 1000000

60000

⇒ vn ≥ 23

⇒ n ≤ − ln 23

ln 1.02= 20.47531885 .

Thus the last payment is #20. The amount of the last payment will then have to be

60, 000 +28, 667.31247

1.02= 88, 105.20830 or 88,105.21.

(c) Let the time of the first payment of 80,000 be n1, and the time of the last paymentof 80,000 be n2. The amount outstanding on the loan immediately after the lastpayment of 60,000 will be

1, 000, 000(1.02)n1−1−60, 000·sn1−1 2% = 1, 000, 000(1.02)n1−1−60, 0000.02

·((1.02)(n1−1) − 1

).

(4)


Thus our constraint is that n1 be the smallest solution of the inequality

1, 000, 000(1.02)n1−1 − 60, 0000.02

·((1.02)(n1−1) − 1

)≤ 500, 000 ,

which reduces to (1.02)n1−1 ≥ 1.25. Taking logarithms yields

n1 − 1 ≥ ln 1.25ln 1.02

= 11.26838111 ,

of which the smallest integer solution is n1 = 13. Just after the 12th payment theoutstanding balance is

1, 000, 000(1.02)12 − 60, 0000.02

·((1.02)12 − 1

)= 463, 516.41 .

Now we embark on the second phase of the problem. The number of paymentsof 80,000 (except possibly the last, which could be less) is the smallest integer n2

satisfying

80, 000 · an274 % ≥ 463, 516.41

⇔ 1 − (1.0175)−n2

0.0175≥ 463, 516.41

80, 000

⇔ (1.0175)−n2 ≤ 1 − (0.0175)(463, 516.41)80, 000

= 0.8986057853

⇔ n2 ≥ − ln 0.8986057853ln 1.0175

= 6.162491959 ,

so n2 = 7. The amount of the unpaid balance just after the (12 + 6)th payment is

463, 516.41(1.0175)6 + 80, 000 · s6 0.0175

= 463, 516.41(1.0175)6 − 80, 0000.0175

·(1.01756 − 1

)

= 12, 868.7759 ,

so the amount of the last payment will be 1.0175 × 12, 868.7759 = 13, 093.98.

3. George buys a home for 300,000, paying a down payment of 120,000, and assuming amortgage to pay off the balance. The mortgage rate is initially a nominal annual rateof 5.25% compounded semi-annually for the first 5 years, but the initial payments arecalculated on the basis of the mortgage being paid off after 20 years at this rate. How-ever, the rate will be adjusted every five years: if George is not willing to accept therate offered after one of these 5-year intervals, he is obliged to pay the mortgage off


immediately. Should the rate be adjusted, a new payment will be calculated, so as tomaintain the plan to pay off the mortgage in 20 years. George has the right to pay, onany payment day, more than the required payment, thereby reducing the principal, butnot the payments which will still be recalculated every five years.

(a) Determine the semi-annual payment George must pay for the first 5 years.

(b) George finds the semi-annual payment onerous, as his salary is paid every half-month. Determine the semi-monthly payment he would have to make to be equiva-lent to the contracted semi-annual payment (for the first five years of the mortgage).

(c) George finds that he has more disposable income at the middle of the month thanat the end of the month, and would prefer that the odd-numbered semi-monthlypayments should be twice as large as those at the ends of the months. Calculate thetwo sizes of payments (for the first five years of the mortgage).

(d) George finds that he has more disposable income at the middle of the month thanat the end of the month, and would prefer that the odd-numbered semi-monthlypayments should be 1000 larger than those at the ends of the months. Calculate thetwo sizes of payments (for the first five years of the mortgage). The 1000 is to beconsidered part of his regular payments, not a voluntary, additional payment.

(e) Suppose that the interest rates for the remaining periods (all nominal annual ratescompounded semi-annually) are: 5.75% for the second 5-year period, 6% for thethird 5-year period, and 10% for the fourth 5-year period. Calculate the semi-annual payment due during each of these periods. All payments are at the end ofthe half-year. In each case the payment you compute is to be based on amortizationfor the number of years remaining from 20, even though it is likely that the paymentwill have to be adjusted for the next 5-year period.

Solution:

(a) Let Y be the amount of the initial semi-annual payment. Then

Y · a40 2.625% = 300, 000 − 120, 000

⇒ Y =180, 000

1−(1.02625)−40

0.02625

= 7322.287 .

(b) Each single payment of 7322.287 will have to be replaced by an annuity-immediatewith 2 × 6 = 12 equal payments. The effective semi-monthly interest rate will be

(1.02625)112 − 1 = 0.002161615 = 0.2161615% .


The amount of each of these payments is

7322.287s12 0.2161615%

= 602.97 .

(c) Let the amount paid at the end of the month be Z, so the amount paid in the middleof the month will be 2Z. The effective interest rate for one half-month has beenfound above to be 0.2161615%, so the rate for a full month will be 0.004327903.We need to know this rate because repetitions of each of the two sizes of paymentsare spaced one month apart. We can now solve the equation

2Z · (1.002161615) · s6 0.004327903 + Z · s6 0.004327903 = 7322.287

⇒ Z =7322.287

(2(1.002161615) + 1) · s6 0.004327903

⇒ Z =7322.287(0.004327903)

(2(1.002161615) + 1)((1.004327903)6 − 1

) = 401.8354085 ,

so the mid-monthly payment is 803.67, and the end-of-the-month payment is 401.84.

(d) Let the amount paid at the end of the month be W, so the amount paid in the middleof the month is 1000 + W. There are several ways of solving this problem. Oneway will be to equate

1000 · (1.002161615) · s6 0.004327903 + W · s12 0.002161615 = 7322.287

⇒ W =7322.287 − 1000 · (1.002161615) · s6 0.004327903

s12 0.002161615

⇒ W =

(0.002161615)(7322.287 − (1002.161615)((1.004327903)6−1)

0.004327903

)

(1.002161615)12 − 1= 102.43 ,

so the payments at the middle and end of the month will be, respectively, 1102.43and 102.43.

(e) i. The semi-annual payment for the first 5 years was computed above in part(3a).

ii. At the end of 5 years the amount that has been paid up is

180, 000 ·s10 2.625%

a40 2.625%

= 180, 000 · (1.02625)10 ·a10 2.625%

a40 2.625%

= 180, 000 · (1.02625)10 · 1 − (1.02625)−10

1 − (1.02625)−40 = 82, 506.49209 .


The amount outstanding on the mortgage is, therefore,

180, 000(1.02625)10(1 − 1 − (1.02625)−10

1 − (1.02625)−40

)= 150, 734.1381, or

180, 000(1.02625)10 − 82, 506.49209 = 150, 734.14 .

The effective semi-annual interest rate for the second 5-year period is 0.05752 =

0.02875. The semi-annual payment for years ##5-10 is therefore

150, 734.141−(1.02875)−30

0.02875

= 7, 566.61 .

iii. At the end of 10 years the amount still owing will have been reduced to

150, 734.14(1.02875)10 − 150, 734.14 ·s10 0.02875

a30 0.02875

= 150, 734.14(1.02875)10 ·(1 − 1 − (1.02875)−10

1 − (1.02875)−30

)

= 113, 884.07 .

For the third 5-year period the effective semi-annual interest rate is 3%, so thesemi-annual payments will each be equal to

113, 884.071−(1.03)−20

0.03

= 7, 654.798

or 7,654.80.iv. At the end of 15 years the amount outstanding will have been reduced to

113, 884.07(1.03)10 − 113, 884.07 ·s10 0.03

a20 0.03

= 113, 884.07(1.03)10 ·(1 − 1 − (1.03)−10

1 − (1.03)−20

)

= 65, 296.98 .

This amount must be repaid by 10 equal semi-annual payments at an effectivesemi-annual interest rate of 5%. The amount of each of these equal paymentswill be

65, 296.98a10 0.05

=(65, 296.98)(0.05)

1 − (1.05)−10 = 8456.258

or 8,456.26.


3.7 Fourth Problem AssignmentDistribution Date: 09 March, 2010

Solutions are to be submitted by the class of Monday, March 22nd, 2010

These problems are to be solved with full solutions, modelled either on the solu-tions to problems in the textbook, presented in class, or in the notes on the Web forthis or previous years. The essence is that the reader should be able to reconstructevery step of the proof from what you have written: getting the right answer isnever enough. You are not being graded for elegance, but simply for the proof be-ing logical and legible, without serious gaps. While the data given may sometimesnot justify a large number of decimal places, you should show your intermediatecalculations in sufficient detail that the grader can determine that your calculationsare correct. The default number of decimal places given by your calculator maynot always be sufficient for this.

1. An annuity pays 25,000 per year continuously for 8 years.

(a) At an effective annual interest rate of 8%, determine the value of the annuity 1 yearbefore the payments begin.

(b) Determine the value of the annuity 1 year before the payments begin if interest iscompounded continuously at a nominal annual rate of 8%.

In each case you should show your answer first as an integral, and then evaluate it.

2. Showing all your work, determine a formula for the value at time t = 0 (the time of thefirst payment) of the following annuity in terms of (Ia)n , an , v, i, and n. At a constanteffective annual rate i of compound interest the annuity pays 2n + 3 at time t = 0, 2n + 1at time t = 1, . . . , 5 at time t = n − 1, 3 at time t = n, 1 at time t = n + 1, 4 at timen + 2, 7 at time n + 3, . . . , 3n − 2 at time t = 2n. You are expected not to use, in yourfinal formula, any symbol for a decreasing annuity, nor any symbol for an annuity-due.You are expected to provide a verbal explanation, as you develop your formula — it isnot sufficient just to write down the answer. You are expected to check your work bycalculating the exact value in terms of powers of v when n = 1, 2.

3. (a) Showing your work, determine, in terms of i only (where we assume i > 0), thelimits of (Ia)n and (Ia)n as n → ∞. You may use a formula in your textbook for(Ia)n , where n is a finite integer.

(b) Annuity #1 is a perpetuity which pays X at the end of every year; annuity #2 is anincreasing perpetuity which pays Y at the end of the first year, 2Y at the end of the


2nd year, . . . , nY at the end of the nth year, . . .; annuity #3 pays (1+ j)n−1Z at the endof the nth year, where j is a non-negative constant, and 0 ≤ j < i. The interest ratewhich is applicable is, in all cases i, and we will be investigating certain optimalvalues for i, always assuming that X,Y,Z are all positive. All three perpetuitiesbegin their payments one year from now. Showing all your work, determine

i. the value(s) of i — if there are such values — for which perpetuities ##1,2 areequal in value;

ii. the value(s) of i — if there are such values — for which the difference ofthe values of perpetuities ##1,2 is as large as possible in magnitude (absolutevalue);

iii. the value(s) of i — if there are such values — for which perpetuities ##1,3 areequal in value.

4. A loan is being repaid by 60 payments spaced 4 months apart. The first 15 instalmentsare 300 each; the next 24 are 400 each; and the last 21 are 600 each. If interest is earnedat a nominal annual rate of 6%, compounded every 4 months,

(a) determine the full amount of the loan, as of 4 months before the first payment;

(b) determine the amount of the loan still outstanding immediately after the 30th pay-ment; and

(c) divide the 33rd payment into principal and interest.

(d) Construct a table with the lines for payments ##14, 15, 16, 39, 40, 59, 60 andcolumn headings as follows: Payment Number, Payment Amount, Interest Com-ponent, Principal Component, Outstanding Loan Balance (after reduction by Prin-cipal component of this Payment).

5. A loan of 300,000 has been negotiated at an effective annual interest rate of 13%. Thelender wishes to receive only the full interest payment each year, with the principal tobe repaid at the end of 10 years. The borrower, however, knows that he will be unableto repay the loan unless he saves progressively over the 10-year period; he requiresthe discipline of a regular, even payment, even if the lender does not wish to receiveit. The borrower arranges to save for the ultimate repayment of the loan by paymentsinto two sinking funds: the first earns 16% effective annually, and the second earnsonly 11% effective annually. He is able to arrange for the 16% sinking fund to receiveannual payments which, at maturity, will be worth one-third of the loan; the 11% sinkingfund will be worth two-thirds. The payments into the sinking funds will be constant.Determine the total annual payment: interest to the lender, and the level contributions tothe two sinking funds.


3.8 Class Tests3.8.1 Class Test, Version 1

McGILL UNIVERSITY, FACULTY OF SCIENCECLASS TEST in MATH 329, THEORY OF INTEREST

EXAMINER: Professor W. G. Brown DATE: Monday, 08 March, 2010.TIME: 45 minutes, 14:35→15:20

FAMILY NAME:GIVEN NAMES:

STUDENT NUMBER:

Instructions

• The time available for writing this test is about 45 minutes.

• This test booklet consists of this cover, Pages 128 through 132 containing questions togetherworth 40 marks; and Page 132, which is blank.

• Show all your work. All solutions are to be written in the space provided on the page wherethe question is printed. When that space is exhausted, you may write on the facing page, onthe blank page, or on the back cover of the booklet, but you must indicate any continuationclearly on the page where the question is printed! (Please inform the instructor if you findthat your booklet is defective.)

• All your writing — even rough work — must be handed in.

• Calculators. While you are permitted to use a calculator to perform arithmetic and/or expo-nential calculations, you must not use the calculator to calculate such actuarial functions asan i, sn i, etc. without first stating a formula for the value of the function in terms of exponen-tials and/or polynomials involving n and the interest rate. You must not use your calculatorin any programmed calculations. If your calculator has memories, you are expected to havecleared them before the test. Your solutions should include sufficient detail that the examinercan conclude that you have not used built-in annuity functions from a financial calculator.

• In your solutions to problems on this test you are expected to show all your work . You areexpected to simplify algebraic and numerical answers as much as you can.

• Your neighbours may be writing a version of this test which is different from yours.

PLEASE DO NOT WRITE INSIDE THIS BOX

1(a) 1(b) 1(c) 2(a) 2(b) 3 4 Total

/4 /4 /4 /4 /4 /6 /14 /40


1. Showing your work in detail, determine each of the following; the rates you determineshould be accurate to at least 4 decimal places, or as a percentage accurate to at least 2decimal places:

(a) [4 MARKS] the nominal annual interest rate, compounded every two months, cor-responding to an effective annual interest rate of i = 5%

(b) [4 MARKS] the effective interest rate per 3-year period corresponding to a nominaldiscount rate, compounded quarterly, of d = 4.%

(c) [4 MARKS] the effective monthly interest rate corresponding to a nominal annualrate of interest of 6%, compounded continuously.


2. Suppose that the force of interest at time t is δt =0.03

1 + 0.03t. Showing all your work,

determine

(a) [4 MARKS] the accumulation function a(t); you are expected to show an integraland evaluate it, not simply to quote a formula;

(b) [4 MARKS] the value at time t = 3 of payments of 100 at time t = 0 and 200 attime t = 1.


3. [6 MARKS] An annuity at an effective annual interest rate of i consists of payments of515 1 year from now, at time t = 1, 490 at the end of two years, 465 at the end of threeyears, decreasing by a constant amount until the last payment in the amount of 140. Theannuity is to be evaluated as of one year before the first payment. Express its value interms of any of symbols (Ia)n , (Is)n , an and/or sn , but do not evaluate. You are notexpected to quote formulæ needed for the evaluation of any of these functions. (Youare expected to show all of your work — it is not sufficient to simply write down theanswer.)


4. [14 MARKS] A loan of 2,000,000 is to be repaid by semi-monthly payments of 20,000to commence half a month from now, and to continue thereafter for as long as neces-sary. Find the time and amount of the final payment if the final payment is to be nosmaller than the regular payments. Assume a nominal annual interest rate of i = 6%compounded every half-month.


continuation page for problem number

You must refer to this continuation page on the page where the problem is printed!


3.8.2 Class Test, Version 2




STUDENT NUMBER:

Instructions









1 2(a) 2(b) 2(c) 3(a) 3(b) 4 Total

/6 /4 /4 /4 /4 /4 /14 /40


1. [6 MARKS] An annuity at an effective annual interest rate of i consists of payments of120, 1 year from now at time t = 1, 135 at the end of two years, 150 at the end of threeyears, increasing by a constant amount until the last payment in the amount of 450. Theannuity is to be evaluated as of time t = 0. Express its value in terms of any of symbols(Ia)n , (Is)n , an and/or sn , but do not evaluate. You are not expected to quote formulæneeded for the evaluation of any of these functions. (You are expected to show all ofyour work — it is not sufficient to simply write down the answer.)



(a) [4 MARKS] the effective interest rate per 3-year period corresponding to a nominaldiscount rate, compounded quarterly, of d = 4.%

(b) [4 MARKS] the nominal annual interest rate, compounded every two months, cor-responding to an effective annual interest rate of i = 5%

(c) [4 MARKS] the effective monthly interest rate corresponding to a nominal annualrate of interest of 6%, compounded continuously.


3. Suppose that the force of interest at time t is δt = 4 ln(1.01). Showing all your work,determine


(b) [4 MARKS] the value at time t = 3 of a continuous annuity s3 that pays 1 contin-uously over each year, starting at time t = 0; you are expected to first express thevalue as an integral, and then to evaluate the integral.


4. [14 MARKS] A loan of 1,000,000 is to be repaid by half-yearly payments of 75,000to commence half a year from now, and to continue thereafter for as long as necessary.Find the time and amount of the final payment if the final payment is to be no larger thanthe regular payments. Assume a nominal annual interest rate of i = 6% compoundedevery half-year.






STUDENT NUMBER:

Instructions









1(a) 1(b) 1(c) 2(a) 2(b) 3 4 Total

/4 /4 /4 /4 /4 /6 /14 /40



(a) [4 MARKS] the effective discount rate per 4-year period corresponding to a nom-inal interest rate, compounded 3 times a year, of d = 6.% [There was an inconsis-tency in the wording of this problem. Students were told to resolve the inconsis-tency in a “reasonable“ way.]

(b) [4 MARKS] the nominal annual interest rate, compounded every 4 months, corre-sponding to an effective semi-annual interest rate of i = 5%

(c) [4 MARKS] the nominal annual interest rate compounded quarterly correspondingto a nominal annual rate of interest of 3%, compounded continuously.




determine


(b) [4 MARKS] the value at time t = 0 of a continuous annuity that pays 100 continu-ously over each year between times t = 0 and t = 3.


3. [6 MARKS] An annuity at an effective annual interest rate of i consists of payments of735, 1 year from now at time t = 1, 710 at the end of two years, 685 at the end of threeyears, decreasing by a constant amount until the last payment in the amount of 35. Theannuity is to be evaluated as of just after the last payment. Express its value in terms ofany of symbols (Ia)n , (Is)n , an and/or sn , but do not evaluate. You are not expected toquote formulæ needed for the evaluation of any of these functions. (You are expected toshow all of your work — it is not sufficient to simply write down the answer.)






STUDENT NUMBER:

Instructions









1 2(a) 2(b) 2(c) 3(a) 3(b) 4 Total

/6 /4 /4 /4 /4 /4 /14 /40


1. [6 MARKS] An annuity at an effective annual interest rate of i consists of payments of210, 1 year from now at time t = 1, 225 at the end of two years, 240 at the end of threeyears, increasing by a constant amount until the last payment in the amount of 570. Theannuity is to be evaluated as of the last payment. Express its value in terms of any ofsymbols (Ia)n , (Is)n , an and/or sn , but do not evaluate. You are not expected to quoteformulæ needed for the evaluation of any of these functions. (You are expected to showall of your work — it is not sufficient to simply write down the answer.)



(a) [4 MARKS] the nominal annual interest rate, compounded every 4 months, corre-sponding to an effective semi-annual interest rate of i = 5%

(b) [4 MARKS] the nominal annual interest rate compounded quarterly correspondingto a nominal annual rate of interest of 3%, compounded continuously.

(c) [4 MARKS] the effective discount rate per 4-year period corresponding to a nom-inal interest rate, compounded 3 times a year, of d = 6.% [There was an inconsis-tency in the wording of this problem. Students were told to resolve the inconsis-tency in a “reasonable“ way.]



(a) [4 MARKS] i(3); you are expected to simplify your answer;

(b) [4 MARKS] the value at time t = 0 of a deferred continuous annuity 1

∣∣∣a4 that pays1 continuously over each year between times t = 1 and t = 5.


3.9 Draft Solutions, 2010 Class Test

3.9.1 Problems on Nominal/Effective Interest/Discount Rates




(c) [4 MARKS] the effective monthly interest rate corresponding to nominal annualrate of interest of 6%, compounded continuously.

Solution:

(a) We are given that d(4) = 4%. Hence the effective quarterly discount rate is 1%. The

effective quarterly interest rate will be0.01

1 − 0.01=

199

, so the effective interest ratefor a 3-year period, i.e., for 3 × 4 = 12 quarters will be

(1 +

199

)12

− 1 =

(10099

)12

− 1 = 0.128178098 = 12.8178098% .

The problem is not asking for i13 , which is the nominal annual interest rate com-

pounded every 3 years.

(b) Since(1 +

i(6)

6

)6

= 1.05,

i(6) = 6(1.05

16 − 1

)= 0.048989076 = 4.8989076% .

(c) A nominal annual rate of interest of 6% compounded instantaneously is equivalentto an effective annual accumulation factor of e0.06, or to an effective annual interestrate of e0.06−1 = 0.061836547 = 6.1836547% , and an accumulation factor of e0.06.The effective monthly accumulation factor is e

0.0612 = e0.005, so the effective monthly

interest rate ise0.005 − 1 = 0.00501252 = 0.501252% , .

so the percentage, accurate to two decimal places, is 0.05%.






Solution:

(a) i. Suppose that the inconsistency in the wording was resolved by replacing thestatement d = 6% by i = 6%: We are given that i(3) = 6%. Hence the effectiveinterest rate for a 4 month period is 2%. The effective quarterly discount rate

will be0.02

1 + 0.02=

2102

, so the effective discount rate for a 4-year period, willbe

1 −(1 − 2

102

)12

= 1 −(1 − 2

102

)12

= 0.2115068241 = 21.15068241% .

ii. Suppose that the inconsistency in the wording was resolved by replacing thewords nominal interest rate by nominal discount rate: We are given that d(3) =

6%. Hence the effective discount rate for a 4 month period is 2%. As a 4-yearperiod contains 3 × 4 = 12 4-month intervals, the discount factor for a 4-yearperiod is (1 − 0.02)3×4 = (0.98)12 = 0.78471672, so the effective discount rateis 1 − 0.78471672 = 0.21528328 = 21.53%.

(b) Since(1 +

i(3)

3

)3

= (1.05)2,

i(3) = 3(1.05

23 − 1

)= 0.099184662 = 9.9184662% .

(c) A nominal annual rate of interest of 3% compounded continuously is equivalent toan effective annual accumulation factor of e0.03 , and to a quarterly accumulation

factor of e0.03

4 = 1 +i(4)

4. It follows that

i(4) = 4(e

0.034 − 1

)= 0.030112780 = 3.0112780% .


3.9.2 Problems on Force of Interest



determine

(a) [4 MARKS] a(t); you are expected to show an integral and evaluate it, not simplyto quote a formula;


Solution:

(a) Since δt =a′(t)a(t)

=ddt

(ln a(t)),

a(t) = e∫ t

0 δs ds

= e∫ t

00.03

1+0.03t dt

= e[ln(1+0.03s)]t0

= eln(1+0.03t)−ln 1 = eln(1+0.03t) = 1 + 0.03t.

(The accumulation can thus be seen to be under simple interest.)

(b) The payment of 100 at time t = 0 accumulates to 100·a(3) = 109 at time t = 3. The

payment of 200 at time t = 1 accumulates to 200 · a(3)a(1)

= 200(1.091.03

)= 211.65, so

the total accumulation is

109 + 211.65 = 320.65 .



determine



Solution:



=ddt

(ln a(t)),

a(t) = e∫ t

0 δs ds

= e∫ t

00.04

1+0.04t dt

= e[ln(1+0.04s)]t0

= eln(1+0.04t)−ln 1 = eln(1+0.04t) = 1 + 0.04t.

(The accumulation can thus be seen to be under simple interest.)(b) It was not correct to quote a formula [5, (4.14), p. 125] from the textbook, as this

formula had been computed on the assumption of compound interest. You had tocalculate the value of the annuity from first principles:

100a3 = 100

3∫

0

dta(t)

=

3∫

0

dt1 + 0.04t

=1000.04

[ln(1 + 0.04t)]30

= 100 · ln 1.120.04

= 283.32




Solution:


=ddt

(ln a(t)),

a(t) = e∫ t

0 δs ds

= e∫ t

0 4 ln(1.01) dt

= e[(4 ln 1.01)s]t0

=(eln 1.01

)4t=

((1.01)4

)t.


(The accumulation can thus be seen to be under compound interest at an effectivequarterly rate of 1%.)

(b)

s3 =

3∫

0

a(3)a(t)

dt

=

3∫

0

1.014(3)

(1.01)4t dt

=

3∫

0

(1.01)4(3−t) dt

=

[(1.01)4(3−t)

−4 ln 1.01

]3

0

=1 − (1.01)12

−4 ln 1.01=

(1.01)12 − 14 ln 1.01

= 3.18645.


(a) [4 MARKS] i(3); you are expected to simplify your answer;

(b) [4 MARKS] the value at time t = 0 of a deferred continuous annuity 1


Solution:


=ddt

(ln a(t)),

a(t) = e∫ t

0 δs ds

= e∫ t

0 6 ln(1.02) dt

= e[(6 ln 1.02)s]t0

=(eln 1.02

)6t= 1.026t =

(1.026

)t.

The accumulation can thus be seen to be under compound interest at an effectivequarterly rate of 2% every 6th of a year. We can solve the equation

(1 +

i(3)

3

)3

= 1 + i = (1.02)6


to infer thati(3) = 3

((1.02)2 − 1

)= 0.1212 = 12.12% .

(b)

1

∣∣∣a4 =

5∫

1

dta(t)

=

5∫

1

dt1.026t

=

[(1.02)−6t

−6 ln 1.02

]5

1

=(1.02)−6 − (1.02)−30

6 ln 1.02= 2.82707

3.9.3 Problems on Increasing/Decreasing Annuities

In these problems there was a restriction on the types of functions that could be usedto express the answer. It was not considered acceptable to express the answer in termsof a summation of an i or sn i since the functions , (Ia)n i, (Is)n i, (Da)n i, (Ds)n i were abasic topic in the syllabus. The instructions restricted which of these functions could beused in the solution, and so students may have had to appeal to an identity relating theincreasing and decreasing functions.

7. [6 MARKS] An annuity at interest rate i consists of payments of 120 1 year from now,at time t = 1, 135 at the end of two years, 150 at the end of three years, increasing bya constant amount until the last payment, in the amount of 450. The annuity is to beevaluated as of time t = 0. Express its value in terms of any of symbols (Ia)n , (Is)n , anand/or sn , but do not evaluate. You are not expected to quote formulæ needed for theevaluation of any of these functions. (You are expected to show all of your work — it isnot sufficient to simply write down the answer.)

Solution: The annual increase is in the amount of 15, so one component of the valuecould be 15(Ia)n, where n is approximately the number of payments. And what is thenumber of payments? Since the difference between last and first is 450−120 = 330, and

since33015

= 22, the number of payments is 23 — note that it is not 22, since paymentshave to be associated with an entire year. Thus one component of the value could be


expressed as 15(Ia)23 . Now the amount not accounted for is 120 − 15 = 105. Thus oneway of expressing the present value of this increasing annuity is as 105a23 + 15(Ia)23 .

8. [6 MARKS] An annuity at interest rate i consists of payments of 515 1 year from now,at time t = 1, 490 at the end of two years, 465 at the end of three years, decreasing bya constant amount until the last payment, in the amount of 140. The annuity is to beevaluated as of one year before the first payment. Express its value in terms of any ofsymbols (Ia)n , (Is)n , an and/or sn , but do not evaluate. You are not expected to quoteformulæ needed for the evaluation of any of these functions. (You are expected to showall of your work — it is not sufficient to simply write down the answer.)

Solution: The annual decrease is in the amount of 490 − 465 = 25, so one componentof the value could be 25(Da)n, where n is approximately the number of payments. Andwhat is the number of payments? Since the difference between last and first is 515 −140 = 375, and since

37525

= 15, the number of payments is 16 — note that it is not15, since payments have to be associated with an entire year. Thus one componentof the value could be expressed as 25(Da)16 . Now the amount not accounted for is140− 25 = 115. Thus one way of expressing the present value of this increasing annuityis as 115a16 + 25(Da)16 . However, the instructions required the use of only severalspecific functions, and (Da)n i is not one of them. Since

(Ia)16 + (Da)16 = (17)an ,

we have

115a16 + 25(Da)16 = 115a16 + 25(17 · a16 − (Ia)16

)

= 540a16 − 25(Ia)16 asoneacceptablesolution.

9. [6 MARKS] An annuity at interest rate i consists of payments of 210 1 year from now,at time t = 1, 225 at the end of two years, 240 at the end of three years, increasing bya constant amount until the last payment, in the amount of 570. The annuity is to beevaluated as of the last payment. Express its value in terms of any of symbols (Ia)n ,(Is)n , an and/or sn , but do not evaluate. You are not expected to quote formulæ neededfor the evaluation of any of these functions. (You are expected to show all of your work— it is not sufficient to simply write down the answer.)

Solution: The annual increase is in the amount of 15, so one component of the valuecould be 15(Is)n, where n is approximately the number of payments. And what is thenumber of payments? Since the difference between last and first is 570−210 = 360, and

since36015

= 24, the number of payments is 25 — note that it is not 24, since paymentshave to be associated with an entire year. Thus one component of the value could be


expressed as 15(Is)25 . Now the amount not accounted for is 210 − 15 = 195. Thus oneway of expressing the present value of this increasing annuity is as 195s25 + 15(Is)25 .This isn’t the only solution; another solution is 210s25 + 15(Is)24 .

10. [6 MARKS] An annuity at interest rate i consists of payments of 735 1 year from now,at time t = 1, 710 at the end of two years, 685 at the end of three years, decreasingby a constant amount until the last payment, in the amount of 35. The annuity is to beevaluated as of just after the last payment. Express its value in terms of any of symbols(Ia)n , (Is)n , an and/or sn , but do not evaluate. You are not expected to quote formulæneeded for the evaluation of any of these functions. (You are expected to show all ofyour work — it is not sufficient to simply write down the answer.)

Solution: The annual decrease is in the amount of 735− 710 = 25, so one component ofthe value could be 25(Ds)n, where n is approximately the number of payments. And whatis the number of payments? Since the difference between last and first is 735−35 = 700,

and since70025

= 28, the number of payments is 29 — note that it is not 28, sincepayments have to be associated with an entire year. Thus one component of the valuecould be expressed as 25(Ds)29 . Now the amount not accounted for is 35−25 = 10. Thusone way of expressing the present value of this increasing annuity is as 10s29 +25(Ds)29 .However, the instructions required that the solution be expressed in terms of a specificset of functions, which did not include (Ds)29 . Since

(Is)29 + (Ds)29 = 30 · s29 ,

another solution would be

10s29 + 25(Ds)29 = 10s29 + 25(30 · s29 − (Is)29

)

= 760s29 − 25(Is)29 .

Another solution would be 735s29 − 25(Is)28 .

3.9.4 Problems on Balloon and Drop Payments


Solution: The effective semi-monthly interest rate is 0.25%. Denote the time of the lastpayment by n (in half-months). Then n is the largest integer solution to the inequality

20, 000 · an 0.0025 < 2, 000, 000


⇔ an < 100⇔ 1 − vn < 100(0.0025) = 0.25⇔ vn > 1 − 0.2 = 0.75

⇔ n <ln 0.75− ln 1.0025

= 115.2166102 .

Thus n = 115: the last payment will be the 115th. The first 20,000 of the last paymentwill leave an unpaid balance of

2, 000, 000(1.0025)115 − 20, 000 · a115 0.25%

= 20, 000(100(1.0025)115 − 1.0025115 − 1

0.0025

)

= 4, 325.62705 .

Thus the balloon payment will be in the amount of 20, 000 + 4, 325.63 = 24, 325.63.


Solution: The effective semi-yearly interest rate is 3%. Denote the time of the lastpayment by n (in half-years). Then n is the smallest integer solution to the inequality

75, 000 · an > 1, 000, 000

⇔ an >1000000

75000=

403

⇔ 1 − vn >403· (0.03) = 0.4

⇔ vn < 1 − 0.4 = 0.6

⇔ n >ln 0.6− ln 1.03

= 17.28167534 .

Thus n = 18: the last payment will be the 18th. A full last payment of 75,000 wouldexceed the amount owing by

1, 000, 000(1.03)18 − 75, 000 · a18 3%

= 1, 000, 000(1.03)18 − 750001.0318 − 1

0.03= 53, 649.592 .

Thus the drop payment will be in the amount of 75, 000 − 53, 649.59 = 21, 350.41.



Solution: The effective semi-monthly interest rate is3%24

= 0.125%. Denote the timeof the last payment by n (in half-months). Then n is the largest integer solution to theinequality

20, 000 · an 0.00125 < 1, 000, 000⇔ an < 50⇔ 1 − vn < 50(0.00125) = 0.0625⇔ vn > 1 − 0.0625 = 0.9375

⇔ n <ln 0.9375− ln 1.00125

= 51.66307947 .

Thus n = 51: the last payment will be the 51st. The first 20,000 of the last payment willleave an unpaid balance of

1, 000, 000(1.00125)51 − 20, 000 · s51 0.125%

= 20, 000(50(1.00125)51 − 1.0012551 − 1

0.00125

)

= 13, 247.81500 .



Solution: The effective semi-yearly interest rate is 2.5%. Denote the time of the lastpayment by n (in half-years). Then n is the smallest integer solution to the inequality

100, 000 · an > 2, 000, 000

⇔ an >2, 000, 000100, 000

= 20

⇔ 1 − vn > 20 · (0.025)⇔ vn < 1 − 0.5 = 0.5

⇔ n >ln 0.5− ln 1.025

= 28.07103453 .



2, 000, 000(1.025)29 − 100, 000 · a29 25%

= 2, 000, 000(1.025)29 − 100, 0001.02529 − 1

0.025= 92, 814.788 .



3.10 Draft Solutions, 2010 Class Test, arranged by version3.10.1 Version 1


(a) [4 MARKS] the nominal annual interest rate, compounded every two months, cor-responding to an effective annual interest rate of i = 5%

(b) [4 MARKS] the effective interest rate per 3-year period corresponding to a nominaldiscount rate, compounded quarterly, of d = 4.%


Solution:

(a) Since(1 +

i(6)

6

)6

= 1.05,

i(6) = 6(1.05

16 − 1

)= 0.048989076 = 4.8989076% .

(b) We are given that d(4) = 4%. Hence the effective quarterly discount rate is 1%. The


1 − 0.01=

199

, so the effective interest ratefor a 3-year period, i.e., for 3 × 4 = 12 quarters will be

(1 +

199

)12

− 1 =

(10099

)12

− 1 = 0.128178098 = 12.8178098% .










determine



Solution:


=ddt

(ln a(t)),

a(t) = e∫ t

0 δs ds

= e∫ t

00.03

1+0.03t dt

= e[ln(1+0.03s)]t0

= eln(1+0.03t)−ln 1 = eln(1+0.03t) = 1 + 0.03t.


(b) The payment of 100 at time t = 0 accumulates to 100·a(3) = 109 at time t = 3. The

payment of 200 at time t = 1 accumulates to 200 · a(3)a(1)

= 200(1.091.03

)= 211.65, so

the total accumulation is

109 + 211.65 = 320.65 .

3. [6 MARKS] In these problems there was a restriction on the types of functions that couldbe used to express the answer. It was not considered acceptable to express the answerin terms of a summation of an i or sn i since the functions , (Ia)n i, (Is)n i, (Da)n i, (Ds)n iwere a basic topic in the syllabus. The instructions restricted which of these functionscould be used in the solution, and so students may have had to appeal to an identityrelating the increasing and decreasing functions.

An annuity at interest rate i consists of payments of 515 1 year from now, at time t = 1,490 at the end of two years, 465 at the end of three years, decreasing by a constantamount until the last payment, in the amount of 140. The annuity is to be evaluated asof one year before the first payment. Express its value in terms of any of symbols (Ia)n ,(Is)n , an and/or sn , but do not evaluate. You are not expected to quote formulæ neededfor the evaluation of any of these functions. (You are expected to show all of your work— it is not sufficient to simply write down the answer.)


Solution: The annual decrease is in the amount of 490 − 465 = 25, so one componentof the value could be 25(Da)n, where n is approximately the number of payments. Andwhat is the number of payments? Since the difference between last and first is 515 −140 = 375, and since

37525

= 15, the number of payments is 16 — note that it is not15, since payments have to be associated with an entire year. Thus one componentof the value could be expressed as 25(Da)16 . Now the amount not accounted for is140− 25 = 115. Thus one way of expressing the present value of this increasing annuityis as 115a16 + 25(Da)16 . However, the instructions required the use of only severalspecific functions, and (Da)n i is not one of them. Since

(Ia)16 + (Da)16 = (17)an ,

we have

115a16 + 25(Da)16 = 115a16 + 25(17 · a16 − (Ia)16

)

= 540a16 − 25(Ia)16 asoneacceptablesolution.


Solution: The effective semi-monthly interest rate is 0.25%. Denote the time of the lastpayment by n (in half-months). Then n is the largest integer solution to the inequality

20, 000 · an 0.0025 < 2, 000, 000⇔ an < 100⇔ 1 − vn < 100(0.0025) = 0.25⇔ vn > 1 − 0.2 = 0.75

⇔ n <ln 0.75− ln 1.0025

= 115.2166102 .

Thus n = 115: the last payment will be the 115th. The first 20,000 of the last paymentwill leave an unpaid balance of

2, 000, 000(1.0025)115 − 20, 000 · a115 0.25%

= 20, 000(100(1.0025)115 − 1.0025115 − 1

0.0025

)

= 4, 325.62705 .



3.10.2 Version 2


An annuity at interest rate i consists of payments of 120 1 year from now, at time t = 1,135 at the end of two years, 150 at the end of three years, increasing by a constantamount until the last payment, in the amount of 450. The annuity is to be evaluated asof time t = 0. Express its value in terms of any of symbols (Ia)n , (Is)n , an and/or sn ,but do not evaluate. You are not expected to quote formulæ needed for the evaluation ofany of these functions. (You are expected to show all of your work — it is not sufficientto simply write down the answer.)

Solution: The annual increase is in the amount of 15, so one component of the valuecould be 15(Ia)n, where n is approximately the number of payments. And what is thenumber of payments? Since the difference between last and first is 450−120 = 330, and

since33015

= 22, the number of payments is 23 — note that it is not 22, since paymentshave to be associated with an entire year. Thus one component of the value could beexpressed as 15(Ia)23 . Now the amount not accounted for is 120 − 15 = 105. Thus oneway of expressing the present value of this increasing annuity is as 105a23 + 15(Ia)23 .





Solution:

(a) We are given that d(4) = 4%. Hence the effective quarterly discount rate is 1%. The


1 − 0.01=

199

, so the effective interest rate


for a 3-year period, i.e., for 3 × 4 = 12 quarters will be(1 +

199

)12

− 1 =

(10099

)12

− 1 = 0.128178098 = 12.8178098% .



(b) Since(1 +

i(6)

6

)6

= 1.05,

i(6) = 6(1.05

16 − 1

)= 0.048989076 = 4.8989076% .








Solution:


=ddt

(ln a(t)),

a(t) = e∫ t

0 δs ds

= e∫ t

0 4 ln(1.01) dt

= e[(4 ln 1.01)s]t0

=(eln 1.01

)4t=

((1.01)4

)t.

(The accumulation can thus be seen to be under compound interest at an effectivequarterly rate of 1%.)


(b)

s3 =

3∫

0

a(3)a(t)

dt

=

3∫

0

1.014(3)

(1.01)4t dt

=

3∫

0

(1.01)4(3−t) dt

=

[(1.01)4(3−t)

−4 ln 1.01

]3

0

=1 − (1.01)12

−4 ln 1.01=

(1.01)12 − 14 ln 1.01

= 3.18645.


Solution: The effective semi-yearly interest rate is 3%. Denote the time of the lastpayment by n (in half-years). Then n is the smallest integer solution to the inequality

75, 000 · an > 1, 000, 000

⇔ an >1000000

75000=

403

⇔ 1 − vn >403· (0.03) = 0.4

⇔ vn < 1 − 0.4 = 0.6

⇔ n >ln 0.6− ln 1.03

= 17.28167534 .


1, 000, 000(1.03)18 − 75, 000 · a18 3%

= 1, 000, 000(1.03)18 − 750001.0318 − 1

0.03= 53, 649.592 .



3.10.3 Version 3





Solution:


will be0.02

1 + 0.02=

2102


1 −(1 − 2

102

)12

= 1 −(1 − 2

102

)12

= 0.2115068241 = 21.15068241% .



(b) Since(1 +

i(3)

3

)3

= (1.05)2,

i(3) = 3(1.05

23 − 1

)= 0.099184662 = 9.9184662% .


factor of e0.03

4 = 1 +i(4)

4. It follows that

i(4) = 4(e

0.034 − 1

)= 0.030112780 = 3.0112780% .




determine



Solution:


=ddt

(ln a(t)),

a(t) = e∫ t

0 δs ds

= e∫ t

00.04

1+0.04t dt

= e[ln(1+0.04s)]t0

= eln(1+0.04t)−ln 1 = eln(1+0.04t) = 1 + 0.04t.


(b) It was not correct to quote a formula [5, (4.14), p. 125] from the textbook, as thisformula had been computed on the assumption of compound interest. You had tocalculate the value of the annuity from first principles:

100a3 = 100

3∫

0

dta(t)

=

3∫

0

dt1 + 0.04t

=1000.04

[ln(1 + 0.04t)]30

= 100 · ln 1.120.04

= 283.32



An annuity at interest rate i consists of payments of 735 1 year from now, at time t = 1,710 at the end of two years, 685 at the end of three years, decreasing by a constantamount until the last payment, in the amount of 35. The annuity is to be evaluated as ofjust after the last payment. Express its value in terms of any of symbols (Ia)n , (Is)n , anand/or sn , but do not evaluate. You are not expected to quote formulæ needed for theevaluation of any of these functions. (You are expected to show all of your work — it isnot sufficient to simply write down the answer.)

Solution: The annual decrease is in the amount of 735− 710 = 25, so one component ofthe value could be 25(Ds)n, where n is approximately the number of payments. And whatis the number of payments? Since the difference between last and first is 735−35 = 700,

and since70025

= 28, the number of payments is 29 — note that it is not 28, sincepayments have to be associated with an entire year. Thus one component of the valuecould be expressed as 25(Ds)29 . Now the amount not accounted for is 35−25 = 10. Thusone way of expressing the present value of this increasing annuity is as 10s29 +25(Ds)29 .However, the instructions required that the solution be expressed in terms of a specificset of functions, which did not include (Ds)29 . Since

(Is)29 + (Ds)29 = 30 · s29 ,

another solution would be

10s29 + 25(Ds)29 = 10s29 + 25(30 · s29 − (Is)29

)

= 760s29 − 25(Is)29 .

Another solution would be 735s29 − 25(Is)28 .


Solution: The effective semi-monthly interest rate is3%24

= 0.125%. Denote the timeof the last payment by n (in half-months). Then n is the largest integer solution to theinequality

20, 000 · an 0.00125 < 1, 000, 000⇔ an < 50⇔ 1 − vn < 50(0.00125) = 0.0625⇔ vn > 1 − 0.0625 = 0.9375

⇔ n <ln 0.9375− ln 1.00125

= 51.66307947 .


Thus n = 51: the last payment will be the 51st. The first 20,000 of the last payment willleave an unpaid balance of

1, 000, 000(1.00125)51 − 20, 000 · s51 0.125%

= 20, 000(50(1.00125)51 − 1.0012551 − 1

0.00125

)

= 13, 247.81500 .



3.10.4 Version 4


An annuity at interest rate i consists of payments of 210 1 year from now, at time t = 1,225 at the end of two years, 240 at the end of three years, increasing by a constantamount until the last payment, in the amount of 570. The annuity is to be evaluated as ofthe last payment. Express its value in terms of any of symbols (Ia)n , (Is)n , an and/or sn ,but do not evaluate. You are not expected to quote formulæ needed for the evaluation ofany of these functions. (You are expected to show all of your work — it is not sufficientto simply write down the answer.)

Solution: The annual increase is in the amount of 15, so one component of the valuecould be 15(Is)n, where n is approximately the number of payments. And what is thenumber of payments? Since the difference between last and first is 570−210 = 360, and

since36015

= 24, the number of payments is 25 — note that it is not 24, since paymentshave to be associated with an entire year. Thus one component of the value could beexpressed as 15(Is)25 . Now the amount not accounted for is 210 − 15 = 195. Thus oneway of expressing the present value of this increasing annuity is as 195s25 + 15(Is)25 .This isn’t the only solution; another solution is 210s25 + 15(Is)24 .





Solution:



will be0.02

1 + 0.02=

2102


1 −(1 − 2

102

)12

= 1 −(1 − 2

102

)12

= 0.2115068241 = 21.15068241% .



(b) Since(1 +

i(3)

3

)3

= (1.05)2,

i(3) = 3(1.05

23 − 1

)= 0.099184662 = 9.9184662% .


factor of e0.03

4 = 1 +i(4)

4. It follows that

i(4) = 4(e

0.034 − 1

)= 0.030112780 = 3.0112780% .


(a) [4 MARKS] i(3); you are expected to simplify your answer;(b) [4 MARKS] the value at time t = 0 of a deferred continuous annuity 1


Solution:


=ddt

(ln a(t)),

a(t) = e∫ t

0 δs ds

= e∫ t

0 6 ln(1.02) dt

= e[(6 ln 1.02)s]t0

=(eln 1.02

)6t= 1.026t =

(1.026

)t.


The accumulation can thus be seen to be under compound interest at an effectivequarterly rate of 2% every 6th of a year. We can solve the equation

(1 +

i(3)

3

)3

= 1 + i = (1.02)6

to infer thati(3) = 3

((1.02)2 − 1

)= 0.1212 = 12.12% .

(b)

1

∣∣∣a4 =

5∫

1

dta(t)

=

5∫

1

dt1.026t

=

[(1.02)−6t

−6 ln 1.02

]5

1

=(1.02)−6 − (1.02)−30

6 ln 1.02= 2.82707

4. [14 MARKS] In these problems there was a restriction on the types of functions thatcould be used to express the answer. It was not considered acceptable to express theanswer in terms of a summation of an i or sn i since the functions , (Ia)n i, (Is)n i, (Da)n i,(Ds)n i were a basic topic in the syllabus. The instructions restricted which of thesefunctions could be used in the solution, and so students may have had to appeal to anidentity relating the increasing and decreasing functions.

A loan of 2,000,000 is to be repaid by half-yearly payments of 100,000 to commencehalf a year from now, and to continue thereafter for as long as necessary. Find the timeand amount of the final payment if the final payment is to be no larger than the regularpayments. Assume a nominal annual interest rate of i = 5% compounded every half-year.

Solution: The effective semi-yearly interest rate is 2.5%. Denote the time of the lastpayment by n (in half-years). Then n is the smallest integer solution to the inequality

100, 000 · an > 2, 000, 000

⇔ an >2, 000, 000100, 000

= 20


⇔ 1 − vn > 20 · (0.025)⇔ vn < 1 − 0.5 = 0.5

⇔ n >ln 0.5− ln 1.025

= 28.07103453 .


2, 000, 000(1.025)29 − 100, 000 · a29 25%

= 2, 000, 000(1.025)29 − 100, 0001.02529 − 1

0.025= 92, 814.788 .



3.11 Draft Solutions, Fourth Problem AssignmentDistribution Date: 29 March, 2010

Solutions were to be submitted by the class of Monday, March 22nd, 2010(subject to correction — these solutions may not have been fully checked!)

These problems were to be solved with full solutions, modelled either on the solu-tions to problems in the textbook, presented in class, or in the notes on the Web forthis or previous years. The essence is that the reader should be able to reconstructevery step of the proof from what you have written: getting the right answer isnever enough. You are not being graded for elegance, but simply for the proof be-ing logical and legible, without serious gaps. While the data given may sometimesnot justify a large number of decimal places, you should show your intermediatecalculations in sufficient detail that the grader can determine that your calculationsare correct. The default number of decimal places given by your calculator maynot always be sufficient for this.

1. An annuity pays 25,000 per year continuously for 8 years.

(a) At an effective annual interest rate of 8%, determine the value of the annuity 1 yearbefore the payments begin.

(b) Determine the value of the annuity 1 year before the payments begin if interest iscompounded continuously at a nominal annual rate of 8%.

In each case you should show your answer first as an integral, and then evaluate it.

Solution: In either case the value now will be 25,000 times

9∫

1

dta(t)

=

9∫

1

vt dt =vt

ln v

]9

1=

v9 − v1

− ln(1 + i).

(a) When i = 8%, the value one year before payment begins is

25, 000 × v9 − v1

− ln(1 + i)=

(1.08)−1 − (1.08)−9

ln 1.08= 138, 276.77 .

(b) The effective annual interest rate is now

limm→∞

(1 +

0.08m

)m

− 1 = e0.08 − 1.


Here, using the notation of the preceding part, we have 1 + i = e0.08, v = e−0.08,δ = 0.08, so the value one year before the payment begins is

25, 000 × e−0.72 − e−0.08

−0.08= 136, 363.78 .

2. Showing all your work, determine a formula for the value at time t = 0 (the time of thefirst payment) of the following annuity in terms of (Ia)n , an , v, i, and n. At a constanteffective annual rate i of compound interest the annuity pays 2n + 3 at time t = 0, 2n + 1at time t = 1, . . . , 5 at time t = n − 1, 3 at time t = n, 1 at time t = n + 1, 4 at timen + 2, 7 at time n + 3, . . . , 3n − 2 at time t = 2n. You are expected not to use, in yourfinal formula, any symbol for a decreasing annuity, nor any symbol for an annuity-due.You are expected to provide a verbal explanation, as you develop your formula — it isnot sufficient just to write down the answer. You are expected to check your work bycalculating the exact value in terms of powers of v when n = 1, 2.

Solution: Even with the restricted symbols that are available for your solution, therewill be more than one way of expressing the answer. For example, the following is onederivation. Since the successive payments at the beginning the annuity for the first n fullyears (n + 1 payments) are decreasing by 2, we know that one component in our formulashould involve 2 times the present value of a decreasing annuity. The present value ofthe 1st n + 1 payments under the annuity is

2(Da)n+1 + an+1 = (1 + i)(2(Da)n+1 + an+1

)

= (1 + i)(2((n + 2)an+1 − (Ia)n+1

)+ an+1

)

= (1 + i)((2n + 5)an+1 − 2(Ia)n+1

).

The final n payments are worth, as of time t = n, −2an + 3(Ia)n ; as of time t = 0 thesepayments are worth vn

(−2an + 3(Ia)n

). The total value is, therefore,

(1 + i)((2n + 5)an+1 − 2(Ia)n+1

)+ vn

(−2an + 3(Ia)n

),

whose terms could be combined in various other ways. We were required to check thiswhen n = 1, 2:

n = 1: The payments would be 5 at time t = 1, 3 at time t = 1, and 1 at time t = 2, sotheir value should be 5 + 3v + v2.

(1 + i)((2 · 1 + 5)a1+1 − 2(Ia)1+1

)+ v1

(−2a1 + 3(Ia)1

)

= (1 + i)(7a2 − 2(Ia)2

)+ v

(−2a1 + 3(Ia)1

)

= (1 + i)(7(v + v2) − 2(v + 2v2)

)+ v(−2v + 3v)


= (1 + i)(5v + 3v2

)+ v2

= 5 + 3v + v2 .

n = 2: The payments would be 7 at time t = 1, 5 at time t = 2, 3 at time t = 2, 1 at timet = 3, and 4 at time t = 2, so their value should be 7 + 5v + 3v2 + v3 + 4v4.

(1 + i)((2 · 2 + 5)a2+1 − 2(Ia)2+1

)+ v2

(−2a2 + 3(Ia)2

)

= (1 + i)(9a3 − 2(Ia)3

)+ v2

(−2a2 + 3(Ia)2

)

= (1 + i)(7v + 5v2 + 3v3

)+ v2

(v + 4v2

)

=(7 + 5v + 3v2

)+

(v3 + 4v4

).

3. (a) Showing your work, determine, in terms of i only (where we assume i > 0), thelimits of (Ia)n and (Ia)n as n → ∞. You may use a formula in your textbook for(Ia)n , where n is a finite integer.

(b) Annuity #1 is a perpetuity which pays X at the end of every year; annuity #2 is anincreasing perpetuity which pays Y at the end of the first year, 2Y at the end of the2nd year, . . . , nY at the end of the nth year, . . .; annuity #3 pays (1+ j)n−1Z at the endof the nth year, where j is a non-negative constant, and 0 ≤ j < i. The interest ratewhich is applicable is, in all cases i, and we will be investigating certain optimalvalues for i, always assuming that X,Y,Z are all positive. All three perpetuitiesbegin their payments one year from now. Showing all your work, determine

i. the value(s) of i — if there are such values — for which perpetuities ##1,2 areequal in value;

ii. the value(s) of i — if there are such values — for which the difference ofthe values of perpetuities ##1,2 is as large as possible in magnitude (absolutevalue);

iii. the value(s) of i — if there are such values — for which perpetuities ##1,3 areequal in value.

Solution:

(a)

limn→∞

(Ia)n = limn→∞

(an − nvn

i

)

= limn→∞

1+i

i · (1 − vn) − n(1+i)n

i


=1i

limn→∞

(1 + i

i· (1 − vn) − n

(1 + i)n

)

=1i

limn→∞

(1 + i

i· (1 − vn)

)− 1

ilimn→∞

n(1 + i)n

=1i

(1 + i

i· 1

)− 1

i ln(1 + i)limn→∞

1(1 + i)n

by l’Hospital’s Rule

=1 + i

i2 − 0 =1 + i

i2 .

(b) Denote the values of the three annuities respectively by P1(i), P2(i), P3(i).

P1(i) = X · a∞ =Xi

P2(i) = Y · (Ia)∞ =(1 + i)Y

i2

P3(i) = vZ∞∑

n=0

((1 + j)v)n

=vZ

1 − (1 + j)v=

Zi − j

.

i.

P1(i) = P2(i)

⇔ Xi

=(1 + i)Y

i2

⇒{

i = YX−Y if X > Y

contradiction if X ≤ Y .

ii. The magnitude of the difference will be maximized when P1(i)−P2(i) is eithermaximized or minimized.

ddi

(P1(i) − P2(i)) = −Xi2 +

(2i3 +

1i3

)

=2Y − (X − Y)i

i3 .

If X = Y , the function has no critical points, since we assumed that Y > 0;since the functions are differentiable throughout their domain i > 0, a maxi-mum point must be a critical point — so there is none when X = Y . If X , Y ,the value

i =2Y

X − Y(5)


is critical; if X > Y , this gives the interest rate where the difference is maxi-mized; if X < Y this point is outside of the values we are permitting for i; thusthe only candidate for an extremum is where

X < Y and i =2Y

X − Y.

At this value of i the second derivative is negative, so the function indeed hasa local extremum there; the First Derivative Test could also be used.

iii. P1(i) = P3(i) ⇔ Xi

=Z

i − j⇔ i =

X jX − Z

For equality it is necessary that

X > Z and that i =X j

X − Z.

4. A loan is being repaid by 60 payments spaced 4 months apart. The first 15 instalmentsare 300 each; the next 24 are 400 each; and the last 21 are 600 each. If interest is earnedat a nominal annual rate of 6%, compounded every 4 months,

(a) determine the full amount of the loan, as of 4 months before the first payment;

(b) determine the amount of the loan still outstanding immediately after the 30th pay-ment; and

(c) divide the 33rd payment into principal and interest.

(d) Construct a table with the lines for payments ##14, 15, 16, 39, 40, 59, 60 andcolumn headings as follows: Payment Number, Payment Amount, Interest Com-ponent, Principal Component, Outstanding Loan Balance (after reduction by Prin-cipal component of this Payment).

Solution:

(a) By the Prospective Method, applied 4 months before the first payment, the amountof the loan is

300 · a15 2% + 400 ·15

∣∣∣a24 2% + 600 ·39

∣∣∣a21 2%

= 600 · a60 2% − 200 · a39 2% − 100 · a15 2%

=600

(1 − (1.02)−60

)− 200

(1 − (1.02)−39

)− 100

(1 − (1.02)−15

)

0.02= 14, 191.09 .

(b) By the Prospective Method, applied just after the 30th payment, the amount of theloan still outstanding is

400 · a9 2% + 600 ·9∣∣∣a21 2%


= 600 · a30 2% − 200 · a9 2%

=600

(1 − (1.02)−30

)− 200

(1 − (1.02)−9

)

0.02= 11, 805.43 .

(c) By the Prospective Method, applied just after the 32nd payment, the amount of theloan still outstanding is

400 · a7 2% + 600 ·7∣∣∣a21 2%

= 600 · a28 2% − 200 · a7 2%

=600

(1 − (1.02)−28

)− 200

(1 − (1.02)−7

)

0.02= 11, 474.37 .

The interest earned by this outstanding balance by the next payment date is

0.02(11474.37) = 229.49 ;

so the remainder of the payment of 400, i.e., 170.51, is the amount by which the33rd payment further reduces the principal outstanding on the loan.

(d) i. By the Retrospective Method, the outstanding principal after the 13th paymentis

14, 191.09(1.02)13 − 300 · s13 2% = 13, 953.58866

so the interest component of the 14th payment is

(0.02)(13953.58866) = 279.0717732,

the component of principal reduction is 300 − 279.0717732 = 20.93, and theoutstanding loan is 13, 953.58866 − 20.93 = 13, 932.66. The interest com-ponent of the 15th payment is (0.02)(13, 932.66) = 278.65, the component ofreduction of principal is 300−278.65 = 21.35, and the outstanding principal is13,911.31. The interest component of the 16th payment is (0.02)(13, 911.31) =

278.23, the principal component is 400−278.23 = 121.77, and the outstandingprincipal is 13, 911.31 − 121.77 = 13, 798.54.

ii. By the Prospective Method, the outstanding principal after the 38th paymentis 400(1.02)−1 + 600(1.02)−1a21 2% = 10, 398.75048; alternatively we couldhave used the Retrospective Method and evaluated

14, 191.09(1.02)38 − 300 · (1.02)23s15 2% − 400 · s23 2% = 10, 398.75496 .


Payment Payment Interest Principal OutstandingNumber Amount Component Component Loan Balance

0 14,191.09. . . . . . . . . . . . . . .14 300.00 279.07 20.93 13,932.6615 300.00 278.65 21.35 13,911.3116 400.00 278.23 121.77 13,789.54. . . . . . . . . . . . . . .39 400.00 207.98 192.02 10,206.7340 600.00 204.13 395.87 9,810.86. . . . . . . . . . . . . . .59 600 23.30 576.70 588.2460 600 11.76 588.24 0.00

Table 2: Problem 4 of Assignment 5

The interest component of the 39th payment is (0.02)(10, 398.75) = 207.98,the component of principal reduction is 400 − 207.98 = 192.02, and the out-standing loan is 10, 398.75 − 192.02 = 10, 206.73 The interest component ofthe 40th payment is (0.02)(10, 206.73) = 204.13, the component of reduc-tion of principal is 600 − 204.13 = 395.87, and the outstanding principal is10, 206.73 − 395.87 = 9, 810.86.

iii. By the Prospective Method, the outstanding principal after the 58th payment

is6001.02

+600

(1.02)2 = 1, 164.94. The interest earned by the time of the next

payment is then 0.02(1, 164.94) = 23.30, so the component of that paymentthat reduces principal will be 600 − 23.30 = 576.70, and the principal willreduce to 1, 164.94 − 576.70 = 588.24 (which we could also have computed

as just600

(1.02) = 588.24. The interest earned by this outstanding loan in one

period is then 2% of 588.24, i.e., 11.76, which is the interest component ofthe last payment of 600; the principal component of the payment is then 600−11.76 = 588.24, and after the payment has been made the balance outstandingwill be 0.

5. A loan of 300,000 has been negotiated at an effective annual interest rate of 13%. Thelender wishes to receive only the full interest payment each year, with the principal tobe repaid at the end of 10 years. The borrower, however, knows that he will be unableto repay the loan unless he saves progressively over the 10-year period; he requiresthe discipline of a regular, even payment, even if the lender does not wish to receive


it. The borrower arranges to save for the ultimate repayment of the loan by paymentsinto two sinking funds: the first earns 16% effective annually, and the second earnsonly 11% effective annually. He is able to arrange for the 16% sinking fund to receiveannual payments which, at maturity, will be worth one-third of the loan; the 11% sinkingfund will be worth two-thirds. The payments into the sinking funds will be constant.Determine the total annual payment: interest to the lender, and the level contributions tothe two sinking funds.

Solution: The annual interest payment is 13% × 300, 000 = 39, 000. The 16% sink-

ing fund has a target amount of 100,000, so the annual contributions will be100, 000s10 16%

=

4, 690.108306; the 11% fund requires annual contributions of200, 000s10 11%

= 11, 960.28542.

Thus the total annual payment is 55, 650.39.


4 About the following references• References to these sources are often given in the notes for completeness. Students are

not expected to look up sources, but may wish to do so out of curiosity.

• The entries in this list may not be in alphabetical order. As the notes are constructed, newentries will be added at the end, so as not to upset the earlier numbering of references.

5 References[1] The Canadian Institute of Actuaries English-French lexicon,

http://www.actuaries.ca/members/lexicon/index.html

[2] J. W. Daniel, L. J. Federer Vaaler, Mathematical Interest Theory, Pearson/Prentice Hall(2007), ISBN 0-13-147285-2.

[3] L. J. Federer Vaaler, J. W. Daniel Mathematical Interest Theory, Mathematical Associ-ation of America (2009), ISBN-13 978-0-88385-754-0.

[4] H. S. Hall and S. R. Knight, Higher Algebra, Fourth Edition, MacMillan & Co. (London,1891).

[5] S. G. Kellison, The Theory of Interest, Third Edition. McGraw Hill/Irwin, Boston, etc.(2009). ISBN-10 0-07-338244-2, ISBN-13 987-0-07-338244-9.

[6] S. G. Kellison, The Theory of Interest, Second Edition. Irwin/McGraw Hill, Inc., Boston,etc. (1991). ISBN 0-256-04051-1.

[7] S. G. Kellison, The Theory of Interest. Richard D. Irwin, Inc., Homewood, Ill. (1970).ISBN ???-083841.

[8] McGill Undergraduate Programs Calendar 2009/2010. Also accessible athttp://coursecalendar.mcgill.ca/ug200910/wwhelp/wwhimpl/js/html/wwhelp.htm

[9] R. Muksian, Mathematics of Interest Rates, Insurance, Social Security, and Pensions.Pearson Education, Inc., Upper Saddle River, NJ. (2003). ISBN 0-13-009425-0.

[10] M. M. Parmenter, Theory of Interest and Life Contingencies, with Pension Applications.A Problem-Solving Approach, 3rd Edition. ACTEX Publications, Winsted CT, (1999).ISBN 1-56698-333-9.

[11] J. Stewart, Single Variable Calculus (Early Transcendentals), 6th Edition. Brooks/Cole(2008). ISBN-13 978-0-495-01169-9.


6 Supplementary Lecture Notes

6.1 These notesThese notes will contain some of the material that I propose to discuss in the lectures, and alsosome material that will not make it to the lectures. I will be following the textbook very closely,sometimes explaining statements that I find require some elaboration. In the first chapter ofthe book my notes will be detailed; I do not know yet whether I will be able to continuethis detailed a set of notes for subsequent chapters. Much of the class time will be spent indiscussing problems from the textbook, for which I will be including sketches of solutionswherever possible.

6.2 Supplementary Notes for the Lecture of January 04th, 2010[12] Distribution Date: Monday, January 04th, 2010

(subject to revision)

Textbook Chapter 1. The measurement of interest.

6.2.1 §1.1 INTRODUCTION

While some of the concepts in this course may be applied in non-financial contexts, the lan-guage of the course involves growth of amounts of money under the passage of time. Mostof our discussions will concern one or two specific ways in which amounts of money are, inpractice, assumed to grow; but, initially, we will define very general concepts.

Definition 6.1 1. Principal is the initial amount of money borrowed, lent, or invested.

2. Interest is the compensation paid by a borrower of capital to the lender for the use of thecapital.

3. Interest is said to be earned by the lender, or to accrue to the lender.

4. A distinction may be made between when and how interest is earned, and when it is paidor credited to the lender. When the word interest is used, without further elaboration,payment is normally at the end of each time period. When interest is paid at the beginningof a time period it is usually called discount; however, the compensation itself may stillbe called interest.

6.2.2 §1.2 THE ACCUMULATION AND AMOUNT FUNCTIONS

Definition 6.2 The magnitude after growth of a given, specific amount of principal — as afunction of time, t — is often denoted by the amount function, denoted by A(t).


We will always assume that the rate of growth does not depend on the amount of money thatis invested.11 This scalability assumption permits us to “normalize” many of our discussions,by speaking about the growth of a fund of 1, and then just scaling the resulting figures bymultiplying by the correct initial or terminal value of the fund. We will take this as our first

Axiom 6.1 If A1(t), A2(t) are two amount functions under a prescribed “interest environment”,then the ratio A1(t)

A2(t) is independent of time.

Under this postulate it is often convenient to study amounts that are “normalized” to have value1 at some convenient time. We will sometimes design our notation around this postulate, bydefining pairs of functions with similar names, one for the actual size of the fund, the other forthe “normalized” account. The first example of this is the normalized amount function, whichwe call the “accumulation” function.

Definition 6.3 Corresponding to a specific amount function, which has value A(0) at timet = 0, we define a normalized amount function or accumulation function a by A(t) = A(0) · a(t)or

a(t) =A(t)A(0)

. (6)

A simple consequence of this equation (6) is

Theorem 6.1 a(0) = 1.

(Most mathematicians don’t like to use the word “theorem” for results as trivial as this; wewould be more likely to call it a “proposition” or use some other term that reserves the word“theorem” for more serious results.)

11This is an assumption that is not completely realistic, since, in the real financial world, a person who has, say$1,000,000 to invest can often obtain a higher interest rate on his investment than the person who has only $100.But this model is used throughout the textbook, and we will follow it in the course.


6.3 Supplementary Notes for the Lecture of January 06th, 2010Distribution Date: Wednesday, January 06th, 2010

(revised on 07 January)

6.3.1 §1.2 THE ACCUMULATION AND AMOUNT FUNCTIONS (conclusion)

In practice certain types of amount functions would be absurd: for example, we usuallywouldn’t normally want to permit funds to get smaller with increasing time, although thereare situations where such shrinkage could be given some sense. But, for the present, we willfollow the textbook in assuming

Axiom 6.2 The function a is non-decreasing; i.e.,

t1 < t2 ⇒ a(t1) ≤ a(t2) .

(The textbook uses the word increasing, but many mathematicians restrict the use of that wordto exclude the possibility that there can be two times t1, t2 such that t1 < t2 but t1 = t2. HereI want to permit the amount to sometimes stay unchanged; I am assuming only that it cannotget smaller as time increases.)

The textbook observes that, in certain situations, one wishes to assume also that the functiona(t) is continuous.

Definition 6.4 1. The nth period of time is defined to be the period of time between t =

n − 1 and t = n. More precisely, the period normally will consist of the time intervaln − 1 < t ≤ n.

2. Where n is a non-negative integer, the interest earned by the amount A(t) in the nth periodof time is denoted by In, and defined by

In = A(n) − A(n − 1) . (7)

We may relate this notation to a “standard” convention in the “finite difference calculus”, whichwould define the increment in A(n) by

4A(n) = A(n + 1) − A(n) .

Thus In = 4A(n − 1).12

Example 6.2 The textbook [5, p. 3] gives four graphical examples of amount functions.12Note that the name of the function is 2 characters long: 4A. In situations like this, where the reader might

be at risk of not knowing where the name of the function begins or ends, one might use parentheses, and write(4A)(n − 1).


1. The first has a straight line graph; the graph is sloping upward because the derivative ofthe function is positive, in order that the function should be increasing at a constant rate.In this case the fund is earning a fixed amount which is proportional to the time that haselapsed, even if the time is not an integer number of time-units; this is the situation wewill later call simple interest.

2. The second example is an exponential curve, where A(t) is a function of the form

A(t) = k · e`t , (8)

` being a real number. If we set t = 0 in (8), we see that k = A(0). Since the functionmust be non-decreasing, we know by the calculus that ` ≥ 0. This is the case that we willbe calling compound interest.

3. The third case has a graph which is a horizontal line. This is simply the special case ofthe first case where ` = 0, and A(t) is constant; the amount neither grows nor shrinks.

4. The fourth case is a “step” function, whose graph is horizontal line segments of constantlength, each a fixed distance above the preceding one. This function is “piecewise con-tinuous”, having points of discontinuity at the integer times. This is a realistic model, forexample, where an amount earns interest only if the amount remains in the account fora full time period. We can describe this situation by saying that interest is accrued onlyfor completed periods, with no credit for fractional periods, where the interest earned iscredited at the end of the period.To be more precise, the function is “continuous from the right” everywhere, but it is“discontinuous from the left” at integer points; this is because the limit from the left isnot equal to the function value there — i.e., because the value you would infer from thesize of the amount preceding an integer time is not equal to the actual size of the accountat integer time.

Summation of Geometric Progressions and Series (These results will be applied on page1007 below in the solution of [6, Exercise 3, p. 30].) The reader is reminded of the easilyderived formula [5, Appendix C, A.2, p. 592], [6, Appendix III, A.2, p. 394]

n−1∑

k=0

ark = a(rn − 1r − 1

)= a

(1 − rn

1 − r

)(r , 1) . (9)

The textbook overlooks the essential restriction that r , 1 for the use of this formula; whenr = 1 the sum is

n−1∑

k=0

a1k = a · n (r , 1) . (10)


We shall be using these formulæ repeatedly. We shall also be interested in the behavior of (9)in the limit as n→ ∞: ∞∑

k=0

ark = a(

11 − r

)(|r| < 1) . (11)

(When |r| ≥ 1, the limit does not exist.)

[5, Exercise 1, p. 42], [6, Exercise 1, p. 29] “Consider the amount function A(t) = t2 + 2t + 3.

1. “Find the corresponding accumulation function a(t).”2. (modified) Show that a(0) = 1, that a is a non-decreasing function of time, and that

a is continuous.3. “Find In.”

Solution:

1.

a(t) =A(t)A(0)

=t2 + 2t + 3

02 + 2(0) + 3=

13

t2 +23

t + 1 .

2. (a) a(0) = 13 · 02 + 2

3 · 0 + 1 = 1(b) If the author intends that we consider this a function of a real variable that may

take any values on the non-negative real axis, then one way to prove the functionis non-decreasing is to show that the derivative is not negative. (This can be usedonly for a function that is differentiable, but the present function is a polynomial,and polynomials have that property.)

a′(t) =13· 2 · t +

23· 1 =

23

(t + 1) .

We see that a(t) is a non-decreasing function provided t ≥ −1. This could alsohave been approached from “first principles”. Suppose that t1 < t2. Then

a(t2) − a(t1) =13

(t22 − t2

1

)+

23

(t2 − t1)

= (t2 − t1)(t2

3+

t1

3+

23

)

and both factors are positive for non-negative t2 and t1 and t1 < t2.(c) Polynomial functions are continuous everywhere.

3. In = A(n) − A(n − 1) =(n2 + 2n + 3

)−

((n − 1)2 + 2(n − 1) + 3

)= 2n + 1.

[5, Exercise 2, p. 42], [6, Exercise 2, p. 30] 1. “Prove that A(n) − A(0) = I1 + I2 + . . . + In.”2. “Verify verbally the result obtained” above.

Solution:


1. Summing the equations A(m) − A(m − 1) = Im for 1 ≤ m ≤ n, we can obtain

A(n) − A(0) =

n∑

m=1

Im .

That, however, is based on the following kind of cancellations:

(A(1) − A(0)) + (A(2) − A(1)) + . . . + (A( j) − A( j − 1)) + . . . + (A(n) − A(n − 1))= (A(1) − A(0)) + (A(2) − A(1)) + . . . + (A( j) − A( j − 1)) + . . . + (A(n) − A(n − 1))= (A(1) − A(0)) + (A(2) − A(1)) + . . . + (A( j) − A( j − 1)) + . . . + (A(n) − A(n − 1))= ...

= −A(0) + . . . + (A(n) − A(n − 1))= −A(0) + A(n) .

This kind of “proof” seems to involve hand-waving. It can be made rigorous, withoutthe use of . . ., but it takes a lot of care, and I wouldn’t expect this type of proof inthis course. Here is an example:

n∑

j=1

(A( j) − A( j − 1))

=

n∑

j=1

A( j) −n∑

j=1

A( j − 1)

=

n∑

j=1

A( j) −n−1∑

k=0

A(k)

changing the variable by defining k = j − 1 and changing limits accordingly

=

n−1∑

j=1

A( j) +

n∑

j=n

A( j) −0∑

k=0

A(k) −n−1∑

k=1

A(k)

breaking each sum up into two parts

=

n−1∑

j=1

A( j) + A(n) − A(0) −n−1∑

k=1

A(k) .

Now observe that the two sums are identical except for the name of the variable; so

they have the same value. (This is analogous to

b∫

a

f (x) dx =

b∫

a

f (u) du.) This proof

is pleasing, in that it has no use of . . .; but it’s a lot of work for that purpose, and Iwouldn’t normally expect it of students in this course.


2. Verbally, the (amount of) interest earned over the concatenation of n periods is thesum of the interest earned in each of the periods separately.

The following exercises were not discussed at the lecture, and will not be discussed at thenext lectures.

[6, Exercise 3, p. 30] “Find the amount of interest earned between time t and time n, wheret < n, if

1. Ir = r2. Ir = 2r.”

Solution: In the preceding problem we showed that the amount earned between time 0and time n was the sum of the values of I. The interest earned between time t and timen will be the total interest from time 0 to time n diminished by the interest earned fromtime 0 to time t, so it must be

n∑

m=1

Im −t∑

m=1

Im =

n∑

m=t+1

Im

When Im = m, this is13

n∑

m=1

m −t∑

m=1

m =n(n + 1)

2− t(t + 1)

2=

(n − t)(n + t + 1)2

.

When Im = 2m, the sum is

n∑

m=t+1

2m = 2t+1(2n−t − 12 − 1

)= 2n+1 − 2t+1 .

[5, Exercise 3, p. 42] “For the 5000 investment given in Example 1.1 [5, p. 4], find the amountof interest earned during the second year of investment, i.e., between times t = 3 andt = 4.”Solution: The data given in this example are first a table for a fund of 10,000 invested for4 years:

t A(t)0 10, 000.001 10, 600.002 11, 130.003 11, 575.204 12, 153.96

13using the familiar formula for the sum of the first natural numbers


In the example 5000 is invested at time t = 2: it appears to be this 5000 to which thepresent problem refers. For the full fund the year beginning at t = 3 starts a balanceof 11,575.20 which, at the very end of the period, increases to 12,153.96. The interestearned is A(4) − A(3) = 12, 153.96 − 11, 575.20 = 578.76. For an amount of 5000 for

that year we can scale the interest payment down to5000

11130.00× 578.76 = 260.

[5, Exercise 3, p. 42] “For the 5,000 investment given in [5, Example 1.1, p. 4], find theamount of interest earned during the second year of investment, i.e., between times t = 3and t = 4.”Solution: The hypothesis that the investment of 5,000 at time t = 2 is “under the sameinterest environment” implicitly requires that the interest be scalable — that the rate ofgrowth is proportional to the rate of growth in the original fund during the times 2 ≤ t ≤ 3(in the example), and 3 ≤ t ≤ 4 in this exercise. In the time interval 2 ≤ t ≤ 3 the 5,000grows to become

11, 575.2011, 130.00

× 5, 000 = 5, 200

at time t = 3; it is this amount whose further growth interests us. In the interval 3 ≤ t ≤ 4this amount grows to

12, 153.9611, 575.20

× 5, 200 = 5, 460

at time t = 4. Consequently the interest earned in the interval 3 ≤ t ≤ 4 is 5, 460−5, 200 =

260.

[5, Exercise 4, p. 42], [6, Exercise 4, p. 30] “It is known that a(t) is of the form ct2 + b. If$100 invested at time 0 accumulates to $172 at time 3, find the accumulated value at time10 of $100 invested at time 5.”Solution: Let’s begin by normalizing the function: a(0) = 1⇔ c·02+b = 1, so b = 1. Thegiven growth condition implies that 100(c32 + 1) = 172, so c = 72

900 = 0.08. Consequently100 invested at time 5 grows to 100((0.08) · 52 + 1) = 300.


6.4 Supplementary Notes for the Lecture of January 08th, 2010Distribution Date: Friday, January 08th, 2010


Change of Lecture Room

Effective beginning on Monday, January 11th, 2010, this class hasbeen moved to LEA 26. (There will be a change of one of my officehours because of this, but this will be announced later.)

Summary of some topics from the previous lectures1. definition of Amount function A(t)

2. definition of Accumulation function a(t): normalized amount function, such that a(0) = 1.

3. nth period = time from t = n − 1 to t = n, i.e., the period ending at time t = n (not theusual mathematical convention)

4. Interest earned in nth period: In = A(n) − A(n − 1)

6.4.1 §1.3 THE EFFECTIVE RATE OF INTEREST

When we speak of a rate of interest, we mean the amount of interest per unit investment. Asthis rate is normalized, we follow the convention we began with the pair A and a, by using alower case letter to represent it.

Definition 6.5 The effective rate of interest during the nth period, in, is the ratio of the amountof interest earned during, and paid at the end of the period to the amount of principal investedat the beginning of the period. When the effective rate is constant over all unit time intervals,we may suppress the subscript, and simply write i.

We may use the term effective for periods of lengths other than one unit; but the mostcommon use will be for terms of one year, where we will speak of the effective annual rate.

Note that, when we use the unqualified word interest alone, the listener may not know whetherwe are referring to the amount of interest, the rate of interest, or the general concept of paymentfor the use of capital. There may also be doubts about the accrual period.

Theorem 6.3 In symbols,

in =In

A(n − 1)=

A(n) − A(n − 1)A(n − 1)

=a(n) − a(n − 1)

a(n − 1)(12)

for integer n ≥ 1. And, when the rate is constant, i, recalling that a(0) = 1, we have

a(n) = (1 + i) · a(n − 1) = (1 + i)2 · a(n − 2) = . . . = (1 + i)n · a(0) = (1 + i)n . (13)


In all of this discussion the interest rate is associated with one time interval; this will be con-trasted later with rates — called “nominal” — that are stated for one time interval, but need tobe applied to another.

[5, Exercise 5, p. 42], [6, Exercise 5, p. 30] “Assume that A(t) = 100 + 5t.” Find i5 and i10.Solution:

i5 =A(5) − A(4)

A(4)=

5 · 1100 + 5 · 4 =

124

.

i10 =A(10) − A(9)

A(9)=

5 · 1100 + 5 · 9 =

129

.

[5, Exercise 6, p. 42], [6, Exercise 6, p. 30] “Assume A(t) = 100(1.1)t.” Find i5 and i10.Solution:

i5 =A(5) − A(4)

A(4)=

100(1.1)4(0.1)100(1.1)4 = 0.1 .

i10 =A(10) − A(9)

A(9)=

100(1.1)9(0.1)100(1.1)9 = 0.1 .

[5, Exercise 6, p. 42], [6, Exercise 7, p. 30] “Show that A(n) = (1 + in) A(n − 1), where n is apositive integer.”Solution: (1 + in)A(n − 1) =

(1 +

A(n)−A(n−1)A(n−1)

)A(n − 1) =

A(n)A(n−1) · A(n − 1) = A(n). �

An equation like this is an ideal opportunity to formulate a “verbal” proof. Here is one:

Since in is defined to be the ratio of interest to the initial amount (at time t =

n − 1) in the account, 1 + in will be the ratio of the terminal amount A(n) to theinitial amount A(n − 1).

[5, Exercise 6, p. 42], [6, Exercise 8, p. 30] “If A(4) = 1000 and in = 0.01n, where n is apositive integer, find A(7).”Solution: I(n) = in · A(n − 1) = (0.01)nA(n − 1). It follows that

A(n) = In + A(n − 1) = (0.01(n) + 1)A(n − 1)A(7) = (0.01(7) + 1)A(6)

= (0.01(7) + 1)(0.01(6) + 1)A(5)= (0.01(7) + 1)(0.01(6) + 1)(0.01(5) + 1)A(4) = (1.07)(1.06)(1.05)1000= 1190.91 .


6.4.2 §1.4 SIMPLE INTEREST

We saw in the preceding discussion that a(0) = 1 and a(1) = 1 + i1. Under simple interest weassume the linear function we saw in the first graphical example above: the amount of interestearned in a time interval of length t and starting at a fixed time, like t = 0, is assumed to beproportional to t; we obtain the formula

a(t) = 1 + i1t

over the 1st time interval, or any longer interval where the so-called simple interest rate remainsequal to i1. Suppose that the rate of interest is i, and that this rate is to remain in effect over aninterval of n or more successive time periods without adjusting the base principal from which itis computed. We can determine the rate of interest during the nth period under this assumption;it is

in =a(n) − a(n − 1)

a(n)=

[1 + i1n] − [1 + i1(n − 1)]1 + i1(n − 1)

=i1

1 + i1(n − 1)(14)

where we see that, even though the amount of interest earned in any period of time is a constantmultiple of the length of the period, the rate of interest per period is not constant — it isdecreasing from each period to the next.14 Thus a constant rate of simple interest implies thatthe effective rates of interest for the successive time intervals form a decreasing sequence.What is constant for simple interest is the absolute amount of interest earned in each timeinterval; under compound interest which we shall meet in the next section, it is the relativeamount of interest that is constant — that is, the ratio of interest earned to principal.

Note that, when we speak of simple interest, we must specify not only the rate, but alsogive enough information so that the reader will know when the principal amount is adjusted toincorporate the accrued interest, and on what base amount the calculations are based.

[5, Exercise 9, p. 42], [6, Exercise 9, p. 30] 1. “At what rate of simple interest will $500accumulate to $615 in 2 1

2 years?2. “In how many years will $500 accumulate to $630 at 7.8% simple interest?”

Solution:

1. Under simple interest at a rate i, 615 = 500a(2.5) = 500(1 + i2.5); solving for i, weobtain i =

615500−1

2.5 = 0.092 = 9.2%.

2. Under a rate of 7.8% simple interest for a period of t years, 630 = 500a(t) = 500(1 +

0.078t); solving for t, we obtain t =630500−10.078 = 10

3 years.

[6, Exercise 10, p. 30] “If ik is the rate of simple interest for period k, where k = 1, 2, . . . , n,show that a(n) − a(0) = i1 + i2 + . . . + in.”

14But the derivative of the amount function will be constant.


Solution: Note that there is an implicit assumption that the earned interest is not incorpo-rated into the principle during the period 0 ≤ t ≤ n. What is denoted by it in this problemis not the effective rate of interest for the tth period, shown in equation (1.6). Rather, it isthe total amount earned during the tth period by the amount that began as 1 at time t = 0.Under simple interest an initial investment of a(0) = 1 grows by the prescribed amountsin each of the following years: a(n) = a(0)+i1+i2+. . .+in, so a(n)−a(0) = i1+i2+. . .+in .

[5, Exercise 10, p. 42], [6, Exercise 11, p. 30] “At a certain rate of simple interest $1000 willaccumulate to $1100 after a certain period of time. Find the accumulated value of $500at a rate of simple interest three fourths as great over twice as long a period of time.”Solution: It is known that, at the rate of interest i, and for a prescribed length of timet, 1000(1 + it) = 1110, so it = 0.11. Then 500

(1 + 3i

4 · 2t)

= 500(1 + 3

2 it)

= 500(1 +

0.165) = 582.50.

[5, Exercise 11, p. 43], [6, Exercise 12, p. 30] “Simple interest of i = 4% is being credited toa fund. In which period is this equivalent to an effective rate of 2 1

2%?”Solution: We have to solve for n the equation

0.025 = in =0.04

1 + (0.04)(n − 1)

and find that n = 16.

[5, Exercise 12, p. 43] “A deposit of 1000 is invested at simple interest at time t = 0. The rateof simple interest during year t is equal to 0.01t for t = 1, 2, 3, 4, and 5. Find the totalaccumulated value of this investment at time t = 5.Solution: There is no compounding. In year t the investment earns interest in the amountof (0.01t) ∗ 1000 = 10t. Thus, by time t = 5, the original investment of 1000 has earnedinterest in the amount of 1000(0.01)(1 + 2 + 3 + 4 + 5) = 150. The accumulated value ofthe investment is, therefore, 1000+150 = 1150.. (Note that, if the interest had been com-pounded, the accumulated value would have been 1000(1.01)(1.02)(1.03)(1.04)(1.05) =

1158.73.)

6.4.3 §1.5 COMPOUND INTEREST

Under compound interest, the earned interest is incorporated into the principal after each ac-crual period, following which the interest in the next period is based on the updated principal.Thus the earned interest is reinvested together with the original principal. This is what we ob-served in equation (13) above, where the interest rate remained constant over successive timeperiods.

The growth of the principal can be analyzed in various ways. Thus, for example, the valuea(2) = (1 + i)2a(0) = (1 + i)2 may be interpreted


• as (1 + i) · a(1) = (1 + i) · (1 + i) as we did in the discussion preceding (13)

• as 1+2i+i2, where the principal of 1 is augmented by 2i, representing the “simple” interestearned by the principal over two successive time periods, augmented by the interest atrate i earned by the interest i from the first period and credited at the end of that period.


6.5 Supplementary Notes for the Lecture of January 11th, 2010Distribution Date: Monday, January 11th, 2010


6.5.1 §1.5 COMPOUND INTEREST (conclusion)

Under compound interest at an effective interest rate of i we assume that an amount of 1 growsto (1 + i)t during a time interval of length t, where t is not necessarily a non-negative integer.In most of the sequel it is compound interest that will be assumed to be acting. Where weassume simple interest, it will often be for fractions of a time period. In some ways thisassumption is obsolete now — a relic of the days when exact computing was difficult becauseof the unavailability of reliable and inexpensive calculators. However, some of the practices ofusing simple interest have become entrenched in the financial system, and so we are obligedto continue using them.

Not all of the following problems were considered in the lecture.

[5, Exercise 13, p. 43], [6, Exercise 14, p. 30] “It is known that $600 invested for two yearswill earn $264 in interest. Find the accumulated value of $2000 invested at the same rateof compound interest for three years.”Solution: What is known is that, at the given interest rate i, 600(1 + i)2 = 600 + 264,so (1 + i)2 = 1.44 and (1 + i)1 = 1.2 (i.e. i = 20%). At the same rate, 2000(1 + i)3 =

2000(1.2)3 = 2000 × 1.728 = 3456.

[5, Exercise 14, p. 43], [6, Exercise 15, p. 31] “Show that the ratio of the accumulated valueof 1 invested at rate i for n periods, to the accumulated value of 1 invested at rate j for nperiods, i > j, is equal to the accumulated value of 1 invested for n periods at rate r. Findan expression for r as a function of i and j.”Solution: The language of this problem can be improved It would be better to include theindefinite article “a” before the words “rate r”, since r is a new symbol that has not beendefined earlier in the problem. Now the objective is just to express the given ratio as annth power, and that’s simply an exercise in algebra.

1(1 + i)n

1(1 + j)n =

(1 +

i − j1 + j

)n

.

Thus r =i − j1 + j

.

[5, Exercise 15, p. 43], [6, Exercise 16, p. 31] “At a certain rate of compound interest, 1 willincrease to 2 in a years, 2 will increase to 3 in b years, and 3 will increase to 14 in c years.If 6 will increase to 10 in n years, express n as a function of a, b, and c.”


Solution: If the common rate is i, the hypotheses are that

1(1 + i)a = 22(1 + i)b = 33(1 + i)c = 156(1 + i)n = 10 .

Taking logarithms and solving, we obtain

a =ln 2

ln(1 + i)(15)

b =ln 3 − ln 2ln(1 + i)

(16)

c =ln 5

ln(1 + i)(17)

n =ln 5 − ln 3ln(1 + i)

= c − a − b (18)

Note that this answer is “overdetermined”, in the sense that a, b and c are not independent:by equations (15), (16), (17), a : b : c :: ln 2 : ln 3

2 : ln 5, that is, a, b, and c are related byproportionality equations of the form

aln 2

=b

ln 32

=c

ln 5.

Put another way, we didn’t need all 3 of these equations to solve the problem — any 2 ofthem would have been sufficient!

[5, Exercise 16 p. 43], [6, Exercise 17 p. 31] “An amount of money is invested for one yearat a rate of interest of 3% per quarter. Let D(i) be the difference between the amount ofinterest earned on a compound interest basis, and on a simple interest basis for quarter k,where k = 1, 2, 3, 4. Find the ratio of D(4) to D(3).”Solution: Let the amount of money invested be x. The interest rate is 3% per quar-ter. (The given information appears, at first, to be ambiguous; for, ‘difference between’doesn’t tell us which of the two numbers is larger. However, we can show that, for properfractions of an interest period, growth under compound interest is less than under simpleinterest; and that the opposite occurs outside of one interest period. In the present prob-lem the interest period is a quarter-year, so all the periods we are considering are at leastas long as one interest period, and the compound interest earned is always at least as large


as the possibility under simple interest; in any case, the worst that would have happenedin our solution if we hadn’t known this would have been a sign error.) We have

D(1) = x((1.03)0((1.03) − 1) − 0.03

)

D(2) = x((1.03)1((1.03) − 1) − 0.03

)

D(3) = x((1.03)2((1.03) − 1) − 0.03

)

D(4) = x((1.03)3((1.03) − 1) − 0.03

)

from which it follows that

D(4)D(3)

=(1.03)3((1.03) − 1) − 0.03(1.03)2((1.03) − 1) − 0.03

=1.033 − 11.032 − 1

= 1.522610837 .

[5, Exercise 17, p. 43] “The two sets of grandparents for a newborn baby wish to invest enoughmoney immediately to pay 10,000 per year for four years toward college costs, startingat age 18. Grandparents A agree to fund the first two payments, while Grandparents Bagree to fund the last two payments. If the effective rate of interest is 6% per annum, findthe difference between the contributions of Grandparents A and B.”Solution: The value at time t = 0 of the first two payments is

10000((1.06)−18 + (1.06)−19

),

while the value at time t = 0 of the payments sponsored by Grandparents B is

10000((1.06)−20 + (1.06)−21

).

The excess of the value of the A payments over the B payments is

10000((1.06)−18 + (1.06)−19

) (1 − (1.06)−2

)

= 10000(1.06)−21(1.06 + 1)((1.06)2 − 1

)

= 10000(1.06)−21(2.06)(0.1236) = 748.96672... = 748.97.

6.5.2 §1.6 PRESENT VALUE

Definition 6.6 1. Under an interest rate of i the sum 1 + i is called the accumulation factorwhich, when applied to an amount, yields the value of the amount at the end of 1 timeperiod.

2. When we view an amount from the past, asking what should be today’s value of anamount receivable some time in the future, we call the value today the present or currentvalue or the value discounted (back) to the present.


3. To obtain the amount which, at the beginning of a time period, would yield, at the interestrate i, an amount of 1 at the end of the period, we may divide by 1 + i or multiply by the

discount factor1

1 + i, which is often denoted by v.

The textbook denotes the reciprocal1

a(t)by a−1(t). This is an unfortunate notation, since it

appears to involve an inverse function, and that is not the author’s intention. You are advised

not to use this notation: write1

a(t)or (a(t))−1.

Definition 6.7 Analogous to the accumulation function a(t), we can call (a(t))−1 the discountfunction. It represents the amount that needs to be invested today to yield an amount of 1 atthe end of t time periods.

In the following problems there will be a gradual transition from the general formulation ofthe preceding definitions to the specific case of compound interest.

[6, Exercise 18 p. 31] “Find an expression for the discount factor during the nth period fromthe date of investment, i.e., (1 + in)−1, in terms of the amount function.”

Solution: A(n) = (1 + in) · A(n − 1)⇒ 11 + in

=A(n − 1)

A(n).

[5, Exercise 18 p. 43], [6, Exercise 19 p. 31] “The sum of the present value of 1 paid at theend of n periods and 1 paid at the end of 2n periods is 1. Find (1 + i)2n.”Solution:

(1 + i)−n + (1 + i)−2n = 1 ⇔ ((1 + i)−n)2

+ (1 + i)−n − 1 = 0

⇔((1 + i)−n +

12

)2

=54

(completing the square)

⇔ (1 + i)−n =−1 ± √5

2But the negative sign would be associated with a negative value of (1 + i)n, which isimpossible. Hence

(1 + i)−n =−1 +

√5

2

(1 + i)−2n = 1 − (1 + i)−n =3 − √5

2

(1 + i)2n =2

3 − √5=

2

3 − √5· 3 +

√5

3 +√

5

=2(3 +

√5)

9 − 5=

3 +√

52


[5, Exercise 19, p. 44], [6, Exercise 21 p. 31] “It is known that an investment of $500 willincrease to $4000 at the end of 30 years. Find the sum of the present values of threepayments of $10,000 each which will occur at the end of 20, 40, and 60 years.”Solution: At the rate i, 500(1 + i)30 = 4000, so (1 + i)30 = 8, and (1 + i)10 = 2. Hence thepresent value of payments of 10,000 to occur at the ends of 20, 40, and 60 years is

10000(v20 + v40 + v60

)= 10000

(14

+1

16+

164

)

=210000

64= 3281.25.

[6, Exercise 20, p. 31] “Show that the current value of a payment of 1 made n periods ago anda payment of 1 to be made n periods in the future is greater than 2, if i > 0.”Solution: The payment from the past is currently worth (1 + i)n, while the future paymentis currently (=presently) worth vn = (1 + i)−n. The excess of this sum over 2 is

(1 + i)n + (1 + i)−n − 2 =((1 + i)

n2 − (1 + i)−

n2)2, (19)

which, being a square, cannot be negative. Hence

(1 + i)n + (1 + i)−n ≥ 2 ,

which equality holding precisely when (1 + i)n2 = (1 + i)−

n2 , i.e., when (1 + i)n − 1 = 0,

i.e., if either i = 0 or n = 0.




6.6.1 §1.7 THE EFFECTIVE RATE OF DISCOUNT

We will try to develop, in parallel to the interpretation of amount and accumulation functionsfrom the point of view of the addition of interest, paid at the ends of time periods, a discountinterpretation under which the amount and accumulation functions are discounted by subtract-ing an amount of discount. This will lead to several symbols analogous to those in the interestmodel. This parallel development will be incomplete, and most of the time we will be workingonly with the interest model. For example, we could interpret In given in equation (7) as anamount of discount. Then we could parallel equation (12) on 1009 of these notes with

In

A(n)=

A(n) − A(n − 1)A(n)

=a(n) − a(n − 1)

a(n)(20)

Definition 6.8 The common amount of the members of equation (20)is denoted by dn, andcalled the effective rate of discount.

Note the differences between the two models:

• Under the interest model, the payment for the use of the money is made at the end of theperiod, based on the balance at the beginning of the period.

• Under the discount model, the “payment” is deducted at the beginning of the period fromthe final amount which will be present at the end of the period.

We will use the word equivalent to describe schemes of interest and/or discount that producethe same ultimate accumulated value from an initially invested amount. In the present appli-cation, which is only the first where we will meet this concept, we can, for example, equate

11 + i

= 1 − d (21)

from which we obtain various other relationships, e.g.,

Theorem 6.4

d =i

1 + i(22)

i =d

1 − d(23)

d = 1 − v (24)i − d = id (25)


You should try to explain any identity “verbally”, i.e., in words, in a way that a person lackingyour technical knowledge can find the equation plausible. I will “prove” this theorem verbally:Proof:

1. (22). Suppose an amount of 1 is loaned for 1 time interval, and the respective equivalentrates of interest and discount are i and d. Under the interest model, the cost of the loanis i, paid at the end of the period; i paid at that time is worth only (1 + i)−1i when theamount is discounted back to the beginning of the interval, and this has to be equal to d,the equivalent cost of the loan when charged at the beginning of the time period.

2. (23). Analogously, the amount of d paid at the beginning of the interval can be accumu-lated to the end of the interval by dividing by 1 − d.

3. (24). The cost now of borrowing the 1, which we have defined to be d, can be interpretedas the initial value of the loan, i.e., 1, diminished by the present value of the 1 which willbe all that remains at the end of the period: that 1, when discounted back to the present,is worth only (1 + i)−1, which we denote by v.

4. (25). The cost of borrowing the 1 may be paid in two ways: by d paid at time t = 0, or byi paid at time t = 1. The excess of the amount of interest at time t = 1 over the amount ofdiscount at time t = 0 is the cost of using d for 1 period, i.e., di.

�If a discount rate d is applied t times to a final amount of 1, the current value will be

1a(t)

= vt = (1 − d)t.

The textbook makes several observations about the use of the discount model, and about theterm discount. For these and other reasons, we shall not be devoting equal time to the discountmodel in this course. But you should be aware that most of what we do could be redevelopedfrom the discount point of view.

Definition 6.9 The accumulation function for simple discount at a discount rate d ≥ 0 is given

by a(t) =1

1 − dt, for t <

1d

.

Note that, unlike the situation for simple interest, we must restrict the length of time over whichwe propose to apply simple discount.

Definition 6.10 The accumulation function for compound discount at a discount rate d is given

by a(t) =1

(1 − d)t .

While we need not restrict t in the compound discount case, we must restrict the effectivediscount rate d, requiring that 0 ≤ d < 1.


[5, Exercise 20, p. 44], [6, Exercise 23 p. 31] 1. “Find d5 if the rate of simple interest is10%.”

2. “Find d5 if the rate of simple discount is 10%.”Solution:

1. Under simple interest at 10%, In = 0.1 × A(0) for all non-negative integers n. So

d5 =I5

A(5)=

(0.1)A(0)(1 + 5(0.1))A(0)

=0.11.5

=1

15.

2. But, if the rate of simple discount is 10%, then A(n) =A(0)

1 − 0.01n.

d5 =A(5) − A(4)

A(5)=

10.05

− 10.06

10.5

=16.

[5, Exercise 21, p. 44] “Find the effective rate of discount at which a payment of 200 imme-diately and 300 one year from today will accumulate to 600 two years from today.”Solution: Denote the effective annual rate of discount by d. The sum of the presentvalues of the two payments is 200 + 300(1 − d). This must be equal to the presentvalue of 600 received 2 years from today, i.e., to 600(1 − d)2. The equation of value is200 + 300(1−d) = 600(1−d)2, which has only one positive solution, 1−d = 0.8791529,from which we see that d = 0.120847, i.e., d = 12.08%.

In both the interest and discount points of view we denote the payment for the use during thenth period of the money in an amount A(t) by In = A(n)− A(n− 1), and (unfortunately) use thesame word “interest” here for In. We define for integers n ≥ 1

in =In

A(n − 1)=

A(n) − A(n − 1)A(n − 1)

=a(n) − a(n − 1)

a(n − 1)(26)

dn =In

A(n)=

A(n) − A(n − 1)A(n)

=a(n) − a(n − 1)

a(n), (27)

soA(n − 1) = (1 + i)−1A(n) = (1 − d)A(n) . (28)

[5, Exercise 22 p. 44], [6, Exercise 22 p. 31] “The amount of interest earned on A for oneyear is $336, while the equivalent amount of discount is $300. Find A.”Solution: Denote the equivalent rates of interest and discount by i and d respectively. Thehypotheses are that

iA = 336dA = 300


subject to the relationship between i and d, which may be expressed in various ways. Wewish to obtain A, not i or d. So let us solve the preceding equations for i and d in termsof A,

i =336A

d =300A

and substitute into one of the equations relating i and d, e.g., into i − d = id:

336A− 300

A=

336 × 300A2 ⇔ 36A = 336(300)

⇔ A = 2800

[5, Exercise 23, p. 44], [6, Exercise 2, p. 53] “Find the present value of 5000 to be paid at theend of 25 months at a rate of discount of 8% convertible quarterly:

1. Assuming compound discount throughout.2. Assuming simple discount during the final fractional period.”

Solution:

1. The effective quarterly discount rate is 8%4 = 2%. The present value of 5000 to

be paid at the end of 2512 × 4 = 8 1

3 quarterly periods from now is 5000(1 − d)253 =

500(0.845053031) = 4225.27.2. The last month is to be interpreted as 1

12 of a period during which simple interestis charged at 8%, so the discounting factor for that month would be 1 − 8

12% =

0.99333333. Further discounting back through 8 quarter-years at a rate of 84% = 2%

per quarter year yields a present value of 5000(0.988

)(0.99333333) = 4225.46 .

[5, Exercise 24 p. 44], [6, Exercise 27 p. 32] “Show that

d3

(1 − d)2 =(i − d)2

1 − v.”

Solution: I give an algebraic proof. In an algebraic proof of an identity we can, in themost uninspired proof, prove both members of the alleged equation are equal to the samequantity. Usually many possible proofs exist, but some may be more interesting thanothers, because they may be easier to explain in words, i.e., verbally. Such proofs aremore satisfying if we can start with the quantity on one side of the equal sign, and, byapplying familiar results, transform it into the quantity on the other side. For example

d3

(1 − d)2 = d(

11 − d

)2

= di2 [5, (1.14), p. 17]

=(id)2

d=

(i − d)2

1 − v[5, (1.16), (1.17), p. 18] .


[5, Exercise 25 p. 44], [6, Exercise 26 p. 32] “If i and d are equivalent rates of simple interestand simple discount over t periods, show that i − d = idt.”Solution: The discounted value (at simple interest) at time 0 of 1 at time t is 1 − dt. Atsimple interest this accumulates to (1−dt)(1+ it) = 1. Expanding and dividing by t yields

i − d = idt . �

6.6.2 §1.8 NOMINAL RATES OF INTEREST AND DISCOUNT

Definition 6.11 [6, p. 14] Two rates of interest or discount are equivalent if an amount ofprincipal invested for the same length of time under either of the rates accumulates to the samevalue.

We have already applied this concept in the preceding section, when we related equivalent ratesof interest and discount over the same time interval. In this section we will relate equivalentrates of interest and/or discount over different time intervals, where one interval is a multipleof the other.

When we speak of a nominal rate of interest, we will normally describe a payment schemethat is more or less frequent than the usual payment interval; the interest rate that we willdescribe will represent the total of all payments made over that usual interval. Summing thepayments does not take into account the interest or discount to shift all payments to one spe-cific time. We will determine relationships between nominal rates and effective rates. Thedefinitions are particularly unintuitive, so they must be memorized.

Definition 6.12 Let m be either a positive integer or the reciprocal of a positive integer.

1. A nominal rate of interest i(m) payable m times per period represents m times the amountof compound interest that will have to be paid at the end of each 1

m th of a period which isequivalent to an effective interest rate of i per period.

2. A nominal rate of discount d(m) charged m times per period represents m times the amountof compound discount that will have to be charged at the beginning of each 1

m th of aperiod which is equivalent to d, the amount charged at the beginning of the period underan effective discount rate of d per period.




6.7.1 §1.8 NOMINAL RATES OF INTEREST AND DISCOUNT (conclusion)

Definition 6.13 [6, p. 14] Two rates of interest or discount are equivalent if an amount ofprincipal invested for the same length of time under either of the rates accumulates to the samevalue.

We have already applied this concept in the preceding section, when we related equivalent ratesof interest and discount over the same time interval. In this section we will relate equivalentrates of interest and/or discount over different time intervals, where one interval is a multipleof the other.

When we speak of a nominal rate of interest, we will normally describe a payment schemethat is more or less frequent than the usual payment interval; the interest rate that we willdescribe will represent the total of all payments made over that usual interval. Summing thepayments does not take into account the interest or discount to shift all payments to one spe-cific time. We will determine relationships between nominal rates and effective rates. Thedefinitions are particularly unintuitive, so they must be memorized.

Definition 6.14 Let m be either a positive integer or the reciprocal of a positive integer.

1. A nominal rate of interest i(m) payable m times per period represents m times the amountof compound interest that will have to be paid at the end of each 1

m th of a period which isequivalent to an effective interest rate of i per period.

2. A nominal rate of discount d(m) charged m times per period represents m times the amountof compound discount that will have to be charged at the beginning of each 1

m th of aperiod which is equivalent to d, the amount charged at the beginning of the period underan effective discount rate of d per period.

Theorem 6.5 1. A nominal rate of interest of i(m) payable and compounded at the end ofevery 1

m th of a year is equivalent to15 an effective rate of interest of i(m)

m payable every mthof a year. Thus a principal of 1 accumulates at the end of a full year to

(1 + i(m)

m

)m. We

have, therefore, the relationship(1 +

i(m)

m

)m

= 1 + i , (29)

15Indeed, it has been defined to be


or, equivalently,i(m) = m

((1 + i)

1m − 1

). (30)

2. A nominal rate of discount of d(m) payable and compounded at the beginning of every 1m th

of a year is equivalent to an effective rate of discount of d(m)

m payable in advance, everymth of a year. Thus an accumulation of 1 at the end of a full year has value

(1 − d(m)

m

)mat

the beginning of the year. We have, therefore, the relationship(1 − d(m)

m

)m

= 1 − d , (31)

or, equivalently,d(m) = m

(1 − (1 − d)

1m)

= m(1 − v

1m). (32)

We can combine equations (29) and (31) into(1 +

i(m)

m

)m

= 1 + i =

(1 − d(p)

p

)−p

; (33)

when m = p, this yields the equation

i(m)

m− d(m)

m=

i(m)

m· d(m)

m(34)

of which (25) is the special case m = 1.

A warning about nominal rates. It is not true that i(m) and d(m) inherit all familiar relation-ships that hold between i and d. An example is equation (34) on page 1025 of these notes,which coincides with the familiar relationship between i and d only when m = 1. Because ofthe definitions we have chosen for i(m) and d(m), the effective rates of interest and discount for1m th of a year are i(m)

m and d(m)

m , which do have properties analogous to those that hold between iand d

[5, Exercise 26 p. 44], [6, Exercise 28 p. 32] 1. “Express d(4) as a function of i(3).”2. “Express i(6) as a function of d(2).”

Solution:

1. Since(1 +

i(3)

3

)3

= 1 + i

and(1 − d(4)

4

)4

= 1 − d ,


(1 + i)(1 − d) = 1 ⇒ 1 − d(4)

4=

(1 +

i(3)

3

)− 34

⇒ d(4) = 4

1 −(1 +

i(3)

3

)− 34 .

2.(1 +

i(6)

6

)6

= 1 + i = (1 − d)−1 =

(1 − d(2)

2

)−2

⇒

i(6) = 6

(1 − d(2)

2

)− 13

− 1

.

[5, Exercise 27 p. 44], [6, Exercise 32 p. 32] 1. “Show that i(m) = d(m) · (1 + i)1m .

2. “Verbally interpret the result obtained above.”Solution:

1.

d(m) · (1 + i)1m = m

(1 − v

1m)· (1 + i)

1m

= m((1 + i)

1m − v

1m · (1 + i)

1m)

= m((1 + i)

1m − (v(1 + i))

1m)

= m((1 + i)

1m − 1

1m)

= m((1 + i)

1m − 1

)= i

1m

2. Consider a loan of m. Then i(m) = m · i(m)

m is the interest payable on that loan at theend of 1

m period. If this interest had been prepaid at the beginning of the time period,it would have been m · d(m)

m = d(m), which accumulates for 1m period at a compound

interest accumulation factor of (1 + i)(m).

[5, Exercise 28 p. 44], [6, Exercise 30 p. 32] “Find the accumulated value of $100 at the endof two years:

1. “if the nominal annual rate of interest is 6%, convertible quarterly; and2. “if the nominal annual rate of discount is 6%, convertible once every four years.”

Solution:

1. 100(1 + 0.06

4

)2×4= 100(1.015)8

2. 100 (1 − 4 × 0.06)−24 = 100(1 − 0.24)−

12


[6, Exercise 29 p. 32] “On occasion, interest is convertible less frequently than once a year.Define i(

1m ) and d( 1

m ) to be the nominal annual rates of interest and discount convertibleonce every m years. Find a formula analogous to formula (1.22a) for this situation.”Solution:

1 +i(

1m )

m

1m

=

1 −d

(1p

)

p

1p

[6, Exercise 31 p. 32] “Derive the formula

i(m)

m− d(m)

m=

i(m)

m· d(m)

m[5, (1.24), p. 25][6, (1.23), p. 20].” (35)

Solution: For 1m th of a year, the effective interest and discount rates are, respectively

i(m)

m and d(m)

m . We apply the following relationship between effective rates of interest anddiscount:

i − d = id [6, (1.17), p. 15]

for 1m th of a year.

Another derivation of this equation would be from (??) with p = m.

[5, Exercise 29 p. 45], [6, Exercise 33 p. 32] “Given that i(m) = 0.1844144, and d(m) = 0.1802608,find m.”Solution: (

1 +i(m)

m

) (1 − d(m)

m

)= 1

⇒ m =i(m)d(m)

−d(m) + i(m) =0.1844144 × 0.1802608

0.0041536= 8.0033434...

[5, Exercise 30 p. 45], [6, Exercise 34 p. 32] “It is known that 1 +i(n)

n=

1 + i(4)

4

1 + i(5)

5

. Find n.”

Solution:

1 +i(n)

n=

1 + i(4)

4

1 + i(5)

5

⇒ (1 + i)1n = (1 + i)

14− 1

5 = (1 + i)120

⇒ n = 20 .

[5, Exercise 31 p. 45], [6, Exercise 35 p. 32] “If r = i(4)

d(4) , express v in terms of r.”


Solution: The following is a valid solution, but the form could be improved:

i(4) = 4((1 + i)

14 − 1

)

d(4) = 4(1 − (1 − d)

14)

r =i(4)

d(4) =1

(1 − d)14

= v−14

Hence v = r−4.Now, having found the appropriate relationships, let’s rebuild the solution in a moreelegant form, beginning with the function v that we wish to evaluate:

v =1

1 + i= 1 − d

=

(1

(1 + i)14

)4

=

(1 + i)

14 − 1(

(1 + i)14 − 1

)· (1 + i)

14

4

=

1 − (1 + i)−14

(1 + i)14 − 1

4

=

1 − (1 − d)14

(1 + i)14 − 1

4

=

4(1 − (1 − d)

14

)

4((1 + i)

14 − 1

)

4

=

(d(4)

i(4)

)4

=1r4 .




6.8.1 §1.9 FORCES OF INTEREST AND DISCOUNT

This section of the textbook begins with the definition of the force of interest, δt at time t:

Definition 6.15δt =

A′(t)A(t)

=a′(t)a(t)

=ddt

ln A(t) =ddt

ln a(t) . (36)

A definition in this generality is appropriate if we are considering very general interest schemes.However, in the context of compound interest at a constant rate we can take the following the-orem as a definition:

Theorem 6.6 When i is constant, δt = ln(1 + i).

Proof: When a(t) = (1 + i)t, a′(t) = (1 + i)t · ln(1 + i), so δt =a′(t)a(t)

= ln(1 + i). �

Theorem 6.7 When i is constant, limm→∞

i(m) = limm→∞

d(m) = δ .

Proof: When i is constant,

limm→∞

i(m) = limm→∞

m((1 + i)

1m − 1

)

= limm→∞

(1 + i)1m − 1

1m

= limm→∞

− 1m2 · ln(1 + i) · (1 + i)

1m

− 1m2


= ln(1 + i) = δ

limm→∞

d(m) = limm→∞

m(1 − (1 − d)

1m)

= limm→∞

1 − (1 − d)1m

1m

= limm→∞

− 1m2 · ln(1 − d) · (1 − d)

1m

− 1m2


= − ln(1 − d) = + ln(1 + i) = δ �


We could avoid the use of l’Hospital’s Rule above by changing the variable to x = 1m . For

example,

limm→∞

(1 + i)1m − 1

1m

= limx→0+

(1 + i)x − 1x

= limx→0+

(1 + i)x − (1 + 0)x

x

=ddx

((1 + i)x)∣∣∣∣∣x=0

= (1 + i)x · ln(1 + i)|x=0 = ln(1 + i) .

For the present, please omit the discussion of force of discount on [5, p. 30].

Force of interest under simple interest and simple discount. While δt is constant undercompound interest and compound discount, it is not constant under simple interest and simplediscount: under the former it decreases with time, and, under the latter, it increases with time.

Theorem 6.8 1. When a(t) = 1 + it, where i is constant, δt = i1+it for t ≥ 0.

2. When a(t) = 11−dt , then δt = d

1−dt , for 0 ≤ t < 1d .

Relations between interest, discount, and force of interest We have already seen that,under constant compound interest and compound discount,

δ = ln(1 + i) .

Since (1 + i)(1 − d) = 1, this relationship implies that

δ = − ln(1 − d) .

These equations could be expressed, equivalently, in the forms

1 + i = eδ (37)1 − d = e−δ (38)

i = eδ − 1 (39)d = 1 − e−δ . (40)

[5, Exercise 34, p. 45], [6, Exercise 42, p. 33] “Fund A accumulates at a simple interest rateof 10%. Fund B accumulates at a simple discount rate of 5%. Find the point in time atwhich the forces of interest on the two funds are equal.”


Solution: The notation here is confusing, because we need the letter A to denote theamount function. I am superscripting the amount functions by a label to indicate whichfund is involved: note that the typography should tell you which letter “A” is being used.)For a unit of capital in Fund A, A(a)(t) = 1 + 0.1t, so the force of interest is

δ(A)t =

a(A)′(t)

a(A)(t)=

0.11 + 0.1t

.

For a unit of capital in Fund B, a(B)(t) = (1 − 0.05t)−1, so the force of interest is

δ(B)t =

a(B)′(t)

a(B)(t)=

0.051 − 0.05t

.

Equating these forces of interest yields

0.1(1 − 0.05t) = 0.05(1 + 0.1t)⇔ t = 5,

so the forces of interest are equal after precisely 5 years.

[5, Exercise 35, p. 45]. [6, Exercise 43, p. 33] “An investment is made for one year in a fundwhose accumulation function is a second degree polynomial. The nominal rate of interestearned during the first half of the year is 5% convertible semiannually. The effective rateof interest earned for the entire year is 7%. Find δ0.5.”Solution: We are to assume that a(t) has the form a(t) = Kt2 + Lt + M; since A(0) = 1,M = 1.

a(0.5) = 1.025 ⇒ 0.25K + 0.5L = 0.025a(1) = 1.07 ⇒ K + L = 0.07

⇒{

K = 0.04L = 0.03

⇒ a(t) = 0.04t2 + 0.03t + 1⇒ a′(t) = 0.08t1 + 0.03

⇒ δ0.5 =a′(0.5)a(0.5)

=0.04 + 0.03

0.01 + 0.015 + 1=

14205

[6, Exercise 36 p. 32] Derive the formula limm→∞

d(m) = δ.

Solution:

limm→∞

i(m) = limm→∞

m((1 + i)

1m − 1

)


= limm→∞

(1 + i)1m − 1

1m

= limx→0+

(1 + i)x − 1x

=ddx

((1 + i)x)∣∣∣∣∣x=0

= (1 + i)x ln(1 + i)|x=0 = ln(1 + i)

Analogously,

limm→∞

d(m) = limm→∞

m(1 − (1 − d)

1m)

= limm→∞

1 − (1 − d)1m

1m

= limx→0+

1 − (1 − d)x

x

= − ddx

((1 − d)x)∣∣∣∣∣x=0

= − ln(1 − d) = ln(1 + i)

(to be continued)




6.9.1 §1.9 FORCES OF INTEREST AND DISCOUNT (conclusion, for now)

Motivation for Definition 36. Under constant compound interest i the growth of the amountfunction A(t) in a small period ∆t is given by A(t + ∆t) = A(t) · (1 + i)∆t, so

A(t+∆t)−A(t)∆t

A(t)=

(1 + i)∆t − 1∆t

.

In the limit, as ∆t → 0, the left side approachesA′(t)A(t)

, while the right side approaches ln(1+i) =

δ. This motivates Definition 6.6 (equation (36)),

δt =A′(t)A(t)

=a′(t)a(t)

=ddt

ln A(t) =ddt

ln a(t)

when the interest rate is now permitted to be time-dependent; the force of interest also becomestime-dependent, and we denote it by δt. In general we can solve differential equation (36) byintegrating:

A(t) = A(0) · e∫ t

0 δr dr ,

which implies thata(t) = e

∫ t0 δr dr. (41)

We can interpret the product A(t) ·δt ·∆t = A′(t) ·∆t ≈ A(t +∆t)−A(t) = ∆A(t) as the incrementin A(t) over the time interval from t to t + ∆t, i.e., as the interest earned in the account in thattime interval. Then we have

Theorem 6.9∫ n

0A(t) · δt dt =

∫ n

0A′(t) dt = A(t)]n

0 = A(n) − A(0) , (42)∫ n

0a(t) · δt dt =

∫ n

0a′(t) dt = a(t)]n

0 = a(n) − 1 , (43)

giving an integral formula for the interest earned over a time interval. These formulæ may beapplied under conditions of variable interest.


6.9.2 §1.10 VARYING INTEREST

[5, Exercise 37, p. 45] “Find the level, effective rate of interest over a three-year period whichis equivalent to an effective rate of discount of 8% the first year, 7% the second year, and6% the third year.”Solution: Suppose the rate of interest sought is i effective per annum. Then (1 + i)−3 =

(1−0.06)(1−0.07)(1−0.08)⇒ 1+ i =3

√1

(0.94)(0.93)(0.92)= 1.0753103⇒ i = 7.53%.

[5, Exercise 39, p. 46] “An investor makes a deposit today, and earns an average continuousreturn (force of interest) of 6% over the next five years. What average continuous returnmust be earned over the subsequent five years in order to double the investment at theend of ten years?”Solution: The accumulation factor for the first 5 years is

e∫ 5

0 δt dt = e∫ 5

0 0.06 dt = e0.06]50 = e0.3 .

Let i be the average continuous return earned over the subsequent five years. Then theaccumulation factor for that period will be

e∫ 10

5 δt dt = e∫ 10

5 i dt = e5i]105 = e5i .

Thus

e0.3+5i = 2⇔ 0.3 + 5i = ln 2⇔ i =ln 2 − 0.3

5= 0.0786294 = 7.86% .

[5, Exercise 40, p. 46] “In Fund X money accumulates at a force of interest δt = 0.01t + 0.1for 0 ≤ t ≤ 20. In Fund Y money accumulates at an annual effective interest rate i. Anamount of 1 is invested in each fund for 20 years. The value of Fund X at the end of 20years is equal to the value of Fund Y at the end of 20 years. Calculate the value of FundY at the end of 1.5 years.”Solution: After 20 years the value of the dollar invested in Fund X is

e∫ 20

0 (0.01t+0.1) dt = e

[0.005t2 + 0.1t

]20

0 = e2+2 = e4 .

The dollar in Fund Y has grown to (1 + i)20, and these two amounts are postulated to beequal. Accordingly, the dollar in Fund Y grows to

(1 + i)1.5 = e1.5×4

20 = 1.3498588

after one and one-half years. While this appears to be a high rate of growth, one must notoverlook that the rate of growth of Fund X is very high; for example, in its final year thatfund grows by a factor of more than e0.19+0.1 = 1.336, i.e., more than 33%.


[5, Exercise 41, p. 46] “If the effective rate of discount in year k is equal to 0.01k + 0.06 fork = 1, 2, 3, find the equivalent rate of simple interest over the three-year period.”Solution: Let the equivalent (annual) rate of simple interest be i. Then a principal of 1grows under simple interest to 1 + 3i. This must be equal to

11 − 0.07

· 11 − 0.08

· 11 − 0.09

= 1.2843631 ,

implying that i = 0.0947877 or 9.48%.

6.9.3 §1.11 SUMMARY OF RESULTS

Table 6.9.3, page 1035 is reproduced from the textbook [5, Table 1.1, p. 38][6, Table 1.1, p.29].

Rate of interest The accumulated value The present value of 1

or discount of 1 at time t = a(t) at time t =1

a(t)Compound interest

i (1 + i)t vt = (1 + i)−t

i(m)

(1 +

i(m)

m

)mt (1 +

i(m)

m

)−mt

d (1 − d)−t (1 − d)t

d(m)

(1 − d(m)

m

)−mt (1 − d(m)

m

)mt

δ eδt e−δt

Simple interest

i 1 + it (1 + it)−1

Simple discount

d (1 − dt)−1 1 − dt

Table 3: Summary of Relationships in Chapter 1

Textbook Chapter 2. Solution of problems in interest



Chapter 2 “discusses general principles to be followed in the solution of problems in interest.The purpose of this chapter is to develop a systematic approach by which the basic principlesfrom Chapter 1 can be applied to more complex financial transactions.”

“Successive chapters have two main purposes:

1. “to familiarize the reader with more complex types of financial transactions, includingdefinitions of terms, which occur in practice;

2. “to provide a systematic analysis of these financial transactions, which will often lead toa more efficient handling of the problem than resorting to basic principles.”

6.9.5 §2.2 THE BASIC PROBLEM

As the textbook observes, most interest problems can be broken down into problems that in-volve 4 parameters:

1. The principal amount originally invested. (In these notes I will normally not show thedollar sign for amounts invested.)

2. The length of time during which the amount will be held in the account.

3. The rate of interest or discount to be applicable; in the limit, this could be expressed asthe force of interest of discount.

4. The amount accumulated in the account at the end of the investment period.

These parameters are not independent: when three of them have been specified, the fourth isdetermined; but finding the value of that 4th parameter may not be trivial.

Numerical results can be obtained by

• direct calculation

• direct calculation by hand, typically using series expansions.16

16Students in MATH 329 do not often have background from Calculus 3, where the computation using seriesis normally introduced. Thus any applications of this type will have to be justified to students whose backgroundcannot be assumed to include more than Calculus 1 and Calculus 2. Examples of series of this type are

(1 + i)k = 1 + ki +k(k − 1)

2!i2 +

k(k − 1)(k − 2)3!

i3 + . . . (44)

ex = 1 + x +x2

2!+

x3

3!+ . . . (45)

ln(1 + x) = x − x2

2+

x3

3− . . . (46)

which are “MacLaurin” series (a special case of “Taylor” series), and have restrictions on the numbers for whichthey are valid. (The first is always valid if |i| < 1, and also for any i if k is a non-negative integer; the second isvalid for all x; and the third is valid for −1 < x ≤ 2.) The error that occurs when one “truncates” a series — i.e.,


• use of compound interest tables, as in [6, Appendix I]; in this case we may interpolatebetween entries in the tables; more information will be provided — you are not expectedto have any prior knowledge about interpolation;17

[6, Exercise 1, p. 53] “10000 is invested for 4 months at 12.6%, where interest is computedusing a quadratic to approximate an exact calculation. Find the accumulated value.”Solution: The exact value using compound interest is

10000(1.126)412 = 10403.50 .

Alternatively, when we approximate by a quadratic, we truncate the MacLaurin seriesafter the 2nd degree term (cf. (44))

(1 + i)13 = 1 +

13· i +

13·(−2

3

)· 1

2!· i2 +

13·(−2

3

)·(−5

3

)· 1

3!· i3 + . . .

≈ 1 +13· i +

13·(−2

3

)· 1

2!· i2

= 1.040236

hence the accumulated value is approximately 10402.36.Note that a linear approximation, which would be equivalent to simple interest, wouldgive 10420: as the period of time is less than one unit, a linear approximation is higherthan the value given by compound interest — equivalently, the line joining the points(x, y) =

(0, (1.043)0

)and (x, y) =

(1, (1.043)1

), passes over the graph of y = (1.043)x.

6.9.6 §2.3 EQUATIONS OF VALUE

We will often express the relationship between different sums in a problem in an “equation ofvalue”, which is simply a statement obtained by bringing all amounts to a particular time (byaccumulating or discounting according to whatever accumulation functions are appropriate)”.Sometimes we will have an inequality rather than an equation. The time where the variousamounts are compared is the comparison date. Under compound interest equations of valuefor the same payments, made a different comparison dates, should be equivalent: you shouldbe able to extract one from the other simply by multiplying or dividing by the appropriateaccumulation factors.

when one stops adding at a particular term — can be estimated.“Using series expansions for calculation purposes is cumbersome and should be unnecessary except in unusual

circumstances.”17The current edition of the textbook does not contain interest tables, although I may discuss some examples

of interpolation and provide specific tables at that time, for historical reasons, even though the technique may beobsolete in these applications.


Time diagrams A time diagram is a one-dimensional diagram where the only variable istime, shown on a single coordinate axis. We may show above or below the coordinate of apoint on the time-axis values of money intended to be associated with different funds. As inany mathematical exercises, the diagram is not a formal part of a solution, but may be veryhelpful in visualizing the solution. Some authors use variants of time diagrams where theremay be several parallel horizontal axes, representing several funds; alternatively, you coulduse the usual type of graph that you have seen in calculus, where the values of the fund willbe shown as points in the plane, with the horizontal axis representing time, and the secondcoordinate giving the function value. (The system I am using for these notes accommodatesfigures with great difficulty; I will usually not attempt to show time diagrams for that reason.)

In a typical problem there will exist more than one way in which to view the various sumsof money involved, at any given point in time (the comparison date). In an Equation of Valuewe equate two such representations, and thereby may obtain an equation which we can solvefor one of the variables in terms of the others.

If the prevailing rate of interest or discount is non-zero, the equations obtained at differ-ent points in time will be different. Under compound interest or compound discount — theenvironments we will be considering most of the time — the information we derive fromequations of value will be the same no matter which comparison date we will be using; undersimple interest and discount, and other environments, the comparison date could be relevant.Sometimes we represent such a situation by a one-dimensional diagram, where time is repre-sented along a horizontal axis, the direction of increase being to the right. Notwithstanding theone-dimensionality, there could be several lines above and below the time-axis on which we in-dicate amounts of money being added or removed, as well as the time labels. The comparisondate is often indicated by a vertical arrow.

[5, Exercise 1, p. 67], [6, Exercise 8, p. 55] “In return for payments of 2000 at the end of fouryears and 5000 at the end of ten years, an investor agrees to pay 3000 immediately and tomake an additional payment at the end of three years. Find the amount of the additionalpayment if i(4) = 0.06.”Solution: If we denote the additional payment by X, then the equation of value at timet = 0 is

3000 + X(1.015)4×(−3) = 2000(1.015)4×(−4) + 5000(1.015)4×(−10)

which we can solve to yield

X = −3000(1.015)12 + 2000(1.015)−4 + 5000(1.015)−28

= −3586.854514 + 1884.368461 + 3295.496247 = 1593.010194

so the additional payment should be 1593.01. Depending on where rounding has beenapplied, the amount could vary slightly from this figure.


[5, Exercise 2, p. 67] “You have an inactive credit card with a 1000 outstanding unpaid bal-ance. This particular credit card charges interest at the rate of 18% compounded monthly.You are able to make a payment of 200 one month from today, and 300 two months fromtoday. Find the amount that you will have to pay three months from today to completelypay off this credit card debt.”

Solution: The effective interest rate per month is18%12

= 1.5%. Denote the payment 3months hence by X. The equation of value at time t = 3 months is

1000(1.015)3 = 200(1.015)2 + 300(1.015)1 + X(1.015)

which we can solve, to yield

X = 1000(1.015)3 − 200(1.015)2 − 300(1.015)= 535.133375

so the final payment must be in the amount of 535.13.


6.10 Supplementary Notes for the Lecture of January 22nd, 2010Distribution Date: Friday, January 22nd, 2010


6.10.1 §2.3 EQUATIONS OF VALUE (conclusion)

[5, Exercise 3, p. 67], [6, Exercise 9, p. 54] “At a certain interest rate the present values ofthe following two payment patterns are equal:

(i) 200 at the end of 5 years plus 500 at the end of 10 years;(ii) 400.94 at the end of 5 years.

At the same interest rate 100 invested now plus 120 invested at the end of 5 years willaccumulate to P at the end of 10 years. Calculate P.Solution: Payment schemes (i) and (ii), both with comparison date t = 0, give rise to thefollowing equations of value:

200(1 + i)−5 + 500(1 + i)−10 = 400.94(1 + i)−5 (47)100 + 120(1 + i)−5 = P(1 + i)−10 (48)

Equation (47) implies that(1 + i)−5 = 0.40188 . (49)

Substituting in (48) yields

P =100

(0.40188)2 +120

0.40188= 917.76.

[5, Exercise 4, p. 67], [6, Exercise 10, p. 54] “An investor makes three deposits into a fund,at the end of 1, 3, and 5 years. The amount of the deposit at time t is 100(1.025)t. Find thesize of the fund at the end of 7 years, if the nominal rate of discount convertible quarterlyis 4

41 .”Solution: The nominal rate of discount of d(4) = 4

41 convertible quarterly produces an

annual accumulation factor of(1 − 1

41

)−4= (1.025)4. At the end of 7 years the 3 payments

accumulate to

100(1.025)1+(4×6) + 100(1.025)3+(4×4) + 100(1.025)5+(4×2) = 483.11 .

[5, Exercise 5, p. 67], [6, Exercise 11, p. 54] “Whereas the choice of a comparison date hasno effect on the answer obtained with compound interest, the same cannot be said ofsimple interest. Find the amount to be paid at the end of 10 years which is equivalent totwo payments of 100 each, the first to be paid immediately, and the second to be paid atthe end of 5 years. Assume 5% simple interest is earned from the date each payment ismade, and use a comparison date of


1. The end of 10 years.2. The end of 15 years.”

Solution:

1. With a comparison date at t = 10, the payment at the end of 10 years will be 100(1 +

10 × 0.05) + 100(1 + 5 × 0.05) = 275.2. With a comparison date at t = 15, the amount P that must be paid at the end of 10

years satisfies the equation of value

100(1 + 15 × 0.05) + 100(1 + 10 × 0.05) = P · (1 + 5 × 0.05) ,

which implies that P = 260.18

6.10.2 §2.4 UNKNOWN TIME

Method of Equated Time Omit

Rule of 72 How long does it take money to double? Solving the equation (1 + i)n = 2 yields

n =ln 2

ln(1 + i)≈ 0.6931471806

i· i

ln(1 + i). (50)

The last ratio can be approximated using the MacLaurin expansion (46):

iln(1 + i)

=i

i − i22 + i3

3 − . . .=

1

1 −(

i2 + i2

3 − . . .)

= 1 +

(i2

+i2

3− . . .

)+

(i2

+i2

3− . . .

)2

+ . . .

= 1 +i2

(1 − 2i

3+

2i2

4− . . .

)+

i2

4

(1 − 2i

3+

2i2

4− . . .

)+ . . .

≈ 1 +i2− i2

12+ . . .

18The question is: What is the meaning of equivalent? The meaning taken here is that all amounts are consid-ered from the point of view of the comparison date, at which time the single amount P is to equal the sum of thevalues of the two payments made at different times and transported to the specified comparison date using simpleinterest.


which is approximately equal to 1.0395 when i = 8%. Applying this approximation to equation(50) yields

n ≈ 72100i

, (51)

which is called the “Rule of 72”. The textbook reports that this rule is “surprisingly accurateover a wide range of interest rates.”

[5, Exercise 6, p. 68], [6, Exercise 13, p. 55] “Find how long 1000 should be left to accumu-late at 6% effective in order that it will amount to twice the accumulated value of another1000 deposited at the same time at 4% effective.”Solution: If we let t denote the number of years, the equation of value at t = n is

1000(1.06)t = 2000(1.04)t

son =

2ln 106 − ln 104

= 36.39

years.

[5, Exercise 7, p. 68] “You invest 3000 today, and plan to invest another 2000 two years fromtoday. You plan to withdraw 5000 in n years, and another 5000 in n + 5 years exactly,liquidating your investment account at that time. If the effective rate of discount is equalto 6%, find n.”Solution: The effective rate of discount is 6%. One version of the equation of value as oftime t = 0 is

3000 + 2000(1 − 0.06)2 = 5000(1 − 0.06)n + 5000(1 − 0.06)n+5 ,

which implies that

(0.94)n =3000 + 2000(0.94)2

5000(1 + (0.94)5) .

This implies that

n =ln

(3000 + 2000(0.94)2

)− ln(5000

(1 + (0.94)5

))

ln 1.06= 9.665461121 .

Thus n is approximately 9 2/3 years.

[5, Exercise 8, p. 68], [6, Exercise 14, p. 55] “The present value of two payments of 100 eachto be made at the end of n years and 2n years is 100. If i = 0.08, find n.”


Solution: Solving the equation of value, 100v2n + 100vn = 100, we obtain vn = −1±√52 , in

which only the + sign is acceptable, since vn > 0. Taking logarithms gives

n =ln −1+

√5

2

− ln(1.08)= 6.2527 years.

[5, Exercise 10, p. 68], [6, Exercise 16, p. 55] “You are asked to develop a rule of n to ap-proximate how long it takes money to triple. Find n.”Solution: On page 1041 of these notes I have considered the “Rule of 72.” Let us considerthe rationale of that rule:

(1 + i)n = 2

⇒ n =ln 2

ln(1 + i)

=ln 2

i· i

ln(1 + i)

=ln 2

i· i

i − i22 + i3

3 − i44 + . . .

=ln 2

i· 1

1 − i2 + i2

3 − i34 + . . .

=ln 2

i· 1

1 −(

i2 − i2

3 + i34 − . . .

)

=ln 2

i·1 +

(i2− i2

3+

i3

4− . . .

)+

(i2− i2

3+

i3

4− . . .

)2

+ . . .

=ln 2

i·(1 +

i2− i2

12+ . . .

)

≈(ln 2)

(1 + i

2 − i212 + . . .

)∣∣∣∣i=0.08

i

The only change if we wish to obtain a rule for the time for money to triple is that weobtain

n ≈(ln 3)

(1 + i

2 − i212 + . . .

)

i.

If we approximate i in the numerator by 8%, we obtain a numerator of approximately1.14; the rule could be called the “Rule of 114”.

[5, Exercise 11, p. 68] “A deposits 10 today and another 30 in five years into a fund payingsimple interest of 11% per year. B will make the same two deposits, but the 10 will be


deposited n years from today, and the 30 will be deposited 2n years from today. B’sdeposits earn an annual effective rate of 9.15%. At the end of 10 years the accumulatedvalue of B’s deposits equals the accumulated value of A’s deposits. Calculate n.”Solution: At time 10 the value of A’s deposits is 10(1 + 10(0.11)) + 30(1 + 5(0.11)) =

21 + 46.5 = 67.5. At the same time the present value of B’s deposits is 10(1.0915)10−n +

30(1.0915)10−2n. Equating the two amounts yields a quadratic equation for X = (1.0915)−n:

3X2 + X − 67.510(1.0915)10 ,

whose only positive solution is

X =−1 +

√1 +

12(67.5)10(1.0915)10

6= 0.8157901067

implying that n = − ln 0.8157901067ln 1.0915

= 2.325430525 or 2.33 years.

[5, Exercise 12, p. 68], [6, Exercise 18, p. 55] “Fund A accumulates at a rate of 12% convert-

ible monthly. Fund B accumulates with a force of interest δt =t6

. At time t = 0 equaldeposits are made in each fund. Find the next time that the two funds are equal.”Solution: Without limiting generality I can take the amount of each deposit to be 1. Theequation of value is

(1.01)12t = e∫ t

0 δr dr = e∫ t

0r6 dr = e

t212 ,

of which one solution is t = 0. Taking logarithms yields 12t ln 1.01 =t2

12, so the funds

are next equal when t = 144 ln 1.01 = 1.432847643 years.

6.10.3 §2.5 UNKNOWN RATE OF INTEREST

The textbook and its previous edition describe four methods for determining an unknown rateof interest:

• Sometimes — particularly if only one payment is involved — it may be possible to solvethe equation of value by using the fact that the logarithm and exponential functions aremutual inverses.

• Sometimes the equation may be solved by algebraic techniques, for example when theequation is equivalent to a polynomial equation that factorizes.

• This method is not even mentioned in the current edition. Sometimes the equation can besolved by interpolation in interest tables.


• When all else fails, it may be necessary to solve by successive approximation or iteration.There are various methods, and some converge much faster than others. The justificationof the better methods is beyond this course, but is the subject matter of courses likeMATH 317 (Numerical Analysis).

[5, Exercise 13, p. 68], [6, Exercise 19, p. 55] “Find the nominal rate of interest convertiblesemiannually at which the accumulated value of 1000 at the end of 15 years is 3000.”

Solution: The equation of value at time t = 15 is 1000(1 + i

2

)2(15)= 3000, implying that

i = 2(3

130 − 1

)= 7.4598394% .

[5, Exercise 14, p. 68], [6, Exercise 20, p. 55] “Find an expression for the exact effective rateof interest at which payments of 300 at the present, 200 at the end of one year, and 100at the end of two years will accumulate to 700 at the end of two years.”Solution: For the unknown interest rate i, which we assume to be non-negative, the equa-tion of value at time 2 is

300(1 + i)2 + 200(1 + i)1 + 100 = 700 ,

yielding a quadratic equation in 1 + i:

3(1 + i)2 + 2(1 + i) − 6 = 0 ,

whose only positive solution is 1 + i = −1+√

193 = 1.119632981, so i = 11.9632981%.

[5, Exercise 15, p. 69] “You can receive one of the following two payment streams:

(i) “100 at time 0, 200 at time n, and 300 at time 2n;(ii) “600 at time 10.

“At an annual effective interest rate of i, the present values of the two streams are equal.Given vn = 0.75941, determine i.”Solution: At time t = 0 an equation of value is

600v10 = 100 + 200vn + 300v2n = 100 + 151.882 + 173.011 = 424.893 ,

implying that v10 = 0.7082, so v = 0.9661, i = 3.51%.

[5, Exercise 16, p. 69], [6, Exercise 21, p. 55] “It is known that an investment of 1000 willaccumulate to 1825 at the end of 10 years. If it is assumed that the investment earnssimple interest at rate i during the 1st year, 2i during the second year, . . . , 10i during the10th year, find i.”Solution: The intended interpretation was that the equation of value is

1 + i + 2i + 3i + ... + 10i = 1.825 ,

which implies that i = 0.15 exactly.


[5, Exercise 17, p. 69] “It is known that an amount of money will double itself in 10 years ata varying force of interest δt = kt. Find an expression for k.”Solution: An equation of value is

2 = e∫ 10

0 kr dr = e50k ,

implying that k =ln 250

= 0.01386.

[5, Exercise 18, p. 69], [6, Exercise 23, p. 55] “The sum of the accumulated value of 1 at theend of three years at a certain effective rate of interest, and the present value of 1 to bepaid at the end of three years at an effective rate of discount numerically equal to i is2.0096. Find the rate.”Solution: The equation of value at time t = 3 of 1 payable at time 0 and 1 payable at time6 is (1 + i)3 + (1 − i)3 = 2.0096, which implies that i = 0.04 (the only positive solution).



(subject to correction or revision)

6.11.1 §2.6 DETERMINING TIME PERIODS

The syllabus now reads “Students should peruse §2.6, but will not be expected to memorizethe definitions contained therein.”

A number of schemes are in common use for calculating interest for fractions of a period:

• Under exact simple interest or “actual/actual” one counts the exact number of days, andassumes the year has 365 days.

• Under ordinary simple interest or “30/360” one assumes that each calendar month has30 days, and the year has 360 days.

• Under Banker’s Rule or “actual/360” one uses the exact number of days but treats a yearas having 360 days. The treatment of February 29th in leap years (like the present) isnot completely standardized. These calculation bases can be used for either simple orcompound interest.

Example 6.10 Here is a statement I once found at the bottom of a bank statement that I re-ceived from a large Canadian bank: “(name of bank) calculates interest daily using a 365-dayyear, including leap years. The interest rate charged in a leap year will be equal to the AnnualInterest Rate in effect on each day in that year multiplied by 366 and divided by 365. Althoughthis results in slightly more interest being charged, the effective annual rate is the same whenrounded to the nearest 1/8th of 1%.”

[5, Exercise 19, p. 69], [6, Exercise 5, p. 53] “If an investment was made on the date the UnitedStates entered World War II, i.e., December 7, 1941, and was terminated at the end of thewar on August 8, 1945, for how many days was the money invested:

1. “on the actual/actual basis?2. “on the 30/360 basis?”

Solution:

1. “From December 7, 1941 to December 7, 1944, 3 years passed, the last of whichwas a leap year. The number of days is 3(365) + 1. From December 7, 1944 toAugust 7, 1945, the numbers of days in the months December through July were31 + 31 + 28 + 31 + 30 + 31 + 30 + 31. Add one day, from 7th to 8th August. Thetotal is 1340.


2. From December 7, 1941 to December 7, 1944, 3 years passed, the last of which wasa leap year. The number of days is 3(360). From December 7, 1944 to August 7,1945, the numbers of days in the months December through July were 8(30). Addone day, from 7th to 8th August. The total is 1321.

[5, Exercise 20, p. 69], [6, Exercise 6, p. 54] “A sum of 10,000 is invested for the months ofJuly and August at 6% simple interest. Find the amount of interest earned:

1. “Assuming exact simple interest.2. “Assuming ordinary simple interest.3. “Assuming the Banker’s Rule.”

Solution:

1. The number of days is 31 + 31 = 62; exact simple interest is 621365 · 10000 · (0.06) =

101.92.2. Ordinary simple interest is 60

360 · 10000 · (0.06) = 100.3. Interest under the Banker’s Rule is 62

360 · 10000 · (0.06) = 103.33.

[5, Exercise 21, p. 69], [6, Exercise 7, p. 54] 1. “Show that the Banker’s Rule is always morefavourable to the lender than is exact simple interest.

2. “Show that the Banker’s Rule is usually more favourable to the lender than is ordi-nary simple interest.

3. “Find a counterexample to (the preceding) for which the opposite relationship holds.”Solution:

1. Under the Banker’s Rule the denominator of the fraction is decreased, so the fractionis increased.

2. Suppose that k months have passed. The Banker’s Rule counts the days as 30k.How does this compare with the actual number of days? To answer this questioncorrectly, we need to compare all terms of lengths under a year. But the question isnot precisely stated, and it’s not clear how we would have to weight the results. Asthe computations are extensive, we will not bother.

3. A loan between February 1 and March 1 provides a counterexample.

6.11.2 §2.7 PRACTICAL EXAMPLES

[5, Exercise 22, p. 70], [6, Exercise 25, p. 56] “A bill for 100 is purchased for 96 three monthsbefore it is due. Find

1. “The nominal rate of discount convertible quarterly earned by the purchaser.2. “The annual effective rate of interest earned by the purchaser.”

Solution:


1. 100(1 − d(4)

4

)= 96⇒ d(4) = 16%.

2. 96(1 + i)14 = 100⇒ i =

(10096

)4 − 1 = 17.7375701%.

[5, Exercise 23 p. 70], [6, Exercise 26 p. 56] “A two-year certificate of deposit pays an an-nual effective rate of 9%. The purchaser is offered two options for prepayment penaltiesin the event of early withdrawal:

A - a reduction in the rate of interest to 7%B - loss of three months’ interest.

In order to assist the purchaser in deciding which option to select, compute the ratioof the proceeds under Option A to those under Option B if the certificate of deposit issurrendered:

1. at the end of 6 months2. at the end of 18 months”

Solution:

1. Under Option A the effective annual interest rate is now 7%, so a principal of 1returns (1.07)

12 . Under Option B a principal of 1 returns (1.09)

12− 1

4 . The ratio of

returns under Option A to those under option B is (1.07)12

(1.09)12 −

14

= 1.012360669.

2. Under Option A the effective annual interest rate is now 7%, so a principal of 1returns (1.07)

32 . Under Option B a principal of 1 returns (1.09)

32− 1

4 . The ratio of

returns under Option A to those under option B is (1.07)32

(1.09)32 −

14

= .9937852442.

[5, Exercise 24 p. 70] “The ABC Bank has an early withdrawal policy for certificates of de-posit (CD’s), which states that interest sill be credited for the entire length the moneyactually stays with the bank, but that the CD nominal interest rate will be reduced by1.8% for the same number of months as the CD is redeemed early. An incoming collegefreshman invests 5000 in a two-year CD with a nominal rate of interest equal to 5.4%compounded monthly on September 1, at the beginning of the freshman year. The stu-dent intended to leave the money on deposit for the full two-year term to help financethe junior and senior years, but finds the need to withdraw it on May 1 of the sophomoreyear. Find the amount that the student will receive for the CD on that date.”Solution: The CD is redeemed after 18 months, although it was issued for 24 months.Since it is being redeemed 6 months early, the nominal interest rate will be reduced after18 − 6 = 12 months: the effective monthly rate is 5.4

12 % for the first 12 months, and5.4−1.8

12 % = 3.612 % for the last 6 months. The student receives

5000(1 +

0.05412

)1

2(1 +

0.03612

)6

= 5000(1.055356752)(1.074496218) = 5372.48 .


[5, Exercise 25 p. 70] “Many banks quote two rates of interest on certificates of deposite(CD’s). If a bank quotes 5.1% compounded daily, find the ratio of the APY (annualpercentage yield) to the quoted rate for this CD.”

Solution: Since(1 +

0.051365

)365

= 1.05231914, the ratio of the APY to the quoted rate is

5.2319145.1

= 1.0259 .

[5, Exercise 26, p. 70], [6, Exercise 27, p. 56] “A savings and loan association pays 7% ef-fective on deposits at the end of each year. At the end of every 3 years a 2% bonus is paidon the balance at that time. Find the effective rate of interest earned by an investor if themoney is left on deposit

1. “Two years.2. “Three years.3. “Four years.”

Solution: Let i denote the effective rate of interest in each case.

1. There being no bonus, i = 7%.2. Here (1 + i)3 = (1.07)3(1.02), so i = (1.07)(1.02)

13 − 1 = 7.7086300%.

3. This time i = (1.07)(1.02)14 − 1 = 7.5410377%.

[5, Exercise 27 p. 71], [6, Exercise 28 p. 56] “A bank offers the following certificates of de-posit:

Nominal annual interest rateTerm in years (convertible semiannually)

1 5%2 6%3 7%4 8%

The bank does not permit early withdrawal. The certificates mature at the end of theterm. During the next six years the bank will continue to offer these certificates of de-posit. An investor deposits 1000 in the bank. Calculate the maximum amount that can bewithdrawn at the end of six years.”Solution: It is instructive to begin first with an interpretation that resulted from a mis-reading of the problem. In this misreading, I overlooked the fact that the rates wereconvertible semi-annually. the terms are correct, but, where I used factors 1.05, 1.06,1.07, 1.08, I should have used (1.025)2, (1.03)2, (1.035)2, and (1.04)2 respectively; butfractional years were still not permitted. “We consider the partitions of 6 into sums ofintegers: 6 = 4 + 2 = 4 + 1 + 1 = 3 + 3 = 3 + 2 + 1 = 3 + 1 + 1 + 1 = 2 + 2 + 2 =


2+2+1+1 = 2+1+1+1+1 = 1+1+1+1+1+1. Because the rates increase for periodsof increased length, there is no advantage to partitioning a summand further; so we neednot consider any partition containing 1 + 1, since 2 will always be better, etc. This leavesthe partitions 6 = 4 + 2 = 3 + 3 = 3 + 2 + 1 to consider. These three partitions correspondto accumulations of 1000(1.08)4(1.06)2, 1000

((1.07)3

)2, and 1000(1.07)3(1.06)2(1.05)1,

which are respectively equal to

1528.645395, 1500.730352, and 1445.281231,

so the best partition is into 4 + 2 years, and the best return is 1528.645395.”Now I repeat the computations, this time correctly. I must compare 1000(1.04)8(1.03)4,1000

((1.035)6

)2, and 1000(1.035)6(1.03)4(1.025)2, which are respectively equal to

1540.336523, 1511.068657, 1453.579295;

thus the best partition is again into 4 + 2 years, and the best return is 1540.336523.

6.11.3 §2.8 MISCELLANEOUS PROBLEMS

[5, Exercise 28, p. 71] “A store is running a promotion during which customers have two op-tions for payment. Option One is to pay 90% of the purchase price two months after thedate of sale. Option Two is to deduct X% off the purchase price and pay cash on the dateof sale. Determine X such that a customer would be indifferent between the two optionswhen valuing them using an effective interest rate of 8%.”Solution: The present value of a payment of 0.9 two months hence is (0.9) (1.08)−

212 =

0.8885296, corresponding to an immediate discount of X = 11.15%.

[5, Exercise 29, p. 71], [6, Exercise 29, p. 56] “A manufacturer sells a product to a retailerwho has the option of paying 30% below the retail price immediately, or 25% below theretail price in six months. Find the annual effective rate of interest at which the retailerwould be indifferent between the two options.

Solution: We solve the equation of value, 70% = 75% · (1 + i)12 . i =

(7570

)2 − 1 =

14.7959183%.

[5, Exercise 30, p. 71] “You deposit 1000 into a bank account. The bank credits interest ata nominal annual rate of i, convertible semiannually for the first 7 years, and a nominalannual rate of 2i, convertible quarterly for all years thereafter. The accumulated amountin the account at the end of 5 years is X. The accumulated amount in the account at theend of 10.5 years is 1980. Calculate X to the nearest dollar.”Solution: After 10.5 years the accumulation is

1000(1 +

i2

)2×7

·(1 +

2i4

)4×(10.5−7)

= 1908


⇒(1 +

i2

)14+14

= 1.980

⇒ 1 +i2

= 1.980128

⇒ 1000(1 +

i2

)2×5

= 1000 × 1.9801028 = 1276.30 .

[5, Exercise 31, p. 71], [6, Exercise 31, p. 56] “Fund A accumulates at 6% effective, and FundB accumulates at 8% effective. At the end of 20 years the total of the two funds is 2000.At the end of 10 years the amount in Fund A is half that in Fund B. What is the total ofthe two funds at the end of 5 years.?”Solution: Denote the principals in funds A and B by A and B respectively. Then we knowthat

A(1.06)20 + B(1.08)20 = 2000

A(1.06)10 =12· B(1.08)10

and wish to determine A(1.06)5 + B(1.08)5. Solving the equations yields

A =2000(1.06)−10

(1.06)10 + 2(1.08)10 = 182.8195812

B =2000(1.08)−10

12 (1.06)10 + (1.08)10

= 303.3009728 .

Hence

A(1.06)5 + B(1.06)5 =2000(1.06)−5

(1.06)10 + 2(1.08)10 +2000(1.08)−5

12 (1.06)10 + (1.08)10

= 182.8195812(1.06)5 + 303.3009728(1.08)5 = 690.3024748.

[5, Exercise 32, p. 71], [6, Exercise 32, p. 57] “An investor deposits 10,000 in a bank. Dur-ing the first year the bank credits an annual effective rate of interest i. During the secondyear the bank credits an annual effective interest i − 0.05%. At the end of two years theaccount balance is 12,093.75. What would the account balance have been at the end ofthree years if the annual effective rate of interest were i+0.09 for each of the three years?”Solution: The equation of value is

10000(1 + i)(1 + i − 0.05) = 12093.75 ,

which we interpret as a quadratic equation in 1 + i:

(1 + i)2 − 0.05(1 + i) − 1.209375 = 0



1 + i =0.05 +

√(0.05)2 + 4(1.209375)

2= 1.125000000

from which we conclude that i = 12.5%, and that the account balance after 3 years wouldbe 10000(1.125 + 0.09)3 = 10000(1.215)3 = 17936.13.

[5, Exercise 33, p. 71], [6, Exercise 33, p. 57] “A signs a one-year note for 1000 and receives920 from the bank. At the end of six months, A makes a payment of 288. Assumingsimple discount, to what amount does this reduce the face amount of the note?”Solution: The rate of simple discount is 80

1000 = 8%. At the due date, 1 year from now,the 288 accumulates to 288

1−0.04 = 300, which is the reduction in the face value of the note,reducing the value to 700.

[6, Exercise 30, p. 56] “If an investment will be doubled in 8 years at a force of interest δ, inhow many years will an investment be tripled at a nominal rate of interest numericallyequal to δ and convertible once every three years?”Solution: The given information implies that

(eδ

)= 2, so δ = ln 2

8 . We need to determinethe number of years n with the property that (1 + 3δ)

n3 = 3, so n = 3 ln 3

ln(1+3δ) = 14.26421450years.

Textbook Chapter 3. Basic Annuities


Definition 6.16 1. An annuity is a series of payments, usually made at equal intervals oftime.

2. The total period during which payments will be made is called the term.

3. The interval between payments is called the payment period; the name “annuity” suggestsa default period of 1 year, but, in some specific applications, the default period may havesome other length; there can, in fact, be two distinct periods involved — one associatedwith the interest calculations, and one associated with the payments under the annuity.

4. An annuity-certain consists of payments that are all certain to be made. Otherwise anannuity is a contingent annuity. An annuity whose payments are contingent on whether


a given individual is alive is a life annuity; the study of such annuities is concerned with“life contingencies”, and is outside of this course; following the textbook, we will usuallysuppress the suffix -certain.

6.11.5 §3.2 ANNUITY-IMMEDIATE

In an annuity-immediate the payments are made at the end of each period. The present valueof an annuity-certain for n periods is denoted by an , or, if the interest rate needs to be statedexplicitly an i . The author uses a diagram to denote annuities, with 2 arrows, one sometimeslabelled t1 and the other sometimes labelled t2. The first shows the beginning of the first period,at the end of which a payment is due under the annuity-certain. The second arrow, labelledt2, indicates the last payment date — more precisely, just after the payment has been made. Ishall usually not include diagrams in these notes, as they are very time-consuming to arrangeusing the text-preparation software that I am using.

The accumulated value of the payments as of time t2 is denoted by sn , or, if the interestrate needs to be stated explicitly sn i . We can prove the following basic formulæ:

an = v + v2 + . . . + vn−1 + vn

= v · 1 − vn

1 − vwhen i , 0

=1 − vn

i(52)

sn = 1 + (1 + i) + (1 + i)2 + . . . + (1 + i)n−1

=(1 + i)n − 1(1 + i) − 1

when i , 0

=(1 + i)n − 1

i(53)




6.12.1 §3.2 ANNUITY-IMMEDIATE (conclusion)

In the last lecture I developed formulæ for an and sn; note that, in deriving these formulæ, Iassumed in both cases — as is usually the case — that i , 0.19 We can also prove the followingidentities, both algebraically and verbally:

1 = ian + vn (54)sn = an · (1 + i)n (55)1an

=1sn

+ i (56)

A verbal proof of the identity sn = an · (1 + i)n. The value at time n of n payments of 1at the ends of the years, the last payment being at time n, is sn. Instead of shifting each of thepayments of 1 immediately to time n, we could first view them at time 0, where they are worthan. That amount at time 0 accumulates at an annual accumulation factor of 1 + i, and is worthan · (1 + i)n at time n.

A verbal proof of the identity 1 = ian + vn. The borrower of 1 today can repay the debtn years from now by paying 1 at that time (worth only vn today, when discounted back to thepresent) after paying i in interest at the end of each year for the use of that 1 unit. Those interestpayments constitute an n-payment annuity-immediate, and are today worth i · an i.

A verbal proof of the identity1an

=1sn

+ i. In the previous two verbal proofs I interpreted

the quantities on one or both of the sides of the equal sign as the accumulation of several pay-ments. In this case I propose to interpret the identity as a decomposition of an annual payment

of1an

at the end of each year for n years. n annual payments of1an

, one at the end of each year,

can be interpreted as being the sole payments to repay a loan of 1 today. Alternatively, we canthink of delaying repayment, and instead paying the lender annual interest payments of i, while

the debt of 1 remains at the same level, unpaid. The remainder of the annual payment,1sn

can

be thought of as being held in a separate fund, also accumulating interest at the annual effective

19When i = 0, an = n = sn. In fact neither of these formulæ would have a meaning if i = 0; (it could havehappened that the formulæ might have been meaningful, but the proofs had been defective.


rate i, accumulating until the end of the nth year. At that time the payments have accumulated

a fund of value1sn· sn = 1, and the principal of the loan can also be repaid. We will return to

this formalization later, when we introduce the concept of a sinking fund [5, §5.4].

Remember to sketch a time diagram as you read each of the following problems.

[5, Exercise 1, p. 107], [6, Exercise 1, p. 88] “A family wishes to accumulate 50,000 in a col-lege education fund (by) the end of 20 years. If they deposit 1000 into the fund at the endof each of the first 10 years, and 1000 + X at the end of each of the second 10 years, findX to the nearest unit if the fund earns 7% effective.”Solution: The equation of value at time t2 = 20 is

1000s20 + X · s10 = 50000 ,

which we may solve to yield

X =50000s10 0.07

− 1000s20 0.07

s10 0.07

=50000s10 0.07

− 1000(1.07)20 − 1(1.07)10 − 1

= 50000s−110 0.07

− 1000((1.07)10 + 1)

= 50000(0.072378) − 1000(2.96715) from the tables in [6]= 651.75.

A more precise computation would yield 651.724; the difference is due to rounding errorsand/or the limited precision of the tables. (The current edition of the textbook does notprovide tables.)

[5, Exercise 2, p. 107], [6, Exercise 2, p. 88] “The cash price of a new automobile is 10,000.The purchaser is willing to finance the car at 18% convertible monthly and to makepayments of 250 at the end of each month for 4 years. Find the down payment which willbe necessary.”Solution: Let X be the down payment. Then the equation of value at time t = 0 isX+250a48 1.5% = 10000, from which we determine the down payment, X = 1489.361590.(Here again, the answer obtained using the tables in the 2nd edition of the text-book isslightly incorrect, at 1489.35.)




6.13.1 §3.3 ANNUITY-DUE

In an annuity-due the payments are made at the beginning of each period. The present valueof an annuity-certain for n periods is denoted by an , or, if the interest rate needs to be statedexplicitly an i . Again the author uses a diagram to denote annuities, with 2 arrows, one labelledt1 and the other labelled t2. The first shows the beginning of the first period, just before apayment is due under the annuity-due. The second arrow, labelled t2, indicates the end of thelast period at the beginning of which a payment was made. The accumulated value of thepayments as of time t2 is denoted by sn , or, if the interest rate needs to be stated explicitly sn i .We can prove the following basic formulæ:

an = 1 + v + v2 + . . . + vn−2 + vn−1

=1 − vn

1 − vwhen i , 0

=1 − vn

d(57)

sn = (1 + i) + (1 + i)2 + . . . + (1 + i)n

=(1 + i)n − 1

dwhen i , 0 . (58)

Note that we have assumed — as is usually the case — that i , 0; when i = 0, we can see thatan = an = 1 + 1 + . . . + 1 = n.20. We can also prove the following identities, both algebraicallyand verbally:

an = (1 + i)an (59)sn = (1 + i)sn (60)an = 1 + an−1 (61)sn = −1 + sn+1 (62)sn = (1 + i)nan (63)

(Discount, Due) vs. (Interest, Immediate). There are analogies between annuities-immediateand annuities-due that can be explained by the two points of view of interest and discount. So

20Since this is the limit of an i as i→ 0, we have proved the right continuity of an i and an i at i = 0.


we have the following analogous formulæ:

an =1 − (1 + i)−n

ian =

1 − (1 − d)n

d

sn =(1 + i)n − 1

isn =

(1 − d)−n − 1d

Note that, in spite of this analogy, the notation is “prejudiced” to expression in terms of interestrather than discount. Thus, for example, in the symbols an k, an k k always represents the interestrate — never the rate of discount.

[5, Exercise 7, p. 107], [6, Exercise 11, p. 89] “Find a8 if the effective rate of discount is 10%.”

Solution: Using the discount version of the formula, i.e., an =1 − (1 − d)n

d, we have

a8 =1 − (0.9)8

0.1= 5.695327900.

[5, Exercise 8, p. 108], [6, Exercise 9, p. 89] “Find the present value of payments of 200 ev-ery six months starting immediately and continuing through four years from the present,and 100 every six months thereafter through ten years from the present, if i(2) = 0.06.”Solution: The reader is likely to make assumptions in a casual reading that were notintended by the author. Reading carefully, one might see that he intends that, though thepayments start immediately, they continue to the end of 4 years from now — that is, thatthere be 9 payments of 200, not 8. Similarly, it is the payments that end after 10 years,not the years for which they are prepaid — so there are payments over 21 half-years inall. With this interpretation, at the effective rate of 1

2 i(2) = 3% per half-year,

Present value = 100a21 3% + 100a9 3%

= 100(1.03)(100a21 3% + 100a9 3%

)

= 2389.716705

Note that my solution here shows my prejudice (or laziness) in defaulting to the interestversion of the formulæ; but as we are given the rate of interest, rather than the rate ofdiscount, this turns out to be the easiest approach.

6.13.2 §3.4 ANNUITY VALUES ON ANY DATE

Three cases are considered: the date is always an integral number of periods from each pay-ment date.


1. present values more than one period before the first payment date

2. accumulated values more than one period after the last payment date;

3. current values at a date strictly between the first and last payment dates.

Present values more than one period before the first payment date An annuity is said tobe deferred by m time units if the first payment is m time units later than the type of annuity inquestion would normally be paid, and the subsequent payments occur at the expected intervals.The value of an annuity that is deferred through k periods is denoted by the earlier symbol,prefixed by k|. Thus k

∣∣∣an is the present value of an annuity-immediate whose first payment hasbeen deferred through k time periods; analogous symbols can be used for an. It can be seenthat

k

∣∣∣an = vkan

= an+k − ak , (64)

k

∣∣∣an = vkan

= an+k − ak . (65)

Accumulated values more than 1 period after the last payment date. Here we can expressthe current values by multiplying by the appropriate power of 1 + i; or, alternatively, by takingthe difference of two annuity values.

Summary. You are urged to follow the textbook’s suggestion, “The reader should not try towork problems by memorizing formulas...” Remember the reasoning that was used to derivethe formulæ, and apply that reasoning “from first principles” in each case.

Some exercises from the textbook. The textbook asks you to prove several formulæ. Thereare two levels at which such exercises should be approached:

• an algebraic proof

• a verbal justification

Usually an algebraic proof should not be difficult; I will not normally include proofs in thesenotes, but you can see me if you have difficulty working through a proof. The issue is notto find an elegant proof — just to show that the two sides of the equation are equal. As fora verbal proof, that will be much harder, and I will spend increasing amounts of time at thelectures discussing problems of this type.

[5, Exercise 12, p. 108], [6, Exercise 17, p. 89] “Payments of 100 per quarter are made fromJune 7, year Z through December 7, year Z + 11, inclusive. If the nominal rate of interest,convertible quarterly, is 6%:


1. find the present value on September 7, year Z − 1;2. find the current value on March 7, year Z + 8;3. find the accumulated value on June 7, year Z + 12”.

Solution:

1. As of September 7, year Z − 1, no payments have yet been made. The present valueis21

(1.015)−2100a44+3 1.5% = 3256.879998 .

2. As of March 7, year Z + 8, the value as of March 7, year Z will accumulate by afactor (1.015)(8×4), for an accumulated value of 5403.152103.

3. As of June 7, year Z + 12, the originally computed value will accumulate by a factor(1.015)(13×4)−1, for an accumulated value of 6959.369761.

[5, Exercise 13, p. 108], [6, Exercise 22, p. 90] “Simplify a15

(1 + v15 + v30

)to one symbol.”

Solution: Here is an algebraic solution:

a15

(1 + v15 + v30

)=

1 − v15

i·(1 + v15 + v30

)

=1 − v45

i= a45

A verbal proof could be based on the equation

a15

(1 + v15 + v30

)= a15 + v15 · a15 + v30 · a15

= a15 + 15

∣∣∣∣a15 + 30

∣∣∣∣a15

In the last member the first summand is the present value of an annuity of 15 annualpayments of 1, the first one year from now and the last 15 years from now; the secondsummand is the present value of a sequence of 15 payments which begin one year afterthe last represented by the first summand and the last being paid 30 years from now; andthe last summand is the present value of a final subseries of 15 payments of 1, the firstto be paid one year after the last of the 30 payments mentioned above, and the last to bepaid 45 years from now. We know that the present value of such a series of 45 paymentsof 1 is a45.

21Why is the exponent −2 and not −3 even though the the evaluation is being made 9 months before the firstpayment? Because I am viewing the payments as an annuity-immediate: the clock starts ticking one period beforethe first payment. And why have 3 periods been added, even though June and December are only 2 quarter-yearsapart? Again because the first payment is associated with the 3-month period ending with the payment, so theactual length of the annuity period is 3 months longer.


6.14 Supplementary Notes for the Lecture of February 01st, 2010Distribution Date: Monday, February 01st, 2010


6.14.1 §3.4 ANNUITY VALUES ON ANY DATE (conclusion)

[5, Exercise 14, p. 108], [6, Exercise 21, p. 90] “It is known that

a7

a11

=a3 + sx

ay + sz. (66)

Find x, y, and z.”Solution: Intuitively we usually expect that the determination of 3 variables requires 3constraints. While this is not always the case, the fact that only one equation has beenpresented here should ring an alarm bell. The variable i has not been mentioned, and wemight also wish to know whether the solution we are asked to find should be dependenton i.22 It appears from the textbook that the values given above are the only solution thetextbook was seeking — but there exist others!.At time t = 0 the present values of annuities-due of 1 per year for respectively 7 and 11

years have value in the ratioa7

a11

. Both of these values should increase by a factor of 1 + i

per year under compound interest. Hence, 4 years later, the ratio will not have changed.

But, at that time, it can be interpreted asa3 + s4

a7 + s4

. Thus one solution to the problem is

(x, y, z) = (4, 7, 4), (67)

and this solution is valid for all i. I would not expect students to be able to generate therest of this solution!Having found one solution, which we can see from [5, p. 605] to be the solution theauthor is seeking, we might be expected to stop. But, to a mathematician, an instructionlike “Find x, y, and z” means, implicitly,

“Find all possible sets of values for x, y, and z.”

So let’s investigate whether we have all solutions. This raises new issues. The solutionwe found above in (67) is valid for all i. I have two subquestions:

• Could there be other solutions for all i?22This can explain the apparent paucity of equations: to assume that the solution holds for all i is equivalent

to assuming an equation for every value of i — infinitely many equations, for the 3 unknowns we are trying todetermine. In such a situation we should not be surprised if there is no solution at all.


• Could there be solutions that hold for specific values of i, but not for all i?

Could there be other solutions for all i? When i = 0, equationa7

a11

=a3 + sx

ay + sz(68)

becomes7

11=

3 + xy + z

,

implying that11x − 7y − 7z = −33 . (69)

As i→ ∞, v→ 0, and the left side of equation (68) approaches

limi→∞

1 − (1 + i)−7

1 − (1 + i)−11 = 1 ;

the limit of the right side will be 1 only if

x = z . (70)

Then (69) becomes4x − 7y = −33 . (71)

Our earlier solution (67) satisfies this last “diophantine” equation; some other solutionsare

(x, y, z) = (11, 11, 11), (18, 15, 18), (−3, 3,−3) .

If we definew =

y − 34

=x + 3

7,

the conditions we have determined on x, y, z have the general solution

(x, y, z) = (7w − 3, 4w + 3, 7w − 3)

and the case (4, 7, 4) corresponds to w = 1. What about the other solutions? All wehave shown is that, if there are any solutions valid for all i, then they have to be of thepreceding form. But, when we substitute these general values into equation (66), we mayreduce the equation to the following condition on w:

1 − v7

1 − v11 =1 − v7w

1 − v11w .

All values of w other than w = 1 produce a polynomial equation which constrains v, sothe equality will not hold for all interest rates. Thus solution (67) is the only solutionvalid for all i.You are not expected to be able to reproduce this detailed argument.


[5, Exercise 15, p. 108], [6, Exercise 19, p. 89] “Annuities X and Y provide the following pay-ments:

End of Year Annuity X Annuity Y1–10 1 K

11–20 2 021–30 1 K

“Annuities X and Y have equal present values at an annual effective interest rate i suchthat v10 = 1

2 . Determine K.”Solution: As of today, the present values of annuities X and Y are respectively 1 · a30 +

1(a20 − a10

)and K · a30 − K

(a20 − a10

). Setting these amounts equal and solving, we

obtain

K =a30 + a20 − a10

a30 − a20 + a10

=1 − v30 − v20 + v10

1 − v30 + v20 − v10

=1 − 1

8 − 14 + 1

2

1 − 18 + 1

4 − 12

=95

= 1.8.

[5, Exercise 16, p. 108] “You are given that 5

∣∣∣∣a10 = 3 ·10

∣∣∣∣a5 . Find (1 + i)5.”

Solution:

Case 1 — i = 0. In this case there is no discounting, and the given equation is equivalentto 10 = 3(5). From this contradiction we conclude that i , 0.

Case 2 — i , 0.

5

∣∣∣∣a10 = 3 ·10

∣∣∣∣a5 ⇒ v5 · 1 − v10

i= 3v10 · 1 − v5

i⇒ 2v15 − 3v10 + v5 = 0⇒ v5

(2v5 − 1

) (v5 − 1

)= 0 ,

provided i , 0. Aside from the trivial solution i = −1, there are two non-trivialsolutions to this equation: but the solution v5 = 1 is not valid, since we assumedearlier that i , 0; the remaining non-trivial solution is equivalent to (1 + i)5 = 2. (Iwill not attempt to provide an interpretation of i = −1.)

[5, Exercise 17, p. 108], [6, Exercise 23, p. 90] “Find the present value to the nearest dollaron January 1 of an annuity which pays 2000 every six months for five years. The firstpayment is due on the next April 1, and the rate of interest is 9% convertible semiannu-ally.”


Solution: On April 1st the annuity is worth 2000 · a10 4.5%. Discounting back to January1st, i.e. through half of a half-year period, we obtain a value of

(1.045)−12 · 2000a10 4.5% = (1.045)

12 · 2000a10 4.5%

= (1.045)12 · 2000 · 1 − (1.045)−10

0.045= 16177.59053

or 16178 to the nearest dollar.

Non-constant annuities. One student asked about annuities where the payments are notconstant. There are two types that we will be investigating in the next chapter:

1. Annuities where the payments are in “geometric progression”. Suppose that the pay-ments begin with 1, and then increase by a constant ratio of 1 + k over the previouspayment. Then the present value of an n-term annuity-immediate is

v + v2(1 + k) + v3(1 + k)2 + . . . + vn(1 + k)n−1

= v ·1 −

(1 + k1 + i

)n

1 − 1 + k1 + i

2. Annuities where the payments are in “arithmetic progression”. This problem will beconsidered in Chapter 5.

6.14.2 §3.5 PERPETUITIES

The symbol a∞ is used for the present value of a perpetual annuity — an infinite sequence ofpayments, the first one period from now; where the interest rate is not clear from the context,we may write a∞ i. When the interest rate i is 0 the present value would be infinite. Otherwisewe sum an infinite geometric series, obtaining

v + v2 + v3 + . . . =v

1 − v=

viv

=1i.

It can be seen that this is the limit of the usual formula for an as n→ ∞. An infinite annuity ofthis type is called a perpetuity-immediate, or usually just a perpetuity.

Analogously we may define a perpetuity-due to be an infinite sequence of equal payments,where the first is made immediately. We use the symbol a∞ , and can prove that its present

value is1d

. While perpetuities do exist in the real world, for example in the bond market, theyare also useful in providing verbal explanations of identities. For example, we can explainthe formula an = 1−vn

i by expressing the quotient on the right as a difference, 1i − vn 1

i and


interpreting the n-payment annuity as the difference of 2 perpetuities, one beginning a yearfrom now, and the other n + 1 years from now; i.e.,

an = a∞ −(

n|a∞)

[5, Exercise 18, p. 109], [6, Exercise 25, p. 90] “Deposits of 1000 are placed into a fund atthe beginning of each year for the next 20 years. After 30 years, annual payments com-mence, and continue forever, with the first payment at the end of the 30th year. Find anexpression for the amount of each payment.”Solution: The deposits accumulate to a fund worth 1000s20 a year after the last payment,and 1000(1 + i)10 s20 10 years after the last payment. A perpetuity-due bought with thisamount will have annual payments of

1000s20(1 + i)10d = 1000(1 + i)10 · (1 + i)21 − (1 + i)i

· iv= 1000

((1 + i)30 − (1 + i)10

).

Alternatively, we could set up an equation of value at any other time, for example attime t = 0. The value of the deposits will be 1000a20. Suppose that the level amount ofthe withdrawals is X. These withdrawals may be interpreted as either a perpetuity-duefor which the clock starts at the time of the first payment, 30 years from now; or as aperpetuity-immediate for which the clock starts 29 years from now in order that the firstpayment occur at the end of that year. This leads to equations, either

1000a20 = Xv30a∞

or1000a20 = Xv29a∞

which may be solved for the same value of X as determined earlier.

[5, Exercise 19, p. 109] “A deferred perpetuity-due begins payments at time n with annualpayments of 1000 per year. If the present value of this perpetuity is equal to 6561, andthe effective rate of interest is i = 1

9 , find n.”Solution:

1000 ·n∣∣∣∣a∞ 1

9= 6561 ⇔ vn

d= 6.561

⇔ vn−1 = 6.561i = 0.729

⇔(109

)n−1

=

(109

)3

⇔ n − 1 = 3 taking logarithms to base109

⇔ n = 4 .


[5, Exercise 20, p. 109] “A woman has an inheritance in a trust fund for family members leftby her recently deceased father that will pay 50,000 at the end of each year indefinitelyinto the future. She has just turned 60, and does not think that this perpetuity-immediatemeets her retirement needs. She wishes to exchange the value of her inheritance in thetrust fund for one which will pay her a 5-year deferred annuity-immediate, providing hera retirement annuity with annual payments at the end of each year for 20 year followingthe 5-year deferral period. She would have no remaining interest in the trust fund after20 payments are made. If the trustee agrees to her proposal, how much annual retirementincome would she receive? The trust fund is earning an annual effective rate of interestequal to 5%. Answer to the nearest dollar.”Let X be the annual retirement income the woman would receive annually for 20 yearsunder the new, deferred annuity-immediate. The interest rate for the new annuity is notstated, so I am assuming it will be the same as the rate for the trust fund.

50000 · a∞5% = X ·5∣∣∣∣a205%

⇔ X = 50000 · 10.05

(1.05)−5 · 1 − (1.05)−20

0.05

⇔ X =50000(1.05)25

1.0520 − 1= 102, 412.1345 ,

so the annual payment under the new scheme will be 102,412.13.

[5, Exercise 21, p. 109], [6, Exercise 26, p. 90] “A benefactor leaves an inheritance to 4 char-ities, A, B, C, and D. The total inheritance is a series of level payments at the end of eachyear forever. During the first n years, A, B, and C share each payment equally. All pay-ments after n years revert to D. If the present values of the shares of A, B, C, and D areall equal, find (1 + i)n.”Solution: Imposing the condition that the sum of the first n payments is equal to 3 timesthe present value of payments ##n + 1, n + 2, . . ., we obtain an = 3vna∞ ⇒ vn = 1

4 ⇒(1 + i)n = 4.

[5, Exercise 22, p. 109], [6, Exercise 27, p. 90] “A level perpetuity-immediate is to be sharedby A, B, C, and D. A receives the first n payments, B the second n payments, C the thirdn payments, and D the payments thereafter. It is known that the ratio of the present valueof C’s share to A’s share is 0.49. Find the ratio of the present value of B’s share to D’sshare.”Solution: The present values of the shares of A, B, C, D are, respectively, an, vnan, v2nan,and v3na∞. The fact that C’s share, divided by A’s share is to equal 0.49 implies thatvn = 0.7. The ratio of B’s share to D’s is then seen to be

vnan

v3na∞=

0.7 × 0.3(0.7)3


=3049.


6.15 Supplementary Notes for the Lecture of February 03rd, 2010Distribution Date: Wednesday, February 03rd, 2010


6.15.1 §3.6 UNKNOWN TIME

In this section the textbook considers situations where n, the number of annuity payments, isnot known.

Fractions of a payment period It is possible to give a reasonable interpretation to a non-integer number of level annuity payments. The textbook observes that, if we take k to lie inthe interval 0 ≤ k ≤ 1, then, using the formula (52) on page 1054 of these notes, we have

an+k =1 − vn+k

i= an + vn · 1 − vk

i= an + vn+k · (1 + i)k − 1

i, (72)

which can be interpreted as either an n − period level annuity of 1 with an additional paymentof 1−vk

i at time n; or with a final payment of (1+i)k−1i at time n + k. In the first case the additional

payment can be shown to equal, for small i, approximately

k(1 − 1 + k

2i + . . .

);

in the second case the final payment is approximately

k(1 − 1 − k

2i + . . .

);

thus the payment is approximately consistent with a payment of 1 per period. But, becauseof the approximations involved, and the possibility that a court might order a specific inter-pretation — such problems are usually resolved in advance by having a final payment in anotherwise regular sequence of payments cover the amount necessary to make up the difference.Where the final payment needed to meet a goal is larger than a regular payment, it is calleda balloon payment; where it is smaller, it is called a drop payment. Solving problems in thissection typically involves two phases:

• Determination of the number of regular payments.

• Determination of the value of the final payment, subject to given constraints (e.g. that itbe a balloon or drop payment).

We can proceed by first solving an inequality. Read [5, Example 3.7, pp. 92-93], where theauthor considers a problem where a planned final payment turns out to be negative.


[5, Exercise 23, p. 110], [6, Exercise 29, p. 90] “Compute a5.25 if i = 5% using the followingdefinitions:

1. Formula [5, (3.20)] ([6, (3.22)]).2. A payment of 0.25 at time 5.25.3. A payment of 0.25 at time 6.

Solution:

1. a5 + v5.25 (1.05)14

0.05 = 4.519457520.

2. a5 + v5.250.25 = 4.522983461.

3. a5 + v60.25 = 4.516030529.

[5, Exercise 24, p. 110], [6, Exercise 32, p. 91] “A loan of 1000 is to be repaid by annual pay-ments of 100 to commence at the end of the 5th year, and to continue thereafter for aslong as necessary. Find the time and amount of the final payment if the final payment isto be larger than the regular payments. Assume i = 4.5%.”Solution: Let the time of the last — balloon — payment be n, and let the amount of thelast payment be X. Then n is the largest integer solution to the inequality

1000 ≥ 100(1.045)−4an−4 = 100(1.045)−4 · 1 − (1.045)−(n−4)

0.045

⇔ (1.045)−(n−4) ≥ 1 − 1000 × 0.045 × (1.045)4

100⇔ −(n − 4) ln 1.045 ≥ ln

(1 − 10 × 0.045 × (1.045)4

)

⇔ −(n − 4) ≥ln

(1 − 0.45(1.045)4

)

ln 1.045

⇔ n ≤ 4 −ln

(1 − 0.45(1.045)4

)

ln 1.045= 21.47594530.

Thus we conclude that the balloon payment is made at time t = 21. The equation of valueat time t = 21 is

1000(1.045)21 = 100s17 + (X − 100)

implying that

X = 100 + 1000(1.045)21 − 1000.045

((1.045)17 − 1

)= 146.070467 .

[6, Exercise 33, p. 91] “A fund of 2000 is to be accumulated by n annual payments of 50,followed by n annual payments of 100, plus a smaller final payment made 1 year after


the last regular payment. If the effective rate of interest is 4.5%, find n and the amount ofthe final irregular payment.”Solution: We shall interpret the payments to be made under two annuities-due: the first,for 2n years, consists of an annual deposit of 50 in advance; the second, for n years,deferred n years after the first, also consists of an annual deposit of 50 in advance. Itis at the end of year 2n that the final, drop payment is to be made, and it is to be under100. (Note that this is the type of problem where the drop payment could turn out to benegative. We seek the smallest n for which

50s2n + 50sn > 2000 − 100

⇔ 50(1.045) · (1.045)2n + (1.045)n − 20.045

> 1900

⇔ (1.045)2n + (1.045)n − 2 >190050· 0.045

1.045

⇔((1.045)n +

12

)2

>1900

50· 0.045

1.045+ 2.25 = 3.886363636

Since the exponential is positive, the preceding inequality is equivalent to (1.045)n >1.471386222, and, in turn, to

n >ln 1.471386222

ln 1.045= 8.774018446 .

Thus the drop payment will be when t = 2× 9, i.e., 18 years after the first payment underthe annuity with payments of 50. Just before the drop payment the accumulated value ofall previous payments is

50(s9 + s18

)= 50(1.045) · (1.045)18 + (1.045)9 − 2

0.045= 1967.588591

so the drop payment at time t = 18 is 2000 − 1967.588591 = 32.411409.Note that there is an error in the answers in the textbook: while n = 9 is correct, thepayment of 32.41 is not “at time n = 9”.(If tables like those in the textbook were available, one could determine the value of n byinspecting the value of s2n + sn. We observe from the 4.5% tables the following values:

n s2n + sn

8 32.09939 37.6572

10 43.6596


We seek the smallest n such that

50s2n + 50sn > 2000 − 100

i.e., such thats2n + sn > 38 ,

equivalently,

s2n + sn >38

1.045= 36.37 ,

and so can conclude that n = 9.)

Suppose now that we change the last problem slightly, as follows:

Example 6.11 (cf. [6, Exercise 33, p. 91]) A fund of 2000 is to be accumulated by n annualpayments of 50, followed by n annual payments of 100, plus a smaller final payment of notmore than 25 made 1 year after the last regular payment. If the effective rate of interest is4.5%, find n and the amount of the final irregular payment.”Solution: I will adapt my solution for the original version. Again I interpret the payments tobe made under two annuities-due: the first, for 2n years, consists of an annual deposit of 50 inadvance; the second, for n years, deferred n years, also consists of an annual deposit of 50 inadvance. It is at the end of year 2n that the final, drop payment is to be made, and it is to beunder 25. We seek the smallest integer n for which

50s2n + 50sn > 2000 − 25

⇔ 50(1.045) · (1.045)2n + (1.045)n − 20.045

> 1975

⇔ (1.045)2n + (1.045)n − 2 >1975

50· 0.045

1.045

⇔((1.045)n +

12

)2

>1975

50· 0.045

1.045+ 2.25 = 3.950956938.

Since the exponential is positive, the preceding inequality is equivalent to (1.045)n > 1.487701421,and, in turn, to

n >ln 1.487701421

ln 1.045= 9.024542616 .

Thus the drop payment will be when t = 2 × 10, i.e., 20 years after the first payment underthe annuity with payments of 50. Just before the drop payment the accumulated value of allprevious payments is

50(s10 + s20

)= 50(1.045) · (1.045)20 + (1.045)10 − 2

0.045= 2281.215780 ,


so the drop payment at time t = 20 is 2000 − 2281.215780 = −281.215780. Thus, by insistingthat the last payment be very small, the payment schedule was extended so that the borroweractually gets money back after the n + n payments as prescribed.


6.16 Supplementary Notes for the Lecture of February 05th, 2010Distribution Date: Friday, February 05th, 2010


6.16.1 §3.7 UNKNOWN RATE OF INTEREST

I do not expect you to be able to use the iteration algorithms built into financial calculators.

Approximation formulæ for i, given sn i or an i In what follows, assume that n > 0 isgiven, as well as either an i or sn i. It is trivial to test whether i = 0, since then an 0 = n = sn 0.Now assume i > 0, and that i is “small”.

Approximation formula for i, given an i

1an i− 1

n=

11 − (1 + i)−n −

1n

=i

1 −(1 − ni +

n(n+1)2 i2 + . . .

) − 1n

(binomial expansion)

=1n

11 − n+1

2 i + . . .− 1

≈ 1n

(1 +

n + 12

i − 1)

(geometric series)

=n + 1

2ni

Solving this approximate equation for i we obtain

i ≈2(n − an i

)

an i · (n + 1). (73)

Approximation formula for i, given sn i

1n− 1

sn i=

1n− 1

(1 + i)n − 1

=1n− i(

1 + ni +n(n−1)

2 i2 + . . .)− 1


(binomial expansion)

=1n

1 − 11 + n−1

2 i + . . .

≈ 1n

(1 − 1 +

n − 12

i)

(geometric series)

=n − 1

2ni

Solving this approximate equation for i we obtain i ≈2(sn i − n

)

sn i · (n − 1).

Approximating compound interest for fractions of a measurement period In the previousedition of the textbook [6, §2.2] the author observed that a method that is often used when afraction of a measurement period is involved is to approximate compound interest by simpleinterest. This amounts to using series (44) above, but stopping after the 1st degree term:

(1 + i)k ≈ 1 + ki .

It can also be interpreted as linear interpolation between values of (1+i)x for integer x. Supposethat we are interested in the accumulation factor for compound interest between t = n and t =

n+k, where 0 ≤ k ≤ 1. If we think of a line joining the points (n, (1 + i)n) and(n + 1, (1 + i)n+1

),

and take as a value approximating (1 + i)n+k, the ordinate of the point where this line meets theline x = n + k, we have similar triangles, leading to the equation

k1

=approximation − (1 + i)n

(1 + i)n+1 − (1 + i)n

from which we determine that the approximation is

k(1 + i)n+1 + (1 − k)(1 + i)n = (1 + i)n (k(1 + i) + (1 − k)) = (1 + i)n(1 + ki).

The textbook remarks that linear approximation between successive integer values of vx isequivalent to simple discount. We can obtain a better approximation by truncating the MacLau-rin expansion after a higher term than the first degree term. In the following example theaccumulation factor is approximated by a quadratic — not a linear — function.

Unknown rate of interest Where it is the interest rate that is not known, there can be severaldifferent approaches:

• Algebraic methods:


– Where possible, one tries to find a solution using algebraic methods. Since the vari-ous formulæ we work with are usually polynomial in i or ratios of polynomials in i,it may be possible to find the particular solution(s) we seek by algebraic means.

– Algebraic means may still be available where solving for i fails, if we can find an-other convenient intermediary variable.

• It may be possible to solve by interpolation on tables, provided the functions we areinterested in are tabulated.

• The favoured method is by successive approximation, to obtain a solution to any desiredaccuracy.

[5, Exercise 29, p. 110], [6, Exercise 38, p. 92] “If a2 = 1.75, find an exact expression for i.”Solution:

a2 = 1.75 ⇔ (1 + i)2 − 1i

= 1.75(1 + i)2

⇔ 1.75i2 + 2.5i − 0.25 = 0 (74)⇔ 7i2 + 10i − 1 = 0

⇔ i =−10 +

√100 + 28

14=−5 − 4

√2

7or−5 + 4

√2

7. (75)

Since i > 0, the second solution is inadmissible, and i = 0.09383632129.

[5, Exercise 30, p. 111], [6, Exercise 42, p. 92] “A beneficiary receives a 10,000 life insur-ance benefit. If the beneficiary uses the proceeds to buy a 10-year annuity-immediate,the annual payout will be 1,538. If a 20-year annuity-immediate is purchased, the annualpayout will be 1,072. Both calculations are based on an annual effective interest rate of i.Find i.”Solution: The equations of value at time 0 are

10000 = 1538 · a10 = 1072 · a20

We know that i , 0, as a10 0 = 1. The last equation alone implies that 1 + v10 =

1.434701492, which implies that i = 8.6878222%. The information about the amount ofthe insurance benefit was redundant.

[5, Exercise 31, p. 111] “The present values of the following three annuities are equal:

(i) “perpetuity-immediate paying 1 each year, calculated at an annual effective interestrate of 7.25%;

(ii) “50-year annuity-immediate paying 1 each year, calculated at an annual effectiveinterest rate of j%;


(iii) “n-year annuity-immediate paying 1 each year, calculated at an annual effective in-terest rate of j − 1%.

“Calculate n.”Solution: We are given that

a∞ 7.25% = a50 j% = an ( j−1)% ,

which is equivalent to

13.79310345 =1

0.0725=

1 − (1 + j)−50

j=

1 − (0.99 + j)−n

j − 0.01.

Equality of (i) and (ii) yields, using the approximation formula in [5, Appendix 3, p. 106](equation (73) on page 1073 of these notes),

j ≈ 2(50 − 13.7931625)13.79310345(51)

= .1029410086 .

For these large values of n, i, the approximation is not very good, since

a50 0.102941 =1 − (1.102941)−50

0.102941= 9.6419 ,

which is quite far from the known value of 13.7931625. However, since the perpetuityis calculated at a rate of 7.25%, and is being truncated, the interest rate for a50 should beless than 7.25%. By trial and error and then by an iterative procedure such as bisection orsomething more sophisticated, we can show that j = 0.07004382216, or approximatelyj = 7.00%.Now we need to solve an equation

an (7.00−1)% = 13.79310345 ⇔ 1.06−n = 1 − (0.06)(13.79310345) = 0.1724137930⇔ n = 30.168

so n is approximately 30.2. (There remains, of course, the issue of what is meant by anwhen n is not an integer.)

6.16.2 §3.8 VARYING INTEREST

I will omit this section, except for stating the following definition.Suppose that the interest rate ik applies only to the kth period of time — from time k − 1

to time k. The textbook considers two different ways of interpreting the applicability of thesedifferent rates. Of course there is a need for clarity when more than one interest rate is stated,as the reader must be able to infer precisely which of these methods is applicable, if not anentirely different method.


Definition 6.17 1. In the Portfolio Rate Method rate ik applies to all interest transactionsthat cross this time interval, even if the actual payment was made later than the kth period.Thus, for example,

an =1

1 + i1+

11 + i1

· 11 + i2

+ . . . +1

1 + i1· 1

1 + i2· . . . · 1

1 + in.

andsn = (1 + in) + (1 + in)(1 + in−1) + . . . + (1 + in)(1 + in−1) · (1 + i1) .

2. In the Yield Curve Method rate ik applies only to the payments made during the kth period,and follow those payments back to the present. Thus, for example,

an =1

1 + i1+

(1

1 + i1

)2

+ . . . +

(1

1 + in

)n

.

andsn = (1 + in) + (1 + in−1)2 + . . . + (1 + i1)n .

6.16.3 §3.9 ANNUITIES NOT INVOLVING COMPOUND INTEREST

I wish to consider only one problem from this section:

[5, Exercise 38, p. 112] “Given that δt = 120−t (t ≥ 0), find s10.”

Solution: Note that the theory of [5, §4.5] does not apply here, as it is for continuousannuities with a constant rate. In this case the payments are discrete — at 1-year intervals,but the interest rate is changing continuously. When we use the symbol sn here we don’thave a constant interest rate, so we can’t use the formula that we have proved to applywhen interest is constant.23 The payment at time n is worth, at time 10,

e

10∫nδt dt

= e−[ln(20−t)]10n =

20 − n10

.

Hence

s10 =1

10

10∑

n=1

(20 − n) =19 · 20

20− 9 · 10

20= 14.5 .

23That isn’t so unusual, as we also can’t use that formula in situations where i = 0.



[5, Exercise 40, p. 112] “At an annual effective interest rate of i, both of the following annu-ities have a present value of X:

(i) a 20-yeear annuity-immediate with annual payments of 55;(ii) a 30-year annuity-immediate with annual payments that pay 30 per year for the first

10 years, 60 per year for the second 10 years, and 90 per year for the final 10 years.

Calculate X.”Solution: There are two unknowns here: i and X. The hypotheses yield two equations forthese unknowns:

X = 55a20 i

X = 90a30 i − 30a20 i − 30a10 i

Could i = 0? If that were the case, the annuities could be evaluated, and the two equationswould reduce to

X = 55(20) = 1100X = 90(30) − 30(20) − 30(10) = 1800 .

Since the equations are contradictory, we conclude that the hypothesis that i = 0 is in-valid.Now, using the property that vn = 1 + ian, we find from the second equation that

iX = 90(1 − v30) − 30(1 − v20) − 30(1 − v10)= 30(1 − v10)(3v20 + 2v10 + 1)

so the second of our original equations may be rewritten in the form

X = 30a10 ·(3v20 + 2v10 + 1

).

The first original equation may be similarly rewritten in the form

X = 55a10 ·(1 + v10

).

Dividing one equation by the other yields

55(v10 + 1

)= 30

(3v20 + 2v10 + 1

),

which is equivalent to the factorized quadratic equation(2v10 − 1

) (9v10 + 5

)= 0 .

Of the two roots, the negative is extraneous, so v10 = 12 and i =

10√2. Hence a20 =1− 1

410√2−1

=

10.44954456, so X = 55a20 = 574.7249508.


[5, Exercise 41, p. 112] “To accumulate 8000 at the end of 3n years, deposits of 98 are madeat the end of the first n years, and 196 at the end of the next 2n years. The annual effectiverate of interest is i. You are given that (1 + i)n = 2. Determine i.”Solution: An equation of value, at time 3n is

8000 = 98(s3n + s2n

)

⇒ 8000 =98i

(((1 + i)3n − 1

)+

((1 + i)2n − 1

))

⇒ 8000 =98i

((8 − 1) + (4 − 1) = 980 ,

which implies that i = 12.5%.

[5, Exercise 42, p. 112], [6, Exercise 51, p. 93] “A loan of 10,000 is to be repaid, with annualpayments at the end of each year for the next 20 years. For the first 5 years the paymentsare k per year; the second 5 years, 2k per year; the third 5 years 3k per year; and thefourth 5 years, 4k per year. Find an expression for k.”Solution: The trick is to interpret the payment scheme in terms of annuities-certain, allbeginning at the present. We find that

10000 = k(4a4k − a3k − a2k − ak

),

sok =

100004a4k − a3k − a2k − ak

.

6.16.5 APPENDIX 3

This section was discussed in connection with [5, §3.7]


6.17 Supplementary Notes for the Lecture of February 07th, 2010Distribution Date: Monday, February 07th, 2010


Prior to the beginning of the lecture, students were asked for an opinion on thetentative date for the Term Test, presently scheduled for 03 March. The consensusappears to be that the test should be on one of the days: Wednesday, 03 March, 2010;Monday, 08 March, 2010; Friday, 10 March, 2010; Monday, 15 March, 2010.

Textbook Chapter 4. More General Annuities.


The textbook considers

• annuities whose payments do not have the same frequency as the interest conversionperiods;

• annuities whose payments are not constant.

For the first of these topics, contained in the following three sections of the textbook, we shallnot study all the material carefully, but shall consider ad hoc solutions to problems.

6.17.2 §4.2 ANNUITIES PAYABLE AT A DIFFERENT FREQUENCY THAN INTER-EST IS CONVERTIBLE

A typographical error in [6] (corrected in [5]). The following example, given in the textbook[6, Example 4.1, p. 96] contains a serious error. “Example 4.1 Find the accumulated value at the end of45 years of an investment fund in which 100 is deposited at the beginning of each quarter for the first2 years, and 200 is deposited at the beginning of each quarter for the second 2 years, if the fund earns12% convertible quarterly.”Solution: If the interest rate were as stated the problem would be of a type studied in Chapter 2. Areading of the solution shows that the author intended the last word to be monthly.

[5, Exercise 1, p. 146], [6, Exercise 1, p. 122] “Find the accumulated value 18 years after thefirst payment is made of an annuity on which there are 8 payments of 2000 each made at2-year intervals. The nominal rate of interest convertible semiannually is 7%. Answer tothe nearest dollar.”Solution:


1. First Solution: Working with 2-year periods. The nominal biennial interest ratei(

12 ) is given by

1 +i(

12 )12

12

= 1 + i =

(1 +

i(2)

2

)2

=

(1 +

0.072

)2

⇒ i(12 ) =

12·((1.035)4 − 1

)

and the effective biennial rate — call it j — is

j =i(

12 )12

= (1.035)4 − 1 .

Using this interest rate and a scale with time unit of a 2 year interval, the payments— 2 intervals after the last — are worth

2000(s10 − s2

)=

2000j

((1 + j)10 − (1 + j)2

)

= 2000((1.035)40 − (1.035)8

(1.035)4 − 1

)

= 35824.25354

which is 35824 to the nearest unit. (Why, then, does the textbook [6, p. 419] give thevalue as 35825? Because the author is using his tables. If we work with the 3.5% tables [6,p. 383], we obtain

2000(s10 − s2

)= 2000

s40 0.035 − s8 0.035

s4 0.035

(76)

= 2000(84.5503 − 9.0517

4.2149

)(77)

= 35824.62217 . (78)

The tables also contain values of the inverse of s4 0.035:

2000

s40 0.035 − s8 0.035

s4 0.035

= 2000s−14 0.035

(s40 0.035 − s8 0.035

)

= 2000(0.237251)(84.5503 − 9.0517)

= 35824.23670 ,

which gives an answer which is closer to the correct one.)


2. Second Solution: Working with half-year periods. The effective interest rate forhalf-year periods is 1

2 (7%) = 3.5%. A payment of 2000 at the end of a 2-year periodcould be replaced by 4 equal payments at the ends of the first, second, third, and 4thhalf year (which is the time of the payment of 2000). If we denote the amount ofthese equal payments by X, we have the equation of value at the end of the 2-yearperiod,

X · s4 3.5% = 2000

so X =2000s4 3.5%

. There will be 4× 8 = 32 such payments of X, the last of them 8 half-

years before the date at which we require the accumulated value. We can, insteadthink of a series of 32 + 8 = 40 payments, and then subtract the accumulated valueof the last 8 payments that we have added. We obtain as the accumulated value

X(s40 3.5% − s8 3.5%

)=

2000s4 3.5%

·(s40 3.5% − s8 3.5%

)

which agrees with equation (76) above.[5, Exercise 2, p. 146], [6, Exercise 2, p. 122] “Find the present value of a 10-year annuity

which pays 400 at the beginning of each quarter for the first 5 years, increasing to 600per quarter thereafter. The annual effective rate of interest is 12%. Answer to the nearestdollar.”Solution: With 1 + i = (1.12)

14 ,

Present Value = 600a40 − 200a20

= (600 − 200) + 600(1 − v39

i

)− 200

(1 − v19

i

)

= 11466.12687

[5, Exercise 3, p. 146], [6, Exercise 3, p. 122] “A sum of 100 is placed into a fund at the be-ginning of every other year for 8 years. If the fund balance at the end of 8 years is 520,find the rate of simple interest earned by the fund.”Solution: The wording of this problem is not as precise as it could be — while it is clearthat the payments into the fund are 2 years apart, it is not clear whether they are at thebeginnings of years ##0, 2, 4, 6 or the beginnings of years ##1, 3, 5, 7. From the author’sanswer we see that he intended the former interpretation.Let i be the annual rate of simple interest. Then the equation of value at time t = 8 is

100[(1 + 8i) + (1 + 6i) + (1 + 4i) + (1 + 2i)] = 520⇒ i = 6%

(The other interpretation mentioned would have given

100[(1 + 7i) + (1 + 5i) + (1 + 3i) + (1 + 1i)] = 520⇒ i = 7.5% .)


[5, Exercise 4, p. 146] “An annuity-immediate that pays 400 quarterly for the next 10 yearscosts 10,000. Calculate the nominal interest rate convertible monthly earned by thisinvestment.”Solution: Let i be the effective quarterly interest rate. Then

a40 i =1 − v40

i=

10000400

= 25,

which can be solved by iteration to yield i = 0.025243849. The effective monthly rateis, therefore, (1.025243849)0.25 − 1 = 0.006252085, and the nominal annual interest rate,convertible monthly, is 12 × 0.006252085 = 7.5025020%, presumably 7.5%.


6.18 Supplementary Notes for the Lecture of February 10th, 2010Distribution Date: Wednesday, February 10th, 2010


6.18.1 §4.3 ANNUITIES PAYABLE LESS FREQUENTLY THAN INTEREST IS CON-VERTIBLE

[5, Exercise 5, p. 146], [6, Exercise 4, p. 122] “Rework [5, Exercise 1, p. 146] using the ap-proach developed in [5, §4.3].”Solution: There is more than one approach in the section. It would appear that the text-book wishes the student to apply the formulæ developed in the section, whereby a scalingfactor is used. Preferring to work from first principles, I interpret this instruction to referto the replacement of the biennial payments by payments spaced according to the inter-est compounding interval — here a half-year. If we let R denote the amount of such apayment under an annuity-immediate paying 2000 every 2 years, we have

R · s4 3.5% = 2000

so R =2000s4 3.5%

. The value 18 years after the first biennial payment is made — i.e., 4 years

after the last biennial payment of 2000 is made will be equal to the value of a semi-annual32-payment annuity-immediate of R per half-year, 8 half-years after the last payment, i.e.

(1.035)8 · R · s32 3.5% = (1.035)8 · 2000s4 3.5%

· s32 3.5%

= 2000(1.035)8 ·s32 3.5%

s4 3.5%

= 2000(1.035)8 · 57.33454.2149

= 35824.61636.

[5, Exercise 6, p. 146] “Give an expression in terms of functions assuming a rate of interestper month for the present value, 3 years before the first payment is made, of an annuityon which there are payments of 200 every 4 months for 12 years:

1. expressed as an annuity-immediate;2. expressed as an annuity-due.”

Solution: (Since any scheme of equal payments can be viewed from the appropriate timeas either an annuity-immediate or annuity-due, the instruction isn’t completely clear.While I obtain the answers the author appears to expect, my solution to each of the partsgives rise to a solution to the other when we recognize that the addition or removal of


the double-dots can be effected in a fraction of the type I obtain as solutions without anyother change.)In order to use these functions, we need to replace the regular payment every 4 monthsby an equivalent monthly payment. Let i represent the effective monthly interest rate,and A and B respectively be the monthly payments at the end of the month/in advanceequivalent to a payment under an annuity-immediate/annuity-due of 200 every 4 months.Then

As4 i = 200Ba4 i = 200

Both replacement annuities will pay 12 × 12 = 144 monthly payments.

1. The annuity-due is deferred 3 years, i.e., 3×12 = 36 months, before the first paymentis made — that means 32 months before time-0 of an equivalent annuity-immediateat 4-month intervals, and also 32 months before the beginning of the replacementannuity-immediate with monthly payments of A.

Value = A ·(

32

∣∣∣a144i

)

= 200 ·v32a144

s4

= 200 ·a176 − a32

s4

= 200 ·a176 − a32

s4

2. This annuity-due is deferred 36 months.

Value = B ·(

36

∣∣∣a144i

)

= 200 · v36a144

a4

= 200 ·a180 − a36

a4

= 200 ·a180 − a36

a4

[5, Exercise 7, p. 146], [6, Exercise 8, p. 122] “Find an expression for the present value of anannuity-due of 600 per annum payable semiannually for 10 years, if d(12) = 0.09.”


Solution: The effective discount rate per month is 0.0912 = 3

4%. The effective discount rateper 6 months is, therefore, d′ = d(2)

2 = 1 − (1 − 0.0075)6. The textbook wishes us tointerpret the statement “...600 per annum payable semiannually” to mean “...300 paid perhalf-year”. With this interpretation the present value of the 20-half-year annuity-due is,therefore,

300 · a20 = 600 · 1 − (1 − d′)20

d′

= 300 · 1 − (1 − 0.0075)120

1 − (1 − 0.0075)6

[5, Exercise 8, p. 146], [6, Exercise 9, p. 122] “The present value of a perpetuity paying 1 at

the end of every 3 years is12591

. Find i.”

Solution: A payment of 1 at the end of the year for 3 years is equivalent to a payment ofs−1

3at the end of every year. The equation of value is

12591

= s−13· 1

i

⇒ 12591

=1

(1 + i)3 − 1

⇒ (1 + i)3 =

(65

)3

⇒ i = 20%.

[5, Exercise 9, p. 146], [6, Exercise 10, p. 123] “Find an expression for the present value ofan annuity on which payments are 100 per quarter for 5 years, just before the first pay-ment is made, if δ = 0.08.”Solution: The effective interest rate per quarter is i = e

0.084 − 1, so v = e−0.02. Accordingly,

the present value of the annuity-due is

100a20 i = 100 ·(1 + i) ·

(1 − v20

)

i

= 100 · 1 − e−0.4

e−0.02 · (−1 + e0.02)

= 100 · 1 − e−0.4

1 − e−0.02

[5, Exercise 10, p. 146], [6, Exercise 12, p. 123] “Find an expression for the present value ofan annuity on which payments are 1 at the beginning of each 4-month period for 12 years,assuming a rate of interest per 3-month period.”


Solution: If the effective interest rate per 3-month period is i, then the equivalent effectiverate per 4-month period will be j = (1 + i)

43 − 1. The present value of the annuity-due at

this rate for 12 × 3 4-month periods will be

a36 j =(1 + j) − (1 + j)−35

j

=(1 + i)

43 − (1 + i)−

1403

(1 + i)43 − 1

=1 − v48

1 − v43



revised 11 February, (subject to further revision)

The data of the class test has still not been finalized. The latest choices appear to beeither March 8th or March 12th.

6.19.1 §4.3 ANNUITIES PAYABLE LESS FREQUENTLY THAN INTEREST IS CON-VERTIBLE

[5, Exercise 5, p. 146], [6, Exercise 4, p. 122] “Rework [5, Exercise 1, p. 146] using the ap-proach developed in [5, §4.3].”Solution: There is more than one approach in the section. It would appear that the text-book wishes the student to apply the formulæ developed in the section, whereby a scalingfactor is used. Preferring to work from first principles, I interpret this instruction to referto the replacement of the biennial payments by payments spaced according to the inter-est compounding interval — here a half-year. If we let R denote the amount of such apayment under an annuity-immediate paying 2000 every 2 years, we have

R · s4 3.5% = 2000

so R = 2000s4 3.5%

. The value 18 years after the first biennial payment is made — i.e., 4 years

after the last biennial payment of 2000 is made will be equal to the value of a semi-annual32-payment annuity-immediate of R per half-year, 8 half-years after the last payment, i.e.

(1.035)8 · R · s32 3.5% = (1.035)8 · 2000s4 3.5%

· s32 3.5%

= 2000(1.035)8 ·s32 3.5%

s4 3.5%

= 2000(1.035)8 · 57.33454.2149

= 35824.61636.

[5, Exercise 6, p. 146] “Give an expression in terms of functions assuming a rate of interestper month for the present value, 3 years before the first payment is made, of an annuityon which there are payments of 200 every 4 months for 12 years:

1. expressed as an annuity-immediate;2. expressed as an annuity-due.”

Solution: (Since any scheme of equal payments can be viewed from the appropriate timeas either an annuity-immediate or annuity-due, the instruction isn’t completely clear.


While I obtain the answers the author appears to expect, my solution to each of the partsgives rise to a solution to the other when we recognize that the addition or removal ofthe double-dots can be effected in a fraction of the type I obtain as solutions without anyother change.)In order to use these functions, we need to replace the regular payment every 4 monthsby an equivalent monthly payment. Let i represent the effective monthly interest rate,and A and B respectively be the monthly payments at the end of the month/in advanceequivalent to a payment under an annuity-immediate/annuity-due of 200 every 4 months.Then

As4 i = 200Ba4 i = 200

Both replacement annuities will pay 12 × 12 = 144 monthly payments.1. The annuity is deferred 3 years, i.e., 3 × 12 = 36 months, before the first pay-

ment is made — that means 32 months before the beginning of annuity-immediateat 4-month intervals, and also 32 months before the beginning of the replacementannuity-immediate with monthly payments of A.

Value = A ·(

32

∣∣∣a144i

)

= 200 ·v32a144

s4

= 200 ·a176 − a32

s4

.

If the intention of part 2 was simply with reference to the symbols used in the answer,then we could simply multiply the numerator and denominator of the preceding ratioby v, and thereby obtain

200 ·a176 − a32

s4

.

2. This annuity is deferred 36 months. Suppose that the author’s intention is that wereplace each 4-monthly payment by 4 monthly payments, the first of which occursat the time of the 4-monthly payment being replaced. Then the new payments willbe in the amount of B, calculated above. It follows that

Value = B ·(

36

∣∣∣a144i

)

= 200 · v36a144

a4

= 200 ·a180 − a36

a4

.


Here also we could multiply numerator and denominator by v and obtain 200 ·a180 − a36

a4

.

[5, Exercise 7, p. 146], [6, Exercise 8, p. 122] “Find an expression for the present value of anannuity-due of 600 per annum payable semiannually for 10 years, if d(12) = 0.09.”Solution: The effective discount rate per month is 0.09

12 = 34%. The effective discount rate

per 6 months is, therefore, d′ = d(2)

2 = 1 − (1 − 0.0075)6. The textbook wishes us tointerpret the statement “...600 per annum payable semiannually” to mean “...300 paid perhalf-year”. With this interpretation the present value of the 20-half-year annuity-due is,therefore,

300 · a20 = 600 · 1 − (1 − d′)20

d′

= 300 · 1 − (1 − 0.0075)120

1 − (1 − 0.0075)6

[5, Exercise 8, p. 146], [6, Exercise 9, p. 122] “The present value of a perpetuity paying 1 at

the end of every 3 years is12591

. Find i.”

Solution: A payment of 1 at the end of the year for 3 years is equivalent to a payment ofs−1


12591

= s−13· 1

i

⇒ 12591

=1

(1 + i)3 − 1

⇒ (1 + i)3 =

(65

)3

⇒ i = 20%.

[5, Exercise 9, p. 146], [6, Exercise 10, p. 123] “Find an expression for the present value ofan annuity on which payments are 100 per quarter for 5 years, just before the first pay-ment is made, if δ = 0.08.”Solution: The effective interest rate per quarter is i = e

0.084 − 1, so v = e−0.02. Accordingly,

the present value of the annuity-due is

100a20 i = 100 ·(1 + i) ·

(1 − v20

)

i

= 100 · 1 − e−0.4

e−0.02 · (−1 + e0.02)


[5, Exercise 10, p. 146], [6, Exercise 12, p. 123] “Find an expression for the present value ofan annuity on which payments are 1 at the beginning of each 4-month period for 12 years,assuming a rate of interest per 3-month period.”Solution: If the effective interest rate per 3-month period is i, then the equivalent effectiverate per 4-month period will be j = (1 + i)

43 − 1. The present value of the annuity-due at

this rate for 12 × 3 4-month periods will be

a36 j =(1 + j) − (1 + j)−35

j

=(1 + i)

43 − (1 + i)−

1403

(1 + i)43 − 1

=1 − v48

1 − v43





6.20.1 §4.3 ANNUITIES PAYABLE LESS FREQUENTLY THAN INTEREST IS CON-VERTIBLE (conclusion)

[6, Exercise 6, p. 122] “Show that the present value at time 0 of 1 payable at times 7, 11, 15,19, 23, and 27 is

a28 − a4

s3 + a1

.

Solution: s3 + a1 is the present value — just after the third payment — of a 4-paymentannuity of 1 for which the first payment was 1 unit ago. Hence a single payment of 1 at

time 3 is equivalent to payments of1

s3 + a1

at times 1, 2, 3, 4. An n-payment annuity

that pays 1 every 4 years with first payment in 3 years, may be replaced by a 4n-payment

annuity which pays1

s3 + a1

at the end of every year, first payment at the end of the year

which starts now, and its value now isa4n

s3 + a1

. But, if we delete the first payment from

the original annuity which pays every 4 years — i.e., if we delete the payment of 1 at theend of year 3, then we have deferred the new annuity by 4 years, and the present value isa4n − a4

s3 + a1

. The problem posed by the textbook is the case n = 7, as there are 6 payments

of 1. Thus the present value of the annuity described is

a28 − a4

s3 + a1

.

�

[6, Exercise 7, p. 122] “A perpetuity of 750 payable at the end of every year, and a perpetuityof 750 payable at the end of every 20 years are to be replaced by an annuity of R payableat the end of every year for 30 years. If i(2) = 0.04, show that

R = 37500 · 1

s2

+v40

a40

·s2

a60

where all functions are evaluated at 2% interest.”Solution: The given nominal annual interest rate of 4%, compounded semi-annually, isequivalent to an effective semi-annual rate of 2%.


1. Each payment of 750 at the end of a year is equivalent to 2 payments of750s2 2%

, one

at the end of 6 months, the other at the end of the year. Thus the present value of the

first perpetuity is750s2 2%

· a∞ 2%.

2. The perpetuity of 750 payable at the end of 20-year intervals, is, analogously, worth

at present750

s40 2%

· a∞ 2%.

3. Hence the amount available to purchase the annuity is

750 · a∞ 2% · 1

s2 2%

+1

s40 2%

= 750 · 10.02

1s2 2%

+v40

v40s40 2%

= 37500 · 1

s2 2%

+v40

a40 2%

(79)

4. A payment of R at the end of a year is equivalent to a payment of Rs2

at the end of

every 6 months. The present value of the 30-year annuity is, therefore,

R ·a60

s2

(80)

since we are replacing it by a 60-half-year annuity-immediate.5. The equation of value equates amounts (79) and (80), implying that

R =

37500 · 1

s2 2%

+v40

a40 2%

a60

s2

= 37500 · 1

s2 2%

+v40

a40 2%

·s2

a60

�

[6, Exercise 11, p. 123] “A perpetuity paying 1 at the beginning of each year has a presentvalue of 20. If this perpetuity is exchanged for another perpetuity paying R at the begin-ning of every 2 years, find R so that the values of the two perpetuities are equal.”Solution: An equation of value now for the perpetuity is 1

d = 20, implying that i =d

1 − d=

119

. At an effective annual discount rate of d, a payment of 1 now is equivalent

to a 2-year annuity-due paying 11+v = 1+i

2+i each year, in advance. The perpetuity-duepaying R at the beginning of each 2-year period is equivalent to a perpetuity-due payingR · 1+i

2+i at the beginning of every year. Equating this to 20 yields R = 3920 = 1.95.




6.21.1 §4.4 FURTHER ANALYSIS OF ANNUITIES PAYABLE MORE FREQUENTLYTHAN INTEREST IS CONVERTIBLE

In §4.4 we have a new notation. A right-superscribed (m) indicates that the annuity is payablem times in a time unit for compounding of interest; more than that, we will assume that the unitpreviously associated with one payment is divided into m parts; this usage is analogous to theuse of the upper right parentheses in i(m). Thus a(m)

n irepresents the present value of an annuity

of 1m payable in m time-subintervals of length 1

m : there will be immediate and due versions ofthe symbol, analogous symbols for perpetuities, and corresponding symbols for s.

We can develop formulæ for functions like a(m)i , where i is the interest rate for a period

which contains m payments of 1m . We find, for example, that

a(m)n i

=1 − vn

i(m) =1 − vn

m((1 + i)

1m − 1

) ;

a(m)n i

=1 − vn

d(m) =1 − vn

m(1 − (1 − d)

1m

) =1 − vn

m(1 − (1 + i)−

1m

) ;

but we can also solve problems involving this function from first principles, by convertinginterest rates or payments so that the intervals of interest and payment coincide; and it is thislatter practice that I shall try to follow, rather than making extensive use of the new functions.We can also generalize these functions to perpetuities.

[5, Exercise 11, p. 147], [6, Exercise 13, p. 123] “Rework [5, Exercise 2, p. 146] using theapproach developed in [5, Section 4.4].”Solution: The annuity can be expressed as the difference

4 × 600a(4)

10 12%− 4 × 200a(4)

5 12%=

4(600)(1 − (1.12)−10

)− 4(200)

(−(1.12)−5

)

4(1 − 1.12−

14

)

= 11466.12691 .

[5, Exercise 12, p. 147], [6, Exercise 19, p. 123] 1. “Show that a(m)n

=1m

m∑

t=1

vtm an .

2. “Verbally interpret the (preceding) result.”Solution:


1.

a(m)n

=1m

(v

1m + v

2m + . . . + v

nm−1m + v

nmm)

=v

1m

m

(1 + v + v2 + . . . + vn−1

)+

v2m

m

(1 + v + v2 + . . . + vn−1

)+ . . .

+v

m−1m

m

(1 + v + v2 + . . . + vn−1

)+

vmm

m

(1 + v + v2 + . . . + vn−1

)

=1m

(v

1m + v

2m + . . . + v

mm) (

1 + v + v2 + . . . + vn−1)

�.

2.1m

a(m)n

is the present value of an annuity-due whose m × n payments of 1m occur 1

m thof an interest conversion period apart, the first payment being immediate. We candecompose this sequence of payments into m subsequences of n payments of 1

m , thefirst occurring t mths of a conversion period from now (t = 1, 2, . . . ,m − 1). For thetth of these subsequences, the payments may be viewed as constituting an annuity-due deferred t m’ths of a conversion period. The present value of that subsequence

is then1m· v t

m an.

[5, Exercise 13, p. 147], [6, Exercise 20, p. 123] “A sum of 10000 is used to buy a deferredperpetuity-due paying 500 every 6 months forever. Find an expression for the deferredperiod expressed as a function of d.”Solution: The effective discount rate per 6 months is 1 − v

12 . If the perpetuity-due has

been deferred for n full years, the equation of value is

10000 = 500 · vn · 11 − √v


n =ln(20(1 − √v))

ln v

=ln(20(1 − √1 − d))

ln(1 − d)

[5, Exercise 14, p. 147], [6, Exercise 21, p. 124] “If 3 · a(2)n

= 2 · a(2)

2n= 45 · s(2)

1, find i.”

Solution:

3 · a(2)n

= 2 · a(2)

2n= 45 · s(2)

1

⇔ 32

(1 − vn) =22

(1 − v2n

)=

452

i


The first equation implies that 1 + vn = 32 ⇒ (1 + i)n = 2. Substitution in the equations

yields i = 130 .

[5, Exercise 15, p. 147], [6, Exercise 22, p. 124] “Find an expression for the present value ofan annuity which pays 1 at the beginning of each 3-month period for 12 years, assuminga rate of interest per 4-month period.”Solution: If i be the effective interest rate per 4-month period, the effective rate per 3-month period will be j = (1 + i)

34 − 1. Accordingly the value of the desired annuity

is

a48 j = (1 + j) · 1 − (1 + j)−48

j

= (1 + i)34 · 1 − (1 + i)−36

(1 + i)34 − 1

=1 − (1 + i)−36

1 − (1 + i)−34

An alternative approach to this problem would be to define A to be the amount of paymentthat would have to be made under an annuity-due every four months for 12 years to givethe same present value as the annuity-due described in the problem. We can look at asingle year in order to determine A. The present value of payments of A now, 4 monthsfrom now, and 8 months from now, is

A(1 + (1 + i)−1 + (1 + i)−2

)= A · (1 + i) · 1 − (1 + i)−3

i

and the value of the 4-payment annuity-due at 3 month intervals is

1(1 + (1 + j) + (1 + j)−2 + (1 + j)−3 = (1 + j) · 1 − (1 + j)−4

j.

Equating the two, and recalling that (1 + i)3 = (1 + j)4, gives A =i(1 + j)j(1 + i)

. The value of

the annuity will then be

a48 j = A · a36 i

=i(1 + j)j(1 + i)

· (1 + i) · 1 − (1 + i)−36

i

=1 − (1 + i)−36

1 − (1 + i)−34

as found previously.


[6, Exercise 14, p. 123] “Derive formula [6, (4.9), p. 122].”

a(m)n

=1m

(1 + v

1m + v

2m + . . . + vn− 1

m)

=1m· 1 − vn

1 −(1 − d(m)

m

)

=1 − vn

d(m) .

Analogously we can prove that

a(m)n

=1m

(v

1m + v

2m + . . . + vn

)

=1m· v 1

m · 1 − vn

1 − v1m

=1m· 1 − vn

(1 + i)1m − 1

=1m· 1 − vn

1m · i(m)

=1 − vn

i(m) .


6.22 Supplementary Notes for the Lecture of February 15th, 2010Distribution Date: Monday, February 15th, 2010

(updated on 16 February, 2010, by comment to solution to [5, Exercise 17, p. 147]subject to further revision)

At this lecture the students present were again consulted on the date for theclass test, tentatively shown in the course outline as 03 March, 2010. Themajority favoured delaying the test until the following Monday, 08 March,2010, and so that will be the date for the 45-minute class test. (Of course,all details of the course — including the date of this test — could be subjectto discretionary change in case of force majeure.)

6.22.1 §4.5 CONTINUOUS ANNUITIES

In the previous section we considered annuities in which a payment of 1 per period was splitinto m payments of 1

m paid regularly during the period. We can pass to the limit as m → ∞and generalize to one unit being paid continuously over the period, where the payment inan infinitesimal subinterval of time is equal in magnitude to the length of the subinterval.We denote this situation by placing a bar above the corresponding symbol, as in an and sn,which we represent by integrals with respect to time, in which we discount or accumulate theinfinitesimal payments respectively to times 0 or n.

Definition 6.18 1. an =

n∫

0

1a(t)

dt =

n∫

0

vt dt

2. sn =

n∫

0

a(t) dt =

n∫

0

(1 + i)t dt

When the force of interest is constant, we can show that

an =

n∫

0

vt dt =vt

ln v

]n

0=

1 − vn

δ

since ln v = − ln(1 + i) = −δ

sn =

n∫

0

(1 + i)t dt =(1 + i)t

ln(1 + i)

]n

0=

(1 + i)n − 1δ

,


both of which formulæ could have been obtained by passing to the limit in analogous formulæin the preceding section of the textbook. We can prove that

an = limm→∞

a(m)n

= limm→∞

1 − vn

i(m) =1 − vn

δ=

1 − e−nδ

δ

an = limm→∞

a(m)n

= limm→∞

1 − vn

d(m) =1 − vn

δ

sn = limm→∞

s(m)n

= limm→∞

(1 + i)n − 1i(m) =

(1 + i)n − 1δ

=enδ − 1δ

sn = limm→∞

s(m)n

= limm→∞

(1 + i)n − 1d(m) =

(1 + i)n − 1δ

.

As always in these situations we need to be careful to distinguish the interest rate which isapplied continuously and the nominal annual rate to which it is equivalent.

[5, Exercise 17, p. 147] “There is 40,000 in a fund which is accumulating at 4% per annum,convertible continuously. If money is withdrawn continuously at the rate of 2400 perannum, how long will the fund last?”Solution: Let the unknown time be n years. The fund is accumulating continuously at4% per annum, so δ = 0.04,24 and v = e−δ = e−0.04. An equation of value is

40000 = 2400an = 2400 · 1 − vn

δ

⇒ e−0.04n = 1 − 40000 × 0.042400

=13

⇒ n =ln 30.04

= 27.46530722 ,

so the fund will last 27.47 years.

[5, Exercise 18, p. 147] “If an = 4 and sn = 12, find δ.”Solution: The hypotheses imply that

1 − vn = 4δ(1 + i)n − 1 = 12δ

which together imply that (1 + i)n = 3. Substitution of this result into the second equation

yields δ =16

.

24I am not pleased with this interpretation of the words “accumulating continuously”: had the intention beenthat the force of interest was to be 4%, I believe the problem should have said precisely that; I am solving theproblem in the present way because, from his answer, I can see that this is the intention of the author.



[5, Exercise 19, p. 147] “Find an expression for an if δt =1

1 + t.”

Solution: We cannot use [5, Formula (4.14), p. 125] here, since it is based on a constantforce of interest. By [5, Example 1.15, pp. 34-35] dt units at time n discount back to thepresent in the value of

dt

e∫ n

0 δs ds=

dt1 + n

,

so dt units at time t is now worthdt

1 + t. It follows that an n-year continuous annuity

which delivers dt units at time t is now worthn∫

0

11 + t

dt = ln(1 + t)]n0 = ln(1 + n) − ln 1 = ln(1 + n) .

[5, Exercise 20, p. 147] “Find the value of t, 0 < t < 1, such that 1 paid at time t is equivalentto 1 paid continuously between time 0 and 1.”Solution: Unlike the preceding problem, the author here is assuming a constant force ofinterest.

a1 = vt ⇔ ivδ

= vt

⇔ vt−1 =iδ

⇔ t = 1 +ln δ − ln i

δ

⇔ t = 1 +1δ· ln δ

i

Review on instantaneous compounding

Nominal and Effective Rates To say that we are considering compound interest is to saythat we are assuming that the function a(t) is exponential. The constant base of the exponentialfunction is the accumulation factor for one interest period, as it represents the factor of increaseover one time interval; we normally require that this base be no less than 1, so that a(t) doesnot decrease with time; the rate of interest is the excess of this accumulation factor over 1.The word nominal alerts the reader to the fact that an associated rate of interest is not the truerate associated with the given function a(t), but is stated as a multiple of the true rate; where anominal rate i is described as compounded m times per period, the intention is that the effectiverate, i.e., the true rate of interest for 1

m th of a time period, is to be im ; in this usage we permit

m to be any positive number. Once the accumulation factor for the effective rate of interest


is established for a given time interval, we can determine the accumulation factors for otherlengths of time, and can determine the corresponding effective rate of interest by subtracting 1.We can compute an effective rate of interest for any time interval, not only for intervals of unitlength or intervals whose lengths are integer multiples and submultiples of 1 — i.e., not onlyfor intervals whose lengths are rational numbers of intervals.

Instantaneous Compounding As we increase the frequency of compounding we obtain,in the limit, instantaneous compounding. We have already seen that the equation

(1 +

i(m)

m

)m

= 1 + i ,

or, equivalently,i(m) = m

((1 + i)

1m − 1

)

leads, in the limit as m→ ∞, to

limm→∞

i(m) = ln(1 + i) = δ .

What this equation states is that δ = ln(1+ i) is the nominal annual rate of interest which, whencompounded instantaneously, produces an effective annual rate of interest of i. Note that wecan recover the nominal rate (compounded instantaneously) from the effective annual rate bysolving the equation δ = ln(1 + i), to obtain i = eδ − 1.




6.23.1 §4.6 PAYMENTS VARYING IN ARITHMETIC PROGRESSION

A general formula will be developed for an annuity-immediate with a term of n periods, withpayments beginning at P and increasing by Q per period thereafter, at interest rate i, showingthat its present value A is

A = P · an + Q · an − nvn

i. (81)

and the accumulated value just after the last payment is

S = P · sn + Q · sn − ni

. (82)

When P = Q = 1, specific symbols are used:

(Ia)n =an − nvn

i(83)

(Is)n =sn − n

i=

sn+1 − (n + 1)

i(84)

Similarly, when P = n and Q = −1, we have a decreasing annuity for which the present andaccumulated values are given by

(Da)n =n − an

i,

(Ds)n =n(1 + i)n − sn

i.

Other symbols of interest (with obvious definitions) are (Ia)n , (I s)n , (Da)n , (Ds)n , (Ia)∞ ,(Ia)∞ . You may omit the discussion of Fn, Gn, Hn in [5, Appendix 4, pp. 143-145], and anyexercises based on these functions. However, you should be able to derive formulæ

(Ia)∞ = limn→∞

an − nvn

i=

1 + ii2

(Ia)∞ = (1 + i)(Ia)∞ =(1 + i)2

i2

using the calculus (l’Hospital’s Rule, or even simpler methods).


The sum of the payments valued by (Ia)n and (Da)n is an annuity-immediate with constantpayments all equal to n + 1, and similar statements may be made about the other pairs offunctions. We have the identities

(Ia)n + (Da)n = (n + 1)an

(Is)n + (Ds)n = (n + 1)sn

(Ia)n + (Da)n = (n + 1)an

(I s)n + (Ds)n = (n + 1)sn

Verbal Interpretations. A number of the identities we can develop in connection with in-creasing and decreasing annuities admit interesting verbal interpretations.

1.an = i(Ia)n + nvn (85)

This equation can be obtained from (83): that provides an algebraic proof. One verbalinterpretation is as follows. The annual payments of 1 under an n-payment annuity-dueof present value an may be each invested for repayment together n years after the first ofthe payments. The interest earned by these investments will be multiples of i: 1i at theend of year 1, 2i at the end of year 2, . . . , ni at the end of year n; the principal of n repaidafter n years is worth nvn at time 0.

2. Equation (83)

(Ia)n =an − nvn

i,

can be interpreted directly as describing a decomposition of the increasing annuity. Thinkof a perpetuity paying an − nvn at the end of every year. That annual payment may beinterpreted as the value of an n-payment annuity-due of regular payments of 1, the firstpayment at that time, followed by a charge of n the year after the last payment. Thesepayments, when summed, yield 0 for years n+1, . . ., and yield an increasing annuity withpayments 1, 2, . . . , n for the first n payments.




6.24.1 §4.6 PAYMENTS VARYING IN ARITHMETIC PROGRESSION (continued)

Verbal Interpretations (continued from last lecture). A number of the identities we candevelop in connection with increasing and decreasing annuities admit interesting verbal inter-pretations.

1.an = i(Ia)n + nvn (86)

This equation can be obtained from (83): that provides an algebraic proof. One verbalinterpretation is as follows. The annual payments of 1 under an n-payment annuity-dueof present value an may be each invested for repayment together n years after the first ofthe payments. The interest earned by these investments will be multiples of i: 1i at theend of year 1, 2i at the end of year 2, . . . , ni at the end of year n; the principal of n repaidafter n years is worth nvn at time 0.

2. Equation (83)

(Ia)n =an − nvn

i,

can be interpreted directly as describing a decomposition of the increasing annuity. Thinkof a perpetuity paying an − nvn at the end of every year. That annual payment may beinterpreted as the value of an n-payment annuity-due of regular payments of 1, the firstpayment at that time, followed by a charge of n the year after the last payment. Thesepayments, when summed, yield 0 for years n+1, . . ., and yield an increasing annuity withpayments 1, 2, . . . , n for the first n payments.

Analogous verbal interpretations can be constructed for the identities related to (Da)n, (Is)n,(Ds)n, etc.

Expressing annuities with payments in Arithmetic Progression in terms of functionswe have defined. Suppose, for example, that you wish to evaluate the present value of anannuity-immediate whose first payment is P, and whose subsequent payments are each ob-tained from its predecessor by adding Q. If we begin with an annuity whose payments are Q,2Q, 3Q, . . ., whose present value is Q(Ia)n, we will be left with residual payments of P−Q oneach payment date. Hence the present value of this annuity-immediate is

Q(Ia)n + (P − Q)an .


[5, Exercise 21, p. 147], [6, Exercise 31, p. 124] “Show algebraically, and by means of a timediagram, the following relationship between (Ia)n and (Da)n :

(Da)n = (n + 1) · an − (Ia)n .”

Solution: In this method of presenting notes it is difficult to present a time diagram;instead, I give a verbal explanation. To simplify, I shall move the subtracted term fromthe right side to an added term on the left side of the identity. Then we are summingtwo variable annuities: as the amount of one decreases by 1 unit, the amount of the othertakes up the slack and increases by one unit, so the sum of the two remains constant forn payments. And that sum begins with value n + 1, so that is the amount that remainsconstant, giving an annuity-immediate of that constant payment for n payments.Algebraically, we have

(n + 1) · an − (Ia)n = (n + 1) · 1 − vn

i− an − nvn

i

=(n + 1) − vn − an

i

=(n + 1) − vn − 1 − an−1

i

=(n + 1 − 1) −

(an−1 + vn

)

i

=n − an

i= (Da)n

[6, Exercise 22, p. 148] “Simplify20∑

t=1

(t + 5)vt.”

Solution: One method for solving this is as follows. Represent the unknown sum by S .Then

S =

20∑

t=1

(t + 5)vt ⇔ vS =

20∑

t=1

(t + 5)vt+1

⇔ vS =

21∑

s=2

(s + 4)vs taking s = t + 1

⇔ vS =

21∑

t=2

(t + 4)vt replacing s by t


⇔ vS =

21∑

t=2

((t + 5) − 1)vt =

21∑

t=2

(t + 5)vt −21∑

t=2

vt

⇔ vS = S + v(26v20 − 6

)−

v(1 − v20

)

i

⇔ dS = v(6 − 26v20

)+

v(1 − v20

)

i⇔ iS =

(6 − 6v20

)+ a20 − 20v20

⇔ iS =6 − 6v20

i+

a20 − 20v20

i= 6a20 +

a20 − 20v20

i.

[5, Exercise 23, p. 148], [6, Exercise 32, p. 124] “The following payments are made underan annuity: 10 at the end of the 5th year, 9 at the end of the 6th year, decreasing by1 each year until nothing is paid. Show that the present value is

10 − a14 + a4 (1 − 10i)

i.”

Solution: The payments decrease until a payment of 1 at the end of the 14th year. Wecan think of a decreasing annuity starting with a payment of 14 at the end of the 1st year,and then make corrections. We can subtract a 4-payment decreasing annuity-immediatebeginning with a payment of 4 at the end of the year, and a 4-payment annuity-immediatewith a constant payment of 10. Thus we have

(Da)14 − (Da)4 − 10a4 =

(14 − a14

)−

(4 − a4

)− 10i · a4

i

=10 − a14 + a4 (1 − 10i)

i


6.25 Supplementary Notes for the Lecture of March 01st, 2010Distribution Date: Monday, March 01st, 2010


Announcements

Due Date for Assignment 3: There was some ambiguity in the due datefor Assignment 3. The class agreed that the due date should now beFriday, March 05th: the assignments must be handed in at the latest atthe lecture of that day (not through the Departmental mailbox on thatday!)

Solutions to Assignment 2: An error on page 115 of these notes, in thesolution to Problem 3c of Assignment 3, has been corrected.

6.25.1 §4.6 PAYMENTS VARYING IN ARITHMETIC PROGRESSION (conclusion)

[5, Exercise 25, p. 148], [6, Exercise 34, p. 124] “A perpetuity-immediate has annual paymentsof 1, 3, 5, 7, . . . . If the present value(s) of the 6th and 7th payments are equal, find thepresent value of the perpetuity.”Solution: We equate the values of the 6th and 7th payments:

(1 + (6 − 1)2)v6 = (1 + (7 − 1)2)v7

⇔ 1 + i =1311

⇔ i =2

11

The perpetuity can be viewed as the sum of an increasing perpetuity with payments in-creasing by 2 each year, i.e., 2(Ia)∞ diminished by a constant perpetuity-immediate a∞ :

2 · (Ia)∞ − a∞ = 2limn→∞

(an − n

(1+i)n

)

i− 1

i

=2a∞ − 1

iby l’Hopital’s Rule

= 66

[5, Exercise 27, p. 148], [6, Exercise 36, p. 125] “An annuity-immediate has semiannual pay-ments of 800, 750, 700, . . . , 350, at i(2) = 0.16. If a10 0.08 = A, find the present value ofthe annuity in terms of A.”


Solution: We will be working with an effective semi-annual interest rate of j = i(2)

2 = 8%.The truncated decreasing annuity-immediate that we wish to evaluate has present value

50(Da)10 − 50v10(Da)6 = 50 ·(16 − a16

)− v10

(6 − a6

)

i

=50i·((

16 − 6v10)−

(a16 − v10a6

))

=50i·(10 + 6ia10 − a10

)

=50i·(10 + (6i − 1)a10

)

= 625(10 − 0.52A) = 6250 − 325A .

Alternatively we could express the payments as a constant annuity of payment size 300,superimposed on a decreasing annuity of payments 500, 450, 400, . . . , 50. This has value

300 · a10 8% + 50(Da)10 8% = 300A + 50 · 10 − A0.08

= 300A + 6250 − 625A = 6250 − 325A.

[6, Exercise 29, p. 124] In [5, Example 4.10, p. 132], [6, Example 4.13, p. 116] it is shownthat the present value of an annuity-immediate such that payments start at 1, increase byannual amounts of 1 to a payment of n, and then decrease by annual amounts of 1 to afinal payment of 1, is an · an . The present exercise is to justify this value verbally.Solution: Consider a sequence of n annuities-due, each of them consisting of n paymentsof 1. The first of these annuities is to make its first payment 1 year from now, the second2 years from now, ..., the nth n years from now. The total payments made will increasefrom 1 to n, then decrease to a payment of 1, 2n − 1 years from now. The value of thepayments under each of these annuities-due is an just before the first payment. Thesevalues, when discounted to the present, have value an · an .


6.26 Supplementary Notes for the Lecture of March 03rd, 2010Distribution Date: Wednesday, March 03rd, 2010


6.26.1 §4.7 PAYMENTS VARYING IN GEOMETRIC PROGRESSION

I shall work some problems to illustrate that problems of this type are not hard to solve, sincethe effect of the geometric progression is equivalent to altering the interest rate.

[5, Exercise 28, p. 148], [6, Exercise 38, p. 125] “Find the present value of a 20-year annuitywith annual payments which pays 600 immediately and each subsequent payment is 5%greater than the preceding payment. The annual effective rate of interest is 10.25%.Answer to the nearest dollar.”Solution:

Present Value =

19∑

n=0

600vn(1.05)n

= 600 ·1 −

(1.05

1.1025

)20

1 − 1.051.1025

(87)

= 7851.1926 ,

or 7,851 to the nearest dollar.

[5, Exercise 29, p. 148] “ In [5, Exercise 28, p. 148]] find the interest rate i′ such that thepresent value would be equal to the present value of the level annuity-due 600a20i′ .”Solution: If we hadn’t seen the preceding solution, we might have to solve for i′ the

equation a20 i′ =7851.1926

600. We would then find by iteration, interpolation, or other

methods that i = 5.0%. However, with the preceding solution available, we can seeimmediately from (87) that the sum represents a20 i′ , where

1 + i′ =1.1025

1.05= 1.05 .

[5, Exercise 30, p. 148], [6, Exercise 37, p. 125] “Annual deposits are made into a fund at thebeginning of each year for 10 years. The first 5 deposits are 1,000 each, and depositsincrease by 5% per year thereafter. If the fund earns 8% effective, find the accumulatedvalue at the end of 10 years.”Solution: The first 5 deposits are today worth

1000a5 8% = 4312.12684 .


I see 2 ways of interpreting the words “increase by 5% per year thereafter”: either theincrease is geometric, by a factor of 1.05 applied repeatedly; or the deposits increase inarithmetic progression (before discounting). If the deposits increase in geometric pro-gression, then the present value will be

1000(v5(1.05) + v6(1.05)2 + . . . + v9(1.05)5

)

= 1000v5(1.05) ·(1 − (v(1.05))5

)

1 − v(1.05)

= 36000(1.05)(1.08)−5 ·1 −

(1.051.08

)5= 3379.996182.

The present value will then be 4312.1268 + 3379.9962 = 7692.1230; the accumulatedvalue at the end of 10 years will be 16606.72. This is the answer given by the textbook,so, presumably, our interpretation is the one the author intended, (which could be ex-pected by the presence of the exercise in this section of the chapter). (But the languageis ambiguous, and the other interpretation is plausible also. If the deposits increase inarithmetic progression, then the present value will be

1000(1.08)−5a5 8% + 50(1.08)−5(Ia)5 8% = 3510.09347

so the present value is 7822.2203, and the value after 10 years is 16,887.59.)

[5, Exercise 31, p. 149] “A perpetuity makes payments starting five years from today. Thefirst payment is 1000, and each payment thereafter increases by k% per year. The presentvalue of this perpetuity is equal to 4096, when compounded at i = 25%. Find k.”Solution: Five years from now the value of the perpetuity is

10001 +

(1 + 0.01k

1.25

)1

+

(1 + 0.01k

1.25

)2

+

(1 + 0.01k

1.25

)3

+ . . .

=1000

1 − 1+0.01k1.25

=12500025 − k

.

An equation of value is, therefore12500025 − k

= 4096(1.25)5 = 4 · 55, implying that k = 15.

[5, Exercise 32, p. 149] “An employee currently is aged 40, earns 40,000 per year, and ex-pects to receive 3% annual raises at the end of each year for the next 25 years. Theemployee decides to contribute 4% of annual salary at the beginning of each year for thenext 25 years into a retirement plan. How much will be available for retirement at age 65if the fund can earn a 5% effective rate of interest? Answer to the nearest dollar.”


Solution: The accumulation into the fund by age 65 is

(0.04)(40000)((1.05)25(1.03)0 + (1.05)24(1.03)1 + . . . + (1.05)1(1.03)24

)

= 1600(1.05)25

1 +1.031.05

+

(1.031.05

)2

+ . . . +

(1.031.05

)24

= 1600(1.05)25 ·1 −

(1.031.05

)25

1 − 1.031.05

= 1600(1.05)25 · 1052·1 −

(103105

)25 = 108576.4687

or 108,576 to the nearest unit.

6.26.2 §4.8 MORE GENERAL VARYING ANNUITIES

We will not formally study this section and its interesting generalizations. If you meet anyproblems of these types, they should be solvable “by first principles”. To repeat: you don’tneed to remember derivations or formulæ, but you have already been exposed to machinerythat can be used to solve problems of these types — no new machinery is required.


[5, Exercise 44, p. 150] “A perpetuity paying 1 at the beginning of each 6-month period hasa present value of 20. A second perpetuity pays X at the beginning of every 2 years.Assuming the same annual effective interest rate, the two present values are equal. De-termine X.”Solution: In terms of the effective annual discount rate d, the rate of discount for one-

half year is 1 − √1 − d, and the value of the first perpetuity is1

1 − √1 − d. The rate of

discount for a 2-year period is 1− (1−d)2 = 2d−d2, so the value of the second perpetuity

isX

2d − d2 . Equating the values of the two perpetuities yields

X =2d − d2

1 − √1 − d= (2 − d)

(1 +√

1 − d).

We are told that the first perpetuity is worth 20, so

1

1 − √1 − d= 20⇒ d =

39400⇒ X =

761400· 39

20= 3.709875 .


[5, Exercise 45, p. 150], [6, Exercise 54, p. 126] “ For a given n it is known that an = n − 4

and δ = 10%. Find

n∫

0

at dt.”

Solution: Note that the exercise does not say that an = n − 4 for all n; in any case, thiswould be absurd for n ≤ 4. (The hypothesis that δ = 10% should not be linked to theprevious statement: it is part of the general background for the exercise.)You are not asked to interpret the integral, but it is enlightening to do so. The following commenthas be be interpreted as beyond the course, since the prerequisites do not include Calculus 3. Sincethe integral can be interpreted as a double integral, i.e., the integral of a function of 2 variablesover a region, we can reverse the order of integration when evaluating it as a the result of twointegrations. We obtain

n∫

0

at dt =

n∫

0

t∫

0

vs ds dt

=

n∫

0

n∫

s

vs dt ds

reversing the order of integration

=

n∫

0

vs

n∫

s

dt ds

=

n∫

0

vs(n − s) ds .

This can be interpreted as an increasing continuous annuity, the continuous analogue of (Ia)n and(Ia)n .

By [5, equation (4.14), p. 125] an = 1−vn

δ, so the first hypothesis is equivalent to

1 − vn

δ=

n − 4. This equation also enables us to find an antiderivative:∫ n

0at dt =

∫ n

0

1 − vn

δdt

=1δ

[t +

1δ· vt

]n

0

=1δ

(n +

vn − 1δ

)

=n − 1−vn

δ

δ


=n − an

δ=

4δ

= 40.

[5, Exercise 46, p. 150], [6, Exercise 56, p. 127] “A family wishes to provide an annuity of100 at the end of each month to their daughter, now entering college. The annuity willbe paid for only nine months each year, for four years. Show that the present value onemonth before the first payment is 1200a4 · a(12)

9/12.”

Solution: The proposed factor a4 represents a 4-payment annual annuity-due. We needonly determine the value of the annuity-immediate of 9 monthly payments. These resultfrom division of the year into 12 months — that’s the message of the superscript (12) —but the payments stop after the 9th — that’s the message of the subscript 9/12 .


6.27 Supplementary Notes for the Lecture of March 05th, 2010Distribution Date: Friday, March 05th, 2010


Textbook Chapter 5. Amortization schedules and sinking funds.


We consider methods of repaying a loan, in particular

The Amortization Method: In this method the borrower makes instalment payments to thelender. Usually these payments are at regularly spaced periodic intervals; the progressivereduction of the amount owed is described as the amortization of the loan.

The Sinking Fund Method: In this method the loan will be repaid by a single lump sumpayment at the end of the term of the loan. However the borrow may prepare himselffor the repayment by making deposits to a fund called a sinking fund to accumulate therepayment amount. (Sometimes the lender may be aware of the existence of the sinkingfund; for example, an institutional borrower that issues a series of bonds may let thepublic know that the accumulation of funds to redeem the bonds may be disciplined by asinking fund.)

6.27.2 §5.2 FINDING THE OUTSTANDING LOAN BALANCE

When a loan is being amortized the outstanding balance is being reduced by the amortizationpayments. Each payment may be analyzed and interpreted as consisting of an interest compo-nent and a component for reduction of principal. An equation of value can be set up at anytime during the amortization, equating

Current value of payments = Accumulated value of Loan

where “Payments” consists of both past and future payments. Decomposing the term andrearranging the equation gives

Present Value of Future Payments=Accumulated Value of Loan − Accumulated Value of Past Payments

Synonymous terms:

• outstanding loan balance

• outstanding principal

• unpaid balance


• remaining loan indebtedness

[5, Exercise 1, p. 185], [6, Exercise 1, p. 195] “A loan of 1,000 is being repaid with quarterlypayments at the end of each quarter for 5 years, at 6% convertible quarterly. Find theoutstanding loan balance at the end of the 2nd year.”

Solution: The level payments under this annuity-immediate will be1000

a20 1.5%

.

Retrospective method: The value of the payments already made is

1000a20 1.5%

· s8 1.015 = 491.1769.

Subtracting this from 1000(1.015)8 yields 635.3157.Prospective method: The value of the 12 remaining payments is

1000a20 1.5%

· a12 1.5% = 635.3157 .

[5, Exercise 2, p. 185], [6, Exercise 2, p. 195] “A loan of 10,000 is being repaid by instal-ments of 2,000 at the end of each year, and a smaller final payment made one year afterthe last regular payment. Interest is at the effective rate of 12%. Find the amount ofoutstanding loan balance remaining when the borrower has made payments equal to theamount of the loan.”Solution: The problem asks for the outstanding loan balance just after payments totalling10,000 have been made; this will be immediately after the 5th payment. We shall usethe retrospective method only here. The accumulated value of the loan at time t = 5 is10000(1.12)5. The accumulated value of the payments made is 2000s5. The outstandingloan balance will, therefore be

10000(1.12)5 − 2000s5 = 10000(1.12)5 − 20000.12

((1.12)5 − 1

)

= 4917.72212

Were we to use the prospective method, we would need to determine the value of the lastdrop payment. This is interesting information, and we could have been asked for it. Butit has not been requested, and so we shall not bother finding it. (But you should knowhow to do that if it is necessary.)

[5, Exercise 3, p. 185], [6, Exercise 3, p. 195] “A loan is being repaid by quarterly instalmentsof 1,500 at the end of each quarter, at 10% convertible quarterly. If the loan balance atthe end of the first year is 12,000, find the original loan balance.


Solution: Denote the original loan balance by L. Here the retrospective method is themost appropriate, since we don’t know how many future payments have to be made. Theequation of value at time 1 year is

L(1.025)4 − 1500s4 2.5% = 12000

which we solve to yield

L = (1.025)−4(1200 + 1500

((1.025)4 − 1

))

= −48000(1.025)−4 + 60000 = 16514.36905.

[5, Exercise 4, p. 185], [6, Exercise 6, p. 196] “A 20,000 loan is to be repaid with annual pay-ments at the end of each year for 12 years. If (1 + i)4 = 2, find the outstanding balanceimmediately after the fourth payment.”

Solution: The annual payment is, by the prospective method,20000a12 i

. Again by the

prospective method, the outstanding balance after the 4th payment is

20000a12 i

· a8 i = 20000(

1 − (1 + i)−8

1 − (1 + i)−12

)=

67· 20000 = 17142.85714.

[5, Exercise 5, p. 185], [6, Exercise 7, p. 196] “A 20,000 mortgage is being repaid with 20annual instalments at the end of each year. The borrower makes 5 payments, and thenis temporarily unable to make payments for the next 2 years. Find an expression for therevised payment to start at the end of the 8th year if the loan is still to be repaid at the endof the original 20 years.”

Solution: The original payments are (by the prospective method)20000

a20

. The outstanding

balance at the end of the 7th year (with no payment then or at the end of the previous year)is the value then of all unpaid payments, i.e.,

20000a20

·(a13 + s2

).

It follows that the level payment needed to repay the loan in 13 payments (under anannuity-immediate) is

20000a20

·(a13 + s2

)· 1

a13

=20000 · (1 + i)2 · a15

a20 · a13

.


[5, Exercise 6, p. 185][6, Exercise 8, p. 196] “A loan of 1 was originally scheduled to be re-paid by 25 equal annual payments at the end of each year. An extra payment K witheach of the 6th through the 10th scheduled payments will be sufficient to repay the loan5 years earlier than under the original schedule. Show that

K =a20 − a15

a25 · a5

.”

Solution: The level payment of this loan of 1 is1

a25

. The extra payments are equal in

value to the value of the last 5 payments, so, at time t = 5,

K · a5 =a20 − a15

a25

which yields the desired value for K.

6.28 Supplementary Notes for the Lecture of March 08th, 2010Distribution Date: Wednesday, March 10th, 2010


There was no lecture this Monday, as the Class Test was administered. The versions of the testare contained in these notes on pages 127–150




Assignment #4 has been posted on pages 125-126 of these notes. It is dueat the latest at the lecture on Monday, March 22nd, 2010.

6.29.1 §5.2 FINDING THE OUTSTANDING LOAN BALANCE (conclusion)

[5, Exercise 7, p. 185] “A husband and wife buy a new home and take out a 150,000 mort-gage loan with level annual payments at the end of each year for 15 years, on which theeffective rate of interest is equal to 6.5%. At the end of 5 years they decide to make amajor addition to the house, and want to borrow an additional 80,000 to finance the newconstruction. They also wish to lengthen the overall length of the loan by 7 years (i.e.,until 22 years after the date of the original loan). In the negotiations the lender agreesto these modifications, but only if the effective interest rate for the remainder of the loanafter the first 5 years (be) raised to 7.5%. Find the revised annual payment which wouldresult for the remainder of the loan. Answer to the nearest dollar.”Solution: The statement “after the first 5 years” is ambiguous — is the interest rate tobegin immediately with the new loan, or is it to begin after 5 years of the new loan? Iwill assume the intention is that the new interest rate begins immediately.

The initial payment is150000a15 6.5%

. Immediately after the 5th payment the amount outstand-

ing on the loan is

150000(1.065)5 −

s5 6.5%

a15 6.5%

.

The principal of the new loan will be

150000(1.065)5 −

s5 6.5%

a15 6.5%

+ 80000 .

The 17 level payments on the new loan will therefore be

150000((1.065)5 −

s5 6.5%

a15 6.5%

)+ 80000

a17 7.5%

=150000

((1.065)5 − (1.065)5−1

1−(1.065)−15

)+ 80000

1−(1.075)−17

0.075

= 20636.38388 ,

so the new level payment should be 20,636.


[6, Exercise 4, p. 195] A loan is being repaid by annual payments at the ends of 15 successiveyears. The first 5 instalments are 4,000 each, the next 5 are 3,000 each, and the final 5are 2,000 each. Find expressions for the outstanding loan balance immediately after thesecond 3,000 instalment.

1. prospectively;2. retrospectively.

Solution:

1. Prospective Method. The value at time 7 of the payments yet to be made is 2000a8+

1000a3.2. Retrospective Method. We have to use the prospective method at some time to

obtain the initial value of the loan. This will be the value of all payments at timet = 0, i.e., 2000a15 + 1000a10 + 1000a5. The value of all payments made before andat time t = 7 is 4000s − 2s2. The outstanding loan balance is, therefore

(2000a15 + 1000a10 + 1000a5

)(1 + i)7 − 4000s7 + 1000s2

[6, Exercise 5, p. 196] “A loan is to be repaid with level instalments payable at the end ofeach half-year for 3 1

2 years, at a nominal rate of interest of 8% convertible semiannually.After the fourth payment the outstanding loan balance is 5,000. Find the initial amountof the loan.”Solution: The effective semiannual rate is 4%. There are to be 7 level payments in all.The amount of the payments is, by the prospective method

5000a3 4%

=5000 × 0.041 − (1.04)−3 .

It follows that the amount of the loan, now by the prospective method, is

5000a3 2%

· a7 2% = 5000(1 − (1.04)−7

1 − (1.04)−3

)= 10814.15817 .

6.29.2 §5.3 AMORTIZATION SCHEDULES

An amortization schedule is a chart showing, for each payment date, the information shown inthe columns of the following beginning of such a schedule with n payments of size x:

Year Payment Interest Principal Outstandingamount paid repaid loan balance

0 x · an i1 x x(1 − (1 + i)−n) x(1 + i)−n x · an−1 i2 x x(1 − (1 + i)−n+1) x(1 + i)1−n x · an−2 i· · · · · · · · · · · · · · ·


In practice there will be rounding errors as the table is generated line by line, and the last linemay not quite balance. “Standard practice is to adjust the last payment so that it is exactlyequal to the amount of interest for the final period plus the outstanding loan balance at thebeginning of the final period,” in order to bring the outstanding loan balance to 0. In order todetermine the entries in one row of the table, we do not need to generate the table line by line;the table is useful when a number of rows are of interest.

Few of the exercises in the textbook require extensive computation of schedules. Studentsshould, however, be capable of completing a full table for a loan when all the informationneeded is available.

[5, Exercise 8, p. 186], [6, Exercise 10, p. 196] “A loan is being repaid with quarterly instal-ments of 1,000 at the end of each quarter for 5 years at 12% convertible quarterly. Findthe amount of principal in the 6th instalment.”Solution: The principal is

1000a20 3% =10000.03

(1 − (1.03)−20

)= 14877.47486 .

After the 5th payment, the outstanding principal is (by the prospective method) 1000a20−5 3%,so the interest component of the next payment is

30a20−5 3% =30

0.03

(1 − (1.03)−15

)= 358.1380526 ,

and the amount of principal reduction is

1000 − 358.1380526 = 641.8619474 .

[5, Exercise 9, p. 186], [6, Exercise 12, p. 196] “A loan of 10,000 is being repaid with 20 in-stalments at the end of each year, at 10% effective. Show that the amount of interest inthe 11th instalment is

10001 + v10 .”

Solution: The amount of each payment is, by the prospective method,

10000a20 10%

.

By the prospective method, the unpaid balance after the 10th payment is

10000a20 10%

· a10 10% =10000

1 + (1.1)10

so the interest component of the 11th payment is

0.01(

100001 + (1.1)10

)=

1001 + (1.1)10 .


[5, Exercise 10, p. 186], [6, Exercise 14, p. 197] “A loan is being repaid with a series of pay-ments at the end of each quarter, for 5 years. If the amount of principal in the 3rd paymentis 100, find the amount of principal in the last 5 payments. Interest is at the rate of 10%convertible quarterly.”Solution: The loan is being repaid in 5 × 4 = 20 quarterly payments, and the effectiveinterest rate per quarter is 1

4 (10%) = 2.5%. Let’s assume that the amount of each levelpayment is x, and begin by compiling the first part of the amortization table.

Payment Payment Interest Principal Outstandingamount paid repaid loan balance

0 x · a20 2.5%1 x x(1 − (1.025)−20) x(1.025)−20 x · a19 2.5%2 x x(1 − (1.025)−19) x(1.025)−19 x · a18 2.5%3 x x(1 − (1.025)−18) x(1.025)−18 x · a17 2.5%

from which we see that x(1.025)−18 = 100, so

x = 100(1.025)18 = 155.9658718 .

(We didn’t need to use the schedule here. By the Prospective Method, the unpaid balancejust after the 2nd payment is x · s18, so the interest component of the 3rd payment is ix · s18and the residue for reduction of principal is

x(1 −

(1 − v18

))= xv18 .)

We could now compile the last lines of the amortization table backwards. Alternatively,if the author is requesting the total amount of principal in the last 5 payments, that is

x(v + v2 + v3 + v4 + v5) = 155.9658718 · a5 0.025 = 724.5906916 .

[5, Exercise 11, p. 186], [6, Exercise 15, p. 197] “A loan is being repaid with instalments of1 at the end of each year for 20 years. Interest is at effective rate i for the first 10 years,and effective rate j for the second 10 years. Find expressions for

“a) the amount of interest paid in the 5th instalment;“b) the amount of principal repaid in the 15th instalment.”

Solution: The present value of the last 10 payments is (1 + i)−10a10 j; the principal of theloan is, therefore,

P = a10 i + (1 + i)−10a10 j .

We compile the first lines of the amortization table:


Year Payment Interest Principal Outstandingamount paid repaid loan balance

0 P1 1 iP 1 − iP P(1 + i) − s1 i12 1 i(P(1 + i) − 1) (1 + i)(1 − iP) P(1 + i)2 − s2 i

3 1 i(1 + i)2((

P − a2 i

))s2 i − i(1 + i)2P P(1 + i)3 − s3 i

4 1 i(1 + i)3((

P − a3 i

))s3 i − i(1 + i)3P P(1 + i)4 − s4 i

5 1 i(1 + i)4((

P − a4 i

))s4 i − i(1 + i)4P P(1 + i)5 − s5 i

1. The amount of interest in the 5th instalment is

i(1 + i)4((a10 i + (1 + i)−10a10 j − a4 i

)= i

(a6 i + (1 + i)−6a10 j

).

More simply, we can observe (using the Prospective Method) that the outstandingbalance just after the 4th instalment is

(1 + i)−6a10 j + a6 i .

The interest component of the 5th payment is obtained by multiplying this amountby i.

2. After the 10th instalment has been paid, we shift to the second interest rate. Theoutstanding balance after the 14th payment is, again by the Prospective Method, a6 j;interest one payment later will be

j · a6 j = 1 − v6

so the payment will reduce principal by 1 − (1 − v6) = v6.

[5, Exercise 12, p. 186], [6, Exercise 17, p. 197] “A borrower has a mortgage which calls forlevel annual payments of 1 at the end of each year for 20 years. At the time of the7th regular payment an additional payment is made equal to the amount of principalthat, according to the original amortization schedule, would have been repaid by the8th regular payment. If payments of 1 continue to be made at the end of the 8th andsucceeding years until the mortgage is fully repaid, show that the amount saved in interestpayments over the full term of the mortgage is 1 − v13.”Solution: At time 0 the amount owing is a20. By the 8th regular payment on the originalschedule, the principal repaid would have been

A − A(1 + i)8 + s8 = (Ai − 1)s8 .

It is intended that this amount is added to the 7th payment. Immediately after the original7th payment the principal owing would have been

A(1 + i)7 − s7 .


The additional payment at time 7 reduces this amount to

A(1 + i)7 − s7 −(Ai − 1)s8

)= (1 + i)7 + A

(1 − i(1 + i)7

)

[5, Exercise 13, p. 186], [6, Exercise 18, p. 197] “A loan of L is being amortized with pay-ments at the end of each year for 10 years. If v5 = 2

3 , find the following:

“a) The amount of principal repaid in the first 5 payments.“b) The amount due at the end of 10 years if the final 5 payments are not made as

scheduled.”

Solution:

1. The annual level payments constitute an annuity-immediate with annual payment of

La10

=Li

1 −(

23

)2 =9iL5.

By the prospective method, the amount of principal remaining to be paid immedi-ately after the 5th annual payment is

9iL5· a5 =

9L5·(1 − 2

3

)=

3L5.

Hence the amount of principal that has already been paid at that time is(1 − 3

5

)L =

2L5

.

2. If no further payments are made, the amount repayable at the end of 10 years is

(1 + i)5 · 3L5

=32· 3L

5=

9L10

.

[5, Exercise 14, p. 187], [6, Exercise 19, p. 197] “A 35-year loan is to be repaid with equalinstalments at the end of each year. The amount of interest paid in the 8th instalment is135. The amount of interest paid in the 22nd instalment is 108. Calculate the amount ofinterest paid in the 29th instalment.”Solution: Suppose the amount of the loan at time 0 was L. The annual instalments are

eachL

a35

. By the prospective method the amount of principal outstanding just after the

7th instalment is La35· a28. This will incur an interest payment of i · L

a35· a28 in the 8th

instalment; we thus have the equation

iL · 1 − v28

1 − v35 = 135 .


A similar computation involving instalments 21 and 22 gives

iL · 1 − v14

1 − v35 = 108 .

Taking the ratio of the two equations, we obtain

1 + v14 =1 − v28

1 − v14 =135108

=54

so v7 = 12 ; substitution in either equation above yields iL =

2792

. The amount of interestin the 29th instalment is

iL ·a7

a35

= iL · 1 − v7

1 − v35 =16iL31

= 72 .

[5, Exercise 15, p. 187] “A 10-year loan of L is repaid by the amortization method, with pay-ments of 1000 at the end of each year. The annual effective interest rate is i. The totalamount of interest repaid during the life of the loan is also equal to L. Calculate theamount of interest paid during the first year of the loan.”Solution: The level payments under the amortized loan are

1000 =L

a10i

;

the total amount of interest paid during the life of the loan is 10000−L = L, so L = 5000,and a10i = 5. Solving this equation (how?) yields i = 0.1509841448 = 15.09841448%,so the amount if interest paid at the end of the first year is 0.1509841448 × 5000 =

754.9207240 or 754.92.

[5, Exercise 16, p. 187] “A bank customer borrows X at an annual effective rate of 12.5%,and makes level payments at the end of each year for n years.

(i) “The interest portion of the final payment is 153.86.(ii) “The total principal repaid as of time n − 1 is 6009.12.

(iii) “The principal repaid in the first payment is Y .

Calculate Y .”

Solution: The level payments are each equal toX

an 12.5%. The interest portion of the final

payment is

153.86 =dX

an 12.5%=

0.1251.125 · Xan 12.5%

=X

9an 12.5%,


implying that

an12.5% =X

153.86 × 9=

X1384.74

,

so the level payments are 1384.74. The unpaid principal at time n−1 — is it clear whether

this is before or after the payment? — 1384.74a1 = 1384.74v =1384.74

1.125= 1230.88, so

the total loan is X = 6009.12 + 1230.88 = 7240.; hence

an12.5% =7240.

1384.74= 5.2284

so 1 − (1.125)−n = (0.125)(5.2284) = 0.65355, and vn = 0.34645. But this implies thatthe principal component of the first payment is Y = 0.34645 × 1384.74 = 479.74.

[6, Exercise 11, p. 196] “Consider a loan which is being repaid with instalments of 1 at theend of each period for n periods. Find an expression at issue for the present value of theinterest which will be paid over the life of the loan.”Solution: We can use the information in [5, Table 5.1, p. 157]. The sum of the interest

portions of the payments is, as shown in the table,n∑

r=1

(1− vr) = n− an . This, however, is

not what the problem requests. The present value of the interest payments — presumablyas of time t = 0, is

n∑

r=1

vn−r+1(1 − vr) = an − nvn+1 .

[6, Exercise 13, p. 196] “A loan is being repaid with 20 instalments at the end of each year at9% effective. In what instalment are the principal and interest portions most nearly equalto each other?”Solution: Without limiting generality, assume the payments are all of size 1, so thatthe loan is for a20 9%, and [5, Table 5.1, p. 157] applies. In that table we see that therepayments of principal range between v20 in the first payment to v1 in the last. The twoportions under consideration sum to a constant, so they cannot both exceed 1

2 at any time.Both sequences — interest payments and reductions of principal — are monotone: therewill be only one pair of values t = a, t = a + 1, where they reverse their positions frombeing greater than/less than 1

2 to the reverse. But it is not clear which of those two willhave the closest values! We see from [5, Table 5.1, p. 157] that the difference betweenprincipal and interest components in the tth payment is

∣∣∣1 − 2vn−t+1∣∣∣. We shall begin by

determining the largest value of t — if any — for which

v20−t+1 ≤ 12,


equivalently, the largest value of t such that

(1 + i)21−t ≥ 2 ,

equivalently, the largest value of t such that

t ≤ 21 − ln 2.ln 1.09

= 12.95676827 ,

i.e., t = 12. This is the “best” choice where the difference 1 − 2vn−t+1 is positive. Weneed also to consider the “best” choice when the difference is negative, i.e., t = 13. Wecompute the differences:

t = 12 1 − 2(1.09)−9 = 0.0791444410t = 13 1 − 2(1.09)−8 = −0.003732559

and see that the values are closest when t = 13.




Negative final payment? I begin this lecture with an example from an earlier chapter whichcan be relevant to sinking fund problems. In [5, Example 3.7, pp. 92–93] a fund is to be accu-mulated “by means of deposits of 1,000 made at the end of every year, as long as necessary; ifthe fund earns an effective rate of interest of 8%, find how many regular deposits will be nec-essary, and the size of a final deposit to be made one year after the last regular deposit”. Theintention is that the final deposit be smaller than the others, if necessary. By finding the largest

integer n such that 1000 · sn 0.08 ≤ 25, 000 ⇔ n ≤ ln 3ln 1.08

= 14.275, we conclude that the lastregular payment is the 14th. The balance in the account after the 14th deposit is 24,214.920,which, after a year, grows to 26,152.11, which exceeds the target of the fund. Hence the 15thpayment is negative, in the amount of −1, 152.11: it even exceeds 1,000 in magnitude! Thistype of situation can occur in other ways, so you should be prepared for it, even though it isnot the most likely outcome.

6.30.1 §5.4 SINKING FUNDS

Differences between the Amortization and Sinking Fund Methods of repayment In theAmortization Method for repaying a loan, the borrower makes regular payments — often levelpayments — directly to the lender. The Outstanding Loan Balance at any time is then the netamount owing on the lender’s books — either the excess of the accumulated value of the loanminus the accumulated value of the payments made Retrospective Method or the present valueof the payments yet to be made Prospective Method.

Where a loan is repaid by a Sinking Fund, portions of the borrower’s payments are nottransmitted to the lender until a later date, usually when they have accumulated in fund —the Sinking Fund — often at an interest rate different from the rate associated with the loan.Where the borrower is required or permitted to make regular payments directly to the lenderin addition to those paid into the fund, those are described as service on the loan; where thoseregular payments cover the interest costs on the loan, the principal outstanding will remainconstant. This should not be confused with the Net Amount of the loan, which will be theexcess of the outstanding principal over the accumulated value of the sinking fund.

Where the interest rate associated with the Sinking Fund is the same rate as is being paidon the loan, the Sinking Fund Method is equivalent to the Amortization Method.

The identity1

an=

1sn

+ i. While the identity can easily be proved algebraically (as was

done in the lecture), it admits an interesting verbal proof in terms of a loan of 1 being repaid


over n periods.1

anis the regular payment necessary under an annuity-immediate.

1sn

is the

portion of that regular payment that will accumulate in a sinking fund to a value of 1 just afterthe nth payment; the complement — i — is the amount necessary to service the loan annuallyuntil the time that the sinking fund matures.

[5, Exercise 17, p. 187], [6, Exercise 20, p. 197] “A has borrowed 10,000 on which interestis charged at 10% effective. A is accumulating a sinking fund at 8% effective to repaythe loan. At the end of 10 years the balance in the sinking fund is 5000. At the end of the11th year A makes a total payment of 1500.

“a) How much of the 1500 pays interest currently on the loan?“b) How much of the 1500 goes into the sinking fund?“c) How much of the 1500 should be considered as interest?“d) How much of the 1500 should be considered as principal?“e) What is the sinking fund balance at the end of the 11th year?”

Solution: This problem requires attention to the terminology used.

“a) Presumably we are to assume that the borrower is servicing the loan so that theoutstanding balance remains constant. The interest payment necessary at the end ofthe 11th year will, therefore, be 0.10 × 10000 = 1000.

“b) The remainder of the payment of 1500 is a contribution of 500 to the sinking fund.“c) The net interest paid is the excess of the interest paid — 1000 — over the 8% ×

5000 = 400 interest earned by the sinking fund, or 600.“d) The excess of the contribution over the net interest payment can be assigned to re-

ducing the principal of the loan. The amount is 1500 − 600 = 900. This can beconsidered as made up of two components: 500 which is paid into the sinking fund,and 8% of 5000, which is 400 — the interest earned by the sinking fund, and whichwill ultimately be paid to the lender to retire the loan.

“e) At the end of the 11th year, the sinking fund balance is

5000(1.08) + 500 = 5900 .

[5, Exercise 18, p. 187], [6, Exercise 21, p. 198] “A loan of 1000 is being repaid with levelannual payments of 120 plus a smaller final payment made one year after the last regularpayment. The effective rate of interest is 8%. Show algebraically and verbally that theoutstanding loan balance after the 5th payment has been made is:

“a) 1000(1.08)5 − 120 · s5

“b) 1000 − 40s5.”

Solution: I give only the verbal explanations.


“a) Interpret the loan as being amortized by the regular payments and the final droppayment. The amount of 1000(1.08)5 − 120 · s5 is that given by the retrospectivemethod.

“b) Now interpret the loan as being repaid by the sinking fund method. As there isno formal sinking fund, one may simply view a portion of the payments as beingcontributed to a sinking fund in the hands of the lender, where the sinking fundearns interest at the same rate as the principal. Each annual payment of 120 may beinterpreted as being the sum of the service cost of 80, plus a contribution of 40 to thesinking fund which will repay the loan at maturity. After the 5th payment the valueof the sinking fund is 40 · s5. The “outstanding loan balance” will be the excess ofthe face value of the loan (which has been serviced annually, so there is no additionalaccumulation of interest) over the value of the sinking fund.

[5, Exercise 19, p. 187], [6, Exercise 23, p. 198] “On a loan of 10,000, interest at 9% effec-tive must be paid at the end of each year. The borrower also deposits X at the beginningof each year into a sinking fund earning 7% effective. At the end of 10 years the sinkingfund is exactly sufficient to pay off the loan. Calculate X.”Solution: An equation of value at time t = 10 is X · s10 7% = 10000, so

X =10000 × 0.07(

(1.07)10 − 1)

1.07= 676.4252593 .

The interest rate of 9% is totally irrelevant.[5, Exercise 20, p. 188], [6, Exercise 24, p. 198] “A borrower is repaying a loan with 10 an-

nual payments of 1,000. Half of the loan is repaid by the amortization method at 5%effective. The other half of the loan is repaid by the sinking fund method, in which thelender receives 5% effective on the investment and the sinking fund accumulates at 4%effective. Find the amount of the loan.”Solution: Let the amount of the loan be L. The amortization of the loan of L

2 entails an

annual payment ofL

2a10 5%

=0.025L

1 − (1.05)−10 . For the other half of the loan the borrower

must pay interest annually in the amount of 0.05 × L2 = 0.025L. These two expenses —

the amortization of half the loan, and the servicing of the other half — leave from hisannual payment a balance of

1000 − 0.025L1 − (1.05)−10 − 0.025L

which must accumulate at 4% in the sinking fund to produce a balance of L2 at maturity.

We have the equation of value

1000 − 0.025L1 − (1.05)−10 − 0.025L =

L2s10 4%

=0.02L

(1.04)10 − 1


⇔ L =40000

11−(1.05)−10 + 1 + 0.8

(1.04)10−1

= 7610.479836


6.31 Supplementary Notes for the Lecture of March 15th, 2010Distribution Date: Monday, March 15h, 2010


6.31.1 §5.4 SINKING FUNDS (continued)

[5, Exercise 21, p. 188], [6, Exercise 25, p. 198] “A borrows 12,000 for 10 years, and agreesto make semiannual payments of 1,000. The lender receives 12% convertible semiannu-ally on the investment each year for the first 5 years and 10% convertible semiannuallyfor the second 5 years. The balance of each payment is invested in a sinking fund earn-ing 8% convertible semiannually. Find the amount by which the sinking fund is short ofrepaying the loan at the end of the 10 years.”Solution: The interest payments for the first 10 half-years are 6% of 12,000, i.e. 720per half-year; and, for the second 10 half-years, 600 per half-year. This leaves 280 atthe end of each of the first 10 half-years, and 400 at the end of each of the second 10half-years to accumulate in the sinking fund, which earns 4% effective every half year.The accumulated balance in the sinking fund at maturity will be

120s10 4% + 280s20 4% =1

0.04

(120

((1.04)10 − 1

)+ 280

((1.04)20 − 1

))

= 25(120(1.04)10 + 280(1.04)20 − 400

)

= 9778.594855

implying that the shortfall to repay the loan will be 12, 000 − 9778.59 = 2221.41.

[5, Exercise 22, p. 188], [6, Exercise 26, p. 198] 1. “A borrower takes out a loan of 3000for 10 years at 8% convertible semiannually. The borrower replaces one-third ofthe principal in a sinking fund earning 5% convertible semiannually, and the othertwo-thirds in a sinking fund earning 7% convertible semiannually. Find the totalsemiannual payment.

2. “Rework (a) if the borrower each year puts one-third of the total sinking fund depositinto the 5% sinking fund and the other two-thirds into the 7% sinking fund.

3. “Justify from general reasoning the relative magnitude of the answers to (a) and (b).”Solution:

1. The semiannual contribution to the sinking funds is

1000s20 2.5%

+2000

s20 3.5%


and the semiannual interest payment is 4% of 3, 000, or 120. Hence the total semi-annual payment is

1000s20 2.5%

+2000

s20 3.5%

+ 120 =25

(1.025)20 − 1+

70(1.035)20 − 1

+ 120

= 229.8692824

2. Let the total sinking fund deposit be D. Then the equation of value at maturity is

D3· s20 2.5% +

2D3· s20 3.5% = 3000 ,

implying that

D =9000

s20 2.5% + 2s20 3.5%

=9000

(1.025)20−10.025 + 2 · (1.035)20−1

0.035

= 109.6170427 ,

so the total semi-annual payment is 109.6170427+120=229.6170427.3. In the original repayment scheme the portion of the payment contributed to the 5%

sinking fund grows more slowly than that to the 7% fund. Thus, while the final ac-cumulations in the funds will be in the ratio of 1:2, the proportion of the contributionto the 5% fund would have been more than 1

3 . By reducing that proportion to 13 we

increased the interest earned by the fund, so a smaller total contribution was requiredfor the sinking fund.

[5, Exercise 23, p. 188], [6, Exercise 27, p. 198] “A payment of 36,000 is made at the end ofeach year for 31 years to repay a loan of 400,000. If the borrower replaces the capital bymeans of a sinking fund earning 3% effective, find the effective rate paid to the lender onthe loan.”Solution: The annual contribution to the sinking fund is

400000s31 3%

=12000

(1.03)31 − 1= 7999.571516.

Hence the annual interest payment is 36, 000 − 7, 999.57 = 28000.43, i.e., 7% of theprincipal of 400,000.

(The graded test were returned to students present at this time.)




6.32.1 §5.4 SINKING FUNDS (conclusion)

[6, Exercise 28, p. 198] “A 20-year annuity-immediate has a present value of 10,000, whereinterest is 8% effective for the first 10 years, and 7% effective for the second 10 years.An investor buys this annuity at a price which, over the entire period, yields 9% on thepurchase price; and, further, allows the replacement of capital by means of a sinking fundearning 6% for the first 10 years and 5% for the second 10 years. Find an expression forthe amount that is placed in the sinking fund each year.”Solution: The level annual payments under the annuity will be

10000a10 8% + (1.08)−10a10 7%

.

It appears to be intended that the sinking fund payments be level also. If their value is S ,and the purchase price is P, then

S((1.05)10s10 6% + s10 5%

)= P .

The purchase price satisfies the following equation of value at time 0: 10000

a10 8% + (1.08)−10a10 7%

− S a20 9% + (1.09)−20P = P ,

which implies that 10000a10 8% + (1.08)−10a10 7%

− S = 0.09P .

Solving with the earlier equation yields

S =10000(

a10 8% + (1.08)−10a10 7%

) (1 + 0.09

((1.05)10s10 6% + s10 5%

)) .

6.32.2 §5.5 DIFFERING PAYMENT PERIODS AND INTEREST CONVERSION PE-RIODS

Omit this section.


6.32.3 §5.6 VARYING SERIES OF PAYMENTS

Omit this section.

6.32.4 §5.7 AMORTIZATION WITH CONTINUOUS PAYMENTS

Omit this section.

6.32.5 §5.8 STEP-RATE AMOUNTS OF PRINCIPAL

Omit this section.

Textbook Chapter 6. Bonds and other securities.


The chapter is concerned with relations between the price of a security and its yield rate, andwith the value of a security at any time after it has been purchased, even at a time that is not aninterest compounding date.

6.32.7 §6.2 TYPES OF SECURITIES

We shall confine our study to bonds, which are a commitment by the issuer to repay a loan ata particular time, with interest payments according to a prescribed rate and schedule. Manyvariations are possible, and this type of instrument is still evolving.

Read the book and become familiar with the following terms concerning bonds:

Definition 6.19 • The word bond originally had a much more general meaning: we areusing the word in the sense of a security that commits a borrower to pay one or more spe-cific sums at specific times, subject to detailed requirements of interest and/or bonuses.

• The term of the bond is the length of time from the date of issue until the date of finalpayment, which date is the maturity date.

• The detailed conditions of the bond may permit the bond to be called at some date priorto the maturity date, at which time the issuer (=the lender) will repay the commitment,possibly with some additional amounts. Such a bond is callable.


• While the calling of a bond is at the initiative of the borrower, the lender (=the purchaser)may possibly have the right to redeem the bond prior to the date of maturity. Or, he mayhave some other type of right, e.g., to exchange the bond for shares of the stock of theissuing company; this is a convertible bond.

• A bond may be supplied with coupons — portions of the paper bond that are to be cutfrom the bond and exchanged for interest payments. Nowadays the coupons may nolonger be printed, but the purchaser may receive regular payments from the lender. Thiscan be the case if the bond is fully registered, so that the issuer has the coordinates ofthe purchaser. (A bond could also be registered only as to principal, in which case thecoupons are of the traditional type.)

• A coupon bond may be “stripped”, separating the coupons from the commitment to re-pay the face value of the bond, following which the two parts may be sold to separatepurchasers.

The preceding is just a brief introduction: this is not the course in which to learn about thevariety of investment vehicles available in today’s financial markets.

Unlike the problems we have been considering in earlier chapters, those here sometimes in-volve technical definitions that are not necessarily intuitive, and need simply to be memorized.I will eventually discuss with the class what definitions need to be absorbed for examinationpurposes.




6.33.1 §6.2 TYPES OF SECURITIES (conclusion)

[5, Exercise 1, p. 240], [6, Exercise 1, p. 240] “Find the price which should be paid for a zerocoupon bond which matures for 1000 in 10 years to yield:

1. 10% effective2. 9% effective3. Thus a 10% reduction in the yield rate causes the price to increase by what percent-

age?”

Solution:

1. The bond is now worth 1000(1.10)−10 = 385.5432894.2. When the interest rate is reduced to 9%, the present value of the bond increases to

422.4108069.3. The 10% decrease in the interest rate thereby increases the price by

422.4108069385.5432894

− 1 = 9.5624846%.

[5, Exercise 2, p. 240], [6, Exercise 2, p. 240] “A 10-year accumulation bond with an initialpar value of 1000 earns interest of 8% compounded semiannually. Find the price to yieldan investor 10% effective.”Solution:

Definition 6.20 [6, p. 205] An accumulation bond is one in which the redemption priceincludes the original loan plus all accumulated interest.

Solution: The only return payment is at maturity. The price to yield 10% interest willtherefore be

(1.10)−10(1000(1.04)20

)= 844.7728240.

[5, Exercise 3, p. 240], [6, Exercise 3, p. 240] “A 26-week (U.S.) T(reasury)-bill is boughtfor 9,600 at issue, and will mature for 10,000. Find the yield rate computed as:

1. A discount25 rate, using the typical method for counting days on a T-bill26.

25simple discount26i.e., “actual/360” [6, p. 39], using the exact number of days, but assuming 360 days in the year.


2. An annual effective rate of interest, assuming the investment period is exactly half ayear.”

Solution: Here is an example of a problem requiring some technical preparation. Whilewe have encountered the use of discount rather than interest in isolated problems, this isthe first time we have met it in a complex transaction.

1. The time is 26 weeks, i.e., 26 × 7 = 182 days, or, under the actual/360 system,182360

of a year. The discount rate will, therefore, be

360182× 400

10000= 7.912087912% .

2. The effective interest rate for half a year is

4009600

=1

24.

The effective rate for a full year will be(1 +

124

)2

− 1 =1

12+

1576

= 8.506944444% .

6.33.2 §6.3 PRICE OF A BOND

Remember that a bond is essentially a contract to pay a large amount “at maturity”, and smalleramounts of interest periodically on “coupon dates” until maturity. It differs from the types ofloans we have studied hitherto in that the contract is often (but not always)27 transferable fromone owner to another, and so it is reasonable to investigate the value of the entire contract underconditions of varying yield.

Further definitions Familiarize yourself with the following terms, defined in the textbook,and with the symbols usually used for them.

Definition 6.21 1. The price P paid for a bond. In practice bond prices are usually quotedin terms of a bond with face value (see next item) of 100.

2. The par value or face value or face amount. This amount is usually printed in the bondcontract, but may not be the amount paid at maturity. Its function is to determine, oncethe coupon rate r has been specified, the magnitude of the coupons.

27Of course, the issuer of a security cannot prevent the purchaser from arranging privately to transfer theproceeds to another person, and to receive payment for that. What the issuer may be able to do is to restrict theestablishment of a secondary market at which the security may be routinely bought and sold, and which wouldcreate a market value for the security. Some securities have restricted transferability conditions, e.g., only on thedeath of the owner. And no security is immune from a court order.


3. The redemption value C is the amount paid when the bond is redeemed. When a bondis “redeemable at par”, C = F. Where the redemption value exceeds the face value, theword premium may be used for the excess; this word premium is also used to denote theexcess of the price paid for a bond over what would have been the value if the yield ratewas the same as the coupon rate.

4. The coupon rate r is the effective rate per coupon payment period, based on which theamount of the coupon is calculated. The default payment period is a half-year.

5. The amount of a coupon is the product Fr.

6. The modified coupon rate g =FrC

is the coupon rate per unit of redemption value, ratherthan per unit of par value.

7. The yield rate or yield to maturity i is the actual interest rate earned by the investor.

8. The number of coupon payment periods from the date of calculation until maturity isdenoted by n.

9. The present value of the redemption value, discounted back to the present by the yieldrate, is denoted by K; so K = C(1 + i)−n.

10. The base amount G isFri

: the amount which, if invested at the yield rate i, would produceperiodic interest payments equal to the coupons.

11. A callable bond is one where the lender has the right to declare that interest paymentswill stop and a bond may be redeemed at certain dates before the maturity date; therecould be a premium paid in addition to the redemption value, to encourage lenders tocash in the bond.

12. The word discount is often used where we have been using the word premium if thepremium is negative: the discount is the negative of the premium.

Four formulæ for price. We shall consider four different ways of determining the presentvalue, or price of a bond. The formulae are derivable one from the other: there are situationswhere one may be more useful than another for specific applications; as with many of theother formula we have met, the relative advantages were often linked to the number of timesthat tables had to be consulted in computing the bond value; such distinctions may no longerbe significant. The most “basic” of the formulæ computes the price of a bond as the presentvalue of the coupons, interpreted as an annuity-immediate, to which is added the present valueK of the redemption value C of the bond, the face value plus any premium that is payable uponredemption. This formula may be applied either at the maturity date, or, if the bond can becalled (by the issuer) or redeemed (by the purchaser), at some earlier date.

Theorem 6.12 (The “Basic” Formula) P = Fr · an i + Cvn = Fr · an i + K


Theorem 6.13 (The Premium/Discount Formula) P = C + (Fr −Ci) · an

Proof: This formula may be derived from the preceding by recalling that 1 = vn + i · an i. �

The Base Amount G was defined above to beFri

.

Theorem 6.14 (The Base Amount Formula) P = G + (C −G)vn

Theorem 6.15 (Makeham’s Formula) P = K +gi(C − K)

There are interesting verbal explanations of the preceding formulæ, to which we may return.

[5, Exercise 5, p. 240], [6, Exercise 8, p. 241] “Two 1000 bonds redeemable at par at the endof the same period are bought to yield 4% convertible semiannually. One bond costs1136.78, and has a coupon rate of 5% payable semiannually. The other bond has a couponrate of 2 1

2% payable semiannually. Find the price of the second bond.”Solution: For the first bond we have F1 = C1 = 1000, i1 = 2%, P1 = 1136.78, r1 = 2.5%.By the Premium/Discount formula,

1136.78 = 1000 + (25 − 20) · an 2% ,

implying that an = 27.356. For the second bond we have F2 = C2 = 1000, i2 = 2%,r2 = 1.25%, n2 = n1,

P2 = 1000 + (12.5 − 20) · an

= 794.83 .


6.34 Supplementary Notes for the Lecture of March 22nd, 2010Distribution Date: Monday, March 22nd, 2010


6.34.1 §6.3 PRICE OF A BOND (conclusion)

Four formulæ for price (conclusion). The first two of the following formulæ were dis-cussed at the last lecture.

Theorem 6.16 (The “Basic” Formula) P = Fr · an i + Cvn = Fr · an i + K

Theorem 6.17 (The Premium/Discount Formula) P = C + (Fr −Ci) · an

Proof: This formula may be derived from the preceding by recalling that 1 = vn + i · an i. �

The Base Amount G was defined above to beFri

.

Theorem 6.18 (The Base Amount Formula) P = G + (C −G)vn

Theorem 6.19 (Makeham’s Formula) P = K +gi(C − K)

There are interesting verbal explanations of the preceding formulæ, to which we may return.

[5, Exercise 6, p. 240], [6, Exercise 9, p. 241] “A 1000 bond with a coupon rate of 9% payablesemiannually is redeemable after an unspecified number of years at 1125. The bond isbought to yield 10% convertible semiannually. If the present value of the redemptionvalue is 225 at this yield rate, find the purchase price.”Solution: We have F = 1000, C = 1125, r = 4.5%, i = 5%, K = 225. From this last fact

we have 225 = 1125(1.05)−n, so (1.05)n = 5, and an i =1− 1

50.05 = 16. By the Basic Formula,

the price is Fr · an + K = 45(16) + 225 = 945.

[5, Exercise 8, p. 241], [6, Exercise 11, p. 241] “An investor owns a 1000 par value 10% bondwith semiannual coupons. The bond will mature at par at the end of 10 years. The in-vestor decides that an 8-year bond would be preferable. Current yield rates are 7% con-vertible semiannually. The investor uses the proceeds from the sale of the 10% bond topurchase a 6% bond with semiannual coupons, maturing at par at the end of 8 years. Findthe par value of the 8-year bond.”Solution: We have C1 = F1 = 1000, r1 = 5%, n1 = 20, i1 = 3.5%. The price of the bondpresently owned is, by the premium/discount formula,

P1 = C1 + (F1r1 −C1i1)a20 3.5% = 1000 + (50 − 35) · 1 − (1.035)−20

0.035= 1, 213.186050 .


With these proceeds the investor buys a bond with par value F2 = C2, n2 = 16, r2 = 3%;by the premium/discount formula,

1213.186050 = F2 + F2(0.03 − 0.035) · 1 − (1.035)−16

0.035,

implying that F2 = 1291.269895.




There will be no 5th assignment.

This year the assignment portion of your grade will be based on the four assign-ments that have already been distributed: there isn’t time to prepare and grade a5th assignment. Grades on the four assignments will count equally in the 10-markportion of the term mark.

6.35.1 §6.4 PREMIUM AND DISCOUNT

Premium = P −C = (Fr −Ci)an i

Discount = C − P = −(Fr −Ci)an i

In practice the word premium is used when P − C > 0, and discount when P − C < 0; whenP = C we speak of a purchase at par.

Book Value For accounting purposes it is necessary to show a gradual progression of thevalue of a bond from purchase to maturity that is in some reasonable relationship with itsmarket value. There are a number of possible methods for doing this in practice, but we shalladhere to the method whereby the value is shown to be what would be the price if the yieldrate does not change after the purchase of the bond. Of course, the yield rate could very wellchange after the purchase, and so this is not a completely realistic assumption. There are othermethods for assigning a value to the bond. In this section we shall consider the value only atcoupon payment dates; generalization to other dates will be considered in the next section.

When a bond is purchased at a premium28 the value of the redemption value and the unpaidcoupons just after the payment of the tth coupon — i.e., of the book value Bt — will decreaseuntil the maturity date. Thus the coupons may be interpreted as consisting of the interest whichis earned at the yield rate i and an amount to amortize the premium. The redemption value Cis C = Bn, and the price is P = B0. We will use this notation even when the bond is purchasedwith some of its coupons already paid (to the vendor). The writing down of the premium maybe shown on a schedule like

28mutatis mutandis, at a discount


Coupon Coupon Interest Amount for Amortization BookNumber amount earned of Premium value

0 . . . . . .1 Fr

. . . Fr . . . . . . . . .

The following example illustrates a problem where the price P of a bond is not determinedexplicitly until an equation for P is determined and solved.

Example 6.20 [5, Example 6.4, p. 212], [6, Example 7.4, p. 219] “Find the price of a 1,000 parvalue 2-year 8% bond with semi-annual coupons bought to yield 6% convertible semiannually,if the investor plans to replace the premium by means of a sinking fund earning 5% convertiblesemi-annually. (Note: The intention is that this plan for a sinking fund has been considered incalculating the yield rate of the bond.)Solution: Note that, following the usual convention, the 8% rate is interpreted as a nominalinterest rate compounded twice a year. The unknown is the price, P. The coupons each havevalue 0.04×1000 = 40, but the interest earned is 3% of the price P, which remains to be deter-mined. The sinking fund is to mature at value P−1000 after 2 years, i.e., after 4 contributions.Thus an equation of value is

(40 − .03P)s4 2.5% = P − 1000 . (88)

This can be solved for P:

P =1000 + 40s4 2.5%

1 + 0.03s4 2.5%

= 1, 036.93 .

Note the assumption used in equation (88): the amount by which each coupon exceeds theinterest “earned” at the yield rate is contributed to the sinking fund; this was not explicitlystated in the problem, but is the author’s interpretation. If, for example, the purchaser haddecided that he would allocate only half of the excess to his sinking fund contributions, thenthe price of the bond would be about 1082.43.

Example 6.21 Let’s return to the Example 6.20 on page 1143 of these notes, considered lastday, [5, Example 6.4, p. 212], [6, Example 7.4, p. 219] “Find the price of a 1,000 par value2-year 8% bond with semi-annual coupons bought to yield 6% convertible semiannually, ifthe investor plans to replace the premium by means of a sinking fund earning 5% convertiblesemi-annually.” We found that the price of this bond would be 1,036.93. Now let’s modify theproblem so that there is no 5% sinking fund. Then, by the Premium/Discount Formula, the


price of the bond would be simply

P = C + (Fr −Ci) · a4 3%

= 1000 + (40.00 − 30.00) · 1 − (1.03)−4

0.03= 1037.17 .

At first glance this appears to be erroneous: how could the price without the sinking fund,which is earning an inferior interest rate, be less than the price with the assumption of thesinking fund? The point is that, when one assumes apriori that part of the coupon will not beavailable, but will be trapped in a sinking fund earning an inferior interest rate for a periodof time ranging from 1.5 years to 0 years, then the value of the bond is less than if the entirecoupon was assumed to be readily available, and that all interest calculations would be madeat a rate of 3% per half-year. You can see this, for example, by tracing the contribution fromthe first coupon to the sinking fund. The coupon is for 40.00, of which (under the sinking fundassumption) the amount that was “earned” was only 0.03 × 1, 036.93 = 31.11, so 8.89 wascontributed to the sinking fund. That amount earns interest for 1.5 years at 2.5% per half-year,and grows to 8.89(1.025)3 = 9.57 at maturity. But, if the prevailing interest rate is to be takento be 3% per half year, then the present value of that amount is 8.89(1.025)3(1.03)−4 = 8.51,and its value on the date the coupon was received was only 8.89(1.025)3(1.03)−3 = 8.76, 13cents less than the amount deposited in the sinking fund at that time.

For the bond purchased at 1037.17 without the sinking fund, let’s set up an amortizationtable at 3% per half-year:

Coupon Coupon Interest Amount for Amortization BookNumber amount earned of Premium value

0 1,037.171 40.00 31.12 8.88 1,028.292 40.00 30.85 9.15 1,019.143 40.00 30.57 9.43 1,009.714 40.00 30.29 9.71 1,000.00

TOTAL 160.00 122.88 37.16

Before leaving this example, let’s work out an analogous table for the sinking fund variationconsidered last day. In this case the premium is not amortized until the date of the last coupon.We have


Coupon Coupon Interest S. Fund S. Fund BookNumber amount earned Contribution Balance Value

0 1,036.931 40.00 31.11 8.89 8.89 1,036.932 40.00 31.11 8.89 18.00 1,036.933 40.00 31.11 8.89 27.34 1,036.934 40.00 31.11 8.89 36.91 1,036.93

where the balance in the Sinking Fund has reduced the book value of the bond to (approxi-mately) the maturity value.

6.35.2 §6.5 VALUATION BETWEEN COUPON PAYMENT DATES (Omit)

6.35.3 §6.6 DETERMINATION OF YIELD RATES (omit)




6.36.1 §6.7 CALLABLE AND PUTABLE BONDS

[5, Exercise 24, p. 243], [6, Exercises 31, 32, p. 243] “A 1000 par value bond has 8% semi-annual coupons, and is callable at the end of the 10th through the 15th years at par.

1. “Find the price to yield 6% convertible semiannually.2. “Find the price to yield 10% convertible semiannually.”3. “If the bond in 2. is actually called at the end of 10 years, find the yield rate.

Solution: The question did not mention the length of term. Does that matter? It doesn’tmatter in the case of 6% yield because, as we will see, the worst value of the bond iswhen it is called as early as possible, and that determines the minimum price to pay toobtain a yield of 6%. However, when the yield is 10% compounded semi-annually, thevalue of the bond decreases as the time to redemption increases; so, if the bond is notcalled, the owner is stuck with a poor investment that he must keep until it matures, andthe price he should pay for the bond depends on knowing that maturity date.

1. Let n be the coupon number at whose date the bond is called matures. Then, by thePremium/Discount Formula

P = 1000 + (40 − 30)an 3% , (n = 20, 22, 24, 26, 28, 30).

Without knowing which will be the date of call, we take the worst possible datein order to minimize the price; since an 3% is an increasing function of n, and ismultiplied by a positive number, 10, we minimize by making n as small as possible,i.e., 2 × 10 = 20:

P = 1000 + (40 − 30)a20 3% = 1000.00 + 148.78 = 1148.78 .

2. When the semi-annual yield rate is 5%, the multiplier is negative, 40 − 50 = −10,and we must choose the largest value of n, i.e., n = 30, for a price of

P = 1000 + (40 − 50)a30 5% = 1000.00 − 153.72 = 846.28 .

But what if the maturity date of the bond is after the n > 30th coupon? Then theprice will be given by

P = 1000 + (40 − 50)an 5%

As n increases, the value decreases. In the limit as n→ ∞ the price approaches

P = 1000 + (40 − 50)a∞ 5% = 100 − 200 = 800 ,

which can be seen to be 40 · a∞ 5%.


3. The yield rate i satisfies the equation

846.28 = 1000 + 1000(0.04 − i)a20 i ,

and can be found by iteration to be 10.52%%.

Example 6.22 [5, Example 6.8, p. 225][6, Example 7.9, p. 231] “Consider a $100 par value4% bond with semiannual coupons callable at $109 on any coupon date starting 5 years afterissue for the next 5 years, at $104.50 starting 10 years after issue for the next 5 years, andmaturing at $100 at the end of 15 years. What is the highest price which an investor can payand still be certain of a yield of: (1) 5% convertible semiannually, and (2) 3% convertiblesemiannually?”Solution: I recommend that you use the Premium/Discount Formula, and remember that thefunction an i is an increasing function of n; this means that, in a problem of this type, the valuesleast favourable to the investor will always be at the end of an interval of payment periods: atthe lower end when Fr − Ci is positive, and at the upper end when Fr − Ci is negative. Theintervals to be considered will be consecutive payment dates when the call premium has thesame value. Each of these intervals need to be treated separately, and then the least favourablefrom each interval taken together and the least favourable of them determined. The previousedition of the textbook stated, in connection with the part of the problem where the yield rateis 5%, that “it is immediately clear that the latest possible redemption date is less favourableto an investor, since the bond will be selling at a discount.” When I was using that version inmy course, I felt the statement may not have been clear, and offered the following explanation;the wording has changed in the current edition, although it may not be any more transparent.The factor that determines which end of the interval is least favourable is Fr − Ci = C(g − i),where g is the modified coupon rate (cf. Definition 6.21.4, p. 1138). The given data lead tofinding the minimum of the following three values:

100 + (2 − 2.50)a30 0.025 = 89.53109 + (2 − 2.725)a29 0.025 = 95.33

104.50 + (2 − 2.6125)a19 0.025 = 94.17

And, indeed, the minimum will be when n = 30. The situation is best viewed through the Baseamount formula:

P = G + (C −G)vn

where G = Fri (cf. Definition 6.21.10). Now vn is a decreasing function of n, and the sign of

C −G = C − Fri

is +: the factor vn is smallest when n is as large as possible; and, because thevalues at which the bond is called or redeemed are not increasing with time, the factor C −Gis also non-increasing. Thus, in this particular case, the bond will be worth least if it is not


called. Thus we did not need to evaluate the prices for the bond being called when n = 19, 29,as they could not be less than the price for n = 30.

When the yield rate is 3% there is no such claim of an obvious solution. In this case thePremium/Discount Formula gives prices

109.00 + (2.0000 − 1.6350)an 0.015 for n = 10, . . . , 19104.50 + (2.0000 − 1.5675)an 0.015 for n = 20, . . . , 29

100.00 + (2.0000 − 1.5000)an 0.015 for n = 30.

We must consider the smallest value of n for each of the intervals associated with the differentredemption values, i.e.,

109.00 + (2.0000 − 1.6350)a10 0.015 = 112.37104.50 + (2.0000 − 1.5675)a20 0.015 = 111.93100.00 + (2.0000 − 1.5000)a30 0.015 = 112.01 ,

so, in this case the lowers price occurs for n = 20.

[5, Exercise 25, p. 243], [6, Exercise 33, p. 243] “A 1,000 par value 8% bond with quarterlycoupons is callable five years after issue. The bond matures for 1,000 at the end of 10years, and is sold to yield a nominal rate of 6% convertible quarterly, under the assump-tion that the bond will not be called. Find the redemption value at the end of 5 years thatwill provide the purchaser the same yield rate.”Solution: If the bond is certain not to be called, its present value (i.e., its price) is

1000 + (20 − 15)a40 1.5% = 1000 + 5(29.9258) = 1149.63.

Let x denote the redemption value after 5 years that will provide the same yield rate.Then

1149.63 = x + (20 − 0.015(x))a20 1.5%

which implies that

x =1149.63 − 20a20 1.5%

1 − 0.05a20 1.5%

= 1085.91 .

[5, Exercise 26, p. 243], [6, Exercise 34, p. 243] “A 1000 par value 4% bond with semian-nual coupons matures at the end of 10 years. The bond is callable at 1050 at the ends ofyears 4 through 6, at 1025 at the ends of years 7 through 9, and at 1000 at the end of year10. Find the maximum price that an investor can pay and still be certain of a yield rate of5% convertible semiannually.”


Solution: We have to find the minimum of the following prices based on the given calldates and premiums:

1050.00 + (20.000 − 26.250)an 2.5% (n = 8, 10, 12) (89)1025.00 + (20.000 − 25.625)an 2.5% (n = 14, 16, 18) (90)

1000.00 + (20.000 − 25.000)a20 2.5% . (91)

In each case the coefficient of an 2.5% is negative, so the lowest value will be when n is aslarge as possible; that is, we have to compare the following three amounts

1050.00 + (20.000 − 26.250)a12 2.5% = 1050 − (6.250)10.2578 = 985.891025.00 + (20.000 − 25.625)a18 2.5% = 1025 − (5.625)14.3534 = 944.261000.00 + (20.000 − 25.000)a20 2.5% = 1000 − (5.000)15.5892 = 922.05 ,

whose minimum is the last, the price of the bond if not called before maturity. That is thehighest price the investor may pay if she wishes to be sure that the yield will not be lessthan 5% convertible semiannually.

[6, Exercise 35, p. 243] “A 1,000 par value 6% bond with semiannual coupons is callable atpar 5 years after issue. It is sold to yield 7% under the assumption that the bond will becalled. The bond is not called, and it matures at the end of 10 years. The bond issuerredeems the bond for 1000 + X without altering the buyer’s yield rate of 7% convertiblesemiannually. Find X.”Solution: Under the assumption that the bond will be called at par 5 years after issue, itsprice, when yielding 3.5% effective semi-annually, would be

1000 + (30 − 35)a10 3.5% = 958.42 .

The premium upon maturity will be given by the equation

958.42 = 1000 + X + (30 − (0.035)(1000 + X))a20 3.5% .

which implies that

X =958.42 − 30a20 3.5%

1 − 0.035a20 3.5%

− 1000 = 58.66 .


6.37 Supplementary Notes for the Lecture of March 29th, 2010Distribution Date: Monday, March 29th, 2010


Textbook Chapter 7. Yield Rates.

6.37.1 §7.1 INTRODUCTION (omit)

6.37.2 §7.2 DISCOUNTED CASH FLOW ANALYSIS (omit)

6.37.3 §7.3 UNIQUENESS OF THE YIELD RATE (part)

Definition 6.22 The yield rate is that rate of interest at which the present value of returns fromthe investment is equal to the present value of contributions into the investment.

Finding the yield rate may require the use of various approximation methods, since the equa-tions that have to be solved may be polynomial of high degree.

Example 6.23 [5, p. 255] [6, p. 133] A person makes payments of 100 immediately and 132at the end of 2 years, in exchange for a payment in return of 230 at the end of 1 year. The yieldrate i can be shown to satisfy the equation

((1 + i) − 1.1)((1 + i) − 1.2) = 0

which has two distinct solutions.

Example 6.24 [5, Example 7.3, p. 258] “A is able to borrow 1000 from B for 1 year at 8%effective, and to lend it to C for 1 year at 10% effective. What is A’s yield rate on this transac-tion.”Solution: The equation of value at time 0 is

1000 − v(1080) = 1000 − v(1100) ,

which has no finite solution. We can say that the yield rate is infinite.

Read [5, Example 7.4, p. 258], in which the borrower cannot possibly earn sufficient in-terest to cover her payments; in this case we might wish to speak of an “imaginary” yieldrate.

6.37.4 §7.4 REINVESTMENT RATES

Single payment with interest reinvested Suppose 1 is invested at time 0, with interest beingpaid at rate i at the ends of n years, but where the interest can be reinvested only at rate j. Atthe end of n years the total accumulated value of the investment is

1 + isn j


Annuity-immediate at rate i, with reinvestment at rate j If an annuity-immediate of 1 perperiod for n periods pays interest at rate i, but the interest can be reinvested only at rate j, theaccumulated value at time n is

n + i · (Is)n−1 j = n +ij·(sn j − n

)

[5, Exercise 11, p. 301], [6, Exercise 10, p. 161] “It is desired to accumulate a fund of 1,000at the end of 10 years by equal deposits at the beginning of each year. If the deposits earninterest at 8% effective, but the interest can only be reinvested at only 4% effective, show

that the deposit necessary is1000

2s11 0.04 − 12.”

Solution: Let x denote the necessary deposit. We will sum the principal payments (x ×10) and treat the interest payments as forming an increasing annuity-immediate withincrements of 0.08x.

x(10 + 0.08 · (Is)10 4%

)= 1000

⇔ x(10 + 0.08 ·

s10 4% − 10

0.04

)= 1000

⇔ x =1000

2s11 0.04 − 12. �

[5, Exercise 12, p. 301][6, Exercise 11, p. 161] “A loan of 10,000 is being repaid with pay-ments of 1,000 at the end of each year for 20 years. If each payment is immediatelyreinvested at 5% effective, find the effective annual rate of interest earned over the 20-year period.”Solution: Let the effective yield rate be i. The payments do not become available untilthe maturity date, after 20 years. Until that time they are locked into a payment-schemethat accumulates to value 1000s20 5%. We are asked for the interest rate that was earned.There are thus just two transactions: the loan at time 0, in the amount of 10,000, and therepayment at time 20, in the amount given above. The equation of value at time t = 0 is

1000s20 5%(1 + i)−20 = 10000

⇔ (1 + i)−20 =10(0.05)

(1.05)20 − 1⇔ i = 6.1619905%.

[5, Exercise 13, p. 301], [6, Exercise 12, p. 161] “An investor purchases a 5-year financial in-strument having the following features:

“(i) The investor receives payments of 1000 at the end of each year for 5 years.


“(ii) These payments earn interest at an effective rate of 4% per annum. At the end of theyear, this interest is reinvested at the effective rate of 3% per annum.

“Find the purchase price to the investor to produce a yield rate of 4%.”Solution: To determine the yield we need to consider when the payment is finally releasedto the investor. The payments of 1000 are to be invested at 4%; they will generate anincreasing annuity whose payments start at 40 at the end of year 2, up to 160 at the endof year 5. But they are not released to the investor; rather, they earn interest at 3%. Atthe end of 5 years — and only then — the investor receives

5(1000) + 40(Is)4 3%

and these amounts have to be discounted to the present at 4%, giving a present value of

(1.04)−5(5(1000) + 40(Is)4 3%

)= 5000 +

400.03

(s5 3% − 5

)

= (1.04)−5(5000 +

400.03

((1.03)5 − 1

0.03− 5

))

= 4448.418326

if the yield is to be 4%.

[6, Exercise 13, p. 161] “An investor deposits 1,000 at the beginning of each year for fiveyears in a fund earning 5% effective. The interest from this fund can be reinvested at only4% effective. Show that the total accumulated value at the end of ten years is

1250(s11 0.04 − s6 0.04 − 1

).”

Solution: The 5 deposits of 1000 would be worth 1000(s10 4% − s5 4%

)at the end of 10

years if they were earning interest together with the reinvested annual interest payments.But they are locked into a fund where they earn 5%, and are not released until time 10,still worth 5,000. The interest payments constitute an increasing annuity to the investor,beginning with 50 at time 1, increasing to 250 at time 5, and then remaining constantuntil time 10. They are available to the investor as they are paid, but she reinvests themat 4%. Their value at time 10 is, therefore 50

((Is)10 4% − (Is)5 4%

). Summing yields

5000 + 50((Is)10 4% − (Is)5 4%

)= 5000 +

500.04

(s11 4% − 11 − s6 4% + 6

)

= (5000 − 6250) + 1250(s11 4% − s6 4%

)

= 1250(s11 0.04 − s6 0.04 − 1

)

= 7316.719914


Using the tables in the textbook, we would find the answer to be

1250(s11 0.04 − s6 0.04 − 1

)= 1250(13.4864 − 6.6330 − 1)= 7316.7500 .

[6, Exercise 14, p. 161] “A invests 2,000 at an effective interest rate of 17% for 10 years.Interest is payable annually and is reinvested at an effective rate of 11%. At the end of 10years the accumulated interest is 5,685.48. B invests 150 at the end of each year for 20years at an effective interest rate of 14%. Interest is payable annually and is reinvested atan effective rate of 11%. Find B’s accumulated interest at the end of 20 years.”Solution: A’s interest payments begin at the end of year 1, in the amount of 17%×2000 =

340 and continue at the same level for 10 years. The equation of value at the end of year10 is

340s10 11% = 5685.48 ,

implying that

s10 11% =5685.48

340= 16.722.

It follows that (1.11)10 = (16.722)(0.11) + 1 = 2.83942, and that

s20 11% =(1.11)20 − 1

0.11= s10 11% ·

((1.11)10 + 1

)

= s10 11% ·(s10 11% + 2

)

= 16.722 × 3.83942 = 64.20278124.

B’s interest payments constitute an increasing annuity-immediate whose first payment isat the end of year 2, in the amount of 21. The accumulated value of B’s accumulatedinterest is

21(Is)19 11% = 21s20 11% − 20

0.11

= 21 · 64.20278124 − 200.11

= 8438.712782.


6.37.5 §7.5 INTEREST MEASUREMENT OF A FUND (Omit)

6.37.6 §7.6 TIME-WEIGHTED RATES OF INTEREST (Omit)

6.37.7 §7.7 PORTFOLIO METHODS AND INVESTMENT YEAR METHODS (Omit)

6.37.8 §7.8 SHORT SALES (Omit)

6.37.9 §7.9 CAPITAL BUDGETING - BASIC TECHNIQUES (Omit)

6.37.10 §7.10 CAPITAL BUDGETING - OTHER TECHNIQUES (Omit)

6.37.11 Appendix 7 (Omit)


6.38 Supplementary Notes for the Lecture of March 31st, 2010Distribution Date: Wednesday, March 31st, 2010


This hour was devoted to discussion of, among other topics, problems on the 5th problemassignments for years 2002/2003 and 2004/2005. Solutions accompany these assignments,elsewhere in these notes.


6.39 Supplementary Notes for the Lecture of April 07st, 2010Distribution Date: Wednesday, April 07th, 2010


This hour was devoted to a review of force of interest, and of general use of the accumulationfunction a(t) in situations where the interest environment may not be compound interest. Noteswill not be posted.


7 Problem Assignments, Tests, and Examinations from Pre-vious Years

7.1 2002/20037.1.1 First 2002/2003 Problem Assignment, with Solutions

In all of the following problems students were are expected to show your work.

1. (a) What principal will earn interest of 100 in 7 years at a simple interest rate of 6%?(b) What simple interest rate is necessary for 10,000 to earn 100 interest in 15 months?(c) How long will it take for money to double at a simple interest rate of 8%?(d) For the rate stated and the period of time computed in the previous part of the ques-

tion, what would 1 grow to if interest were compounded annually?Solution:

(a) Let P denote the unknown principal. Equating the interest earned, P · (0.06) · 7 to100 and solving for P, we obtain that P = 100

7×0.06 = 238.10.(b) If the interest rate is i, the amount of interest earned will be i · 15

12 · 10000 = 100.Solving yields i = 4

5 · 10010000 = 0.008 or 0.8%.

(c) Let the number of years for money to double be denoted by t. We solve 1 + (0.08)t =

2: t = 2−10.08 = 12.5. Money doubles in 12 1

2 years.(d) (1.08)12.5 = 2.62.

2. The total amount of a loan to which interest has been added is 20,000. The term of theloan was four and one-half years.

(a) If money accumulated at simple interest at a rate of 6%, what was the amount of theloan?

(b) If the nominal annual rate of interest was 6% and interest was compounded semi-annually, what was the amount of the loan?

(c) If the rate of interest was 6%, interest was compounded annually for full years, butsimple interest was paid for the last half-year, what was the amount of the loan?

(d) If the rate of interest was 6%, interest was compounded annually for full and partyears, what was the amount of the loan?

(e) If the effective annual rate of interest was 6%, but interest was compounded semian-nually, what was the amount of the loan?

(f) If the nominal annual rate of interest was 6%, but interest was compounded contin-uously, what was the amount of the loan?

(g) If interest was compounded continuously, and the force of interest was 6%, whatwas the amount of the loan?


Solution:

(a) If the amount of the loan is P, then 20000 = (1 + 0.06 × 4.5)P = 1.27P, so P =200001.27 = 15, 748.03.

(b) If the amount of the loan is P, then 20000 = (1 + 0.062 )2×4.5P,

P = 20000(1.03)−9 = 15, 328.33 .

(c) If the amount of the loan is P, then 20000 = (1 + 0.06)4 ·(1 + 0.06

2

)P,

P = 20000(1.06)−4(1.03)−1 = 15, 380.46 . (92)

(d) If the amount of the loan is P, then 20000 = (1 + 0.06)4.5P, P = 20000(1.06)−4.5 =

15, 386.99.(e) If the semi-annual rate of interest is denoted by i, then (1 + i)2 = 1.06, so 1 + i =

(1.06)12 . If the amount of the loan is P, then 20000 = (1 + i)9P = (1.06)4.5P, so

P = 20000 × (1.06)−4.5 = 15, 386.99.Note that this is exactly the same principal as in the preceding version of the problem,since it is precisely the same problem! Note also that the amount of the principal isslightly more than in the version in part 2c, since simple interest for a fraction of ayear in that case will have a higher yield than compound interest in this case.

(f) We are told that the nominal rate of interest, compounded instantaneously, is 6%;thus, if i is the effective annual rate, 0.06 = δ = ln(1 + i); equivalently, e0.06 = 1 + i,so i = 6.18365%. In a full year an amount of 1 will grow by a factor 1.0618365; ina half year, by a factor

√1.0618365. In 4 1

2 years we have P(1.0618365)4.5 = 20000,so P = 20000(1.0618365)−4.5 = 20000(0.763379) = 15267.59.

(g) The force of interest is the nominal rate of interest which is convertible continuously[10, p. 17]. We are told that interest was compounded continuously. The effectiveannual rate of interest, which we shall denote by i, has the property that 0.06 =

ln(1+ i), or that 1+ i = e0.06. If the amount of the loan is P, then 20000 = (1+ i)4.5P =

e4.5×0.06 = e0.27P, so P = 20000e−0.27 = 15267.59, the same result as in the precedingpart.

3. (cf. [10, Exercise 1-13, p. 24]) Henry plans to have an investment of 10,000 on January1, 2006, at a compound annual rate of discount d = 0.11.

(a) Find the value that he would have to invest on January 1, 2003.(b) Find the value of i corresponding to d.(c) Using your answer to part (b), rework part (a) using i instead of d. Do you get the

same answer?


Solution:

(a) The accumulation of 10,000 will have to be discounted by a factor of (1 − 0.11)three times to reduce it by compound discount to January 1, 2003. The amount to beinvested is, accordingly, 10000(1 − 0.11)3 = 7049.69.

(b) The relationship between d and i is given, for example, by (1 + i)(1 − d) = 1, whichimplies that i = d

1−d . Here

i =0.11

1 − 0.11=

1189

= 0.1233596.. = 12.36..%.

(c) When i = 12.36%, v = 11+i = 1

1.1236 = 0.89. The value on January 1, 2003 of the10,000 expected on January 1, 2006 will then be 10000(0.89)3 = 7049.69, as before.

4. (cf. [10, Exercise 1-24, p. 26]) Recall that (cf. [10, (1.21)])[1 +

i(m)

m

]m

= 1 + i =1

1 − d=

[1 − d(m)

m

]−m

. (93)

(a) Determine whether there is an integer n such that

1 +i(n)

n=

1 + i(2)

2

1 + i(3)

3

(94)

and, if there is such an integer, find it.(b) Replace the right member of (94) by a product

(1 +

i(2)

2

) (1 − d(3)

3

)

and then interpret this product verbally to show that it must be equal to 1 + i(n)

n if asuitable n exists.

Solution:

(a) i(2) is the nominal annual interest rate which, when compounded semi-annually,yields an effective annual rate of i. Thus 1 + i(2)

2 =√

1 + i; similarly, 1 + i(3)

3 =3√1 + i.

The ratio 1+ i(2)2

1+ i(3)3

is, therefore, equal to (1 + i)12− 1

3 = (1 + i)16 , which is the accumulation

of 1 after a period of 16 of a year; this is, by definition, equal to 1 + i(6)

6 .

(b) Under an effective annual interest rate of i, the factor(1 + i(2)

2

)is the value of 1 after

122 = 6 months. If this amount is discounted back 12

3 = 4 months, it decreases by areduction factor of

(1 − d(3)

3

). The result is equivalent to a net accumulation period of

6 − 4 = 2 months, i.e. 16 of a year, under which it would grow by a factor 1 + i(6)

6 .


5. (cf. [10, Exercise 1-30, p. 27]) Show that f (t) = (1+i)t−(1+it) is minimized at t = ln i−ln δδ

.Solution:

If only elementary calculus is used, this problem is more difficult than it looks.Students were accorded a full grade for showing that the point claimed is, in-deed, a local minimum; the proof that it is a global = absolute minimum, ismore difficult; one possible solution is given below. No attempt has been madeto produce a compact solution.

Applying elementary calculus, we find that

f ′ = (1 + i)t ln(1 + i) − i = (1 + i)tδ − if ′′ = (1 + i)tδ2

To find the critical points of the function, we solve for t the equation f ′(t) = 0. Takingnatural logarithms yields

t ln(1 + i) + ln δ = ln i

which is satisfied only for

t0 =ln i − ln δ

δ=

ln iδ

δ. (95)

The second derivative, f ′′(t0) is positive everywhere, since it is the product of an expo-nential — always positive — and the square of a real number; this tells us that the pointt = t0 (95) is a local minimum.Does this completely solve the problem? Not yet! To solve an extremum problem weneed to interpret local extremum information with reference to the domain of the func-tion. For example, if the domain is infinite, then the function might not even have a globalor absolute minimum, even though it has a local minimum.29 And, if the domain of thefunction is a closed interval, we need to investigate the behavior at the end points of thatinterval. The function f is meaningful for all real values of t. One interpretation wouldbe to take the domain to be t ≥ 0; another interpretation would be to take the domain tobe −∞ ≤ t ≤ +∞.What follows is just one possible way of completing this problem. We observe thatf (0) = f (1) = 0. Could f (t) = 0 for t different from 0, 1? Rolle’s theorem impliesthe existence of a point with zero slope between any two zeros of the function; as wehave seen that there is only one such point with zero slope, there cannot exist more thantwo zeros of the function: and thus the point t0 is the only local extremum. Thus, bythe Intermediate Value Theorem, f has the same sign throughout each of the intervals−∞ < t < 0, 0 < t < 1, 1 < t. As t → ∞, lim f (t) → ∞; hence f (t) > 0 for all t > 1;

29Consider, for example, the function t3 − t, which has a local minimum at t = 1, a local maximum at t = −1,but has neither a global maximum nor a global minimum over its entire domain −∞ < t < +∞.


as t → −∞, lim f (t) → ∞; so the function is positive in the interval −∞ < t < 0 also.Thus the global minimum is in the interval 0 ≤ t ≤ 1; and, from our investigation of thecritical point, we know that the minimum is attained at one (or more) of t = 0, t = 1 ort = t0. We can complete this investigation if we can argue that f (t0) < 0. As we know thesign of the function will be the same throughout the interval 0 < t < 1, we can take anyconvenient value of t in that interval.

f(12

)=√

1 + i −(1 +

i2

)

=

(√1 + i − (1 + i

2 ))·(√

1 + i + (1 + i2 )

)√

1 + i + (1 + i2)

=(1 + i) −

(1 + i + i2

4

)√

1 + i +(1 + i

2

)

= − i2

4· 1√

1 + i +(1 + i

2

) < 0

Thus the global minimum is attained at t0.

6. (cf. [10, Exercise 1-33, p. 27]) Find the accumulation function a(t) if it is known thatδt = 0.04(1 + t)−1 for t > 0.Solution: Applying definition [10, (1.28), p. 19], we solve the differential equation

ddt

ln(a(t)) = 0.04(1 + t)−1 :

Integration gives

ln(a(t)) =

∫0.041 + t

dt = 0.04 ln(1 + t) + C

where C is the constant of integration. Setting t = 0, where we know, by definition, thata(0) = 1, we have

0 = ln 1 = 0.04 ln 1 + C

so C = 0, anda(t) = e0.04 ln(1+t) =

(eln(1+t)

)0.04= (1 + t)0.04 .

7. Let φ(λ) denote the value of 1 at the end of 3 years, accumulated at an effective rate ofinterest λ; let ψ(λ) denote the present value of 1, to be paid at the end of 3 years at aneffective rate of discount numerically equal to λ. Suppose it is known that φ(λ) + ψ(λ) =

2.0294. Determine λ.


Solution: (cf. [7, Exercise 52, p. 30]) φ(λ) = (1 + λ)3; ψ(λ) = (1 − λ)3. Summingyields φ(λ) + ψ(λ) = 2 + 6λ2, which we equate to 2.0294, and from which we infer that

λ =

√0.0294

6 = .07 = 7%.

8. Showing your work, determine a formula — in terms of the force of interest, δ, for thenumber of years that are needed for a sum of money to double itself. Verify your answerby determining the value of δ when the annual interest rate is 100%.Solution: Let the number of years needed be t, the interest rate be i, and the force ofinterest δ. We solve the equation (1+i)t = 2 by taking logarithms of both sides: t ln(1+i) =

ln 2, so

t =ln 2

ln(1 + i)=

ln 2δ.

When the interest rate is 100% money doubles in one year; here δ = ln 2.

7.1.2 Second 2002/2003 Problem Assignment, with Solutions

1. (cf. Exercise 2-5, p. 36) A vendor has three offers for a house:

(a) three equal payments — one now, one 1 year from now, and the other 2 years fromnow;

(b) a single cash payment now of 120,000;(c) two payments, 45,000 a year from now, and 90,000 two years from now.

He makes the remark that “one offer is just as good as another”. Determine the interestrate and the sizes of the equal payments that will make this statement correct.Solution: Let the interest rate be i, and the equal payments be k. Equating the presentvalue of the payments of 45,000 and 90,000 to 120,000 yields

45000v + 90000v2 = 120000

which we solve for v, obtaining

v =− 1

2 ±√

14 + 16

3

2

=−1 ± √22.33333

4

The lower sign yields a negative value of v, and so that solution is extraneous. We obtainv = 0.931454, so i = 0.0736 = 7.36%.The present value of the equal payments is then k(1 + v + v2) = 2.79906k = 120, 000, sothe equal payments will each be 42,871.53.


2. (cf. [10, Exercise 2-9, p. 37]) Fund A accumulates at 9% effective, and Fund B at 8%effective. At the end of 12 years the total of the two funds is 50,000. At the end of 6years the amount in Fund B is 4 times that in Fund A. How much is in Fund A after 15years?Solution: Let a and b denote the initial amounts in funds A and B. We have two con-straints relating a and b:

a(1.09)12 + b(1.08)12 = 50000b(1.08)6 = 4a(1.09)6

From the second of these we can determine the relative sizes of a and b; substituting inthe first equation and solving gives

a =50000

(1.09)12 + 4(

1.091.08

)6(1.08)12

=50000

(1.09)6 ((1.09)6 + 4(1.08)6) = 3, 715.25.

Hence b = 4(

109108

)6a = 15, 705.96. The value of Fund A after 15 years is, therefore,

3, 715.25(1.09)15 = 13, 532.73.

3. The initial balance in an investment fund was 100,000. At the end of 3 months it hadincreased to 105,000; at that time 25,000 was added to the fund. Six months later thefund had increased to 143,000, and this time 30,000 was removed. Finally, at the end ofa year, the fund had a balance of 120,000. What was the time-weighted rate of return?Solution: [9, Example 2.3.2, p. 39] The balances and withdrawals are respectively B0 =

100, 000, B1 = 105, 000, B2 = 143, 000, B3 = 120, 000; W0 = 0, W1 = 25, 000, W2 =

−30, 000. Hence the rates of interest in the successive time periods are given by

1 + i1 =B1

B0 + W0=

105, 000100, 000 + 0

= 1.05

1 + i2 =B2

B1 + W1=

143, 000105, 000 + 25, 000

= 1.10

1 + i3 =B3

B2 + W2=

120, 000143, 000 − 30, 000

= 1.062

The time-weighted rate of return i is, by definition, given by the product

1 + i = (1.05)(1.10)(1.062) = 1.227

so i = 22.7%.


4. (cf. [10, Exercise 2-15, p. 38]) A trust company pays 5% effective on deposits at the endof each year. At the end of every 3 years a 2% bonus is paid on the balance at the time.Find the effective rate of interest earned by an investor if she leaves her money on deposit

(a) for 2 years;(b) for 3 years (until after the bonus payment is made);(c) for 4 years;(d) forever — take a limit!

Solution:

(a) Since there are no bonus payments, the effective rate of interest is 5%.(b) A deposit of 1 grows to 1.05 at the end of the first year, (1.05)2 at the end of the

second year, and, after the bonus, (1.05)3(1.02) at the end of the 3rd year. If theeffective rate of interest is i, then we must solve the equation

(1 + i)3 = (1.05)3(1.02) .

Taking logarithms, we obtain 3 ln(1 + i) = 3 ln(1.05) + ln(1.02), so

ln(1 + i) =13

(3 ln(1.05) + ln(1.02))

1 + i = e13 (3 ln(1.05)+ln(1.02))

i = e13 (3 ln(1.05)+ln(1.02)) − 1

= 1.053√1.02 − 1 = .056953846 = 5.70%.

(c) Analogously to the preceding,

i = e14 (4 ln(1.05)+ln(1.02)) − 1

= 1.054√1.02 − 1 = .0552 = 5.52%.

(d) The effects of the bonuses depend on whether the remainder of the number of yearsis, upon division by 3, 0, 1, or 2. We have, for any non-negative integer n,

(1 + i)3n = (1.05)3n(1.02)n

(1 + i)3n+1 = (1.05)3n+1(1.02)n

(1 + i)3n+2 = (1.05)3n+2(1.02)n

By taking logarithms and dividing, or, equivalently, by taking the appropriate (posi-tive) root of both sides of the equation, we obtain

1 + i = (1.05)(1.02)1/3

1 + i = (1.05)(1.02)n

3n+1

1 + i = (1.05)(1.02)n

3n+2


As n → ∞, the exponent of 1.02 approaches 13 , and so the interest rate approaches

the value of 5.70% we obtained in case of 3 years.

5. Alice borrows 5000 from The Friendly Finance Company, at an annual rate of interestof 18% per year, where the company compounds interest annually, but charges simpleinterest for fractions of a year.

(a) She plans to pay the company 5000 at the end of 2 years.i. How much will she continue to owe the company at that time?

ii. What is the present value of that residual amount, assuming the same 18% inter-est rate?

(b) Alice discovers that she doesn’t need the loan, so she offers to lend the money to herbrother, at an effective annual rate of 18%. When her brother pays off his loan, Alicewill pay off hers. How much will Alice still owe Friendly if

i. Her brother pays off his loan exactly 3 years from now?ii. Her brother pays off his loan 3.5 years from now?

Solution:

(a) i. The residual amount immediately after the payment will be

5000((1.18)2 − 1

)= 1962 .

ii. The present value of the residual amount is 5000((1.18)2 − 1

)(1.18)−2 = 5000

(1 − (1.18)−2

)=

1409.08.(b) i. If her brother pays off his loan after an integer number of years, and Alice im-

mediately repays her loan, she will owe nothing to Friendly.ii. After 3.5 years Alice will receive 5000(1.18)3.5, but will be owing

5000(1.18)3(1.09); after making her payment, she will continue to owe

−5000(1.18)3.5 + 5000(1.18)3(1.09) = 5000(1.18)3(1.09 −√

1.18) = 30.58 .

6. (cf. [10, Exercise 2-12, p. 37])

(a) Find an equation that gives information about the effective rate of interest i if pay-ments of 200 at the present, 300 at the end of 1 year, and 400 at the end of 3 years,are to accumulate to 1000 at the end of 4 years.

(b) Use the Intermediate Value Theorem to argue that there exists a positive rate ofinterest less than 100% which can solve this problem. Then use the Mean ValueTheorem to show that there is just one solution to the problem (by showing that acertain derivative is positive).


(c) While there exist more efficient algorithms for solving problems like this, a naivesolution could be found by successively subdividing an interval at whose ends acertain function would have values with opposite signs. Apply this idea to find theinterest rate i to within an error of 0.1%.

Solution:

(a) The equation of value at the end of 4 years30 is

200(1 + i)4 + 300(1 + i)3 + 400(1 + i) = 1000

Define f (x) = 200x4 + 300x3 + 400x − 1000. Then f (1) = −100 < 0 < 5400 =

3200 + 2400 + 800 − 1000 = f (2). By the Intermediate Value Theorem functionf , which, being a polynomial, is continuous, has a zero somewhere in 1 < x < 2.If there were 2 or more zeroes, then, by Rolle’s Theorem, (since f , being a poly-nomial, is differentiable), there would be a point between them where f ′ wouldbe 0. But f ′(x) = 800x3 + 900x2 + 400 > 0 for 1 < x < 2. From this con-tradiction we know that there is at most one zero for f , hence exactly one solu-tion i for our effective interest rate. We can apply the Intermediate Value Theo-rem between any two points in the domain. The most naive solution would beto repeatedly halve the interval. We find that f (1.5) = 1625, so we may con-fine ourselves to the interval 1 < x < 1.5; then f (1.25) = 574.22, f (1.125) =

197.51, f (1.0625) = 39.72, f (1.03125) = −32.29. We evaluate f at the midpointof the interval [1.03125, 1.0625]: f (1.046875) = 3.17, so we next use the interval[1.046875, 1.0625], whose midpoint is 1.0546875, where f (1.0546875) = 21.30. Asthere will is a sign change in the interval [1.03125, 1.0546875], we next evaluate fat its mid-point: f (1.042968750) = −5.80.In the course of these calculations we have not bothered to round the decimal ex-pansions of the midpoints. There is nothing to be gained by this persistence, as theprocedure will work even if we do not take the precise midpoints. Having now con-fined the root to the interval [1.043, 1.055]. f (1.049) = 8.07, we try f

(1.043+1.049

2

)=

f (1.046) = 1.15, f(

1.043+1.0462

)= f (1.0445) = −2.29, f (1.045) = −1.15, f (1.0455) =

0.002, f (1.04549913) = −0.0000013. Thus the rate is approximately 4.55%.

7.1.3 Third 2002/2003 Problem Assignment, with Solutions

1. (a) Find the sum of the positive integers 1, 2, . . . , N.30Non-trivial equations are never unique; also, we could have found an equation of value at another time. For

example, an equation of value at the present could be

200 + 300v + 400v3 = 1000v4 .


(b) Find the sum of the odd integers 1, 3, 5, . . . , 2N + 1.(c) In an arithmetic progression x1, x2, . . ., xn, . . . the third term is 4 times the first term,

and the sixth term is 17. Find the general term xn.(d) The sum of n terms of the arithmetic series 2, 5, 8, . . . is 950. Find n.(e) The sum of the first 6 terms of a geometric progression is equal to 9 times the sum

of the first 3 terms. Find the common ratio. Is it possible to determine the sequencefrom this information?

(f) Use your knowledge of the sum of geometric series to determine a “vulgar” fractionof integers m

n which is equal to the repeating decimal number

3.157157157157...

Do not use a calculator for this problem.Solution: These topics were once part of the standard high school curriculum. Theseproblems were adapted from [4].

(a) The common difference is 1 and the first term is also 1; the sum of N terms is,therefore, N

2 (1 + N) =N(N+1)

2 .31

(b) The common difference is 2, the first term is 1, and the N + 1st term is 2N + 1. Thesum of N + 1 terms is N+1

2 (2 · 1 + N · 2) = (N + 1)2.[Many students failed to notice that the number of summands was N + 1 — not N.An error of this type might have been detected by checking one’s computations forsmall values of N, e.g. N = 0 or N = 1. Carry out the summation mechanically, thencompare the sum that you obtain with the value of the formula you have derived; ifthe values are different, you need to check every step of your work carefully.]

(c) Let the first term be a and the common difference be d. We have to solve the equa-tions:

x3 = 4x1 ⇔ a + 2d = 4ax6 = 17 ⇔ a + 5d = 17

which yield a = 2, d = 3. Hence xn = 2 + 3(n − 1) = 3n − 1.(d) We solve the equation n

2 (2 · 2 + (n − 1) · 3) = 950, which reduces to 3n2 +n−1900 =

0, whose only positive solution is n = 25.(e) If the first term is a and the common ratio is r, the given information implies that

a · r6 − 1r − 1

= 9a · r3 − 1r − 1

if r , 1 , (96)

6a = 9a if r = 1 . (97)31This expression is known, from other considerations, to be the number of ways of choosing 2 objects from a

set of N distinct objects; it is often denoted by(

N2

).


When a = 0, both equations are satisfied: the sequence is 0, 0, 0, . . . ; the commonratio is indeterminate. When r , 1, (96) yields r6 − 1 = 9(r3 − 1), so r3 = 1 (whichcontradicts the hypothesis) or r3 = 8, hence r = 2 and the sequence is then

a, 2a, 4a, ..., 2n−1a, ...

But the sequence is not completely determined, since any value of a is acceptable —including the value 0 which we already saw as the solution to (97).

(f)

3.157157157157... = 3 +

(1

10+

5100

+7

1000

)+

11000

(1

10+

5100

+7

1000

)

+1

10002

(1

10+

5100

+7

1000

)+ . . .

= 3 +157

1000+

11000

· 1571000

+1

10002 ·1571000

+ . . .

= 3 +157

1000· 1

1 − 11000

= 3 +157999

=3154999

.

2. (cf. [10, Exercise 3-6, p. 66] An annuity pays 1000 per year for 8 years. If i = 0.05, findeach of the following

(a) The value of the annuity one year before the first payment.(b) The value of the annuity one year after the last payment.(c) The value of the annuity at the time of the 4th payment.(d) If possible, the number of years an annuity-immediate would have to run in order

that its value, viewed one year before the first payment should be twice that of the8-payment annuity whose value at the same time was determined above.

(e) If possible, the number of years an annuity-immediate would have to run in orderthat its value, viewed one year before the first payment, should be three times that ofthe 8-payment annuity whose value at the same time was determined above.

(f) If possible, the number of years an annuity-immediate would have to run in orderthat its value, viewed one year before the first payment, should be four times that ofthe 8-payment annuity whose value at the same time was determined above.32

(g) (cf. [10, Exercise 3-58, p. 73]) Redo part (a), assuming now that the annuity is con-tinuous. (The effective annual interest rate remains 5%, and the time is still 8 years.)

Solution:

32Note that the wording of the cited questions in the textbook required a number of assumptions that have beenmade more explicit in the present questions.


(a) 1000a8 5% = 1000 · 1−(1.05)−8

0.05 = 20000(1 − (1.05)−8

)= 6463.21.

(b) 1000s8 5% = 1000 · (1.058−1)(1.05)0.05 = 10026.56.

(c)

1000s4 5% + 1000a4 5% = 1000(1.05)4a45%

= (1.05)4(6463.21) = 7856.07 .

(d) We have to solve for n:

an 5% = 2a8 5%

⇒ 1 − vn = 2 − 2v8

⇒ vn = 2v8 − 1

⇒ n =ln(2v8 − 1)

ln v= 21.30 years.

(e) We have to solve for n:

an 5% = 3a8 5%

⇒ 1 − vn = 3 − 3v8

⇒ vn = 3v8 − 2

⇒ n =ln(3v8 − 2)

ln v= 71.52 years.

(f) In this case we observe that the value of a perpetuity of 1000 per year at 5% is only10000.05 = 20000 < 4(6463.21); so the problem will have no solution. If we attempt tosolve as in the preceding case, we will obtain the equation 1 − vn = 4 − 4v8 ⇒ vn =

4v8 − 3 = −0.29, which has no solution.(g)

1000a8 5% = 1000

8∫

0

vt dt

= 1000 · 1 − (1.05)−8

ln(1.05)= 6623.48.

3. [10, Exercise 3.15] Prove each of the following identities

• algebraically; and• verbally

(a) an = an + 1 − vn


(b) sn = sn − 1 + (1 + i)n

Solution:

(a) an differs from an in that it has an immediate payment of 1 but lacks the final paymentof 1 n years hence, whose value now is vn.Algebraically,

an = an(1 + i)= an + ian

= an + i · 1 − vn

i= an + (1 − vn)

(b) sn differs from sn in that is lacks a payment of 1 at time t = 0, but has a payment of1 that has accumulated interest over n years, so that its present value is (1 + i)n.Algebraically,

sn = sn(1 + i)= sn + isn

= sn + i · (1 + i)n − 1i

= sn + ((1 + i)n − 1)

4. [10, Exercise 3-45, p. 71] Wilbur leaves an inheritance to four charities: A, B, C, D. Thetotal inheritance is a series of level payments at the end of each year, payable forever.During the first 20 years, A, B, C share each payment equally. All payments after 20years are to revert to charity D. The present value of the shares of A, B, C, and D are allequal. Showing all your work, prove that i = 0.07177.Solution: It does not limit generality to assume that the level payments are all of 1. Thepresent value of the payments to each of A, B, C is 1

3a20 i%. The payments to D constitutea perpetuity-immediate of 1 deferred 20 years; its value is v20 · 1

i . Accordingly we haveto solve the following equation for i:

13· a20 i% = v20 · 1

i

⇔ 1 − v20

3= v20

⇔ v20 =14

⇔ 1 + i = 4120 = 1.0717735

so i = 7.177%.


5. Find the present value at i effective of a perpetuity whose annual payments of 1000 beginwith a payment of 1000 after one year, with the property that each payment thereafter isreduced by 10% from the preceding payment. In particular, determine the present valuewhen i = 2.5%.Solution: [STUDENTS WERE ASKED NOT TO SUBMIT A SOLUTION TO THISPROBLEM.] The present value is

1000(v + 0.9v2 + 0.92v3 + 0.93v4 + ... + 0.9n−1vn + ...)

= 1000v∞∑

n=0

(0.9

1 + i

)n

=1000v

1 − 0.9v

=1000

(1 + i) − 0.9=

10000.1 + i

When i = 2.5%, the present value is 10000.125 = 8000.

6. A fund of 10,000 is to be accumulated by means of deposits of 1000 made at the end ofevery year, as long as necessary. If the fund earns an effective rate of interest of 2 1

2%, findhow many regular deposits will be necessary, and the size of a final deposit to be madeone year after the last regular deposit.Solution: [7, Example 6.5, pp. 59-60] There are often tacit assumptions in interest prob-lems; usually there is an “obvious” intended interpretation, while other interpretationsmight be justified by some unusual reading of the wording. In the present problem oneis to assume that the deposits are not permitted to exceed 1000, and that all deposits (the“regular” deposits) but the last are to be exactly 1000. The last deposit can be smaller,but not larger.Let n be the number of regular deposits required. Then n is the largest integer that satisfiesthe “inequality of value”

1000sn 2.5% ≤ 10000 ;

equivalently, sn 2.5% ≤ 10 . Computing the values of sn 2.5%, we find that s8 2.5% = 8.73612,s9 2.5% = 9.95452, s10 2.5% = 11.20338. Hence n = 9. The final partial deposit will haveto be the excess of 10,000 over the accumulated value of s9 2.5% after one year, i.e., theexcess of 10,000 over (1.025)s9 2.5% = 10, 000 − (1.025)(9954.52) = −203.38. So ratherthan a final deposit, there will be a final refund of 203.38. This situation could also havebeen seen from the value of 10,000s10 2.5% = 11203.38, which would be the value after a10th deposit; since this exceeds 11,000, no 10th deposit would be required.[ADDED March 10th, 2003] Another possible interpretation of the instructions in thisproblem is to treat the 8th as the last “regular” deposit, and to reduce the 9th deposit sothat, when the time arrives for a possible 10th deposit, the balance in the fund is exactly10000. If we define the value of the 9th deposit to be x, then

(1 + i)((1 + i)s8 + x

)= 1000



⇔ x =100001.025

− 1025s8

= 9756.0976 − 8954.5188 = 801.58.

We can verify the correctness of this computation by observing that the excess paymentof 1000 − 801.58 = 198.42 accumulated at 2.5% to 1.025 × 198.42 = 203.38, which wascomputed earlier as the amount refunded one later.[This assignment was intended as a learning exercise, rather than a testing exercise. Stu-dents were not expected to have seen an example of this type before.]

7. A deferred annuity is one that begins its payments later than might otherwise have beenexpected. We define

m

∣∣∣an = vman (98)

m

∣∣∣an = vman (99)

Prove, both algebraically and verbally, that, for non-negative integers m and n,

m

∣∣∣an = am+n − am (100)(1 + i)m · sn = sm+n − sm (101)

1

∣∣∣an = an (102)

Solution:

(a)

m

∣∣∣an = vm(v + v2 + . . . + vn)

= (vm+1 + vm+2 + . . . + vm+n

= (v1 + v2 + . . . + vm+n − (v1 + v2 + . . . + vm

= am+n − am

In deferring an n-payment annuity-immediate by m years we are planning for thefirst payment to be made m + 1 years from now, and the last m + n years from now.These can be viewed as the last n payments of an m + n-payment annuity-immediatewhose first payment begins one year hence; thus we obtain the value of m

∣∣∣an bysubtracting am from am+n.

(b)

(1 + i)m · sn = (1 + i)m(1 + (1 + i)1 + . . . + (1 + i)n−1

)


= (1 + i)m + (1 + i)m+1 + . . . + (1 + i)m+n−1

=(1 + (1 + i)1 + . . . + (1 + i)m+n−1

)

−(1 + (1 + i)1 + . . . + (1 + i)m−1

)

= sm+n − sm

The payments associated with (1 + i)m · sn can be interpreted as the first m paymentsof an m + n-payment annuity whose last payment has just been made. If we subtractfrom sm+n the value of the last n payments as viewed from the day of the last payment,we obtain the value of those first m payments.

(c)

1

∣∣∣an = v(1 + v + v2 + . . . + vn−1

)

= v + v2 + . . . + vn = an

When we defer an annuity-due one year it becomes an annuity-immediate.

7.1.4 Fourth 2002/2003 Problem Assignment, with Solutions

1. (cf. [10, Exercise 3-60, p. 73]) A student attending engineering school has increasingamounts of income as she advances through her programme. Accordingly she agrees toborrow a decreasing annual amount from her parents during her 5 training years, andto repay the loan with increasing amounts for 15 years after graduation. She receivesamounts 5X, 4X, 3X, 2X and X at the beginning of each of 5 years, where the last paymentis paid at the beginning of her final year. At the end of her first year after graduation shepays 500, and then increases the amount by 200 each year until a final payment of 3300.33

If the interest rate is 5%, determine X.Solution: An equation of value at the time of graduation is

(1 + i)X(Ds)5 = 300a15 + 200(Ia)15

implying that

X =300a15 + 200(Ia)15

(1 + i)(Ds)5

=300(1 − v15) + 200((1 + i)a15 − 15v15)

5(1 + i)6 − (1 + i)6a5

= 993.11.33The original version of this problem gave the final payment as 3500, which would have required 16 years of

payments. A correction was announced at the lecture of March 3rd, 2003.


For students who corrected the error in the problem by increasing the number of years by1, here is a solution:Solution: An equation of value at the time of graduation is

(1 + i)X(Ds)5 = 300a16 + 200(Ia)16

implying that

X =300a16 + 200(Ia)16

(1 + i)(Ds)5

=300(1 − v16) + 200((1 + i)a16 − 16v16)

5(1 + i)6 − (1 + i)6a5

= 1082.33.

2. (cf. [10, Exercise 4-2, p. 85]) A loan is being repaid by 36 monthly payments. The first12 installments are 250 each; the next 18 are 300 each; and the last 6 are 500 each.Assuming a nominal annual interest rate of 12% compounded monthly,

(a) Find the principal, A(0), of the loan.(b) Using the Prospective Method, find the loan balance immediately after the 6th pay-

ment.(c) Using the Retrospective Method, find the loan balance immediately after the 6th

payment.(d) Divide the 7th and 8th payments into principal and interest.

Solution:

(a) Using the Prospective Method, the principal is seen to be

A(0) = 250a12 + 300v12 · a18 + 500v30 · a6

= 9, 329.46

(b) Immediately after the 6th payment, the value of the remaining 30 payments (at aninterest rate of 1% per period) is

250a6 + 300v6 · a18 + 500v24 · a6

= 2501 − v6

i+ 300v6 · 1 − v18

i+ 500v24 · 1 − v6

i

=1i·((250 + 500v24)(1 − v6) + 300v6(1 − v18)

)

=1i·(250 + 50v6 + 200v24 − 500v30

)

= 8365.40


(c) The principal of the loan has been determined above. The outstanding principal isthe accumulated value of this principal decreased by the accumulated values of thepayments that have been made, i.e.

(1.01)6A(0) − 250s6 = (1.016)(9, 329.46) − 250 · (1.01)6 − 10.01

= 9, 805.38 − 1, 538.00 = 8, 365.40.

(d) The principal owing immediately after the 6th payment is known to be 8,365.40. Atthe time of the 7th payment, this will have accumulated interest of 1%, or 83.65;the balance of the payment, i.e. 250 − 83.65 = 166.35, will be applied to reductionof principal. The reduced balance of 8365.40 − 166.35 = 8199.05 will accumulateinterest in the amount of 0.01× 8199.05 = 81.99 in the 8th month. The 8th paymentwill include, in addition to this amount of interest, an amount of 250−81.99 = 168.01for the reduction of principle; the outstanding principal after the 8th payment will be8199.05 − 168.01 = 8031.04.

3. (a) [10, Exercise 4-16, p. 87] Harriet is repaying a car loan with payments of 2,000 everythree months and a final payment 3 months after the last full payment of 2,000. Ifthe amount of interest in the 4th installment (paid at the end of the first year) is1,100, find the principal of the loan, the time and amount of the final payment, andthe amounts of principal and interest in that final payment. Assume that interest iscompounded monthly, at a nominal annual rate of 18%.

(b) Construct an amortization schedule for the first year of this loan.Solution:

(a) The interest rate being charged monthly is 0.18/12 = 1.5%. Let the principal ofthe loan be A. The Retrospective Method shows that the amount owing immediatelyafter the 3rd installment (paid at 9 months) is

−2000((1.015)6 + (1.015)3 + (1.015)0

)+ (1.015)9A .

This unpaid balance will, in 3 months, earn the lender interest in the amount of

1100 =((1.015)3 − 1

) (−2000

((1.015)6 + (1.015)3 + (1.015)0

)+ (1.015)9A

).

Thus

A =1100

(1.015)12 − (1.015)9

+2000((1.015)−9 + (1.015)−6 + (1.015)−3

)(103)

= 26552.32


We can consider the payments as constituting an annuity, with time interval 3 months,and interest rate per 3 months of (1.015)3 − 1 = 0.045678375. Using the ProspectiveMethod, we see that the value of n installments, as of the day of the loan, is

2000an0.045678375 = 2000 · 1 − (1.015)−3n

0.045678375.


1 − (1.015)−3n ≥ 26552.32 × 0.0456783752000

= 0.6064334

i.e., such that1.015−3n ≤ 0.3935666

orn ≥ − ln 0.3935666

3 ln 1.015= 20.88

Thus there will be 20 full payments of 2,000, the last full payment being made 5years after the beginning of the loan. At that time the amount outstanding will be

26552.32(1.015)60 − 2000s200.045678375

= 64873.15 − 2000 ·(1.0153

)20 − 1

(1.015)3 − 1= 64873.15 − 63190.50 = 1682.65

The last payment will be 1682.65(1.015)3 = 1759.51; of this, the interest compo-nent will be 1682.65

((1.015)3 − 1

)= 76.86, and the balance will be the outstanding

principal of 1682.65.(b)

Duration Payment Interest Principal Repaid Outstanding Principal(Months)

0 26552.323 2000.00 1212.87 787.13 25765.196 2000.00 1176.91 823.09 24942.109 2000.00 1139.31 860.69 24081.41

12 2000.00 1100.00 900.00 23181.414. John has borrowed 10000, on which he is paying interest at 10% effective per year. He is

required to pay the interest on the loan annually, and is permitted to repay only the entireloan, and only on an anniversary. He decides to accumulate a sinking fund to accumulatethe funds to repay the loan. Suppose that John has 2400 available at the end of eachyear, out of which to pay both the interest on the loan and an annual contribution to hissinking fund. If the sinking fund accumulates at 6%, complete a table under the followingheadings to determine when John will be able to repay the loan.


Duration Contribution Interest Interest Earned Balance of(Years) to Sinking Fund on Loan in Sinking Fund Sinking Fund

0 0 0 0 0. . . . . . . . . . . . . . .

Solution: The instructions asked that the student “complete a table...to determine whenJohn will be able to repay the loan”. The information could have been obtained withoutthe table, however, by finding the smallest value of n for which 1400sn ≥ 10000; this canbe seen to be n = 6, where

10000 − 1400sn = 234.55 . (104)

From (104) we see that the shortfall in the balance of the sinking fund after the lastpayment of 1400 is 234.55. The value of the sinking fund is not yet sufficient to repay theloan. Even without a 7th payment the sinking fund will exceed 10000 by the time whenthat payment is due. It will, however, be necessary to pay the interest charge of 1000 onthe loan. If a full 6th payment of 1400 was made into the sinking fund, there would be arefund of (1.06)(1400)s6 − 1000 = 351.38. However, a better solution would have beenfor John to make a smaller 6th deposit into the sinking fund — just sufficient to bringthe fund up to the level of 10000 at the time of the 7th interest payment. The balancejust after such a 6th payment would need to be 10000

1.06 , and the balance just prior to the 6thdeposit would be (1.06)1400s5; so the appropriate 6th deposit would be

100001.06

− (1.06)1400s5 = 9433.963 − 8365.446 = 1068.52 .

The first table below shows what would happen if John made a full 6th contribution:


0 0.00 0. 0.00 0.001 1400.00 1000. 0.00 1400.002 1400.00 1000. 84.00 2884.003 1400.00 1000. 173.04 4457.044 1400.00 1000. 267.42 6124.465 1400.00 1000. 367.47 7891.936 1400.00 1000. 473.52 9765.457 -351.38 1000. 585.93 10351.38

The following table shows the result of a reduced 6th contribution:



0 0.00 0. 0.00 0.001 1400.00 1000. 0.00 1400.002 1400.00 1000. 84.00 2884.003 1400.00 1000. 173.04 4457.044 1400.00 1000. 267.42 6124.465 1400.00 1000. 367.47 7891.936 1068.51 1000. 473.52 9433.967 0.00 1000. 566.04 10000.00

NOTE TO THE GRADER: PLEASE ACCEPT EITHER OF THESE TABLES.

5. (This is a complicated variant of [10, Exercise 4-6, p. 86]. It requires considerable per-sistence, but is a very thorough exercise. Don’t panic! This is not a typical examinationquestion.) Garfield is repaying a debt with 25 annual payments of 1000 each, at an annualinterest rate of i = 10%. The terms of his loan permit him to make additional paymentson the date of any regular payment. After any such additional payment, the terms of theloan require the borrower to continue with payments of 1000 until a last payment of 1000or less which settles the debt completely.

(a) At the end of the 7th year Garfield proposes to make, in addition to his regular annualpayment of 1000, an extra payment of 5000. At that time he also proposes to reducehis remaining payment period by 4 years, and to make level payments over that time(replacing the originally agreed payments of 1000). Find the revised annual levelpayment, computed using the interest rate i = 10%.

(b) The lender is obliged to accept Garfield’s extra payment. But he is not obliged toaccept Garfield’s proposed method to repay the loan in fewer payments. If, at thetime of the change in the payment scheme, the lender insists on charging an interestrate of i = 12% when the remainder of the loan will be repaid over 14 equal annualpayments, what will be the revised annual level payment that will have to be paid atthe end of the each of the next 14 years?

(c) Determine the premium Garfield is being asked to pay as a result of the increasedinterest rate in part 5b. Express the amount as of the date of the proposed change inthe payment scheme. Make two sets of calculations:

i. when the cost of money34 is 10% per annum;ii. when the cost of money is 12% per annum.

34By the statement The cost of money is i we intend that Garfield is able — as of this particular date — to eitherborrow or lend money in any amount and for any period of time commencing immediately — at the interest ratei.


(d) As the loan is repaid, the lender is able to put his money to work. Suppose that moneynow costs 12%, instead of the 10% that prevailed when the loan was written. Onemight have expected the lender to encourage the borrower to repay the loan faster.Faced with the lender’s intransigence, Garfield makes the supplementary paymentof 5000, but decides to abandon his plans to change the payment size; his paymentswill be 1000 per year until possibly the last payment. Determine whether the lenderhas suffered from his own stubbornness: express his loss (or gain) as of the time ofthe supplementary payment made with the 7th payment.

(e) Suppose that, learning that the cost of money is 12% when he is about to make hissupplementary payment, Garfield changes his plans. He makes no change to his loan,but invests his 5000 elsewhere in an annuity which will provide him with paymentsof 1000 to apply to as many of the final payments under his loan as possible. Asof the beginning of the 8th year of the loan (immediately following the 7th paymentand any supplementary payment) compare the cost of this scheme with

i. his commitment under the original loan contract;ii. his proposed scheme, whereby he would pay 5000 immediately and pay the rest

of the loan over 14 years at 10%;iii. the lender’s proposal, where an immediate payment of 5000 would be followed

by equal payments for 14 years, computed at a rate of 12%.(f) Determine the yield earned by the lender under each of the following repayment

schemes:i. the loan as originally written — 25 annual payments of 1000;

ii. the repayment scheme proposed by Garfield: 1000 per year for 7 years, 5000additional at the end of the 7th year; level payments for 14 years thereafter,amount as computed in part 5a above;

iii. the repayment scheme proposed by the lender, in part 5b, where the level pay-ments are recomputed at 12% charged from the time of the 7th payment; (in thiscase it suffices to write down an equation that must be satisfied by the yield);

iv. the repayment scheme finally followed by Garfield in part 5d, where he investsin an annuity to provide him with payments of 1000 for the final payments, andpays the rest annually from his savings.

Solution:

(a) Using the Prospective Method, we find that the unpaid balance immediately after the7th payment, but prior to the extra payment, is 1000a180.1; after the extra paymentthe amount owed is

1000a180.1 − 5000 = 8201.41 − 5000 = 3, 201.41 .


The level payment to repay this principal in 18 − 4 = 14 years is (with i = 10% andv = 1

1.1 )

1000a180.1 − 5000

a140.1

=1000(1 − v18) − 5000i

1 − v14 (105)

= 434.58.

(b) We will have to evaluate the same ratio as in (105), but where numerator and denom-inator involve different interest rates.

1000a180.1 − 5000

a140.12

=1000(1 − (1.1)−18) − 5000(0.1)

1 − (1.12)−14 · 0.120.1

= 483.

(c) Let’s first determine the nature of Garfield’s commitment under the loan after hemakes his supplementary payment. We have determined that the loan balance is3,201.41. We note that 1, 000a4.10 = 3, 169.87, while 1, 000a5.10 = 3, 790.79.Garfield’s loan contract requires him to make 4 payments of 1000; and, at the end ofthe 5th year, to pay the balance of principal that would be owing at that time. Thatbalance would be

(1.1)5(3201.41) − 1000(s5.1 − 1

)= 5155.90 − 5105.10 = 50.80 .

These 5 payments are prescribed under his contract, and the calculation of theirvalues is not affected by the cost of money today. What is affected is the way inwhich Garfield finances these payments; or, equivalently, the present value of thesepayments, which may not be equal to the loan balance.

i. If the cost of money is 10%, the present value of the 14 payments Garfield wouldhave to make would be 483a14.1 = 3558.11; the present value of the paymentsrequired under the loan contract is the outstanding principal, 3201.41; so thepremium would be 356.70.

ii. If the cost of money is 12%, the present value of the 14 payments would be theoutstanding principal, 3,201.41. The value of the 4 payments of 1000 and onefinal payment (i.e. the cost of financing them at 12%) is

1000a4.12 + (1.12)−550.80 = 3037.35 + 28.83 = 3066.18 .

In this case he would be paying a premium of

3201.41 − 3066.18 = 135.23 .


(d) After his supplementary payment, Garfield owes an unpaid balance of 3201.41. Hadhe been permitted to repay this with 14 annual payments of 434.58, the present valueof those payments would be 2880.47. But Garfield has now been driven to repay theloan by continuing the planned payments of 1000 until a final payment. In part 5c wehave determined that the number of payments of 1000 is 4, and these are followed bya payment one year later of 50.80. The value of these payments today, when moneycosts 12%, is

1000a40.12 + (1.12)−550.80 = 3037.35 + 28.83= 3066.18 .

While neither of these repayment schemes yields the full amount owed — becauseinterest rates are higher than at the outset — the lender was wise to be unwillingto accept Garfield’s offer: he has reduced his losses under the loan by 3066.18 −2880.47 = 185.71.

(e) i. Garfield’s commitment under the original contract is for payments of 1000 for18 more years. At a rate of 12%, Garfield could buy an annuity to cover hispayments at a present cost of 1000a18.12 = 7249.67. We are asked to comparethis cost with the use of the 5000 to purchase a deferred annuity to cover thelast payments due under the contract. We will answer this question naively andthen, when the answer looks “interesting”, observe that there is a much simplersolution.The present value of an annuity that will cover the payments due in years ##k +

1, k + 2, ..., 18 is1000

(a18.12 − ak.12

).

Since

1000a18.12 = 7249.67 − 5000 = 2249.671000a3.12 = 2401.831000a2.12 = 1690.05 ,

Garfield’s 5000 will buy him a deferred annuity paying 1000 per year, startingat the end of 4 years from now until 18 years from now, costing him 1000s18.12−1000s3.12 = 7249.67 − 2401.83 = 4847.84 and he will have 152.16 left over.The cost of the payments not covered by his 4847.84 is 2401.83; the total of hiscommitments today is therefore 7249.67, precisely the same as computed above.This should be no surprise, as both sets of computations are being made with aninterest rate of 12%. Thus the excess of one over the other is zero.

ii. An annuity to cover the payments of 434.58 per year for 14 years would costGarfield today 434.58a14.12 = 2880.47; under this scheme he would also be


making a payment of 5000, for a total of 7880.47: the deferred annuity methodwould cost 7880.47 − 7249.67 = 630.80 less.

iii. The payments of 483 per year for 14 years are worth today 483a14.12 = 1000a180.1−5000 = 3201.41; the sum of the value of these payments and the supplementarypayment is 1000a180.1 = 8201.41 : the deferred annuity method would cost8201.41 − 7249.67 = 951.74 less.

(f) i. The loan was written to provide a yield of 10%. The fact that the cost of moneymay have changed does not affect the yield, which is influenced only by thelender’s payments and receipts under the loan.

ii. Since Garfield’s computation of the new level payment is based on an interestrate of 10%, there has been no change in the yield to the lender: it remains 10%.

iii. When the lender demands that the computation of the replacement level paymentbe based on an interest rate of 12%, he effects a partial improvement of the yield;but it cannot affect those funds that were already repaid. Setting up an equationof value at time 7, just after the supplementary payment and the 7th payment of1000, we find that the yield rate, i, will satisfy the equation:

−(1 + i)71000a2510% + 1000s7i + 5000 + 483a14i = 0


−9077.04(1 + i)7 + 1000((1 + i)7 − 1

i

)+ 5000 + 483

((1 + i)14 − 1

i(1 + i)14

)= 0 .

It can be shown that i = 0.10345 approximately. Thus even the increase in theinterest rate for the final payments does not effect a marked increase in the yieldrate.

iv. In this case the yield rate is 10%, as there are, from the lender’s perspective, nochanges.

7.1.5 Fifth 2002/2003 Problem Assignment, with Solutions

1. (a) (cf. [10, Exercise 5.1, p. 105]) A 15-year bond with face value 20000, redeemableat par, earns interest at 7.5%, convertible semiannually. Find the price to yield aninvestor 8% convertible semiannually.

(b) What is the premium or discount at which the bond will be purchased?(c) (cf. [10, Exercise 5-13, p. 109]) For the bond in part 1a find the market price35 and

flat price at each of the following dates and times:35Market price=amortized value [10, p. 98] is obtained by interpolating linearly between book values on

coupon dates. Thus the market price is a continuous function of time.


i. Just after the 7th coupon has been paid.ii. 3 months after the 7th coupon has been paid. (Use simple interest for fractions

of a period.)iii. Just before the 8th coupon is paid.iv. Just after the 8th coupon is paid.

(d) What would the market price and flat price have been just before and just after pay-ment of the 8th coupon if the bond had been purchased at par?

Solution:

(a) Both of the interest rates are nominal annual rates convertible semiannually; we mustdivide each by 2. Using the “general formula”, we find the price of the bond to be

(20000)(1.04)−30 + (0.0375 × 20000)a30.04 (106)

= (20000)(1.04)−30 +

(750.04

) (1 − (1.04)−30

)(107)

= (20000)(1.04)−30 + 18750(1 − (1.04)−30

)(108)

= 18750 + (20000 − 18750)(1.04)−30 (109)= 18750.00 − 385.40 = 19135.40. (110)

Alternatively, using the ”alternate” formula, we find it to be

200 + (750 − 800)a304%

= 2000 − 500.04

(1 − (1.04)−30

)= 19135.40.

(b) The bond is selling at a discount of 20000.00 − 19135.40 = 864.60 less than itsredemption value.

(c) The book value at the time of an interest payment is the present value of the unpaidportions of the bond; the coupon payments will enter into the accounting in someother way, e.g., as income. These computations make use of the yield rate associatedwith the owner’s acquisition of the bond. The market price at these times will equalthe book value.

i. The remaining 30 − 7 coupons are worth 750a234% = 11142.53; the principal isworth 20000(1.04)−30+7 = 8114.53. The market price is the book value, i.e., thesum, 19257.16.

ii. The book value immediately after the payment of the 8th coupon is the sum ofthe value of the unpaid coupons, 750a224% = 10838.34 and the present value ofthe principal, 20000(1.04)−30+8 = 8439.11; together, 19277.45. The average ofthis book value and that after the payment of the 7th coupon is 19267.30, andthis is what we define to be the market price.


iii. By our definition, market price is a continuous function of time: the market priceimmediately before a coupon payment will be equal to that after the payment —here, 19277.45.

iv. As seen above, the book value is 19277.45. This could also have been computedby subtracting the value of the coupon from 1.04 times the book value after thepayment of the 7th coupon:

(1.04 × 19257.16) − 750.00 = 20027.45 − 750.00 = 19277.45 .

We compute the flat prices:i. The flat price associated with yield rate 4% per interest period, just after payment

of the 7th coupon, is the same as the market price, as there is no accrued interest.Here the value is, as above, 19257.16.

ii. The flat price is the book value at the time of the preceding coupon payment plusaccrued simple interest. That is,

(1 +

12· 0.04

)19257.16 = 19642.30 .

In practice this is often quoted as the market price of 19267.30 plus accruedinterest of 375.00, (half of the next coupon).

iii. The flat price just before the payment of the 8th coupon can be determined inseveral different ways:• Viewed as book value plus accrued interest,

(1 + 0.04) 19257.16 = 20027.45 .

• Viewed as the book value just after the payment of the coupon, plus the valueof the coupon, it is

19277.45 + 750.00 = 20027.45 .

iv. The flat price just after payment of a coupon is the book value, here 19277.45.(The flat price is discontinuous at such points in time: the limit as time ap-proaches the point from the right is different from the limit from the left: theydiffer by the value of the coupon.)

(d) The flat price just after payment of the 8th coupon would have been

20000(1.0375)−22 + 750a223.75%

= 8897.99 + 11102.01 = 20000.00 ;


are you surprised by this result? The flat price just before payment of the couponwould have been 20000.00 increased by the interest that had been earned but notpaid, i.e. 20000.00 + 750.00 = 20750.00.The market price would remain constant at 20000 throughout.

2. (a) [10, Exercise 5-3, p. 106] Prove the “Alternate Price Formula”:

P = C + (Fr −Ci)an

algebraically.(b) (cf. [10, Exercise 5.5, p. 106]) Two bonds with face value 10000 each, redeemable

at par at the end of the same period, are bought to yield 10%, convertible semian-nually. The first bond costs 8246.56, and pays coupons at 7% per year, convertiblesemiannually. The second bond pays coupons at 6% per half-year. Find

i. the price of the second bond;ii. the number of coupons remaining on each of the bonds.

Solution:(a) We can derive the Alternate Price Formula from the “General Formula” [10, (5.1)]

as follows:

P = (Fr)an + Cvn

= (Fr)an + C(1 − ian) [10, (3.6), p. 45]= C + (Fr −Ci)an �

(b) We apply the Alternate Price Formula proved above to the two bonds. Denote theprice of the second bond by P2, and the number of coupons remaining by n. Then

8246.56 = 10000 + (10000(0.035) − 10000(0.05))an (111)P2 = 10000 + (10000(0.06) − 10000(0.05))an (112)

From (111) we find that

an5% =8246.56 − 10000

10000(0.035 − 0.05)= 11.6896 . (113)

i. Substituting in (112) yields P2 = 11168.96 as the price of the second bond.ii. We solve (113) for n:

1 − vn

0.05= 11.6896

⇒ 1 − vn = 0.58448⇒ vn = 0.41552⇒ −n ln(1.05) = ln(0.41552)⇒ n = 18



There are 18 coupons remaining: the bonds mature in 9 years.

3. (cf. [10, Exercise 5-16, p. 108])

(a) Construct a bond amortization schedule for a 3 year bond of face amount 5000,redeemable at 5250 with semiannual coupons, if the coupon rate is 5% and the yieldrate is 6% — both converted semiannually. Use the format

Time Coupon Interest Principal BookValue Adjustment Value

0...

(b) Construct a bond amortization schedule for a 3 year bond of face amount 5000,redeemable at 5250 with semiannual coupons, if the coupon rate is 6% and the yieldrate is 5% — both converted semiannually.

Solution:

(a) The purchase price of the bond will be 5250(1.03)−6 +125a63% = 4396.79+677.15 =

5073.94.Time Coupon Interest Principal Book

Value Adjustment Value0 5073.941 125.00 152.22 -27.22 5101.162 125.00 153.03 -28.03 5129.193 125.00 153.88 -28.88 5158.074 125.00 154.74 -29.74 5187.815 125.00 155.63 -30.63 5218.446 125.00 156.55 -31.55 5249.99

(b) The purchase price of the bond will be 5250(1.025)−6 + 150a62.5% = 4527.06 +

826.22 = 5353.28.Time Coupon Interest Principal Book

Value Adjustment Value0 5353.281 150.00 133.83 16.17 5337.112 150.00 133.43 16.57 5320.543 150.00 133.01 16.99 5303.554 150.00 132.59 17.41 5286.145 150.00 132.15 17.85 5268.296 150.00 131.71 18.29 5250.00

4. (cf. [10, Exercise 5-22, p. 109]) A 10-year bond of face value 12000 with semiannualcoupons, redeemable at par, is purchased at a premium to yield 10% convertible semian-nually.


(a) If the book value (just after the payment of the coupon) six months before the re-demption date is 11828.57, find the total amount of premium or discount in theoriginal purchase price.

(b) Determine the nominal annual coupon rate of the bond, compounded semiannually.(c) Give the amortization table for the last one and one-half years.

Solution:

(a) The book value just after the pænultimate36 coupon is

11828.57 = 12000v + Fr · a10.05

= 12000v + Fr · v =12000 + Fr

1.05

soFr = 1.05 × 11828.57 − 12000 = 420 .

Knowing the amount of each coupon we can now evaluate the purchase price of thebond to have been

12000(1.05)−20 + 420a200.05 = 4522.67 + 5234.13= 9756.80 .

The bond was purchased at a discount of 12000 − 9756.80 = 2243.20.(b) The rate per period was 420

12000 = 3.5%; hence the nominal rate compounded semi-annually, is 2 × 3.5% = 7%.

(c) For convenience we will compile this table backwards, beginning with Time=20.We were given that B19 = 11828.57. Hence the Principal Adjustment contained inthe 20th coupon is

11, 828.57 − 12, 000 = −171.43 .

book value at Time=18 will be (12000)(1.05)−2 +420(1.05−1 +(1.05)−2) = 11665.31;the book value at Time=17 will be (12000)(1.05)−3+420(1.05−1+(1.05)−2+(1.05)−3) =

11509.81.Time Coupon Interest Principal Book

Value Adjustment Value20 420.00 591.43 -171.43 12000.0019 420.00 583.26 -163.26 11828.5718 420.00 575.50 -155.50 11665.3117 420.00 . . . . . . 11509.81

362nd last


5. A 4.5% bond37 with par value of 100 and semiannual coupons is issued on July 1, 2003.It is callable at 110 on any coupon date from July 1, 2008 through January 1, 2011; at105 on any coupon date from July 1, 2011 through January 1, 2013; and at 102.50 on anycoupon date from July 1, 2013 through January 1, 2015; thereafter it is callable withoutpremium on any coupon date up to January 1, 2018 inclusive; its maturity date is July 1,2018. Determine the highest price that an investor can pay and still be certain of a yieldof

(a) 5% convertible semiannually;(b) 4% convertible semiannually.(c) 3% convertible semiannually.

[Hint: For each interest rate, and each range of payments for a given premium, expressthe price of the bond as a function of the payment number.]Solution: As a first step towards organizing data, the student should determine the pay-ment numbers being referred to. If we label the payment dates with natural numbers, anddefine the issue date to be (non)-payment #0, the July dates will have even numbers, andthe January dates odd numbers. The premium of 10 is payable when the calling date is##10-15; the premium of 5 when the calling date is ##16-19; the premium of 2.5 whenthe calling date is ##20-23; and no premium is payable when the calling date is ##24-29nor on the maturity date, which is payment #30.

(a) We tabulate the applicable price formulæ, based on the call or maturity date:

First Date Last Date Price10 15 110.00 · (1.025)−n + 2.25an2.5% = 90 + 20.00(1.025)−n

16 19 105.00 · (1.025)−n + 2.25an2.5% = 90 + 15.00(1.025)−n

20 23 102.50 · (1.025)−n + 2.25an2.5% = 90 + 12.50(1.025)−n

24 30 100.00 · (1.025)−n + 2.25an2.5% = 90 + 10.00(1.025)−n

One way to solve the problem would be to laboriously compute the price for everypossible call date, and then take the minimum. However, as the above formulæexpress the value in terms of a decreasing function vn, it suffices to consider thesmallest value in each interval, i.e. the largest value of n. So we have to compare the

37The convention in bonds is that, lacking any indication to the contrary, the term an r% bond refers to abond whose coupon rate is a nominal rate of r%; the rate is compounded (or converted) as often as indicatedin the description of the bond, with the default being half-yearly if there is no indication to the contrary. Underthis convention the coupon rate for this bond is 2.25%. This convention is stated in the textbook [10, p. 93, 1stparagraph]; however, the author usually supplies additional, redundant, information in his problems.


following four prices:Call Date Price

15 103.8119 99.3823 97.0830 94.77

Thus the highest price that the investor may safely pay is 94.77. Because the priceswere expressible in the form 90 + A(1.025)−n, where A is a non-increasing functionof n and (1.025)−n also a non-increasing function of n, we could have stated imme-diately that the lowest price would be that for the bond held to maturity: it was notnecessary to carry out all these computations. The situation is not so clear when theyield rate is less than the coupon rate.

(b) Again we tabulate the applicable price formulæ, based on the call or maturity date:

First Date Last Date Price10 15 110.00 · (1.02)−n + 2.25an2% = 112.50 − 2.50(1.02)−n

16 19 105.00 · (1.02)−n + 2.25an2% = 112.50 − 7.50(1.02)−n

20 23 102.50 · (1.02)−n + 2.25an2% = 112.50 − 10.00(1.02)−n

24 30 100.00 · (1.02)−n + 2.25an2% = 112.50 − 12.50(1.02)−n

These formulæ express the value in terms of an increasing function −vn, it sufficesto consider the largest value in each interval, i.e. the smallest value of n. So we haveto compare the following four prices:

Call Date Price10 110.4516 107.0420 105.7724 104.73

Thus the highest price that the investor may safely pay is 104.73.(c) As before, we tabulate the applicable price formulæ, based on the call or maturity

date:

First Date Last Date Price10 15 110.00 · (1.015)−n + 2.25an1.5% = 150.00 − 40.00(1.015)−n

16 19 105.00 · (1.015)−n + 2.25an1.5% = 150.00 − 45.00(1.015)−n

20 23 102.50 · (1.015)−n + 2.25an1.5% = 150.00 − 47.50(1.015)−n

24 30 100.00 · (1.015)−n + 2.25an1.5% = 150.00 − 50.00(1.015)−n


As in the case of 4% we have to compare four prices:

Call Date Price10 115.5316 114.5420 114.7324 115.02

This time the highest price that the investor may safely pay is 114.54.

7.1.6 2002/2003 Class Tests, with Solutions

Versions 2 and 4 appear to have been slightly more difficult than Versions 1 and 3,and the grades were adjusted to compensate for this.

Versions 1 (white) and 3 (yellow)1. Showing your work, solve each of the following problems:

(a) [2 MARKS] Determine the nominal annual interest rate, i1, compounded every 3months, which is equivalent to a nominal annual interest rate of 12% compoundedevery 4 months.

(b) [4 MARKS] Determine the nominal annual interest rate compounded semi-annually,i2, which is equivalent to an effective annual discount rate of 6%.

(c) [4 MARKS] Determine the nominal annual interest rate, i3, compounded instan-taneously (=convertible continuously), which is equivalent to an effective monthlydiscount rate of 1%.

Solution:

(a) In one year a sum of 1 will grow, under the first rate, to(1 + i1

4

)4, and under the

second to(1 + 0.12

3

)3. Equating these two yields 1 + i1

4 = (1.04)34 , so

i1 = 4((1.04)

34 − 1

)= 0.1194 = 11.94% .

(b) We know several relationships between i and the corresponding d. For example,d = iv = i

1+i = 1 − 11+i . Solving these equations for i when d = 0.06, we obtain,

corresponding to an effective annual discount rate of 6%, an effective annual interestrate of 1

0.94−1 = 694 = 0.06383. The effective semi-annual interest rate corresponding

to this effective annual rate will be (1+ 694 )

12−1 =

√10094 −1 = 0.03142. Corresponding

to this semi-annual rate, the nominal annual interest rate compounded semi-annuallywill be twice this rate, i.e. 6.284 %.


(c) Since d + v = 1, the value of v corresponding to d = 1% is 99100 , so 1 + i = 100

99 = 1 + 199

and i = 199 ; hence the corresponding effective annual rate of interest is

(1 + 1

99

)12 −1 = 12.81781%. The “nominal annual interest rate, i3, compounded instantaneously(=convertible continuously)” will be the force of interest

i3 = δ = ln(1 +

199

)12

= 12 ln(1 +

199

)= 12 ln

10099

= 0.1206 = 12.06%.

2. (a) [2 MARKS] Define the sequence of payments whose value is represented by thesymbol sni, using a time diagram showing the payments, and indicating the point intime where the value of the various payments is being calculated.

(b) [4 MARKS] Derive a formula for sni by using formulæ known to you for the summa-tion of arithmetic or geometric progressions. Your final formula should be expressedin closed form, i.e., without using summation symbols (

∑) or dots (. . .),

(c) [4 MARKS] Define what is meant by sni. Give, without proof, a formula whichexpresses the value of sni in terms of i and n.

Solution:

(a)

1 1 1 1 1

0 21 3 · · · n↑

(b)

sni = 1 + (1 + i) + (1 + i)2 + . . . + (1 + i)n−1

= 1 · (1 + i)n − 1(1 + i) − 1

=(1 + i)n − 1(1 + i) − 1

=(1 + i)n − 1

i


(c) sni is the value of the sum of n payments of 1 at the beginning of each year, evaluatedone year after the last payment. Its value is sn+1i − 1 =

(1+i)n+1−(1+i)i ; other formulæ

would also have been acceptable.

3. A loan of 10,000 at i = 10% is to be repaid by ten equal annual payments.

(a) [5 MARKS] Determine the annual payment.(b) [10 MARKS] Determine an amortization schedule for the first 5 payments, showing,

for each payment, the interest portion and the portion for reduction of principal. Usethe following format for your table.

Duration Payment Interest Principal Outstanding(Years) Repaid Principal

. . .(c) [5 MARKS] If the loan is sold to an investor immediately after the 5th payment at a

price to yield 12% effective annual interest, determine the price paid by the investor.

Solution: (Source = Deferred/Supplemental Examination in Math 329, August, 2000,Problem 3.)

(a) If the annual payment is denoted by X, it must satisfy the equation X · a1010% =

10, 000. Solving this equation yields X = 10001−(1.1)−10 = 1000

0.61445671 = 1627.45 as the levelannual payment.

(b) If we were interested only in the interest portion of the 5th payment, we might recallhaving proved that [10, p. 79] to be

X(1 − (1.1)10−5+1) = 1627.45 × (1 − 0.56447393) = 708.80.

The portion for reduction of principal would then be

X − 708.80 = 1627.45 − 708.80 = 918.65 .

However, the problem required the construction of an amortization table, so thesedata can be used only to verify our computations in the table:

Duration Payment Interest Principal Outstanding(Years) Repaid Principal

0 10000.001 1627.45 1000.00 627.45 9372.552 1627.45 937.26 690.19 8682.363 1627.45 868.24 759.21 7923.154 1627.45 792.31 835.14 7088.015 1627.45 708.80 918.65 6169.36


(c) While the outstanding principal is shown as 6169.35, that will equal the present valueof the remaining 5 payments of 1627.45 each only if the interest rate remains at 10%.If the interest rate changes to 12%, the present value of the remaining 5 paymentsfalls to 1627.45 · a512% = 1627.45 × 1−(1.12)−5

0.12 = 5866.59. This will be the price paidby an investor who expects the 5 remaining payments to yield 12% effective interest.

Versions 2 (blue) and 4 (green)1. [20 MARKS] A borrower takes out a loan of 2000 to be paid by one payment with full

interest at the end of two years. Construct a sinking fund schedule using the headingsDuration Contribution to Interest Interest Earned Balance of Balance of(Years) Sinking Fund on Loan in Sinking Fund Sinking Fund Principal

. . .assuming that the lender receives 10% convertible semi-annually on the loan, and theborrower replaces the amount of the loan with equal semi-annual deposits in a sinkingfund to mature when the loan becomes due, where the sinking fund earns 8% convertiblesemi-annually.Solution: (Source = Final Examination in Math 329, April, 2000, Problem 3; the presentproblem is simplified from that on the examination.)

(a) [7 MARKS] The sinking fund must attain the value of 2000(1.05)4; if we denote thevalue of the semi-annual payments into this fund, than X · s40.04 = 2000(1.05)4, so

X =2000 × (1.05)4 × 0.04

(1.04)4 − 1=

97.24050.16986

= 572.48.

(b) [13 MARKS] The schedule is as follows:Duration Contribution to Interest Interest Earned Balance of Balance of(Years) Sinking Fund on Loan in Sinking Fund Sinking Fund Principal

0.0 0.00 0.00 0.00 0.00 2000.000.5 572.48 100.00 0.00 572.48 2100.001.0 572.48 105.00 22.90 1167.86 2205.001.5 572.48 110.25 46.71 1787.05 2315.252.0 572.48 115.76 71.48 2431.01 2431.01

2. Showing your work, solve each of the following problems:

(a) [4 MARKS] Determine the nominal annual interest rate compounded semi-annually,i1, which is equivalent to an effective annual discount rate of 4%.

(b) [2 MARKS] Determine the nominal annual interest rate, i2, compounded every 6months, which is equivalent to a nominal annual interest rate of 24% compoundedevery 3 months.


(c) [4 MARKS] Determine the nominal annual interest rate, i3, compounded instanta-neously (=convertible continuously), which is equivalent to an effective quarterlydiscount rate of 2%.

Solution:

(a) We know several relationships between i and the corresponding d. For example,d = iv = i

1+i = 1 − 11+i . Solving these equations for i when d = 0.04, we obtain,

corresponding to an effective annual discount rate of 4%, an effective annual interestrate of 1

0.96−1 = 496 = 0.04167. The effective semi-annual interest rate corresponding

to this effective annual rate will be (1+ 496 )

12−1 =

√10096 −1 = 0.02062. Corresponding

to this semi-annual rate, the nominal annual interest rate compounded semi-annuallywill be twice this rate, i.e. i1 = 4.124%.

(b) In one year a sum of 1 will grow, under the first rate, to(1 + i2

2

)2, and under the

second to(1 + 0.24

4

)4. Equating these two yields 1 + i1

2 = (1.06)2, so

i2 = 2((1.06)2 − 1

)= 0.0472 = 24.72% .

(c) Since d + v = 1, the value of v corresponding to d = 2% is 98100 , so 1 + i = 100

98 = 1 + 298

and i = 298 ; hence the corresponding effective annual rate of interest is

(1 + 2

98

)4 −1 = 8.4166%. The “nominal annual interest rate, i3, compounded instantaneously(=convertible continuously)” will be the force of interest

i3 = δ = ln(1 +

298

)4

= 4 ln(1 +

298

)= 4 ln

10098

= 0.08081 = 8.081%.

3. (a) [2 MARKS] Define the sequence of payments whose value is represented by thesymbol ani, using a time diagram showing the payments, and indicating the point intime where the value of the various payments is being calculated.

(b) [4 MARKS] Derive a formula for ani by using formulæ known to you for the summa-tion of arithmetic or geometric progressions. Your final formula should be expressedin closed form, i.e., without using summation symbols (

∑) or dots (. . .).

(c) [4 MARKS] By allowing n to approach infinity, determine a (closed form) formulafor the value of a∞i.

Solution:


(a)

1 1 1 1 1

0 21 3 · · · n↑

(b)

an = v + v2 + . . . + vn

= v · 1 − vn

1 − v

= (1 + i)v · 1 − vn

(1 + i) − (1 + i)v

=1 − vn

(1 + i) − 1=

1 − vn

i

(c) Since 1 + i > 1, 0 < 11+i < 1, so a∞i = lim

n→∞1−vn

i =lim

n→∞ 1− limn→∞ vn

i = 1−0i = 1

i .

7.1.7 Final Examination, 2002/2003

1. (a) [3 MARKS] The total amount of a loan to which interest has been added is 5,000.The term of the loan was 4 years. If the nominal annual rate of interest was 6% andinterest was compounded semi-annually, determine the original amount of the loan,showing all your work.

(b) [3 MARKS] Showing all your work, determine the simple interest rate under whicha sum of money will double in 5 years.

(c) [4 MARKS] Showing all your work, determine the effective annual compound dis-count rate under which a sum of money will double in 8 years.

(d) [5 MARKS] Showing all your work, determine the rate of interest, convertible con-tinuously, that is equivalent to an effective interest rate of 1% per month.

2. (a) [8 MARKS] To repay a loan, X is obliged to pay Y 1,000 at the end of December,2004, and 1,200 at the end of December, 2006. He proposes to replace these twopayments by a single payment of 2,196 at the end of December, 2005. If Y acceptsthis proposal, what yield rate will he be earning on his loan? Show all your work.

(b) [7 MARKS] Showing all your work, determine the value at time t = 0 of a continu-ous annuity that pays 10,000 per year for 2 years, at an effective annual interest rateof 5%.


3. Express each of the following only in terms of `x, and v.

(a) [2 MARKS] d27

(b) [2 MARKS] 4q24

(c) [2 MARKS] a20:25

(d) [2 MARKS] A120:25

(e) [2 MARKS] A20:25

(f) [2 MARKS] The probability that a 25-year old will survive 40 years, but will diebefore reaching age 75.

(g) [3 MARKS] 12|a20:25

4. The Wallace Widget Company is planning to borrow 150,000 from the Bank of Antigo-nish, and to undertake to pay interest annually at a rate of 12%; they plan to contributeequal annual payments to a sinking fund that earns interest at the rate of 9%. The sinkingfund will repay the principal at the end of 10 years. Showing all your work, determine

(a) [2 MARKS] the annual interest payment,(b) [3 MARKS] the annual payment into the sinking fund

At the end of 4 years, when Wallace has made its annual interest payment and its 4thpayment to the sinking fund, it proposes that this should be the last payment to the sink-ing fund. It will apply the balance X accumulated to date in the sinking fund to repayprincipal, and it will amortize the remainder of the principal by equal annual paymentsover the next 5 years, at a rate of 10%.

(c) [4 MARKS] Determine the annual level payment Y under this proposal.(d) [6 MARKS] Construct an amortization table for this proposal, under the following

headings, beginning immediately after the 4th and last payment to the sinking fund;assume also that all outstanding interest on the loan has been made annually to date:

Duration Payment Interest Principal OutstandingRepaid Principal

4 0.00 0.00 0.00 150000.00−X5 Y =...

......

......

9 0.00

5. Consider a 100 par-value 15-year bond, with semi-annual coupons at the nominal annualinterest rate of 4%, convertible every six months. Let t represent time in half-years;assume that the bond is callable at 109.00 on any coupon date from t = 10 to t = 20inclusive, at 104.50 from t = 21 to t = 29 inclusive, but matures at 100.00 at t = 30.In each of the following cases, determine what price an investor should pay to guaranteehimself


(a) [7 MARKS] a nominal annual yield rate of 5%, convertible semi-annually;(b) [8 MARKS] an effective annual yield rate of 3%.

6. In addition to her down payment, Mary’s purchase of her new home is financed by amortgage of 60,000 payable to the vendor; the mortgage is amortized over 20 years, witha level payment at the end of each month, at a nominal annual rate of 6% compoundedmonthly.

(a) [3 MARKS] Determine the monthly payments under this mortgage.(b) [2 MARKS] Divide the first payment into principal and interest.(c) [3 MARKS] Determine the outstanding principal immediately after the 60th pay-

ment.(d) [4 MARKS] Divide the 60th payment into principal and interest.(e) [3 MARKS] Determine the payment that Mary could make at the end of each year

which would be equivalent to the year’s 12 monthly payments.

7. (a) [5 MARKS] Define what is meant by (Da)n and (Ia)n, and explain verbally why

(Da)30 + (Ia)30 = 31a30 .

(b) [10 MARKS] Showing all your work, find the present value (using effective annualinterest rate i = 6%) of a perpetuity which pays 100 after 1 year, 200 after 2 years,increasing until a payment of 2000 is made, after which payments are level at 2000per year forever. [For this problem you may assume that

(Ia)n =an − nvn

i(114)

(Ia)∞ =a∞i

(115)

(Is)n =sn − n

i.] (116)

8. In order to complete the sale of his home in Vancouver, John accepted, in partial pay-ment, a 200,000 mortgage amortized over 15 years with level semi-annual payments ata nominal annual rate of 5% compounded semi-annually. Fred has cash available, and isprepared to buy the mortgage from John and to invest a fixed portion of the semi-annualpayments he receives in a sinking fund that will replace his purchase capital in 15 years.The sinking fund will earn interest at only 4%, compounded semi-annually. Showing allyour work, determine the following:

(a) [3 MARKS] the amount of the semi-annual mortgage payments(b) [4 MARKS] as a fraction of the purchase price Fred pays for the mortgage, the semi-

annual payment into the sinking fund


(c) [8 MARKS] the amount that Fred should pay for the mortgage in order to obtain anoverall yield rate of 6%, compounded semi-annually on his investment. (Note thatthe sinking fund earns 4% compounded semi-annually.)

7.2 2003/20047.2.1 First 2003/2004 Problem Assignment, with Solutions

Distribution Date: Solutions mounted on the Web on Wednesday, 4 February, 2004Assignment was mounted on the Web on Thursday, January 15th, 2005

Hard copy was distributed on Monday, January 19th, 2004Solutions were to be submitted by Monday, January 26th, 2004

(This is a short assignment. Subsequent assignments can be expected to be longer.)

1. It is known that the accumulation function a(t) is of the form b · (1.1)t + ct2, where b andc are constants to be determined.

(a) If $100 invested at time t = 0 accumulates to $170 at time t = 3, find the accumulatedvalue at time t = 12 of $100 invested at time t = 1.

(b) Show that this function satisfies the requirement [6, p. 2, #2] that it be non-decreasing.(c) Determine a general formula for in, and show that lim

n→∞in = 10%. (Use L’Hopital’s

Rule.)

Solution: [6, Exercise 4, p. 30] Denote the corresponding amount function by A(t).

(a) An accumulation function must have the property that a(0) = 1; this implies that1 = a(0) = b + 0, so b = 1.The given data imply that

170 = 100(a(3)) = 100(1(1.331) + c · 32) (117)

which implies that c = 0.041. We conclude that

A(t)A(0)

= a(t) = (1.1)t + 0.041t2 , (118)

implying that a(1) = 1.141, a(12) = 9.042428377. Then

A(12) = A(1) · A(12)A(1)

= A(1) · a(12)a(1)

= A(1) · 9.0424283771.141

= A(1)(7.925002959) = 792.5002959

so $100 at time t = 1 grows to $792.50 at time t = 12.


(b) It follows from (118) that a′(t) = (1.1)t ln 1.1 + 0.082t, which is positive for positivet; thus a(t) is an increasing function of t for positive t.(This property may also be proved “from first principles”. Let t1 ≥ t2. Then

a(t2) − a(t1) = (1.1)t2 + 0.041t22 − (1.1)t1 + 0.041t2

1

= (1.1)t1 ((1.1)t2−t1 − 1)

+ (0.041) (t2 − t1) (t2 + t1)

where both of the summands are non-negative for 0 ≤ t1 ≤ t1.(c)

in =a(n) − a(n − 1)

a(n − 1)=

(0.1)(1.1)n−1 + (0.041)(2n − 1)(1.1)n−1 − (0.041)(n − 1)2

=0.1 + 0.041

(2n−1

(1.1)n−1

)

1 − 0.041(

(n−1)2

(1.1)n−1

)

By L’Hopital’s Rule

limx→∞

2x − 1(1.1)x−1 = lim

x→∞2

(1.1)x−1 ln 1.1= 0

limx→∞

(n − 1)2

(1.1)n−1 = limx→∞

2(x − 1)(1.1)x−1 ln 1.1

= limx→∞

2(1.1)x−1(ln 1.1)2 = 0

Hencelimn→∞

in =0.1 + 0.041 (0)1 − 0.041 (0)

= 0.1 = 10%.

2. It is known that 1000 invested for 4 years will earn 250.61 in interest, i.e., that the valueof the fund after 4 years will be 1250.61. Determine the accumulated value of 3500invested at the same rate of compound interest for 13 years.Solution: [6, Exercise 14, p. 30] Let i be the rate of compound interest. Then 1000(1 +

i)4 = 1250.61. The accumulated value of 3500 after 13 years will be

3500(1 + i)13 = 3500(1250.61

1000

) 134

= 7239.57 .

3. It is known that an investment of 750 will increase to 2097.75 at the end of 25 years. Findthe sum of the present values of payments of 5000 each which will occur at the ends of10, 15, and 25 years.


Solution: [6, Exercise 21, p. 31] Let i be the interest rate. The known fact is that 750(1 +

i)25 = 2097.75. Hence (1 + i)25 = 2.797 , so v25 = 0.357535924. The present value ofthree payments of 5000 after 10, 15, and 25 years will, therefore, be

5000(v10 + v15 + v25)= 5000

((0.357535924)

1025 + (0.357535924)

1525 + (0.357535924)

2525)

= 5000(0.662709221 + 0.539500449 + 0.357535924)= 7798.73.

4. Find the accumulated value of 1000 at the end of 10 years:

(a) if the nominal annual rate of interest is 6% convertible monthly;(b) if the nominal annual rate of discount is 5% convertible every 2 years.

Solution: [6, Exercise 32, p. 31]

(a) The accumulation factor for each month is 1 + 6%12 = 1.005. After 10 years 1000

grows to1000(1.005)10×12 = 1819.40.

(b) The discount factor for each 2 years is 1 − 2 × 5% = 0.09 (moving backwards),corresponding to an accumulation factor of 1

0.9 . After 10 years 1000 grows to

1000(0.09)−102 = 1693.51 .

5. Given that i(m) = 5√

66 − 2 and d(m) = 2− 8

√0.06, find m, the equivalent annual compound

interest rate, and the equivalent annual compound discount rate.Solution: [6, Exercise 30, p. 32] For an mth of a year the relationship between i(m) andd(m) is given by (

1 +i(m)

m

) (1 − d(m)

m

)= 1

which is equivalent to(m + i(m))(m − d(m)) = m2

or

m =i(m)d(m)

i(m) − d(m) .

Substituting the given values

i(m) = 0.041241452d(m) = 0.040408205


gives m = 2. It follows that

i =

(1 +

i(2)

2

)2

− 1

= (1.020620726)2 − 1 = 4.1666667% = 416

% =1

24

d = 1 −(1 − d(2)

2

)2

= 1 − 0.96 = 4% =125

7.2.2 Second 2003/2004 Problem Assignment, with Solutions

Distribution Date: Mounted on the Web on Friday, February 20th, 2004Assignment was mounted on the Web on January 19th, 2004.Hard copy was distributed on Wednesday, January 28th, 2004

Solutions were due by Monday, February 9th, 2004(Solutions presented subject to correction of errors and omissions.)

1. Find the present value of 1000, to be paid at the end of 37 months under each of thefollowing scenarios:

(a) Assume compound interest throughout, and a (nominal) rate of discount of 6%payable quarterly.

(b) Assume compound interest for whole years only at a (nominal) rate of discount of6% payable quarterly, and simple discount at the rate of 1.5% per 3 months duringthe final fractional period.

(c) Assume compound interest throughout, and a nominal rate of interest of 8% payablesemi-annually.

Solution: (cf. [6, Exercise 2, p. 53])

(a) Present value = 1000(1 − 0.06

4

) 373

= 829.94.

(b) Present value = 1000(1 − 0.06

4

) 363(1 − 0.015

3

)= 1000 × 0.872823 × 0.9995 =

829.96.(c) Present value = (1.04)−

376 = 1000 × 0.785165257 = 785.17.

2. The sum of 5,000 is invested for the months of April, May, and June at 7% simple interest.Find the amount of interest earned


(a) assuming exact simple interest in a non-leap year(b) assuming exact simple interest in a leap year (with 366 days);(c) assuming ordinary simple interest;(d) assuming the “Bankers’ Rule”.


(a) The number of days is 30 + 31 + 30 = 91; exact simple interest is 91365 ·5000 · (0.07) =

87.26.(b) Exact simple interest is 91

366 · 5000 · (0.07) = 87.02(c) Ordinary simple interest is 30+30+30

360 · 5000 · (0.07) = 87.50(d) Interest under the Banker’s Rule is 91

360 · 5000 · (0.07) = 88.47.

3. Find how long 4,000 should be left to accumulate at 5% effective in order that it willamount to 1.25 times the accumulated value of another 4,000 deposited at the same timeat a nominal interest rate of 4% compounded quarterly.Solution: (cf. [6, Exercise 13, p. 55]) The equation of value at n years is

4000(1.05)n = (1.25)(4000)(1.01)4n

son =

ln 1.25ln 105 − 4 ln 101

= 24.82450822 .

4. The present value of two payments of 100 each, to be made at the end of n years and 2nyears is 63.57. If i = 6.25%, find n.Solution: (cf. [6, Exercise 14, p. 55]) Solving the equation of value, 100v2n + 100vn =

63.57, we obtain

vn =−1 ± √3.5428

2,

in which only the + sign is acceptable, since vn > 0. Taking logarithms gives

n =0.818446587

ln 1.0625= 13.50023411.

We conclude, to the precision of the problem, that n = 13.5 years.

5. (a) Find the nominal rate of interest convertible quarterly at which the accumulatedvalue of 1000 at the end of 12 years is 3000.

(b) Find the nominal rate of discount convertible semi-annually at which a payment of3000 12 years from now is presently worth 1000.

(c) Find the effective annual rate of interest at which the accumulated value of 1000 atthe end of 12 years is 3000.

Solution:



(a) (cf. [6, Exercise 19, p. 55]) The equation of value at time t = 12 is

1000(1 +

i(4)

4

)4×12

= 3000 ,

implying thati(4) = 4

(3

148 − 1

)= 9.260676%.

(b) The equation of value at time t = 12 is

3000(1 − d(2)

2

)2×12

= 1000 ,

implying thatd(2) = 2

(1 − 3

124)

= 9.3678763%.

(c) The equation of value at time t = 12 is

1000 (1 + i)12 = 3000 ,

implying thati = 3

112 − 1 = 9.5872691%.

6. An investor deposits 20,000 in a bank. During the first 4 years the bank credits an annualeffective rate of interest of i. During the next 4 years the bank credits an annual effectiverate of interest of i − 0.02. At the end of 8 years the balance in the account is 22,081.10.What would the account balance have been at the end of 10 years if the annual effectiverate of interest were i + 0.01 for each of the 10 years?Solution: (cf. [6, Exercise 32, p. 57]) The equation of value is

20000(1 + i)4(1 + (i − 0.02))4 = 22081.10 ,

which we interpret as a polynomial equation. The equation is of degree 8, and we don’thave a simple algebraic method for solving such equations in general. But this equationhas the left side a pure 4th power, so we can extract the 4th roots of both sides, obtaining

(1 + i)(1 + (i − 0.02)) = (1.104055)14 = 1.025056201 ,

which may be expressed as a quadratic equation in 1 + i:

(1 + i)2 − 0.02(1 + i) − 1.025056 = 0


1 + i =0.02 +

√(0.02)2 + 4(1.025056)

2= 1.0225

from which we conclude that i = 2.25%, and that the account balance after 10 yearswould be 20000(1.0225 + 0.01)10 = 20000(1.0325)10 = 27, 737.89.


7. A bill for 1000 is purchased for 950 4 months before it is due. Find

(a) the nominal rate of discount convertible monthly earned by the purchaser;(b) the annual effective rate of interest earned by the purchaser.


(a) If d(12) be the nominal discount rate, then

1000(1 − d(12)

12

)4

= 950

implying that

1 − d(12)

12= 0.9872585

so d(12) = 15.29%.(b) Let i be the effective annual interest rate. Then

950(1 + i)13 = 1000

implies that

i =

(1000950

)3

− 1 = 16.635% .

8. A signs a 2-year note for 4000, and receives 3168.40 from the bank. At the end of 6months, a year, and 18 months A makes a payment of 1000. If interest is compoundedsemi-annually, what is the amount outstanding on the note at the time if falls due?Solution: If i′ be the rate of interest charged semi-annually, then

3168.40(1 + i′)4 = 4000

so i′ = 6.00%; that is i(2) = 12.00%. The value of the 3 payments at the time the notematures is

1000((1.06)3 + (1.06)2 + (1.06)1

)= 3374.62

so the amount outstanding before the final payment is 625.38.

9. The Intermediate Value Theorem for continuous functions tells us that such a functionf (x) whose value at x = a has the opposite sign from its value at x = b will assume thevalue 0 somewhere between a and b. By computing the value of f at the point 1

2 (a+b), wecan infer that there is a 0 of f in an interval half as long as [a, b], and this procedure maybe repeated indefinitely to determine a zero of f to any desired accuracy. Assuming thatpolynomials are continuous, use this idea to determine the nominal quarterly compoundinterest rate under which the following payments will accumulate to 1000 at the end of 4years:


• 300 today

• 200 at the end of 1 year

• 300 at the end of 2 yearsYour answer should be accurate to 3 decimal places, i.e., expressed as a percentage to 1decimal place.Solution: (We will carry the computations to an accuracy greater than requested in theproblem.)(a) Let the effective annual interest rate be i. The equation of value at the end of 4 years

is300(1 + i)4 + 200(1 + i)3 + 300(1 + i)2 = 1000 . (119)

(b) We need a continuous function to which to apply the Intermediate Value Theorem.Some choices may be better than others. We will choose

f (x) = 3x4 + 2x3 + 3x2 − 10 .

We observe that f (0) = −10, that f (2) = 48 + 16 + 12 − 10 = 66 > 0, and thatf (−2) = 48− 16 + 12− 10 = 34 > 0. This tells us that there is a solution to equation(119) for −2 ≤ x ≤ 0, equivalently for −3 ≤ i ≤ −1: such a solution is of no interestto us, as it does not fit the constraints of this problem. But the function is a cubicpolynomial, and has 2 other zeros. We see that it also has a solution in the interval0 ≤ x ≤ 2, and we proceed to progressively halve intervals.

(c) The midpoint of interval [0, 2] is 1;

f (1) = 3 + 2 + 3 − 10 = −2 < 0 < 66 = f (2) ,

so there must be a root in the interval [1, 2].(d) The midpoint of [1, 2] is 1.5;

f (1.5) = 3(1.5)4 + 2(1.5)3 + 3(1.5)2 − 10= 15.1875 + 6.75 + 6.75 − 10 = 18.6875> 0

so there must be a zero in the interval [1, 1.5], whose midpoint is 1.25.(e)

f (1.25) = 3(1.25)4 + 2(1.25)3 + 3(1.25)2 − 10= 7.3242 + 3.9063 + 4.6875 − 10 = 5.918> 0

so there must be a zero in the interval [1, 1.25], whose midpoint is 1.125.


(f)

f (1.125) = 3(1.125)4 + 2(1.125)3 + 3(1.125)2 − 10= 4.8054 + 2.8477 + 3.7969 − 10 = 1.45> 0

so there must be a zero in the interval [1, 1.125], whose midpoint is 1.0625.(g)

f (1.0625) = 3(1.0625)4 + 2(1.0625)3 + 3(1.0625)2 − 10= 3.8233 + 2.3989 + 3.3867 − 10 = −0.3911< 0

so there must be a zero in the interval [1.0625, 1.125], whose midpoint is 1.09375.(h)

f (1.09375) = 3(1.09375)4 + 2(1.09375)3 + 3(1.09375)2 − 10= 4.2933 + 2.6169 + 3.5889 − 10 = 0.4991> 0

so there must be a zero in the interval [1.0625, 1.09375], whose midpoint is 1.078125.(i)

f (1.078125) = 3(1.078125)4 + 2(1.078125)3 + 3(1.078125)2 − 10= 4.053197 + 2.506325 + 3.487061 − 10 = 0.046582> 0

so there must be a zero in the interval [1.0625, 1.078125], whose midpoint is 1.0703125.(j)

f (1.0703125) = 3(1.0703125)4 + 2(1.0703125)3 + 3(1.0703125)2 − 10= 3.936984 + 2.452233 + 3.436707 − 10 = −0.174076< 0

so there must be a zero in the interval [1.0703125, 1.078125], whose midpoint is1.07421875.

(k) f (1.07421875) = −.064207958 < 0 so there must be a zero in the interval [1.07421875, 1.078125],whose midpoint is 1.076172.

(l) f (1.076172) = −.008925 < 0 so there must be a zero in the interval[1.076172, 1.078125], whose midpoint is 1.077149.


(m) f (1.077149) = .018814 > 0 so there must be a zero in the interval[1.076172, 1.077149], whose midpoint is 1.076661.

(n) f (1.076661) = .004952 > 0 so there must be a zero in the interval[1.076172, 1.076661], whose midpoint is 1.076417.

(o) f (1.076417) = −.001974 < 0 so there must be a zero in the interval[1.076417, 1.076661], whose midpoint is 1.076539.

(p) f (1.076539) = .001488 > 0 so there must be a zero in the interval[1.076417, 1.076539], whose midpoint is 1.076478.

(q) f (1.076478) = −.000243 < 0 so there must be a zero in the interval[1.076478, 1.076539], whose midpoint is 1.076509.

(r) f (1.076509) = .000637 > 0 so there must be a zero in the interval[1.076478, 1.076509], whose midpoint is 1.076494.

(s) f (1.076494) = .000211 > 0 so there must be a zero in the interval[1.076478, 1.076494], whose midpoint is 1.076486.

(t) f (1.076486) = −.000016 > 0 so there must be a zero in the interval[1.076486, 1.076494], whose midpoint is 1.07649.

(u) f (1.07649) = .000097 > 0 so there must be a zero in the interval[1.076486, 1.07649], whose midpoint is 1.076488.

(v) f (1.076488) = .000041 > 0 so there must be a zero in the interval[1.076486, 1.076488], whose midpoint is 1.076487.

(w) f (1.076487) = .000012. One zero will be approximately x = 1.07649. Thus theeffective annual interest rate is approximately 7.649%. This, however, is not what theproblem asked for. The accumulation function for 3-months will then be (1.0749)

14 =

1.01822, so the effective interest rate for a 3-month period will be 1.822%, and thenominal annual interest rate, compounded quarterly, will be 7.288, or 7.3% to theaccuracy requested.

THE FOLLOWING PROBLEM WAS CONSIDERED FOR INCLUSION IN THE ASSIGN-MENT, BUT WAS (FORTUNATELY) NOT INCLUDED.

10. [6, Exercise 6, p. 88]

(a) Show thatam−n = am − vmsn = (1 + i)nam − sn

where 0 < n < m.(b) Show that

sm−n = sm − (1 + i)man = vnsm − an

where 0 < n < m.(c) Interpret the results in (a) and (b) verbally.


Solution:

(a) We prove the first of these identities by technical substitutions in sums, analogousto changes of variables in a definite integral. For the second identity we give a lessformal proof.

am−n =

m−n∑

r=1

vr

=

m∑

r=1

vr −m∑

r=m−n+1

vr

=

m∑

r=1

vr − vmm∑

r=m−n+1

vr−m

=

m∑

r=1

vr − vm1∑

s=n

v1−s

under the change of variable s = m − r + 1

=

m∑

r=1

vr − vmn∑

s=1

v1−s

reversing the order of the 2nd summation

=

m∑

r=1

vr − vmn∑

s=1

(1 + i)s−1

= am − vm · sn

am−n = v + v2 + . . . + vm−n

= v−n+1 + v−n+2 + . . . + v0 + v1 + v2 + . . . + vm−n

−(v−n+1 + v−n+2 + . . . + v0

)

= v−n(v1 + v2 + . . . + vn + vn+1 + vn+2 + . . . + vm

)

−((1 + i)n−1 + (1 + i)n−2 + . . . + (1 + i)v0

)

= (1 + i)n(v1 + v2 + . . . + vn + vn+1 + vn+2 + . . . + vm

)

−((1 + i)0 + (1 + i)1 + . . . + (1 + i)vn−1

)

= (1 + i)nam − sn

(b) These identities could be proved in similar ways to those used above. Instead, weshall show that these identities can be obtained from the preceding simply by multi-plying the equations by (1 + i)m−n:

sm−n = (1 + i)m−nam−n [6, (3.5), p. 60]


= (1 + i)m−n((1 + i)nam − sn

)

= (1 + i)mam − (1 + i)m · (1 + i)−nsn

= sm − (1 + i)man

sm−n = (1 + i)m−nam−n

= (1 + i)m−n(am − vmsn

)

= (1 + i)−n · (1 + i)mam − (1 + i)−nsn

= (1 + i)−nsm − an

(c) i. An (m − n)-payment annuity-immediate of 1 has the same present value as anannuity for a total term of m = (m − n) + n years minus a correction paid todayequal to the value of the deferred n payments. Those n payments are worth smat time t = m, which amount can be discounted to the present by multiplying byvm.The preceding explanation was based on values at the commencement of the firstyear of an m-year annuity-immediate. Let us now interpret the m−n payments asbeing the last payments of an m-year annuity whose mth payment has just beenmade. That m-payment annuity was worth am, n years ago — a year before itsfirst payment; today it is worth (1 + i)nan, including the payments we attached atthe beginning. Those payments are worth sn today, for a net value as claimed.

ii. Consider an value of the first m−n payments of an m-payment annuity-immediateof 1, just after the (m−n)th payment. Since these could be considered simply theaccumulated value of an (m − n)-payment annuity, they are worth sm−n. But thepayments of the m-payment annuity not yet made are worth an, and the entireannuity is worth sm at termination, hence vnsm today; hence vmsm −an is also thevalue today. This gives the equality between the extreme members of the allegedinequality.Now let’s evaluate the same m − n payments, but this time consider them to bethe last m − n payments of an m-payment annuity-certain; again, the m − nthpayment has just been made. From first principles, the accumulated value of thepayments actually received is sm−n , and we are viewing them from the context ofan annuity-certain of m payments that would be worth sm today: let’s determinethe amount that would have to be paid out today to correct for that expandedannuity. The value of the n payments we have tacked on in the past was an oneyear before the payments began, and (1 + i)msn today; so we can also view thevalue today as sm − (1 + i)man .

7.2.3 Third 2003/2004 Problem Assignment, with Solutions

Distribution Date: Mounted on the Web on Wednesday, March 3rd, 2004.


Assignment was mounted on the Web on February 8th, 2004,hard copy of assignment was distributed on Wednesday, February 11th, 2004.

Solutions were to be submitted by 9 a.m., Monday, March 3rd, 2004SUBJECT TO CORRECTION OF TYPO’S AND OTHER ERRORS

Sketch a time diagram to accompany your solution of all problems except the last.

1. A skier wishes to accumulate 30,000 in a chalet purchase fund by the end of 8 years. Ifshe deposits 200 into the fund at the end of each month for the first 4 years, and 200 + Xat the end of each month for the next 4 years, find X if the fund earns a nominal (annual)rate of 6% compounded monthly.Solution: The equation of value at the end of 8 × 12 = 96 months is

200s96 + X · s48 = 30000 ,

which we may solve to yield

X =30000s48 0.005

− 200s96 0.005

s48 0.005

=30000s48 0.005

− 200(1.005)96 − 1(1.005)48 − 1

= 30000s−148 0.005

− 200((1.005)48 + 1)

= 30000(0.018485) − 200(2.27049) from the tables= 100.452.

2. A fund of 2500 is to be accumulated by n annual payments of 50, followed by n + 1annual payments of 75, plus a smaller final payment of not more than 75 made 1 yearafter the last regular payment. If the effective annual rate of interest is 5%, find n and theamount of the final irregular payment.Solution: We shall interpret the payments to be made under two annuities-due: the first,for 2n + 1 years, consists of an annual deposit of 50 in advance; the second, for n + 1years, deferred n years after the first, consists of an annual deposit of 25 in advance. It isat the end of year 2n + 1 that the final — drop — payment is to be made, and it is to beunder 75. (Note that this is the type of problem where the drop payment could turn outto be negative.) We seek the smallest n for which

50 · s2n+1 + 25 · sn+1 > 2500 − 75 = 2425

⇔ 2(1.05) ((1.05)n)2 + (1.05)n −(3 +

0.05(2425)25(1.05)

)


⇔ ((1.05)n)2+

12.1

(1.05)n − 12.1

(3 +

0.05(2425)25(1.05)

)

⇔((1.05)n +

14.2

)2

>1

(4.2)2 +1

2.1

(3 +

0.05(2425)25(1.05)

)

⇒ (1.05)n +1

4.2>

(1

(4.2)2 +1

2.1

(3 +

0.05(2425)25(1.05)

)) 12

= 1.919585178

⇒ n >ln 1.919585178

ln 1.05= 10.65133267

Thus the drop payment will be when t = 11 + 12, i.e., 23 years after the first paymentunder the annuity with payments of 50. Just before the drop payment the accumulatedvalue of all previous payments is

50s23 + 25s12 =1.050.05

(50 · (1.05)23 + 25 · (1.05)12 − 75

)= 2592.924516

so the drop payment at time t = 23 is 2500 − 2592.924516 = −92.92.Thus we have an example here of a negative drop payment. Could this mean that weshould have taken n = 10? No. In that case we would find that the final payment wouldbe larger than the permitted 75. (If tables like those in the textbook were available, onecould determine the value of n by inspecting the value of 2s2n+1 + sn+1 . We observe fromthe 5% tables the following values:

n 2s2n+1 + sn

10 84.016511 97.067812 111.3713


50s2n+1 + 25sn > 2500 − 75

i.e., such that2s2n+1 + sn > 97 ,

equivalently,

2s2n+1 + sn >97

1.05= 92.38 ,

and so can conclude that n = 11.)

3. On his 30th birthday, a teacher begins to accumulate a fund for early retirement by de-positing 5,000 on that day and at the beginnings of the next 24 years as well. Since he


expects that his official pension will begin at age 65, he plans that, starting at age 55 hewill make an annual level withdrawal at the beginning of each of 10 years. Assumingthat all payments are certain to be made, find the amount of these annual withdrawals, ifthe effective rate of interest is 6% during the first 25 years, and 7% thereafter.Solution: Let the constant amount of the withdrawals beginning at age 55 be X. Theequation of value at age 55, just before the first withdrawal, is

5000 · s25 6% = X · a10 7%

⇒ Annual Withdrawal X =5000 · s25 6%

a10 7%

= 5000 · 1.061.07

· 0.070.06

· (1.06)25 − 11 − (1.07)−10

= 33477.74

4. At an effective annual interest rate of i it is known that

(a) The present value of 5 at the end of each year for 2n years, plus an additional 3 atthe end of each of the first n years, is 64.6720.

(b) The present value of an n-year deferred annuity-immediate paying 10 per year for nyears is 34.2642.

Find i.Solution: It is convenient to distinguish two cases.

Case i , 0: From (4a) we have an equation of value

64.6720 = 5 · a2n + 3 · an ; (120)

from (4b) we have the equation of value

34.2642 = vn · 10 · an = 10(a2n − an

). (121)

Solving these equations, we obtain

a2n = 9.3689 (122)an = 5.9425 , (123)

implying that

1 − v2n

1 − vn =9.36895.9425

⇒ 1 + vn = 1.5766⇒ vn = 0.5766.



We can substitute in equation (123) to obtain

i = 0.07125 = 7.125%

which implies that

n = − ln 0.5766ln 1.07125

= 8.000 years.

Case i = 0: Here Equations (120) and (121) become

64.6720 = 5(2n) + 3n = 13n34.2642 = 10n = 10(2n − n)

which are inconsistent. Thus this case is impossible.

5. (a) Find a12 if the effective rate of discount is 5%.(b) Charles has inherited an annuity-due on which there remain 12 payments of 10,000

per year at an effective discount rate of 5%; the first payment is due immediately.He wishes to convert this to a 25-year annuity-immediate at the same effective ratesof interest or discount, with first payment due one year from now. What will be thesize of the payments under the new annuity?

Solution:

(a) Since (1 − d)(1 + i) = 1, v = 1 − d = 0.95 when d = 0.05.

a12 =1 − (0.95)12

0.05= 9.19279825.

(b) i = d(1 + i) = dv = 0.05

0.95 = 119 . Let X be the size of the new payments. We must solve

the equation of value,91927.9825 = X · a25i

where v = 0.95. Hence

X =91927.9825

a25i

=91927.9825i

1 − v25

=91927.9825(

1 − (0.95)25) 19= 6695.606220

So the new annuity-immediate will pay 25 annual payments of 6695.61, beginningone year from now.


6. Give an algebraic proof and a verbal explanation for the formula

m

∣∣∣an = a∞ − am − vm+na∞ .

Solution:

(a)

a∞ − am − vm+na∞ =1i− 1 − vm

i− vm+n · 1

i

=1 − (1 − vm) − vm+n

i

=vm (1 − vn)

i= vm · 1 − vn

i= vm · an = m

∣∣∣an

(b) a∞ i is the present value of a perpetuity at rate i of 1 per year, payments starting ayear from now. am is the present value of the first m payments of that perpetuity; ifwe subtract this we have the present value of a perpetuity-immediate that starts myears from now, i.e., where the first payment is m + 1 years from now. vm+na∞ is thevalue of a perpetuity-immediate of 1 starting m + n years from now, i.e., where thefirst payment is m + n + 1 years from now; if we subtract this term as well, we areleft with the present value of payments at the ends of years m + 1, m + 2, . . ., m + n,i.e., with the present value of an n-payment annuity-certain of 1, deferred m years,i.e., of m

∣∣∣an

7. A level perpetuity-immediate is to be shared by A, B, C, and D. A receives the first npayments, B the next 2n payments, C payments ##3n + 1, . . . , 5n, and D the paymentsthereafter. It is known that the present values of B’s and D’s shares are equal. Find theratio of the present value of the shares of A, B, C, D.Solution: The present values of the shares of A, B, C, D are, respectively, an, vna2n =

a3n − an, v3na2n = a5n − a3n, and v5na∞ = a∞ − a5n. The fact that B’s and D’s shares areequal implies that

vn

i· (1 − v2n) =

v5n

iwhich is equivalent to (v2n)2 + v2n − 1 = 0, implying that

v2n =−1 ± √5

2.

Since v is positive, only the + sign is admissible:

v2n =−1 +

√5

2= 0.618033988...


sovn = 0.7861513777... .

Then the shares of A, B, C, D will be in the ratio

1 − vn : vn − v3n : v3n − v5n : 1 − v5n

i.e.0.2138486221 : 0.3002831059 : 0.1855851657 : 0.6997168937

8. (a) Find the present value of an annuity which pays 4,000 at the beginning of each 3-month period for 12 years, assuming an effective rate of 2% interest per 4-monthperiod.

(b) Suppose that the owner of the annuity wishes to pay now so that payments under hisannuity will continue for an additional 10 years. How much should he pay?

(c) How much should he pay now to extend the annuity from the present 12 years to aperpetuity?

(It is intended that you solve this problem “from first principles”, not by substitution intoformulæ in [6, Chapter 4].)Solution:

(a) If i be the effective interest rate per 4-month period, the effective rate per 3-monthperiod will be j = (1 + i)

34 − 1. Accordingly the value of the desired annuity is

4000a48 j = 4000(1 + j) · 1 − (1 + j)−48

j

= 4000(1 + i)34 · 1 − (1 + i)−36

(1 + i)34 − 1

= 40001 − (1.02)−36

1 − (1.02)−34

= 138, 317.4894.

(b) Repeating the calculations above for 48 + 40 = 88 payments, we obtain

4000a88 j = 4000(1 + j) · 1 − (1 + j)−88

j

= 4000(1 + i)34 · 1 − (1 + i)−66

(1 + i)34 − 1

= 40001 − (1.02)−66

1 − (1.02)−34

= 197, 897.4338,


so the additional payments will cost 197, 897.4338 − 138, 317.4894 = 59, 579.9444today.

(c) The cost of the perpetuity-due today would be

4000a88 j = 4000(1 + j) · 1j

= 4000(1 + i)34 · 1

(1 + i)34 − 1

=4000

1 − (1.02)−34

= 271, 329.4837,

so the additional payments will cost

271, 329.4837 − 138, 317.4894 = 133, 011.9943

today.

9. (No time diagram is needed for the solution to this problem.) In Problem 9 of Assignment2 you were asked to apply the Bisection Method to determine the solution to an interestproblem to 3 decimal places. The equation in question was (119):

300(1 + i)4 + 200(1 + i)3 + 300(1 + i)2 = 1000 .

and the solution given began with the values of

f (x) = 3x4 + 2x3 + 3x2 − 10 .

at x = 0 ( f (0) = −10), x = 2 ( f (2) = 66 > 0), and x = −2 ( f (−2) = 34 > 0),and we were interested in the solution between 0 and 2 — a solution that is uniquebecause f ′ is positive in this interval. Apply Linear Interpolation 4 times in an attempt todetermine the solution we seek. (You are not expected to know the general theory of errorestimation.) The intention is that you apply linear interpolation unintelligently, using itto determine a point where you find the function value and thereby confine the zero toa smaller subinterval: the point that you find will replace the midpoint in the bisectionmethod. In some situations, as in the present one, the procedure may not be better than thebisection method. Indeed, in the present example, it could take many more applicationsthan the bisection method to obtain the accuracy you obtained with that method.Solution: We take x1 = 0, x2 = 2. Then

x3 = 0 + (2 − 0) · −10−10 − 66


= 0.2631578947f (x3) = −9.741407754

x4 = 0.2631578947 + (2 − 0.2631578947) · −9.741407754−9.741407754 − 66

= 0.4865401588f (x4) = −8.891376200

x5 = 0.4865401588 + (2 − 0.4865401588) · −8.891376200−8.891376200 − 66

= 0.6662236083f (x5) = −7.486007579

x6 = 0.6662236083 + (2 − 0.6662236083) · −7.486007579−7.486007579 − 66

= 0.8020951913f (x6) = −5.796139773

x7 = 0.8020951913 + (2 − 0.8020951913) · −5.796139773−5.796139773 − 66

= 0.8988026707f (x7) = −4.166425912

x8 = 0.8988026707 + (2 − 0.8988026707) · −4.166425912−4.166425912 − 66

= 0.9641908820f (x8) = −2.825434839

x9 = 0.9641908820 + (2 − 0.9641908820) · −2.825434839−2.825434839 − 66

= 1.006713115f (x9) = −1.837664218

x10 = 1.006713115 + (2 − 1.006713115) · −1.837664218−1.837664218 − 66

= 1.033620406f (x10) = −1.162055435

x11 = 1.033620406 + (2 − 1.033620406) · −1.162055435−1.162055435 − 66

= 1.050340958f (x11) = −0.721587971

x12 = 1.050340958 + (2 − 1.050340958) · −0.721587971−0.721587971 − 66

= 1.060611435f (x12) = −0.442976533


x13 = 1.060611435 + (2 − 1.060611435) · −0.442976533−0.442976533 − 66

= 1.066874356f (x13) = −0.270019571

x14 = 1.066874356 + (2 − 1.066874356) · −0.270019571−0.270019571 − 66

= 1.070676410f (x14) = −0.163879567

x15 = 1.070676410 + (2 − 1.070676410) · −0.163879567−0.163879567 − 66

= 1.072978227f (x15) = −0.099198908

x16 = 1.072978227 + (2 − 1.072978227) · −0.099198908−0.099198908 − 66

= 1.074369462f (x16) = −0.059950550

x17 = 1.074369462 + (2 − 1.074369462) · −0.059950550−0.059950550 − 66

= 1.075209488f (x17) = −0.036195811

x18 = 1.075209488 + (2 − 1.075209488) · −0.036195811−0.036195811 − 66

= 1.075716385f (x18) = −0.021840825

x19 = 1.075716385 + (2 − 1.075716385) · −0.021840825−0.021840825 − 66

= 1.076022149f (x19) = −0.013174269

x20 = 1.076022149 + (2 − 1.076022149) · −0.013174269−0.013174269 − 66

= 1.076206548f (x20) = −0.007944937

x21 = 1.076206548 + (2 − 1.076206548) · −0.007944937−0.007944937 − 66

= 1.076317739f (x21) = −0.004790698

x22 = 1.076317739 + (2 − 1.076317739) · −0.004790698−0.004790698 − 66


= 1.076384781f (x22) = −0.002888504

x23 = 1.076384781 + (2 − 1.076384781) · −0.002888504−0.002888504 − 66

= 1.076425201f (x23) = −0.001741533

x24 = 1.076425201 + (2 − 1.076425201) · −0.001741533−0.001741533 − 66

= 1.076449571f (x24) = −0.001049952

x25 = 1.076449571 + (2 − 1.076449571) · −0.001049952−0.001049952 − 66

= 1.076464263f (x25) = −0.000632996

x26 = 1.076464263 + (2 − 1.076464263) · −0.000632996−0.000632996 − 66

= 1.076473120f (x26) = −0.000381633

x27 = 1.076473120 + (2 − 1.076473120) · −0.000381633−0.000381633 − 66

= 1.076478460f (x27) = −0.000230079

x28 = 1.076478460 + (2 − 1.076478460) · −0.000230079−0.000230079 − 66

= 1.076481679f (x28) = −0.000138723

x29 = 1.076481679 + (2 − 1.076481679) · −0.0001387233−0.0001387233 − 66

= 1.076483620f (x29) = −0.000083635

x30 = 1.076483620 + (2 − 1.076483620) · −0.000083635−0.000083635 − 66

= 1.076484790f (x30) = −0.000050430

x31 = 1.076484790 + (2 − 1.076484790) · −0.000050430−0.000050430 − 66

= 1.076485496f (x31) = −0.000030391


The reason that the method is not efficient here is that the graph of the function is farfrom linear in the interval under consideration.38

7.2.4 Fourth 2003/2004 Problem Assignment, with Solutions

Distribution Date: Mounted on the Web on Wednesday, March 31st, 2004Solutions were due by Wednesday, March 17th, 2004

Corrected as of April 29th, 2004.(Solutions presented subject to further correction of errors and omissions.)

1. (a) Find, to the nearest unit, the accumulated value 19 years after the first payment ismade of an annuity on which there are 7 payments of 3000 each made at 1 1

2 -yearintervals. The nominal rate of interest convertible semiannually is 6%.

(b) Find, to the nearest unit, the present value of a 20-year annuity-due which pays 200at the beginning of each half-year for the first 8 years, increasing to 250 per half-yearthereafter. The effective annual rate of interest is 6%.

Solution:

(a) We will interpret the payments as being made under an annuity-immediate with time-intervals of 1 1

2 years. The clock starts ticking (i.e. t = 0) 1 12 years before the first

payment; the last payment is made at time t = 7 × 1.5 = 10.5. The evaluation is tobe made at time t = 19 − 1.5 = 17.5, i.e., 7 years after the last payment; this is 14

3intervals of length 1 1

2 years; it is simpler to view this as 14 intervals of length 12 year,

for each of which the effective interest rate is 3%. The effective interest rate — callit j — per 1 1

2 years is j = (1.03)3 − 1 . The accumulated value is, therefore,

3000(1.03)14 · s7 =3000

j· (1.03)14 ·

((1 + j)7 − 1

)

=3000

(1.03)3 − 1· (1.03)14 ·

((1.03)21 − 1

)

= 42100.12386

which is 42,100 to the nearest unit.(b) With 1 + i = (1.06)

12 , v = (1.06)−

12 ,

Present Value = 250a40 − 50a16

= (250 − 50) + 250(1 − v39

i

)− 50

(1 − v15

i

)

38Issues of this type are beyond MATH 329, and are not covered adequately in the current textbook; if you areinterested in Numerical Anaylsis, consider taking MATH 317.


= 200 +250

(1 − v39

)− 50

(1 − v15

)

i

= 200 +200 − 250v39 + 50v15

i= 5342.993032.

To the nearest unit the present value is 5343.

2. (a) The present value of a perpetuity-immediate paying 1 at the end of every 5 years is1.637975. Find i and d.

(b) The present value of a perpetuity-due paying 1 at the beginning of every 5 years is1.637975. Find d and i.

Solution:

(a) A payment of 1 at the end of every 5th year for 5 years is equivalent to a payment ofs−1


1.637975 = s−15· 1

i

⇒ 1.637975 =1

(1 + i)5 − 1⇒ (1 + i)5 = 1.610500⇒ i = 10.0000%,

⇒ d =0.11.1

=1

11.

(b) A payment of 1 at the beginning of every 5th year is equivalent to a payment of a−15

at the beginning of every year. The equation of value is

1.637975 = s−15· 1

d

⇒ 1.637975 =1

1 − (1 + i)−5

⇒ (1 + i)−5 = 0.389490⇒ i = 20.7538%,⇒ d = 15.1703%.

3. Determine the present value, at a nominal interest rate of 6% compounded quarterly, ofthe following payments made under an annuity: 120 at the end of the 3rd year, 110 at theend of the 4th year, decreasing by 10 each year until nothing is paid.


Solution: The payments decrease until a payment of 10 at the end of the 14th year. Wecan think of a decreasing annuity-immediate of value 10 (Da)14 starting with a paymentof 140 at the end of the 1st year, and then make corrections. The effective annual interestrate i is given by

1 + i =

(1 +

0.064

)4

= (1.015)4 .

We need only subtract the present values of the payments due at the end of the first andsecond years.

Present Value = 10(Da)14 − 140v − 130v2

=10

(14 − a14

)

i− 140(1.015)−4 − 130(1.015)−8

=10

(14 − 1−(1.015)−56

(1.015)4−1

)

(1.015)4 − 1− 140(1.015)−4 − 130(1.015)−8

= 532.1438213

4. Find the present value, at an effective annual interest rate of 5.75%, of a perpetuity-immediate under which a payment of 100 is made at the end of the 1st year, 300 at theend of the 2nd year, increasing until a payment of 2500 is made, which level is maintainedfor exactly a total of 10 payments of 2500 (including the first of them in the count of 10),after which the payments fall by 400 each year until they reach a level of 100, which ismaintained in perpetuity. (Note: You are expected to show explicitly how you decomposethe payments; it is not sufficient to simply show a few numbers and a sum.)Solution: The payments reach the level of 2500 at the end of year 13, and continue at thatlevel until the end of year 22, after which they fall by 400 annually until they reach 100at the end of year 28.Since the steady state is a constant perpetuity, we can begin with a perpetuity-immediate

of 100 per year, whose present value is100

0.0575. The remaining non-zero portions of

payments are finite in number: there are additional amounts of 200 at the end of year 2,400 at the end of year 3, . . . , 2400 at the ends of years 13 through 22, 2000 at the end ofyear 23, . . . , and a final amount of 400 at the end of year 27.The value at time t = 22 of the remaining amounts for years 23 through 27 is 400 · (Da)5,so the value at time t = 0 is (1.0575)−22 · 400 · (Da)6. The present value of the remaindersof the payments at the ends of years 2, 3, ..., 12 is (1.0575)−1 · 200 · (Ia)11. Thus the sum

1000.0575

+ (1.0575)−22 · 400 · (Da)5 + (1.0575)−1 · 200 · (Ia)11


covers all the payments except for a level annuity of 2400 payable at the ends of years##13. . . 22, whose present value is

2400(a22 − a12

).

Thus the present value of the entire scheme of payments is

1000.0575

+ (1.0575)−22 · 400 · (Da)5 + (1.0575)−1 · 200 · (Ia)11

+2400(a22 − a12

)

=100

0.0575+ (1.0575)−22 · 400 ·

5 − a5

0.0575+ (1.0575)−1 · 200 ·

a11 − 11v11

0.0575+2400

(a22 − a12

)

=100

0.0575+ (1.0575)−22 · 400 · 5 − 1−(1.0575)−5

0.0575

0.0575

+(1.0575)−1 · 200 ·1.0575−(1.0575)−10

0.0575 − 11(1.0575)−11

0.0575

+2400 · (1.0575)−12 − (1.0575)−22

0.0575= 1739.130435 + 1543.013920 + 8225.826240 + 9138.807264 = 20646.77786 .

5. Find, to the nearest unit, the present value of a 25-year annuity-due which pays 100immediately, 104 at the end of the 1st year, 108.16 at the end of the 2nd year, where eachsubsequent payment is obtained from its predecessor by multiplying by a factor of 1.04.The annual effective rate of interest is 8.%.Solution:

Present Value =

24∑

n=0

100vn(1.04)n

= 100 ·1 −

(1.041.08

)25

1 − 1.041.08

= 1648.998444

or 1649 to the nearest unit.

6. (a) A loan of 15,000 is being repaid with payments of 1,500 at the end of each yearfor 20 years. If each payment is immediately reinvested at 6% effective, find theeffective annual rate of interest earned over the 20-year period.



(b) A loan of 15,000 is being repaid with payments of 1,500 at the end of each year for10 years. Determine the yield rate to the investor.

Solution:

(a) Let the effective yield rate be i. The payments do not become available until thematurity date, after 20 years. Until that time they are locked into a payment-schemethat accumulates to value 1500s20 6%. We are asked for the interest rate that wasearned. There are thus just two transactions: the loan at time 0, in the amount of15,000, and the repayment at time 20, in the amount given above. The equation ofvalue at time t = 0 is

1500s20 6%(1 + i)−20 = 15000

⇔ (1 + i)−20 =10(0.06)

(1.06)20 − 1⇔ i = 6.7293555%.

(b) We have to determine i such that 1500 · a10 i = 15000. This may appear to be adifficult problem. But remember that

a10 = v + v2 + v3 + . . . + v10

and that v ≤ 1. That means that the sum cannot be more than 10; in order for it toequal 10, each of the summands must equal 1, so v = 1. 1 + i = 1, and i = 0.

7.2.5 Fifth 2003/2004 Problem Assignment, with Solutions

Distribution Date: Mounted on the Web on Monday, April 7th, 2004.Assignment was mounted on the Web on Wednesday, March 17th, 2004,hard copy of the assignment was distributed on Friday, March 19th, 2004.

Solutions were to be submitted by Friday, April 2nd, 2004(SUBJECT TO CORRECTION OF TYPO’S AND OTHER ERRORS)

1. A loan is being repaid with instalments of 1000 at the end of each year for 15 years,followed by payments of 2000 at the end of each year for 10 years. Interest is at aneffective rate of 4% for the first 10 years, and an effective rate 6% for the next 15 years.

(a) Showing all your work, find the numeric value of the amount of interest paid in the4th instalment without making use of a schedule.

(b) Showing all your work, find the amount of principal repaid in the 20th instalment,without making use of a schedule.

(c) Then use the information you have computed to compile the lines of a schedulecorresponding to the payments at the ends of years 20, 21, . . . , 25.


(d) Now solve (a), (b), (c) again, this time assuming that all payments are at the begin-nings of the years: the interest rates remain precisely the same.

Solution:

(a) By the prospective method, the unpaid balance of the loan at time t = 0 is

1000 · a10 4% + (1.04)−10(2000 · a15 6% − 1000 · a5 6%

)

= 1000 · 1 − (1.04)−10

0.04

+2000 · (1.04)−10 · 1 − (1.06)−15

0.06− 1000 · (1.04)−10 · 1 − (1.06)−5

0.06= 18387.66857 .

By the retrospective method, the unpaid balance just after the 3rd payment of 1000is, therefore,

18387.66857(1.04)3 − 1000 · s3 4% = 18387.66857(1.04)3 − 1000 · (1.04)3 − 10.04

= 17562.02642 .

The interest component of the 4th instalment is, therefore,

0.04(17562.02642) = 702.48 .

(b) By the prospective method, the unpaid balance just after the 19th instalment is

2000 · a6 6% = 2000 · 1 − (1.06)−6

0.06= 9834.648653 .


0.06(9834.648653) = 590.0789192 ,

so the component for reduction of principal is

2000 − 590.0789192 = 1409.921081 .

(c)

Payment Payment Interest Principal OutstandingNumber amount paid repaid loan balance

19 2000 . . . . . . 9834.6520 2000 590.08 1409.92 8424.7321 2000 505.48 1494.52 6930.2122 2000 415.81 1584.19 5346.0223 2000 320.76 1679.23 3666.7824 2000 220.01 1779.99 1886.7925 2000 113.21 1886.79 0


(d) Since the only interest rate affecting the last payments has not changed, there will beno changes at all in (b) or (c).By the prospective method, the unpaid balance of the loan at time t = 0, just beforethe first payment, is

1000 · a10 4% + (1.04)−10(2000 · a15 6% − 1000 · a5 6%

)

= 1000(1.04) · 1 − (1.04)−10

0.04+ 2000(1.06) · (1.04)−10 · 1 − (1.06)−15

0.06

−1000(1.06) · (1.04)−10 · 1 − (1.06)−5

0.06= 19328.71077 .

By the retrospective method, the unpaid balance just after the 3rd payment of 1000is, therefore,

19328.71077(1.04)2 − 1000 · s3 4%

= 19328.71077(1.04)2 − 1000 · (1.04)3 − 10.04

= 17784.33357 .


0.04(17784.33357) = 711.37 .

2. (This problem is modelled on [6, Exercise 23, p. 198].)On a loan of 30,000, the borrower has agreed to pay interest at 7% effective at the endof each year until the loan is repaid. The borrower has decided to deposit a fixed amountat the beginning of each year into a sinking fund earning 4% effective. At the end of11 years the sinking fund is exactly sufficient to pay off exactly two-thirds of the loan.He plans to continue accumulating the sinking fund until a year when a deposit of notmore than this fixed amount will bring the fund balance up to 30,000 and the loan can beimmediately repaid.

(a) Calculate the total amount the borrower has to pay out each year (at the beginning,and at the end), except possibly in the year when the loan is repaid.

(b) Complete the following table to show how the sinking fund attains the target valueof 30,000, and the net amount of the loan after payments ##10, 11, . . . until the loanis paid off.


Payment Interest Sinking Interest earned Amount in Net amountNumber paid fund deposit on sinking fund sinking fund of loan

101112. . .

Solution:

(a) The borrower is paying a constant amount each year of 7% of 30,000, or 2,100 tocover the interest costs of servicing the loan; denote by X the constant amount theborrower spends each year; thus the amount she contributes to the sinking fund isX − 2100. An equation of value at time t = 11 is (X − 2100) · s11 4% = 20000, so

X = 2100 +20000 × 0.04(

(1.04)11 − 1)

1.04= 3525.943064 .

Thus the level contribution to the sinking fund at the beginning of each year exceptthe last is 3525.94 − 2100.00 = 1425.94.

(b) The intention is that the column labelled “Amount in sinking fund” shows the amountjust after the payment with the given number.


10 2,100.00 1,425.94 603.62 17,120.03 12,879.9711 2,100.00 1,425.94 684.80 19,230.77 10,769.2312 2,100.00 1,425.94 769.23 21,425.64 8,574.3613 2,100.00 1,425.94 857.03 23,708.61 6,291.3914 2,100.00 1,425.94 948.34 26,082.89 3,917.1115 2,100.00 1,425.94 1,043.32 28,552.15 1,447.8516 2,100.00 305.76 1,142.09 30,000.00 0.00

3. A borrows 12,000 for 10 years, and agrees to make semiannual payments of 1,000, plus afinal payment. The lender receives 12% convertible semiannually on the investment eachyear for the first 5 years and 10% convertible semiannually for the second 5 years. Thebalance of each payment is invested in a sinking fund earning 8% convertible semiannu-ally.

(a) Find the amount by which the sinking fund is short of repaying the loan at the endof the 10 years.

(b) Complete the following table to show how the sinking fund attains its maximumvalue, and the net amount of the loan after payments.



09101112

Solution:

(a) The interest payments for the first 10 half-years are 6% of 12,000, i.e. 720 per half-year; and, for the second 10 half-years, 600 per half-year. This leaves 280 at theend of each of the first 10 half-years, and 400 at the end of each of the second 10half-years to accumulate in the sinking fund, which earns 4% effective every halfyear. The accumulated balance in the sinking fund at maturity will be

120s10 4% + 280s20 4% =1

0.04

(120

((1.04)10 − 1

)+ 280

((1.04)20 − 1

))

= 25(120(1.04)10 + 280(1.04)20 − 400

)

= 9778.594855

implying that the shortfall to repay the loan will be 12, 000 − 9778.59 = 2221.41.(b) Just after the 8th payment the balance in the fund is

280 · s8 4% =280

((1.04)8 − 1

)

0.04= 2579.98 .

Interest earned by the 9th payments is 103.20. We can now begin to fill in theschedule:


09 720.00 280.00 103.20 2963.18 9036.8210 720.00 280.00 118.53 3361.71 8638.2911 600.00 400.00 134.47 3896.18 8103.8212 600.00 400.00 155.85 4452.03 7547.97

4. (a) A borrower takes out a loan of 3000 for 10 years at 8% convertible semiannually. Theborrower replaces one-third of the principal in a sinking fund earning 5% convertiblesemiannually, and the other two-thirds in a sinking fund earning 7% convertiblesemiannually. Find the total semiannual payment.

(b) Rework (a) if the borrower each year puts one-third of the total sinking fund depositinto the 5% sinking fund and the other two-thirds into the 7% sinking fund.

Solution:


(a) The semiannual contribution to the sinking funds is

1000s20 2.5%

+2000

s20 3.5%

and the semiannual interest payment is 4% of 3, 000, or 120. Hence the total semi-annual payment is

1000s20 2.5%

+2000

s20 3.5%

+ 120 =25

(1.025)20 − 1+

70(1.035)20 − 1

+ 120

= 229.8692824

(b) Let the total sinking fund deposit be D. Then the equation of value at maturity is

D3· s20 2.5% +

2D3· s20 3.5% = 3000 ,

implying that

D =9000

s20 2.5% + 2s20 3.5%

=9000

(1.025)20−10.025 + 2 · (1.035)20−1

0.035

= 109.6170427 ,

so the total semi-annual payment is 109.6170427+120=229.6170427.

5. A payment of 800 is made at the end of each month for 10 years to repay a loan of 30,000.The borrower replaces the capital by means of a sinking fund earning a nominal annualrate of 6% compounded monthly.

(a) Find the effective annual rate i paid to the lender on the loan.(b) Suppose that the lender had elected to amortize the loan by equal monthly payments,

at the same rate as he is now paying to the lender. What would be the amount of thoseequal payments?

Solution:

(a) The monthly contribution to the sinking fund is

30000s120 0.5%

=600

(1.005)120 − 1= 183.0615058 .

Hence the monthly interest payment is 800 − 183.0615058 = 616.9384942, i.e.,2.056461647% of the principal of 30,000. The effective annual rate is, therefore,

(1.02056461647)12 − 1 = 27.6691838%.


(b) The level monthly payment necessary to amortize a loan of 30000 over 120 paymentsat an effective rate of 2.056461647% is

30000a120 2.056461647%

=(30000)(0.02056461647)1 − (1.02056461647)−120 = 675.6703969 .

(The amount is lower than the borrower is now paying because he is holding fundsin his sinking fund and earning less there than he his paying the lender for the use ofthe capital.)

7.2.6 2003/2004 Class Test, Version 1

Instructions


• This test booklet consists of this cover, Pages 3074 through 3075 containing questionstogether worth 60 marks; and Page 3075, which is blank.

• Show all your work. All solutions are to be written in the space provided on the pagewhere the question is printed. When that space is exhausted, you may write on the facingpage, on the blank page, or on the back cover of the booklet, but you must indicateany continuation clearly on the page where the question is printed! (Please inform theinstructor if you find that your booklet is defective.)


• Calculators. While you are permitted to use a calculator to perform arithmetic and/orexponential calculations, you must not use the calculator to calculate such actuarial func-tions as ani, sni, (Ia)ni, (Is)ni, (Da)ni, (Ds)ni, etc. without first stating a formula for thevalue of the function in terms of exponentials and/or polynomials involving n and theinterest rate. You must not use your calculator in any programmed calculations. If yourcalculator has memories, you are expected to have cleared them before the test.

• In your solutions to problems on this test you are expected to show all your work. Youare expected to simplify algebraic and numerical answers as much as you can.


1. (a) [5 MARKS] Suppose that the nominal annual rate of interest, compounded 8 timesper year, is 5%. Showing all your work, determine the equivalent effective annualrate of discount.

(b) [5 MARKS] Suppose that the nominal annual rate of discount, compounded 3 timesper year, is 7%. Showing all your work, determine the equivalent annual rate ofinterest.


(c) [5 MARKS] State the nominal annual interest rate, compounded instantaneously,which is equivalent to an effective annual interest rate of 4%.

(d) [5 MARKS] Suppose that the effective interest rate for14

year is 3%. Determine theequivalent nominal interest rate, compounded every 2 years.

2. [15 MARKS] The accumulated value just after the last payment under a 12-year annuityof 1000 per year, paying interest at the rate of 5% per annum effective, is to be usedto purchase a perpetuity at an interest rate of 6%, first payment to be made 1 year afterthe last payment under the annuity. Showing all your work, determine the size of thepayments under the perpetuity.

3. [25 MARKS] A loan of 5000 is to be repaid by annual payments of 250 to commenceat the end of the 6th year, and to continue thereafter for as long as necessary. Find thetime and amount of the final payment if the final payment is to be larger than the regularpayments. Assume i = 4%.




Instructions




• All your writing — even rough work — must be handed in.• Calculators. While you are permitted to use a calculator to perform arithmetic and/or

exponential calculations, you must not use the calculator to calculate such actuarial func-tions as ani, sni, (Ia)ni, (Is)ni, (Da)ni, (Ds)ni, etc. without first stating a formula for thevalue of the function in terms of exponentials and/or polynomials involving n and theinterest rate. You must not use your calculator in any programmed calculations. If yourcalculator has memories, you are expected to have cleared them before the test.




1. [25 MARKS] A loan of 1000 is to be repaid by annual payments of 100 to commence atthe end of the 5th year, and to continue thereafter for as long as necessary. Find the timeand amount of the final payment if the final payment is to be NO larger than the regularpayments. Assume i = 4.5%.

2. (a) [5 MARKS] Suppose that the effective interest rate for15

year is 0.02. Determinethe equivalent nominal interest rate, compounded every 3 years.

(b) [5 MARKS] Suppose that the nominal annual rate of interest, compounded 9 timesper year, is 6%. Showing all your work, determine the equivalent effective annualrate of discount.

(c) [5 MARKS] Suppose that the nominal annual rate of discount, compounded 6 timesper year, is 5%. Showing all your work, determine the equivalent annual rate ofinterest.

(d) [5 MARKS] State the nominal annual interest rate, compounded instantaneously,which is equivalent to an effective annual interest rate of 8%.

3. [15 MARKS] The accumulated value just after the last payment under a 12-year annuityof 1000 per year, paying interest at the rate of 5% per annum effective, is to be used topurchase a perpetuity of 500 per annum forever, first payment to be made 1 year after thelast payment under the annuity. Showing all your work, determine the effective interestrate of the perpetuity, assuming it comes into effect just after the last payment under theannuity.


Instructions









1. [15 MARKS] The accumulated value just after the last payment under a 9-year annuityof 2000 per year, paying interest at the rate of 8% per annum effective, is to be usedto purchase a perpetuity at an interest rate of 4%, first payment to be made 1 year afterthe last payment under the annuity. Showing all your work, determine the size of thepayments under the perpetuity.

2. [25 MARKS] A loan of 1000 is to be repaid by annual payments of 200 to commenceat the end of the 4th year, and to continue thereafter for as long as necessary. Find thetime and amount of the final payment if the final payment is to be larger than the regularpayments. Assume i = 5%.

3. (a) [5 MARKS] State the nominal annual interest rate, compounded instantaneously,which is equivalent to an effective annual interest rate of 6%.

(b) [5 MARKS] Suppose that the effective interest rate for16

year is 0.015. Determinethe equivalent nominal interest rate, compounded every 4 years.

(c) [5 MARKS] Suppose that the nominal annual rate of interest, compounded 3 timesper year, is 8%. Showing all your work, determine the equivalent effective annualrate of discount.

(d) [5 MARKS] Suppose that the nominal annual rate of discount, compounded 12 timesper year, is 6%. Showing all your work, determine the equivalent annual rate ofinterest.





Instructions








1. (a) [5 MARKS] Suppose that the nominal annual rate of discount, compounded 6 timesper year, is 0.5%. Showing all your work, determine the equivalent annual rate ofinterest.

(b) [5 MARKS] State the nominal annual interest rate, compounded instantaneously,which is equivalent to an effective annual interest rate of 10%.

(c) [5 MARKS] Suppose that the effective interest rate for15

year is 2%. Determine theequivalent nominal interest rate, compounded every 6 years.

(d) [5 MARKS] Suppose that the nominal annual rate of interest, compounded 7 timesper year, is 2%. Showing all your work, determine the equivalent effective annualrate of discount.

2. [25 MARKS] A loan of 1000 is to be repaid by annual payments of 200 to commence atthe end of the 4th year, and to continue thereafter for as long as necessary. Find the timeand amount of the final payment if the final payment is to be NO larger than the regularpayments. Assume i = 5%.


3. [15 MARKS] The accumulated value just after the last payment under a 10-year annuityof 1000 per year, paying interest at the rate of 6% per annum effective, is to be used topurchase a perpetuity of 800 per annum forever, first payment to be made 1 year after thelast payment under the annuity. Showing all your work, determine the effective interestrate of the perpetuity, assuming it comes into effect just after the last payment under theannuity.



7.2.10 Solutions to Problems on the 2003/2004 Class Tests

The first four problems listed are concerned with equivalent rates of interest and discount (eachin 4 parts); the next four concern annuities and perpetuities; and the last four are concernedwith unknown type and final balloon or drop payments.

1. (a) [5 MARKS] [VERSION 1 #1(a)] Suppose that the nominal annual rate of interest,compounded 8 times per year, is 5%. Showing all your work, determine the equiva-lent effective annual rate of discount.Solution: We are given that i(8) = 0.05, and asked to determine d.

(1 +

i(8)

8

)8

= 1 + i

and (1 − d)(1 + i) = 1

⇒ d = 1 −(1 +

i(8)

8

)−8

= 0.0486225508 = 4.86%

(b) [5 MARKS] [VERSION 1 #1(b)] Suppose that the nominal annual rate of discount,compounded 3 times per year, is 7%. Showing all your work, determine the equiva-lent annual rate of interest.Solution: We are given that d(3) = 0.07, and asked to determine i.

(1 − d(3)

3

)3

= 1 − d

and (1 − d)(1 + i) = 1

⇒ i =

(1 − d(3)

3

)−3

− 1 = 0.073398300 = 7.34%


(c) [5 MARKS] [VERSION 1 #1(c)] State the nominal annual interest rate, compoundedinstantaneously, which is equivalent to an effective annual interest rate of 4%.Solution: The rate we seek is the solution i to the equation

limn→∞

(1 +

in

)n

= 1.04

which is equivalent to ei = 1.04. Solving by taking logarithms, we obtain i =

ln(1.04) = 0.039220713151 = 3.92%.

(d) [5 MARKS] [VERSION 1 #1(d)] Suppose that the effective interest rate for14

yearis 3%. Determine the equivalent nominal interest rate, compounded every 2 years.

Solution: We are given the value ofi(4)

4= 3%, and asked to determine i( 1

2 ). Theequation we have to solve is

(1 +

i(4)

4

)4

= 1 + i =

1 +i(

12 )12

12

.

This equation implies that

i(12 ) =

12

(1 +

i(4)

4

)8

− 1

= 0.1333850405 = 13.3%.

2. (a) [5 MARKS] [VERSION 2 #2(b)] Suppose that the nominal annual rate of interest,compounded 9 times per year, is 6%. Showing all your work, determine the equiva-lent effective annual rate of discount.Solution: We are given that i(9) = 0.06, and asked to determine d.

(1 +

i(9)

9

)9

= 1 + i

and (1 − d)(1 + i) = 1

⇒ d = 1 −(1 +

i(9)

9

)−9

= 0.0580479306 = 5.80%

(b) [5 MARKS] [VERSION 2 #2(c)] Suppose that the nominal annual rate of discount,compounded 6 times per year, is 5%. Showing all your work, determine the equiva-lent annual rate of interest.Solution: We are given that d(6) = 0.05, and asked to determine i.

(1 − d(6)

6

)6

= 1 − d

and (1 − d)(1 + i) = 1

⇒ i =

(1 − d(6)

6

)−6

− 1 = 0.051491358 = 5.15%


(c) [5 MARKS] [VERSION 2 #2(d)] State the nominal annual interest rate, compoundedinstantaneously, which is equivalent to an effective annual interest rate of 8%.Solution: The rate we seek is the solution to the equation

limn→∞

(1 +

in

)n

= 1.08


ln(1.08) = 0.07696104114 = 7.70%.

(d) [5 MARKS] [VERSION 2 #2(a)] Suppose that the effective interest rate for15

yearis 0.02. Determine the equivalent nominal interest rate, compounded every 3 years.




(1 +

i(5)

5

)5

= 1 + i =

1 +i(

13 )13

13

.


i(13 ) =

13

(1 +

i(5)

5

)15

− 1

= 0.1152894460 = 11.5%

3. (a) [5 MARKS] [VERSION 3 #3(c)] Suppose that the nominal annual rate of interest,compounded 3 times per year, is 8%. Showing all your work, determine the equiva-lent effective annual rate of discount.Solution: We are given that i(3) = 0.08, and asked to determine d.

(1 +

i(3)

3

)3

= 1 + i

and (1 − d)(1 + i) = 1

⇒ d = 1 −(1 +

i(3)

3

)−3

= 0.0759156521 = 7.59%

(b) [5 MARKS] [VERSION 3 #3(d)] Suppose that the nominal annual rate of discount,compounded 12 times per year, is 6%. Showing all your work, determine the equiv-alent annual rate of interest.Solution: We are given that d(12) = 0.06, and asked to determine i.

(1 − d(12)

12

)12

= 1 − d


and (1 − d)(1 + i) = 1

⇒ i =

(1 − d(12)

12

)−12

− 1 = 0.061996367 = 6.20%

(c) [5 MARKS] [VERSION 3 #3(a)] State the nominal annual interest rate, compoundedinstantaneously, which is equivalent to an effective annual interest rate of 6%.Solution: The rate we seek is the solution to the equation

limn→∞

(1 +

in

)n

= 1.06


ln(1.06) = 0.05826890812 = 5.83%.

(d) [5 MARKS] [VERSION 3 #3(b)] Suppose that the effective interest rate for16

yearis 0.015. Determine the equivalent nominal interest rate, compounded every 4 years.


6= 1.5%, and asked to determine i( 1


(1 +

i(6)

6

)6

= 1 + i =

1 +i(

14 )14

14

.


i(14 ) =

14

(1 +

i(6)

6

)24

− 1

= .1073757030 = 10.7%

4. (a) [5 MARKS] [VERSION 4 #1(d)] Suppose that the nominal annual rate of interest,compounded 7 times per year, is 2%. Showing all your work, determine the equiva-lent effective annual rate of discount.Solution: We are given that i(7) = 0.02, and asked to determine d.

(1 +

i(7)

7

)7

= 1 + i

and (1 − d)(1 + i) = 1

⇒ d = 1 −(1 +

i(7)

7

)−7

= 0.197733746 = 1.98%

(b) [5 MARKS] [VERSION 4 #1(a)] Suppose that the nominal annual rate of discount,compounded 6 times per year, is 0.5%. Showing all your work, determine the equiv-alent annual rate of interest.


Solution: We are given that d(6) = 0.005, and asked to determine i.

(1 − d(6)

6

)6

= 1 − d

and (1 − d)(1 + i) = 1

⇒ i =

(1 − d(6)

6

)−6

− 1 = 0.005014616 = 0.5014616%

(c) [5 MARKS] [VERSION 4 #1(b)] State the nominal annual interest rate, compoundedinstantaneously, which is equivalent to an effective annual interest rate of 10%.Solution: The rate we seek is the solution to the equation

limn→∞

(1 +

in

)n

= 1.10


ln(1.10) = 0.09531017980 = 9.53%.

(d) [5 MARKS] [VERSION 4 #1(c)] Suppose that the effective interest rate for15

yearis 2%. Determine the equivalent nominal interest rate, compounded every 6 years.




(1 +

i(5)

5

)5

= 1 + i =

1 +i(

16 )16

16

.


i(16 ) =

16

(1 +

i(5)

5

)30

− 1

= 0.1352269307 = 13.5%.

5. [15 MARKS] [VERSION 1 #2] The accumulated value just after the last payment undera 12-year annuity of 1000 per year, paying interest at the rate of 5% per annum effective,is to be used to purchase a perpetuity at an interest rate of 6%, first payment to be made 1year after the last payment under the annuity. Showing all your work, determine the sizeof the payments under the perpetuity.Solution: Let X be the level payment under the perpetuity. The equation of value justafter the last annuity payment is

1000 · s12 5% = X · a∞ 6%


implying that

X = 1000 ·s12 5%

a∞ 6%

= 1000 · 65·((1.05)12 − 1

)= 955.03.

6. [15 MARKS] [VERSION 2 #3] The accumulated value just after the last payment undera 12-year annuity of 1000 per year, paying interest at the rate of 5% per annum effective,is to be used to purchase a perpetuity of 500 per annum forever, first payment to be made1 year after the last payment under the annuity. Showing all your work, determine theeffective interest rate of the perpetuity, assuming it comes into effect just after the lastpayment under the annuity.Solution: Let i be the interest rate of the perpetuity. The equation of value just after thelast annuity payment is

1000 · s12 5% = 500 · a∞ i =500

iimplying that

i =500

1000· 1

s12 5%

=500

1000· 0.05(

(1.05)12 − 1)

= 3.14%.

7. [15 MARKS] [VERSION 3 #1] The accumulated value just after the last payment undera 9-year annuity of 2000 per year, paying interest at the rate of 8% per annum effective,is to be used to purchase a perpetuity at an interest rate of 4%, first payment to be made 1year after the last payment under the annuity. Showing all your work, determine the sizeof the payments under the perpetuity.Solution: Let X be the level payment under the perpetuity. The equation of value justafter the last annuity payment is

2000 · s9 8% = X · a∞ 4%

implying that

X = 2000 ·s9 8%

a∞ 4%

= 2000 · 48·((1.08)9 − 1

)= 999.00


8. [15 MARKS] [VERSION 4 #3] The accumulated value just after the last payment undera 10-year annuity of 1000 per year, paying interest at the rate of 6% per annum effective,is to be used to purchase a perpetuity of 800 per annum forever, first payment to be made1 year after the last payment under the annuity. Showing all your work, determine theeffective interest rate of the perpetuity, assuming it comes into effect just after the lastpayment under the annuity.Solution: Let i be the interest rate of the perpetuity. The equation of value just after thelast annuity payment is

1000 · s10 6% = 800 · a∞ i =800

iimplying that

i =800

1000· 1

s10 6%

=800

1000· 0.06(

(1.06)10 − 1)

= 6.07%.

9. [25 MARKS] [VERSION 1 #3] A loan of 5000 is to be repaid by annual payments of250 to commence at the end of the 6th year, and to continue thereafter for as long asnecessary. Find the time and amount of the final payment if the final payment is to belarger than the regular payments. Assume i = 4%.Solution: (cf. [6, Exercise 32, p. 91]) Let the time of the last — balloon — payment ben, and let the amount of the last payment be X. Then n is the largest integer solution tothe inequality

5000 ≥ 250(1.04)−5an−5 = 250(1.04)−5 · 1 − (1.04)−(n−5)

0.04

⇔ (1.04)−(n−5) ≥ 1 − 5000 × 0.04 × (1.04)5

250⇔ −(n − 5) ln 1.04 ≥ ln

(1 − 20 × 0.04 × (1.04)5

)

⇔ −(n − 5) ≥ln

(1 − 0.8(1.04)5

)

ln 1.04

⇔ n ≤ 5 −ln

(1 − 0.8(1.04)5

)

ln 1.04= 97.39832188.


5000(1.04)97 = 250s92 + (X − 250)


implying that

X = 250 + 5000(1.04)97 − 2500.04

((1.04)92 − 1

)= 346.8818 .

10. [25 MARKS] [VERSION 2 #1] A loan of 1000 is to be repaid by annual payments of100 to commence at the end of the 5th year, and to continue thereafter for as long asnecessary. Find the time and amount of the final payment if the final payment is to beNO larger than the regular payments. Assume i = 4.5%.Solution: Let the time of the last — drop — payment be n, and let the amount of the lastpayment be X. Then n is the smallest integer solution to the inequality

1000 ≤ 100(1.045)−4an−4 = 100(1.045)−4 · 1 − (1.045)−(n−4)

0.045

⇔ (1.045)−(n−4) ≤ 1 − 1000 × 0.045 × (1.045)4

100⇔ −(n − 4) ln 1.045 ≤ ln

(1 − 10 × 0.045 × (1.045)4

)

⇔ −(n − 4) ≤ln

(1 − 0.45(1.045)4

)

ln 1.045

⇔ n ≥ 4 −ln

(1 − 0.45(1.045)4

)

ln 1.045= 21.47594530.

Thus we conclude that the drop payment is made at time t = 22. The equation of valueat time t = 22 is

1000(1.045)22 = 100s17 + X

implying that

X = 1000(1.045)22 − 100 × 1.0450.045

((1.045)17 − 1

)= 48.143638 .

11. [25 MARKS] [VERSION 3 #2] A loan of 1000 is to be repaid by annual payments of200 to commence at the end of the 4th year, and to continue thereafter for as long asnecessary. Find the time and amount of the final payment if the final payment is to belarger than the regular payments. Assume i = 5%.Solution: (cf. [6, Exercise 32, p. 91]) Let the time of the last — balloon — payment ben, and let the amount of the last payment be X. Then n is the largest integer solution tothe inequality

1000 ≥ 200(1.05)−3an−3 = 200(1.05)−3 · 1 − (1.05)−(n−3)

0.05


⇔ (1.05)−(n−3) ≥ 1 − 1000 × 0.05 × (1.05)3

200

⇔ −(n − 3) ln 1.05 ≥ ln(1 − 1000 × 0.05 × (1.05)3

200

)

⇔ −(n − 3) ≥ln

(1 − 0.25(1.05)3

)

ln 1.05

⇔ n ≤ 3 −ln

(1 − 0.25(1.05)3

)

ln 1.05= 10.00252595


1000(1.05)10 = 200s7 + (X − 200)

implying that

X = 200 + 1000(1.05)10 − 2000.05

((1.05)7 − 1

)= 200.492935 .

12. [25 MARKS] [VERSION 4 #2] A loan of 1000 is to be repaid by annual payments of200 to commence at the end of the 4th year, and to continue thereafter for as long asnecessary. Find the time and amount of the final payment if the final payment is to beNO larger than the regular payments. Assume i = 5%.Solution: Let the time of the last — drop — payment be n, and let the amount of the lastpayment be X. Then n is the smallest integer solution to the inequality

1000 ≤ 200(1.05)−3an−3 = 200(1.05)−3 · 1 − (1.05)−(n−3)

0.05

⇔ (1.05)−(n−3) ≤ 1 − 1000 × 0.05 × (1.05)3

200⇔ −(n − 3) ln 1.05 ≤ ln

(1 − 5 × 0.05 × (1.05)3

)

⇔ −(n − 3) ≤ln

(1 − 0.25(1.05)3

)

ln 1.05

⇔ n ≥ 3 −ln

(1 − 0.25(1.05)3

)

ln 1.05= 10.00252595.

Thus we conclude that the drop payment is made at time t = 11. The equation of valueat time t = 11 is

1000(1.05)11 = 200s7 + X

implying that

X = 1000(1.05)11 − 200 × 1.050.05

((1.05)7 − 1

)= 0.517581.


7.2.11 Final Examination, 2003/2004

1. In each of the following problems you are expected to show all your work.

(a) [3 MARKS] If v = 0.97, determine the value of d(4).(b) [3 MARKS] If i(12) = 6%, determine the value of i( 1

2 ).(c) [3 MARKS] Showing all your work, determine the nominal interest rate, com-

pounded semi-annually, under which a sum of money will triple in 12 years.(d) [3 MARKS] Showing all your work, determine the rate of interest, convertible con-

tinuously, that is equivalent to a nominal interest rate of 8% per annum, convertiblemonthly.

2. In each of the following problems, give a formula in terms of i alone; then evaluate theformula and determine the numerical value.

(a) [5 MARKS] Determine the present value of a perpetuity-due of 100 payable everythree months, at an effective annual interest rate of 6%.

(b) [5 MARKS] The present value of a perpetuity-immediate paying 1000 at the end of

every 3 years is125, 000

91.Determine the effective annual interest rate.

3. In each of the following problems, give a formula in terms of i alone; then evaluate theformula and determine its numerical value.

(a) [7 MARKS] At a nominal annual interest rate of 8% compounded quarterly, deter-mine the value, 2 years after the last payment, of a decreasing annuity paying 5,000at the end of the first half-year, 4,500 at the end of the 2nd half-year, and continuingto decrease at 500 per half-year until the final payment of 500.

(b) [8 MARKS] Three years before the first payment, determine the value of an annuitythat pays 4,000 the first year, 3,900 the second year, with payments continuing todecrease by 100 until it pays 2,000 per year, after which it pays 2,000 forever. Theinterest rate is 5% effective per year until the payment of 3,000, after which theinterest rate becomes 4% effective forever.

4. One of the following equations is always true, and one is true only when i = 0.

I.1

an i=

1sn i

+ i

II.1

sn i=

1an i

+1i

III.1

an i=

1sn i

+1i

IV.1

sn i=

1an i

+ i


Table 4: Several Useful Formulas that you were not expected to memorize

(Ia)n i =an i−nvn

i

(Is)n i =sn i−n

i

(Is)n i =sn+1 i

−(n+1)

i

(Da)n i =n−an i

i

(Ds)n i =n(1+i)n−sn i

i

(a) Explain which is always true, and prove iti. [4 MARKS] algebraically; and

ii. [4 MARKS] by a verbal argument, referring to a sinking fund.(b) [4 MARKS] Prove algebraically that one of the other equations is true for i = 0.

5. The purchase of a new condominium is partially financed by a mortgage of 120,000payable to the vendor; the mortgage is amortized over 25 years, with a level payment atthe end of each half-month, at a nominal annual rate of 6.6% compounded every half-month.

(a) [3 MARKS] Determine the half-monthly payments under this mortgage.(b) [2 MARKS] Divide the 1st payment into principal and interest.(c) [3 MARKS] Determine the outstanding principal immediately after the 50th pay-

ment.(d) [4 MARKS] Divide the 52nd payment into principal and interest.(e) [3 MARKS] The amortization by half-monthly payments was designed to accom-

modate the purchase, whose salary was being deposited automatically to his bankaccount every half-month. The purchase changes his profession, after 2 years, andnow would prefer to make a single payment once every half-year. Determine theamount of that payment if the interest rates are unchanged, but if the mortgage isnow amortized to be paid off 5 years earlier than previously.

6. (a) [8 MARKS] Find the price of the following bond, which is purchased to yield 6%convertible semi-annually: the bond has face value of 10,000, matures in 15 yearsat a maturity value of 11,500, and has a nominal coupon rate of 9% per annum,


compounded semi-annually; the investor is replacing the principal by means of asinking fund earning 7% convertible semi-annually.

(b) [7 MARKS] Suppose that the bond is callable when t = 13 at a premium of 1, 000above the maturity value. Explain what price the investor should pay if he is nolonger plans to deposit any of the interest in a sinking fund.

7. A loan is being repaid with 15 annual payments of 1,000 each. At the time of the 5thpayment the borrower is permitted to pay an extra 2000, and then to repay the balanceover 5 years with a revised annual payment.

(a) [5 MARKS] If the effective annual rate of interest is 6%, find the amount of therevised annual payment.

(b) [8 MARKS] Complete an amortization table for the full 10 years of the loan, withthe following columns:

Payment Payment Interest Principal Outstandingnumber amount paid repaid loan balance

01

. . . . . . . . . . . . . . .10

8. [8 MARKS] It was n years ago when James deposited 10,000 in a bank paying 2.4%interest compounded monthly. If he had, instead, placed his deposit in a syndicate payinginterest by cheque annually at the rate of 5% per annum, and he had invested only thisinterest with the bank, how much more interest would he have earned altogether? Showall your reasoning, and express your answer in terms of n.

7.2.12 Supplemental/Deferred Examination, 2003/2004

1. In each of the following problems you are expected to show all your work.

(a) [3 MARKS] If v = 0.95, determine the value of d(3).(b) [3 MARKS] Showing all your work, determine the nominal interest rate, com-

pounded quarterly, under which a sum of money will double in 10 years.(c) [3 MARKS] Showing all your work, determine the rate of discount, convertible con-

tinuously, that is equivalent to a nominal discount rate of 8% per annum, convertiblesemi-annually.

(d) [3 MARKS] If i(12 ) = 1

30 , determine the value of i(12).

2. You must show all your work in solving the following problems:

(a) [5 MARKS] Determine the present value of a perpetuity-immediate of 1000 payableevery three months, at an effective annual interest rate of 8%.


Table 5: Several Useful Formulas that you were not expected to memorize

(Ia)n i =an i−nvn

i

(Is)n i =sn i−n

i

(Is)n i =sn+1 i

−(n+1)

i

(Da)n i =n−an i

i

(Ds)n i =n(1+i)n−sn i

i

(b) [5 MARKS] The present value of a perpetuity-immediate paying 100 at the end of

every 4 years is129, 6001, 105

. Determine the effective annual interest rate.

3. Show detailed work in your solutions to each of these problems.

(a) [7 MARKS] At a nominal annual interest rate of 6% compounded quarterly, deter-mine the value — 5 years after the last payment — of a decreasing annuity paying6,000 at the end of the first half-year, 5,500 at the end of the 2nd half-year, andcontinuing to decrease at 500 per half-year until the final payment of 500.

(b) [8 MARKS] Three years before the first payment, determine the present value of anannuity that pays 6,000 the first year, 5,900 the second year, with payments continu-ing to decrease by 100 until it pays 4,000 per year, after which it pays 4,000 forever.The interest rate is 8% effective per year until the first payment of 4,000, after whichthe interest rate becomes 5% effective forever.

4. The purchase of a new condominium is partially financed by a mortgage of 120,000payable to the vendor; the mortgage is amortized over 35 years, with a level payment atthe end of each half-month, at a nominal annual rate of 9.6% compounded every half-month.

(a) [3 MARKS] Determine the half-monthly payments under this mortgage.(b) [2 MARKS] Divide the 1st payment into principal and interest.(c) [3 MARKS] Determine the outstanding principal immediately after the 60th pay-

ment.(d) [4 MARKS] Divide the 62nd payment into principal and interest.


(e) [3 MARKS] The amortization by half-monthly payments was designed to accom-modate the purchaser, whose salary was being deposited automatically to his bankaccount every half-month. The purchaser changes his profession 4 years after themortgage is executed, and now wishes to make a single payment once every half-year. Determine the amount of that payment if the interest rates are unchanged, butif the mortgage is now amortized to be paid off after 25 years.

5. One of the following equations is always true, and one is true only when i = 0.

I.1

sn i=

1an i

+ i

II.1

sn i=

1an i

+1i

III.1

an i=

1sn i

+ i

IV.1

an i=

1sn i

+1i

(a) Explain which is always true, and prove iti. [4 MARKS] algebraically; and

ii. [4 MARKS] by a verbal argument, referring to a sinking fund. (A detailed ex-planation is expected.)

(b) [4 MARKS] Prove algebraically that one of the other equations is true for i = 0.

6. (a) [8 MARKS] Find the price of the following bond, which is purchased at a premiumto yield 5% convertible semi-annually: the bond has face value of 10,000, maturesin 12 years at a maturity value of 11,500, and has a nominal coupon rate of 8% perannum, compounded semi-annually; the investor is replacing the premium by meansof a sinking fund earning 4% convertible semi-annually.

(b) [7 MARKS] Suppose that the bond is callable at the end of 9, 10, or 11 years at apremium of 1, 000 above the maturity value. Explain what price the investor shouldpay if she is no longer plans to deposit any of the interest in a sinking fund.

7. [8 MARKS] It was n years ago when James deposited 10,000 in a bank paying 1.8%interest compounded monthly. If he had, instead, placed his deposit in a syndicate payinginterest by cheque annually at the rate of 6% per annum, and he had invested only thisinterest with the bank, how much more interest would he have earned altogether? Showall your reasoning, and express your answer in terms of n.

8. A loan is being repaid with 16 annual payments of 1,000 each. At the time of the 4thpayment the borrower requests permission, and is permitted to pay an extra 3000, andthen to repay the balance over 8 years with a revised annual payment.


(a) [5 MARKS] If the effective annual rate of interest is 6%, find the amount of therevised annual payment.

(b) [8 MARKS] Complete an amortization table for the last 8 payments, with the fol-lowing columns:

Payment Payment Interest Principal Outstandingnumber amount paid repaid loan balance

5. . . . . . . . . . . . . . .12

7.3 2004/20057.3.1 First Problem Assignment, with Solutions

Mounted on the Web on Thursday, February 10th, 2005(Caveat lector!,39 There could be some undetected misprints or errors.) Full solutions were to

be submitted by Monday, January 31st, 2005

Students were advised that “These problems are to be solved with full solutions, modelledeither on the solutions to problems in the textbook, or in the notes on the Web for this orprevious years. The essence is that the reader should be able to reconstruct every step of theproof from what you have written: getting the right answer is never enough. You are not beinggraded for elegance, but simply for the proof being logical, without serious gaps.”

1. (a) Show that the function 225 − (t − 10)2 cannot be used as an amount function fort ≥ 10.

(b) For the interval 0 ≤ t ≤ 10, determine the accumulation function a(t) that corre-sponds to A(t) = 225 − (t − 10)2.

(c) Determine, for the above functions, I3, i6, d7.(d) Suppose that A(0) is invested at time t = 0 and accumulates to A(10) at time t = 10.

What is the equivalent effective annual rate of compound interest, compounded everyyear, which would yield the same accumulation after 10 years.

(e) Suppose that A(0) is invested at time t = 0 and accumulates to A(10) at time t = 10.What is the equivalent annual rate of simple interest which would yield the sameaccumulation after 10 years.

(f) Suppose that A(0) is invested at time t = 0 and accumulates to A(10) at time t = 10.What is the equivalent annual rate of simple discount.

(g) Suppose that A(0) is invested at time t = 0 and accumulates to A(10) at time t = 10.What is the equivalent annual rate of compound discount.

39Let the reader beware!


(h) Verify that the rates i and d of compound interest and discount have the property that(1 + i)(1 − d) = 1.

Solution:

(a) By condition [6, 2, p. 2], an accumulation function must be increasing (or at leastnon-decreasing); the same property must hold for an amount function, since it is apositive multiple of its corresponding accumulation function. But A(t) = 225 − (t −10)2 has the property that

ddt

A(t) = 2(10 − t)

which is increasing only when 20 − t ≥ 0, i.e., when t ≤ 10.

(b) a(t) =A(t)A(0)

=125 + 20t − t2

125=

(5 + t)(25 − t)125

.

(c)

I3 = A(3) − A(2) = 15

i6 = a(6) − a(5) =A(6) − A(5)

A(5)=

9200

d7 =I7

A(7)=

A(7) − A(6)A(7)

=7

216

(d) Let i be the equivalent annual rate of compound interest. Since A(10) = 225, A(0) =

125,

125(1 + i)10 = 225 ⇒ 1 + i =

(225125

) 110

⇒ i =10√

1.8 − 1 = 6.0540482%.

(e) Let i be the equivalent annual rate of simple interest. Then

125(1 + 10i) = 225 ⇒ 1 + 10i =225125

⇒ i = 8%.

(f) Let d be the equivalent annual rate of simple discount. Then

125 = (1 − 10d)225⇒ d =245

= 4.4444%.

(g) Let d be the equivalent annual rate of simple discount. Then

125 = 225(1 − d)10 ⇒ d = 1 − 10

√59

= 5.70845551%.


(h) (1 + 0.060540482)(1 − 0.0570845551) = 1 .

2. Determine the accumulated value of 100 at the end of 2 years if

(a) the nominal annual rate of interest is 8% convertible quarterly(b) the nominal annual rate of discount is 8% convertible once every 6 years(c) interest is compounded instantaneously at the rate of 8%(d) an annual effective compound discount rate of 8% is applied(e) an annual simple discount rate of 8% is applied(f) interest is earned during the first year at a nominal annual rate of 8% convertible

quarterly, and during the second year at a nominal annual rate of discount of 8%convertible quarterly.

Solution:

(a) A nominal annual interest rate of 8% convertible quarterly is equivalent to a 3-monthrate of 8

4% = 2%. The accumulated value of 100 after 214

= 8 quarter years is

100(1.02)8 = 117.17 .

(b) We are given thatd( 1

6 ) = 8% .

The accumulated value of 100 after 26 = 1

3 6-year periods

100 (1 − (6 × 0.08))−26 = 124.36 .

(c) Interest compounded instantaneously at the rate of 8% yields an accumulation factorof

limm→∞

(1 +

0.08m

)m

= e0.08 .

After 2 years 100 accumulates to

100e2×0.08 = 117.35 .

(d) At an annual effective compound discount rate of 8%, 100 accumulates after 2 yearsto

100(1 − 0.08)−2 = 118.15 .

(e) At an annual simple discount rate of 8%, 100 accumulates after 2 years to

1001 − 2(0.08)

= 119.05 .


(f) The accumulated value of 100 after the first year will be 100(1.02)4. During thesecond year this amount will accumulate to a final amount of

100(1.02)4(1 − 0.08

4

)−4

= 100(1.020.98

)4

= 117.36 .

3. Today is New Year’s Day. In return for payments of 1500 at the end of January, February,and March, and of 3000 at the end of May, July, and September, an investor agrees topay now the total value of the 6 payments, and to either make or receive an additionalpayment at the end of December. Find the amount of that additional payment if it isknown that the nominal annual interest rate is 6%, compounded monthly. (First set up anequation of value.)Solution: Let us denote the final payment by the investor by X. The effective monthlyinterest rate is 6

12% = 12%. We will take as the comparison date the end of December

(although any date chosen would yield the same information). The value on 31 Decemberof the payments paid to the investor is

1500((1.005)11 + (1.005)10 + (1.005)9

)

+3000((1.005)7 + (1.005)5 + (1.005)3

)= 13, 957.74 .

The value on 31 December (just after she has paid the last payment) of the payments paidout by the investor is

(3(1500) + 3(3000))(1.005)12 + X = 14332.65 + X .

Equating the two values yields X = −374.91. Thus the investor is entitled to receive afinal payment of 374.91 at the end of December. (The fact that the final payment wouldbe received by the investor, rather than paid by her could have been reasoned withoutcalculation; we didn’t need the mathematical calculation to tell us that from the sign ofthe answer.)

4. Analogously to the “rule of 72”, you are asked to develop a rule of n to approximate howlong it takes for money to increase to 1 1

2 times its initial value. (That is, to determine a2-digit integer N = 10n1 + n0 (where n0, n1 are decimal digits), for which (0.01) · N

i isa good approximation to the number of years required.) Your approximation should bebest around 8%.Solution: We have to approximate a solution to the equation

(1 + i)n = 1.5,


n =ln 1.5

i· i

ln(1 + i).

We find that


i (ln 1.5) · iln(1 + i)

7.0% 0.4198.0% 0.4219.0% 0.423

so we take N = 42.

5. (a) Find the smallest nominal rate of interest convertible monthly at which the accumu-lated value of 15,000 at the end of 3 years is at least 24,000.

(b) Find the smallest nominal rate of discount convertible semi-annually at which theaccumulated value of 15,000 at the end of 3 years is at least 24,000.

Solution:

(a) Let i denote the nominal interest rate sought. Then

15000(1 +

i12

)36

≥ 24000

⇔(1 +

i12

)36

≥ 1.6000

⇔ 1 +i

12≥ (1.6)

136 = 1.013141254

so i =≥ 15.7695048%. The smallest interest rate is 15.8%.(b) This could be considered a trick question; but the fact is that the word largest was not

intended. I “solve” it both as written, and with the intended wording. Let d denotethe nominal discount rate.As written, “largest” Then

15000(1 − d

2

)−6

≥ 24000

⇔(1 − d

2

)6

≤ 58

⇔ 1 − d2≤ 6

√58

= .9246555971

implying that d ≥ 15.068880%. Thus any discount rate d ≥ 15.1% has the de-sired property. There is no largest rate, since, as the rate d(2) approaches 100%from below, the accumulated value approaches∞. We haven’t attached a mean-ing to discount rates above 100%. (As for a discount rate of 100% — that doesmake sense, but would not permit the accumulation of funds: if we discounta sum X of money back from time 1 to time 0 at an effective discount rate of100%, we obtain a present value of 0.)


Corrected to “smallest”

15000(1 − d

2

)−6

≤ 24000

⇔(1 − d

2

)6

≤ 58

⇔ 1 − d2≤ 6

√58

= .9246555971

implying that d ≥ 15.068880%. Thus the smallest discount rate is 15.1%.

6. Let x be a positive real number, 0 ≤ x < 1. Prove that

11 − x

≥ 1 + x

for all such x. Conclude that the accumulated value of 100 after 10 years at an interestrate of x% is always less than or equal to the accumulated value of 100 after 10 years ata discount rate of x%, with equality holding only when x = 0.Solution: Since x2 is a square, it cannot be negative, and can be 0 only when x = 0; hence1 − x2 ≤ 1, with equality holding precisely when x = 0. But this inequality is equivalentto

1 ≥ (1 − x)(1 + x)

or to1

1 − x≥ 1 + x

since the inequality is preserved when we divide both sides by the positive number 1− x;again, equality holds when x = 0.

The accumulated value of 100 at the interest rate of x% is 100(1 + x

100

)10. We apply the

preceding argument 10 times, after replacing x by x100 : this cannot exceed 100

(1 − x

100

)−10,

which is the accumulated value of 100 at the discount rate of x%. Equality holds whenx

100 = 0, i.e., when x = 0.

7.3.2 Second Problem Assignment, with Solutions

Mounted on the Web on March 3rd, 2005Full solutions were to be submitted by February 14th, 2005.

(Subject to Correction)

These problems were to be solved with full solutions, modelled either on the solutions toproblems in the textbook, or in the notes on the Web for this or previous years. The essence


is that the reader should be able to reconstruct every step of the proof from what you havewritten: getting the right answer is never enough. You are not being graded for elegance, butsimply for the proof being logical, without serious gaps.

1. (a) At a certain rate of compound interest an investment of 1000 will grow to 1500 atthe end of 12 years. Determine its value at the end of 5 years.

(b) At a certain rate of compound interest an investment of 1000 will grow to 1500 atthe end of 12 years. Determine precisely when its value is exactly 1200.

(c) A debt of 7000 is due at the end of 5 years. If 2000 is paid at the end of 1 year,what single payment should be made at the end of the 2nd year to liquidate the debt,assuming interest at the rate of 6.5% per year, compounded quarterly.

(d) George agrees to buy his brother’s car for 7000. He makes a down payment of 4000,and agrees to pay two equal payments, one at the end of 6 months, and the other atthe end of a year. If interest is being charged at 5% per annum effective, how largeshould each of the equal payments be?

(e) A bill for 1500 is purchased for 1000 15 months before it is due. Determine thenominal rate of discount, compounded monthly, which the purchaser is paying.

(f) Bills for 1500 are regularly purchased for 1000 15 months before they are due. Thepurchaser knows that, among 10 such bills, he will be unable to collect anything onone of them, will have to pay his lawyers 500 each to effect collection on 2 others,and will collect the others without any impediment. Lumping all of these together,what is the effective annual interest rate earned by the purchaser on his investment,if it is assumed that the lawyers’ accounts are due at the same time as the bills?

Solution:

(a) Let I be the effective annual rate of compound interest. Then

1000(1 + i)12 = 1500 (124)

implies that

(1 + i)5 =

(15001000

) 512

= 1.184053587

so the investment is worth 1184.05 after 5 years.(b) Let t be the time in years when the investment is worth exactly 1200. Then

1000(1 + i)t = 1200 ,

where i is the effective annual rate of compound interest. By equation (124),

ln(1 + i) =ln 1.5

12.


Hencet =

ln 1.2ln(1 + i)

=12 × ln 1.2

ln 1.5= 5.395923442 :

the investment has the desired value after 5.40 years.(c) Let x denote the payment that must be made at the end of the 2nd year. The equation

of value at that time is

x + 2000(1 +

0.0654

)4

= 7000(1 +

0.0654

)−12

implying that

x = −2000(1 +

0.0654

)4

+ 7000(1 +

0.0654

)−12

= 3635.67

should be paid at the end of the second year to liquidate the debt.(d) Let x denote the amount of each of the payments that should be made at the ends of

6 months and 1 year. Then the equation of value at time 0 is

x((1.05)−

12 + (1.05)−1

)= 7000 − 4000 = 3000 ,

implying that

x =3000

(1.05)−12 + (1.05)−1

= 1555.79 .

(e) Let d be the effective monthly rate of discount. Then

1000 = (1 − d)151500 ,

implying that

d = 1 −(10001500

) 115

= 2.66689392% .

This is the effective monthly rate. The nominal annual rate of discount, compoundedmonthly, is, therefore 12d = 32%.

(f) Of 10 bills which mature in 15 months, the total return that will enure to the pur-chaser at that time is

7(1500) + 2(1500 − 500) + 0(1500) = 12500 .

If we denote the effective annual interest rate by i, then the discounted value of 12500at time t = 0 will be 12500(1+ i)−

1512 . The equation of value at time t = 0 is, therefore,

12500(1 + i)−1512 = 10(1000) = 10000 ,

implying that (1 + i)54 = 1.25, so i = 1.250.8 − 1 = 19.5440625%.


2. A government offers savings bonds in multiples of $1000, which mature in 10 years at$2000, but pay no interest until they are redeemed.

(a) Assuming interest compounded semi-annually, what nominal annual rate of interestdoes the bond holder earn?

(b) Suppose that the bonds earn interest at the nominal rate of i − 0.01 for the first 5years, compounded semi-annually; and that they earn interest at the nominal rate ofi + 0.01 compounded semi-annually for the last 5 years. Determine i.

(c) The government has contracted with a bank to market the bonds, at a cost of $40per $1000 bond. What interest rate, compounded semi-annually, is the governmentpaying for the net proceeds it receives for each $1000 bond?

Solution:

(a) Let i be the nominal interest rate, compounded semi-annually, earned by the pur-chaser. The equation of value at time 10 is then

1000(1 +

i2

)20

= 2000

implying thati = 2

(2

120 − 1

)= 7.0529848%.

(b) The equation of value is now

1000(1 +

i − 12

)10 (1 +

i + 12

)10

= 2000

implying that(1 +

i − 0.012

) (1 +

i + 0.012

)= 2

110

⇒(1 +

i2

)2

− (0.01)2

4= 2

110

⇒ 1 +i2

=

(2

110 +

(0.01)2

4

) 12

⇒ i = 2(2

110 +

(0.01)2

4

) 12

− 2

= 7.0553996%

(c) Let i be the nominal interest rate, compounded semi-annually. From the govern-ment’s point of view the equation of value at maturity is

960(1 +

i2

)20

= 2000


implying that

i = 2

(2000960

) 120

− 1

= 7.4760322%

3. The cash price of a new automobile is 18,000 plus 15.025% tax. The purchaser is pre-pared to finance the car and taxes at 18% convertible semi-monthly, and to make pay-ments of 230 at the end of every half-month for 3.5 years, with the first payment to bemade one half-month after delivery. The dealer requires a down payment upon delivery,both to make up the gap in the financing, and from which to pay his immediate costs(commission to the salesperson, preparation costs, sales taxes).(a) Determine the value of this down payment.(b) Determine the value of the down payment if the purchaser decides, instead of semi-

monthly payments, to make a payment of 460 at the end of every month, first pay-ment a month after delivery, last payment to be made 3.5 years after delivery. Theinterest rate and compounding period do not change.

(c) Determine the value of the down payment if the original conditions are changedso that the first payment of 230 is made 1.5 months after delivery, but the samenumber of payments of 230 are made as originally planned. The interest rate andcompounding period do not change.

Solution:(a) The present value of the future payments is

230 · a(3.5)(24) 0.1824

= 230(1 − (1.0075)−84

0.0075

)

= 14295.41185

The excess of the purchase price and taxes over the present value of the future pay-ments is, therefore,

(1.0015025)(18000) − 14295.41185 = 6409.08815 .

Thus the down payment will be 6,409.09.(b) We can still interpret the payments as constituting an annuity-immediate. But the

interval of payments has paid, while the compounding interval has not. So we needto determine the interest rate that corresponds to one month. This is

(1.0075)2 − 1 = 0.01505625 .

The present value of the future payments is

460a(3.5)(12) 0.01505625 = 460 · 1 − (1.01505625)−42

0.01505625= 14242.00433


The excess of the purchase price and taxes over the present value of the future pay-ments is, therefore,

(1.0015025)(18000) − 14242.00433 = 6462.49567 .

Thus the down payment will be 6,462.50.(c) Since the repayment schedule is delayed by 2 payments, which are added at the end,

the present value of the future payments is

230a(3.5)(24)+2 0.1824− 230a2 0.18

24= 230

(1 − (1.0075)−86

0.0075− 1 − (1.0075)−2

0.0075

)

= 230(63.20976257 − 1.977722907)= 14083.36912

The excess of the purchase price and taxes over the present value of the future pay-ments is now

(1.0015025)(18000) − 14083.36912 = 6621.13088 .

Thus the down payment will now be 6,621.13.4. An employee aged exactly 40 decides to accumulate a fund for retirement at age 65 by

depositing 200 at the beginning of each month for 25 years. When she reaches age 65, sheplans to withdraw a fixed amount at the beginning of each year for 15 years. Assumingthat all payments are made, determine the amount of the annual payments that she willbe able to withdraw:(a) if the interest rate is always taken to be (a nominal annual rate of) 6% per annum,

compounded monthly.(b) if the interest rate is taken to be 6% per annum compounded monthly during the time

when the fund is being built up, and 8% per annum effective when she reaches age65.

(c) if the interest rate is taken to be 4% per annum compounded monthly for the coming5 years, then 6% per annum compounded monthly for the next 20 years while thefund is being built up, then 8% per annum effective when the fund is paying outannual payments.

Solution:(a) The effective interest rate per month is 6%

12 = 12%. The number of contributions will

be 25× 12 = 300. The value of the fund at maturity, when the employee reaches age65, will be

200s25×12 12 % = 200

(1.005)300 − 1

(0.005)(

11.005

)

= 139291.7864


The withdrawals from the fund will be once a year. If we wish to interpret themas payments under an annuity-due, we need to determine the rate of interest thatcorresponds to the period between the payments, i.e. to a 1-year period; that is, weneed to determine i, knowing i(12). This gives us an annual rate of (1.005)12 − 1 =

6.1677812%. If the annual withdrawal from the fund, at the beginning of each year,be denoted by X, then the equation of value just before the first withdrawal is

X · a15 6.1677812% = 139291.7864

which we can solve to determine

X =139291.78641−(1.061677812)−15

0.061677812( 11.061677812 )

= 15393.80469

so the withdrawals will be 15393.80, at the beginning of each of 15 years.(b) Proceeding as in the previous case, we replace the interest rate of 6.1677812% by

8%. Now we solveX · a15 8% = 139291.7864

to determine

X =139291.7864

1−(1.08)−15

0.08( 11.08 )

= 15067.95928

so the withdrawals will be 15,067.96, at the beginning of each of 15 years.(c) We need to recompute the value at the time of maturity of the fund of the contribu-

tions. The last 20 years of payments accumulate at maturity to

200s20×12 12 % = 200

(1.005)240 − 1

(0.005)(

11.005

)

= 92870.21994

The first five years of payments accumulate to

200 · s60 412 %

at the end of 5 years; at maturity these payments have accumulated to

((1.005)20×12

)200s60 4

12 % =((1.005)20×12

)200

(1.003333333)60 − 1

(0.003333333)(

11.00333333333

)

= 44, 038.93886


The sum of the two components of the fund is

44, 038.93886 + 92, 870.21994 = 136, 909.1588

so the annual withdrawal will be

136909.1588a15 8%

=136909.1588(0.08)(1.08)−1

1 − (1.08)−15

= 14810.21733

or 14,810.22.

5. In his will, a benefactor of McGill contributes a large sum of money to establish a fund,whose proceeds are to be used to provide annual bursaries of 5000 to 6 actuarial students.The principal of the fund is to remain constant, and the bursaries are funded by the interestearned, forever.

(a) If the effective annual interest rate is assumed to be 6% forever, determine the lumpsum that the benefactor’s estate needs to contribute to McGill a year before the firstpayment of bursaries.

(b) If the effective annual interest rate is assumed to be 6% forever, determine the lumpsum that the benefactor’s estate needs to contribute to McGill if the first bursariesare to be issued immediately.

(c) Suppose that each student receives not a bursary of 5000, but an annual annuity-dueof 1250 for 4 years. Under these changed conditions, determine the lump sum pay-ment needed now if the first bursaries are to be awarded immediately or 1 year hencerespectively. (The number of awards will remain the same — each year 6 studentsare awarded a 4-year sequence of bursary payments of 1250, the first payment to bemade immediately.)

Solution:

(a) The lump sum payment must equal the present value of the perpetuity-immediate ofannual payments of 30,000, i.e.

6(5000)a∞ 6% =300000.06

= 500, 000.

(b) The lump sum payment must equal the present value of the perpetuity-due of annualpayments of 30,000, i.e.

6(5000)a∞ 6% =30000

0.06 × 11.06

= 530, 000.

Alternatively, this may be viewed as the cost of funding the bursaries of the precedingquestion, to which must be added the cost of the bursaries that must be paid outimmediately in the amount of 6 × 5, 000.


(c) We must replace each payment of 5000 by

1250 · a5 = (1250)(1.06) · 1 − 1.06−40.06 = 4591.264936.

i. The lump sum payment when the bursaries are awarded beginning one yearhence is

4591.2649365000

× 500, 000 = 459126.4936

or 459,126.49.ii. The lump sum payment when the bursaries are awarded beginning immediately

is4591.264936

5000× 530, 000 = 486674.0832

or 486,674.08.

6. Let m and n be positive integers. Consider an annuity-immediate which pays 1 at theend of every period for mn periods. Explain in words why each of the following formulærepresents the value of this annuity m + 1 years before the first payment is made. (If youhave doubts about the truth of this claim, you may wish to verify algebraically that theclaim is correct before you attempt to explain it in words.)

(a) vm · amn

(b) am(n+1) − am

(c) vm · am(n+1) − vm(n+2) · sm

Solution:

(a) One period before the first payment, the value of an annuity of 1 per period payablefor mn periods is amn. At a time m periods earlier the value must be discounted by afactor of 1

1+i per year, or vm for m years.(b) am(n+1) is the value of an annuity-immediate of 1 payable for mn + n periods. At

a time m periods ago, this represents the value then of the annuity under presentconsideration, augmented by an annuity of m payments of 1, the last to be madenow. If we subtract the value of these m payments as of m periods ago, we are leftwith the value of the mn payments under consideration.

(c) Consider an annuity that consists of all the payments under present consideration,extended by an additional m payments. Such a scheme is worth am(n+1) now, andvm · am(n+1) m periods ago. The m payments added at the end are worth sm at the timeof the last of them, which is mn + m periods from now. We may discount that backto a time m periods ago by multiplying by vmn+m+m = vm(n+2).


7.3.3 Third Problem Assignment, with Solutions

Mounted on the Web on March 7th, 2005Full solutions were to be submitted by March 7th, 2005.

(Caveat lector! There could be misprints or other errors!)

These problems were to be solved with full solutions, modelled either on the solutions toproblems in the textbook, or in the notes on the Web for this or previous years. The essenceis that the reader should be able to reconstruct every step of the proof from what you havewritten: getting the right answer is never enough. You are not being graded for elegance, butsimply for the proof being logical, without serious gaps.

1. McGill plans to create a scholarship fund that will eternally pay 50 students a monthlystipend of $200 at the beginning of months September through April, plus an amount of$300 on the following May 1st.

(a) If interest is assumed to be at a nominal annual rate of 6% per annum, compoundedmonthly, determine the amount that is needed in this fund on September 1st justbefore the fund begins making payments.

(b) Determine the amount that will be in the fund just after the December scholarshippayments in the first year, and on September 1st of the following year, just beforethe September payments.

(c) Suppose that at the beginning of September, 8 years after the fund is established, itis decided to increase the capital in the fund because the interest rate has changed to4% per annum compounded monthly. Determine how much additional capital needsto be added to the fund.

(d) Suppose that 2 years after the interest rate is changed to 4%, it changes again, thistime to 8%. This time it is decided to leave the capital unchanged, but to increase thepayments to students by a lump sum of $M to each student, payable on December1st, together with the regular December payment under the scholarship. Determinethe amount of that lump sum payment.

Solution:

(a) Since monthly payments are not completely regular, we cannot interpret this fundas a monthly perpetuity with regular payments. We could still develop its propertiesfrom first principles, but it is easier to interpret it as an annual perpetuity-due. Thepayments made to each student in an academic year are worth

200a8 0.5% + 300(1.005)−8 = 200(1.005)(1 − (1.005)−8

0.005

)+ 300(1.005)−8

= 1860.680368 .


Alternatively, we can think of an annuity-due of 9 payments of 200 together with anadditional payment of 100 alongside the last payment:

200a9 0.5% + 100(1.005)−8 = 200(1.005)(1 − (1.005)−9

0.005

)+ 100(1.005)−8

= 1860.680368 .

The effective annual interest rate is

(1.005)12 − 1 = 6.1677812% .

It follows that the amount in the fund at the beginning of September, just before theSeptember payments, must be

50(1860.680368)a∞ (1.005)12−1 = 50(1860.680368)(1.005)12

(1.005)12 − 1= 1, 601, 421.16 .

(b) We could evaluate the fund at the beginning of each month, determining the growthin principal, and then subtracting the payments that would need to be made on thatday. Instead, we will shift all payments to the dates under examination. On Decem-ber 1st, just after the December payments, the fund will be worth

1, 601, 421.16(1.005)3 − 50(200)s4 0.005

= 1, 601, 421.16(1.005)3 − 50(200)((1.005)4 − 1

0.005

)

= 1, 585, 261.78

On the following September 1st, just before the first payments of the new academicyear, the value of the fund must be

1, 585, 261.78(1.005)9 − 50(200)s4 0.005(1.005)5 − 50(300)(1.005)4

= 1, 585, 261.78(1.005)9 − 50(200)((1.005)4 − 1

0.005

)(1.005)5 − 50(300)(1.005)4

= 1, 601, 421.16

This is no surprise — we constructed this fund so that the amount just before theSeptember 1st payments would be constant.

(c) The amount in the fund on September 1st, just before payments are made, has beenconstant. Because of the revised interest rate, the annual costs per participant willbe, as of the time of the September payment,

200a8 13 % + 300(1.0033333333)−8

= 200(1.0033333333)(1 − (1.0033333333)−8

0.00333333333

)+ 300(1.0033333333)−8

= 1873.636976 .


The effective annual interest rate is now

(1.0033333333)12 − 1 = 4.0741539% .

The amount of the fund will, therefore, need to be

50 × 1873.636976 × 1.0407415390.040741539

= 2, 393, 100.36 .

The additional capital required is, therefore,

2, 393, 100.36 − 1, 601, 421.16 = 791, 679.20 .

(d) Since this is a perpetuity, the amount in the account is periodic. Unless the interestrate or other conditions change, the amount is always the same on 1 September, andit does not matter how many years have passed. We repeat the calculations of thepreceding part, this time for a nominal rate of 8%. Because of the revised interestrate, the annual costs per participant will be, as of the time of the September payment,

200a8 23 % + 300(1.0066666667)−8

= 200(1.0066666667)(1 − (1.0066666667)−8

0.0066666667

)+ 300(1.0066666667)−8

= 1847.870679 .

The effective annual interest rate is now

(1.0066666667)12 − 1 = 8.2999511% .

The amount of the fund to support the previous obligations will, therefore, need tobe

50 × 1847.870679 × 1.0829995110.082999511

= 1, 205, 575.19 .

The capital no longer required for the previously defined purposes is, therefore,

2, 393, 100.36 − 1, 205, 575.19 = 1, 187, 525.17

on September 1st, prior to the payments due on that date. On December 1st thisexcess is worth

1, 187, 525.17(1.0066666667)3 = 1, 211, 434.36

in total, or 24,228.68728 for each of the 50 participants at any time. This can pur-chase a perpetuity-due of

M =24228.68728a∞ 8.2999511%

=24228.68728 × 0.082999511

1.082999511= 1856.851435 .


(The problem was somewhat ambiguous: some students thought the intention wasthat there would be a single lump sum payment in just one December. If that hadbeen the intention, the value would have been 24,228.69 determined above.)

2. A loan of 10,000 is to be repaid by regular, half-yearly payments of 1,000, the first to bemade at the end of the 3rd year. The loan will be paid off by a final payment which mustbe at least 1,000.

(a) If the interest rate is to be 8% per annum, compounded semi-annually, determine theamount of the final payment, and when it is made.

(b) Suppose that the final payment is now not more than 1,000. and the interest rateremains 8% per annum compounded semi-annually. Determine the amount of thefinal payment, and when it is made.

(c) Suppose that, after signing his original commitment, the borrower decides that hewould like to make the first payment 6 months from the date of borrowing, and thatall the payments under this loan should be exactly equal. If the interest rate is 6%per annum, compounded semi-annually, determine that payment level that would beclosest to 1000 per half-year, and determine exactly when the loan will be paid off.

Solution:

(a) Let n be the number of the last payment (counting in half-years from the date of theloan). Then n will be the largest integer such that

10000(1.04)n ≥ 1000sn−5 4%

since the first 5 opportunities for payments are missed. The inequality is equivalentto

10000(1.04)5

1000≥

sn−5

(1.04)n−5 = an−5

and to(1.04)−(n−5) ≥ 1 − 0.4(1.04)5

which implies that

n − 5 ≤− ln

(1 − 0.04(1.04)5

)

− ln(1.04)= 17.00170887

so the last payment is made at n = 22. The excess of the amount of the payment over1000 will be

10000(1.04)22 − 1000s17 = 1.67552

rounding to 1.68.


(b) This time the final payment will be #23; the amount of the payment will be theaccumulation of the residue remaining unpaid after the payment at time n = 22, i.e.,(1.04)1.67552 = 1.74254 rounding to 1.74.

(c) If the last payment is n half-years after the date of borrowing, the payments will bein the amount

10000an

.

The object is to minimize ∣∣∣∣∣∣10000

an− 1000

∣∣∣∣∣∣or to minimize ∣∣∣∣∣∣

10an− 1

∣∣∣∣∣∣Since an is an increasing function of n, its reciprocal is decreasing. We need to findthe value of n such that

an 3% ≤ 10 < an+1 3%

i.e., the largest n such that an 3% ≤ 10, i.e., such that vn ≥ 0.7, i.e., such that n ≤− ln(0.7)ln(1.03)

= 12.06662371. So the only candidates are n = 12 and n = 13, and the

corresponding semiannual payments are

10000a12

= 1004.620855

and10000

a13

= 940.2954396

Taking the payment closest to 1000, we find that the last payment will be at the endof 6 years, and the amount of the regular payments will be 1004.62.

3. (An acknowledgement of the source of this problem will be contained in the solutions,when published.) Katherine, 25 years old, deposits 10,000 at the beginning of every 4-year period into an RSSP account. The account pays compound interest annually, at theeffective annual rate i. The accumulated amount in the account at the end of 40 yearsis X, which is 6 times the accumulated amount in the account at the end of 20 years.Determine X.Solution: This problem was modelled on Problem 8 of Course 2, May, 2003, of theSociety of Actuaries.There are several ways of attacking a problem like this. One method is to determine fromi the interest rate that would apply to the period of payments, i.e. 4 years; another is to


determine the annual payment in advance that would correspond to a payment of 10,000in advance every 4 years.The problem on which this is modelled was posed as a multiple-choice problem, with5 possible numerical answers. None of the answers was expressed in terms of i; so, infact, there was no need for the question to even mention i. You may judge for yourselfwhether that inclusion made the problem harder or easier.Denote by Y the annual payment in advance that is equivalent to a payment every 4-yearsin advance of 10,000. Then

Y · a4 i = 10000 ,

so Y =10000

a4 i

. The values of the account at the ends of 20 years and 40 years are,

respectively

X = Y · s40 i = Y · (1 + i) · (1 + i)40 − 1i

and Y · s20 i = Y · (1 + i) · (1 + i)20 − 1i

From the hypothesis that these amounts are in the ratio of 6 : 1, we conclude that

6 =Y · (1 + i) · (1+i)40−1

i

Y · (1 + i) · (1+i)20−1i

= (1 + i)20 + 1

which implies that (1 + i)20 = 5. We were not asked to compute the interest rate, anddon’t actually require it. (If you did require it, you could solve the preceding equation toshow that 1 + i = 5

120 = 1.083798, so i = 8.38%.) Then

X = 10000 · (1 + i) · (1+i)40−1i

(1 + i) · 1−v4

i

= 10000 · (1 + i)40 − 11 − v4 =

10000(25 − 1)

1 − 5−15

=240000

1 − 0.72478=

2400000.275220336

= 872, 028.58

4. For each of the following sequences of payments as of the time stated,

• express the value using standard symbols, as simple as possible;• give a formula for the value (in terms of i, v, d, etc.);• evaluate using a calculator or computer — not with tables.

(a) a perpetuity-immediate paying 1 per year at effective annual interest rate 4.25%,evaluated 1 year before the first payment;


(b) a perpetuity paying 1 per year at effective annual interest rate 4.25%, evaluated 2years before the first payment;

(c) a perpetuity-due paying 1 per half-year at nominal interest rate 5%, compoundedsemi-annually, evaluated just before the first payment;

(d) a 20-year increasing annuity paying 1 the first year, 2 the second year, 3 the 3rd year,etc., at an interest rate of 7% per year effective, evaluated just after the last payment;

(e) a 20-year increasing annuity paying 1 the first year, 2 the second year, 3 the 3rdyear, etc., at an interest rate of 7% per year effective, evaluated at the time of the 2ndpayment;

(f) a 10-year decreasing annuity paying 1 less each quarter-year, evaluated just after thelast payment, where the nominal interest rate is a 8% annual, compounded quarterly;

Solution:

(a) a∞ 4.25% =1i

=1

0.00425= 23.52941176 .

(b) v · a∞ 4.25% =vi

=1

1.0425(0.0425)= 22.57017915 .

(c) a∞ 0.025 =1iv

=1

0.025(1.025)−1 = 41.

(d) (Is)20 0.07 =s21 0.07 − 21

0.07=

(1.07)21−10.07 − 21

0.07= 340.9310971 .

(e) v18 · (Is)20 0.07 = (1.07)−18 · (340.9310971) = 100.8692096 .

(f) (Ds)40 0.02 =n(1 + i)n − sn

i=

40(1.02)40 − (1.02)40−10.02

0.02= 1395.980168.

5. X and Y have sold their home for 400,000, and wish to purchase an annuity-immediateso that they can spread the proceeds over the next 10 years. The first payment will be onemonth from the date of purchase of the annuity.

(a) Determine the level monthly payments they will receive if the effective annual inter-est rate is 6%.

(b) X and Y live frugally, and don’t think they need as much income now as they will astime passes. Accordingly they plan to receive monthly payments which will gradu-ally increase. If the first payment is 3,000, and the payments increase each month bythe same dollar amount K, determine K. The effective annual interest rate remains6%. What is the final monthly payment?

(c) Suppose that the payments remain constant in any year. The first year’s monthlypayments are all 3000, the next year’s 3000+L, the next years’ 3000+2L, 3000+3L,. . . , 3000 + 9L. Determine L if the nominal annual interest rate is 6%, compoundedmonthly.


Solution:

(a) The effective monthly interest rate is (1.06)112 − 1 = 0.4867551%, which we shall

denote by j below. The payments will be

400000a

120 (1.06)1

12 −1

=400000((1.06)

112 − 1)

1 − (1.06)−120

12

=400000((1.06)

112 − 1)

1 − (1.06)−10 = 4408.961439

(b) We solve for K the equation of value

400000 = K · (Ia)120 j + (3000 − K) · a120 j

⇔ 400000 = K

a120 j − 120(1 + j)−120

j

+ (3000 − K) · a120 j

= K

a120 j − 120(1 + j)−120

j− a120 j

+ 3000 · a120 j

Hence

K =j(400000 − 3000 · a120 j

)

a120 j − 120(1 + j)−120 − j · a120 j

=j(400000 − 3000 · a120 j

)

a120 j − 120(1 + j)−120

= 26.23459805

so the final payment will be

3, 000.00 + 119(26.23459805) = 6, 121.92 .

(c) Consider payments of L at the end of every month for a year. Discounted back to thebeginning of the year, these payments are worth

L · a12 0.005 = L(1 − 1.005−12

0.005

)= L(11.61893206) .

The value of all of the payments as of the date of purchase of the annuity is, therefore,

3000 · a1200.005 + L · a12 0.005 · (Ia)9 (1.005)12−1 = 400000 .


Now

(Ia)9 (1.005)12−1 =a9 (1.005)12−1 − 9(1.005)−9(12)

(1.005)12 − 1

=(1.005)12 · a9 (1.005)12−1 − 9(1.005)−9(12)

(1.005)12 − 1

=a9 (1.005)12−1 − 9(1.005)−10(12)

1 − (1.005)−12

=

1−(1.005)−9(12)

(1.005)12−1 − 9(1.005)−10(12)

1 − (1.005)−12

= 31.08041107 .

Solving for L, we obtain

L =400000 − 3000 · a120 1

2 %

11.61893206(31.08041107)= 0.002769153521

(400000 − 3000 · a120 0.005

)

= 0.002769153521(400000 − 3000 · 1 − (1.005)−120

0.005

)

= 359.3797471.

The monthly payments in the first year will be 3,000; those in the last year will be

3, 000 + 9(359.3797471) = 6, 234.42 .

7.3.4 Fourth Problem Assignment, with Solutions

Mounted on the Web on March 31st, 2005Full solutions were to be submitted by March 21st, 2005.

These problems were to be solved with full solutions, modelled either on the solutions toproblems in the textbook, or in the notes on the Web for this or previous years. The essenceis that the reader should be able to reconstruct every step of the proof from what you havewritten: getting the right answer is never enough. You are not being graded for elegance, butsimply for the proof being logical, without serious gaps.

1. A loan of 20,000 is to be repaid by 15 annual payments beginning one year after the loanwas received, payments which increase by a factor of 1.03, each over the preceding. Ifthe effective annual interest rate is 7%, determine


(a) the amount of the first payment(b) the amount of the loan which is outstanding immediately after the 10th payment,

determined using the Retrospective Method(c) the amount of the loan which is outstanding immediately after the 10th payment,

determined using the Prospective Method

Solution:

(a) Denote the amount of the first payment by X. The value at time 0 of all futurepayments is

(1.07)−1(1.03)0X + . . . + (1.07)−r(1.03)r−1X + . . . + (1.07)−15(1.03)14X

=X

1.07

1 +

(1.031.07

)+

(1.031.07

)2

+ . . . +

(1.031.07

)14

=X

1.07·

1 −(

1.031.07

)15

1 −(

1.031.07

)1

=X

0.04

1 −(1.031.07

)15 = 10.8830229X .

Equating to 20,000, we find that X = 1837.72472 is the first payment.(b) By the Retrospective Method, the amount outstanding after the 10th payment is

10, 000(1.07)10 − X((1.03)9 + (1.03)8(1.07) + . . . + (1.03)0(1.07)9

)

= (1.07)10(20, 000)

1 −1 −

(103107

)10

1 −(

103107

)15

= 10, 709.669 .

(c) By the Prospective Method, the amount outstanding after the 10th payment is

(1.03)10 X1.07

+ (1.03)11 X(1.07)2 + . . . + (1.03)14 X

(1.07)5

=(1.03)10

1.07· X ·

1 +103107

+

(103107

)2

+ . . . +

(103107

)4= 10, 709.669 .

2. Mary always dreamed of owning a Maserati, and now had the chance. Her employer wasbuying a new Ferrari, and was willing to part with his used Maserati for a mere 20,000.Mary agreed to repay the loan by equal monthly payments over 3 years at 4% interest,


compounded monthly, first payment one month after the sale. Two and one-half yearsafter the sale Mary’s employer declared bankruptcy, and Mary’s future payments becameone of the assets to be administered by the trustee.

(a) Determine Mary’s regular monthly payment.(b) The trustee in bankruptcy needs to know the present value of Mary’s future pay-

ments. Determine this value just after the payment made on the date of bankruptcyif

i. the interest rate remains a nominal 4% compounded monthly;ii. the interest rate is now 6% compounded monthly.

(c) Complete an amortization table beginning on the date of bankruptcy, when the rateis 4% compounded monthly, under the headings

Payment Payment Interest Principal OutstandingNumber Amount paid repaid loan balance

3031

(d) Suppose that Mary’s future payments, discounted at 6%, have been assigned to oneof the employer’s creditors. You have computed the present value of these payments.Now set up an amortization table to show how his outstanding debt in this amountwill be amortized at 6%.

Solution:

(a) The effective interest rate per month is 4%12 = 1

3%. Let X be the regular monthlypayment. Then

X· a36 13 % = 20000

⇒ X =20000

1300

1 −(1 +

1300

)−36

=20000

300(1 −

(300301

)36) = 590.4797061

so the monthly payment is 590.48.(b) i. By the prospective method, the unpaid balance is

X · a6 13 % = 20, 000 ·

a6 13 %

a36 13 %

= 20, 000 ·1 −

(300301

)6

1 −(

300301

)36 = 3501.909296

so the unpaid balance is 3,501.91.


ii. By the prospective method, the unpaid balance is

X · a6 13 % = 20, 000 ·

a6 12 %

a36 13 %

= 20, 000 ·1 −

(200201

)6

1 −(

300301

)36 ·23

= 3481.695351

so the unpaid balance is 3,481.70.(c) The unpaid balance before the 30th payment was

3, 501.91 + 590.48 = 4, 092.39 .

One month earlier, this was worth4, 092.39

1.00333333= 4, 078.79, so interest earned in the

30th period was the difference,

4, 092.39 − 4, 078.79 = 13.60 .

Thus the principal reduction in the 30th payment was

590.48 − 13.60 = 576.88 .


30 590.48 13.60 576.88 3,501.9131 590.48 11.67 578.81 2,923.1032 590.48 9.74 580.74 2,342.3633 590.48 7.81 582.67 1,759.6934 590.48 5.87 584.61 1,175.0835 590.48 3.92 586.56 585.5236 590.48 1.96 588.52 0.00

(d) The new interest rate takes effect after the 30th payment.


30 590.48 13.60 576.88 3,481.7031 590.48 17.41 573.07 2,908.6332 590.48 14.54 575.94 2,332.6933 590.48 11.66 578.81 1,753.8734 590.48 8.77 581.71 1,172.1635 590.48 5.86 584.62 587.5436 590.48 2.94 587.54 0.00


3. On January 1st, 2005, John receives a loan of 50,000 from his brother, and agrees torepay it by semi-annual payments: 2,000 every July 1st, and 2,500 every January 1st; thepayments begin on July 1st, 2005, and continue until one last, possibly irregular paymentwhich will be either less than 2,000 if it is on July 1st, or under 2,500 if it is on January1st. The effective interest rate is 6% per annum.

(a) Determine the outstanding principal just after the payment on January 1st, 2015.(b) Complete an amortization schedule beginning with January 1st, 2020, under the fol-

lowing headings

Date Payment Interest Principal OutstandingAmount paid repaid loan balance

January 1, 2020

Solution:

(a) Just after a January 1st payment of 2,500, that payment and the payment of 2,000from the preceding July 1st are together worth

2, 500 + 2, 000√

1.06 = 4, 559.26028 .

By the retrospective method, the outstanding principal just after the payment of Jan-uary 1st, 2015, is

50, 000(1.06)10 −(2, 500 + 2, 000

√1.06

)· s10 6%

= 50, 000(1.06)10 − (4559.126028) · (1.06)10 − 10.06

= 29, 449.47952

so the outstanding balance is 29,449.48.(b) We need to begin the table with January 1st, 2020. It is convenient to repeat the

preceding computation: the outstanding balance as of January 1st, 2019, is

50, 000(1.06)14 − (4, 559.126028) · (1.06)14 − 10.06

= 17, 234.86372 .

or 17,234.86. The effective interest rate per half year is 1.0612 − 1 = 2.9563%.


Date Payment Interest Principal OutstandingAmount paid repaid loan balance

January 1, 2019 2,500.00 17,234.86July 1, 2019 2,000.00 509.51 1,490.49 15,744.37

January 1, 2020 2,500.00 465.45 2,034.55 13,709.82July 1, 2020 2,000.00 405.30 1,594.70 12115.12

January 1, 2021 2,500.00 358.16 2,141.84 9,973.28July 1, 2021 2,000.00 294.84 1,705.16 8,268.12

January 1, 2022 2,500.00 244.43 2,255.57 6,012.55July 1, 2022 2,000.00 177.75 1,822.26 4,190.29

January 1, 2023 2,500.00 123.88 2,376.12 1,814.17July 1, 2023 1,867.80 53.63 1,814.17 0.00

4. A 30-year loan is amortized by level payments at the end of every month, at a nominalinterest rate of 16.8%, compounded monthly. The amount of interest paid in the 120thpayment is 486.72. Determine

(a) the amount of interest in the 300th payment;(b) the original amount of the loan;(c) the amount due at the end of 30 years if the lender defaults on40 the last 6 payments.

Solution:

(a) Suppose that the original amount of the loan was X. The level payments are thenX

a360 1.4%

, which I will denote by Y . By the prospective method, the amount owing

just after the rth payment is Y · a360−r 1.4% , so the interest component of the r + 1stpayment is

(0.014)Y · a360−r 1.4% = Y(1 − (1.014)r−360

).

The data tell us thatY

(1 − (1.014)119−360

)= 486.72

so Y = 504.4061464: the monthly payments are 504.41. It follows that the interestcomponent of the 300th payment is

Y(1 − (1.014)299−360

)= 285.8812253

or 288.40.(b) The original amount of the loan was

504.4061464 · a360 1.4% =504.410.014

(1 − (1.014)−360

)= 35, 787.47073

or 35,787.47.40=fails to pay


(c) At the due date of the last payment, the value of the last 6 payments is 504.4061464 ·s6 1.4% = 3, 134.360302 or 3,134.36.

5. A loan of 100,000 is to be maintained by a semi-annual interest payment, with the prin-cipal being repaid in a single payment at the end of 25 years. The borrower’s total cost is5,000 at the end of each 6-month period, being made up of the interest and a contributionto a sinking fund maintained by a 3rd party earning 4% effective: the proceeds will bepaid to the lender upon maturity. Find

(a) the effective semi-annual interest rate received by the lender of the loan.(b) the effective interest rate paid out by the borrower. (For this purpose it will suffice

to determine an equation for the interest rate, and to use the tables in your book tolocate the rate between two tabulated values; you are not expected to interpolate.)

Solution:

(a) The sinking fund earns interest at an effective annual rate of 4%; the equivalentsemi-annual rate is √

1.04 − 1 = 1.9803903%.

The semi-annual payments into the sinking fund will each be in the amount of

100000s50

√1.04−1

=100000(

√1.04 − 1)

(1.04)25 − 1= 1188.826454

or 1,188.83. This leaves a residue of 5,000-1,188.83=3,811.17. The interest rate perhalf year is, therefore

3, 811.17100, 000

= 3.811% .

(b) Let i be the effective semi-annual interest rate being paid by the borrower. Then

5, 000 · a50 i = 100, 000

so a50 i = 20. We see from the tables that

a50 4.5% = 19.7620a50 4% = 21.4822

(i is approximately 4.427%.)NOTE TO THE GRADER: STUDENTS WERE NOT EXPECTED TO INTERPO-LATE OR ITERATE TO OBTAIN A GOOD APPROXIMATION.

6. Pierre borrows 18,000 for 8 years, and agrees to make level annual payments. The lenderreceives 6% on the investment for each of the first 5 years, and 8% for the last 3 years.Throughout the 8 years, the balance of each payment is invested in a sinking fund earning7%.


(a) Determine the amount of the lender’s total annual payment.(b) Complete the following table.

Period Total Interest Interest Earned Contribution to Balance inPayment to Lender in Sinking Fund Sinking Fund Sinking Fund

0Solution:

(a) Let Pierre’s annual payment be X. The interest component in this payment will be.06 × 18, 000 = 1, 080 during the first 5 years.08 × 18, 000 = 1, 440 during the last 3 years

Hence the annual contributions in arrears to the sinking fund will beX-1,080 during the first 5 yearsX-1,440 during the last 3 years

This information enables us to infer an equation that determines X:

(X − 1, 080)s8 7% − 360 · s3 7% = 18, 000


X = 1, 080 +18, 000 + 360 · s3 7%

s8 7%

= 1, 080 +1, 260 + 360

((1.07)3 − 1

)

(1.07)8 − 1= 2, 947.2254

(b) Growth of the Sinking Fund:Period Total Interest Interest Earned Contribution to Balance in

Payment to Lender in Sinking Fund Sinking Fund Sinking Fund0 0.001 2,947.23 1,080.00 0.00 1,867.23 1,867.232 2,947.23 1,080.00 130.71 1,867.23 3,865.173 2,947.23 1,080.00 270.56 1,867.23 6,002.964 2,947.23 1,080.00 420.21 1,867.23 8,290.405 2,947.23 1,080.00 580.33 1,867.23 10,737.966 2,947.23 1,440.00 751.66 1,507.23 12,996.857 2,947.23 1,440.00 909.78 1,507.23 15,413.868 2,947.23 1,440.00 1,078.96 1,507.23 18,000.06

Is it surprising that the balance is 18,000? Of course not — the repayment schemewas designed so that Pierre would accumulate in the fund just the amount needed torepay the loan after 8 years.


7.3.5 Fifth Problem Assignment, with Solutions

Mounted on the Web on April 12th, 2005Full solutions were to be submitted by April 4th, 2005.

1. (a) A 25,000 7% bond matures on June 30th, 2020. Interest is payable semiannually onJune 30th and December 31st. Determine the price to be paid for the bond on June30th, 2005, in order to earn the investor a yield of 6%. (Remember the conventionfor bonds — interest is normally interpreted as a nominal (annual) rate, compoundedsemi-annually, even when not explicitly stated.)

(b) Repeat problem 1a if the purchase date is December 31st, 2012.(c) A 10,000 bond has coupon rate 8% payable semiannually, and is redeemable after a

certain number of years at 11,250. The bond is purchased to yield 7% convertiblesemiannually. If the present value of the redemption value of the bond is 4,927 atthe given yield rate, determine the purchase price.

(d) A 1,000 bond matures after a m years at par, and has a coupon rate of 10% convertiblesemiannually. It is purchased at a price to yield 7% convertible semi-annually. If theterm of the bond is 2m years, the price of the bond will increase by 107. Determinethe price of the m-year bond to the nearest dollar.

Solution:

(a) F = C = 25, 000, r = 12 × 7% = 3.5%, i = 3%, n = 2 × 15 = 30. By the Basic

Formula

P = (25, 000)(0.035) · a15 3% + (25, 000)(1.03)−30

= 25, 000((0.035)

(1 − (1.03)−30

0.03

)+ (1.03)−30

)

= 27, 450.06

(b) F = C = 25, 000, r = 3.5%, i = 3%, n = 15.

P = (25, 000)(0.035) · a15 3% + (25, 000)(1.03)−15

= 25, 000((0.035)

(1 − (1.03)−15

0.03

)+ (1.03)−15

)

= 26, 492.24

(c) F = 10, 000, C = 11, 250, r = 4%, i = 3.5%, K = 4927. This last value of K impliesthat K = Cvn = 11, 250(1.025)−n, so (1.035)n = 11,250

4,927 , where n is the number of


half-years remaining before maturity. By the Basic Formula,

P = Fr · an 0.035 + K

= 400(1 − (1.035)−n

0.035

)+ 4, 927

= 400

1 − 4,927

11,250

0.035

+ 4, 927 = 11, 350.37

(d) For the m-year bond, F = C = 1, 000, n = 2m, r = 5%, i = 3.5%. For the 2m-year bond, n = 4m. Denote the price of the m-year bond by P. Then, using thePremium/Discount formula, we have

P = 1, 000 + (50 − 35) · a2m 3.5% (125)P + 107 = 1, 000 + (50 − 35) · a4m 3.5% (126)

Subtracting yields107 = 15

(a4m 3.5% − a2m 3.5%

)

which implies that

0.2496666 =1.035−2m − (1.035)−4m

0.035yielding, approximately,

(1.035)−2m = 0.5182574184 or 0.4817425816

so, by (125),

P = 1, 000 + 15(1 − 0.5182574184

0.035

)= 1206.46

or

P = 1, 000 + 15(1 − 0.4817425816

0.035

)= 1, 222.11;

to the nearest dollar there are two answers, 1,206 and 1,222.

2. An investor is considering the purchase of two 1,000 face value bonds which are re-deemable at the end of the same number of years, and are both to be bought to yield7.5% convertible semiannually. One bond costs 907, and pays coupons at the rate of 6%per year, convertible semiannually. The second bond pays coupons at 7% per half-year.

(a) Determine the price he should pay for the second bond so that it is an equivalentinvestment to the first.


(b) Determine the price the investor should pay for the second bond if the first bondcosts 960.45, and pays a premium of 100 upon redemption, while the second pays apremium of 150 upon redemption.

Solution: The earlier references in the problem to interest rates always stated that the ratewas “convertible semiannually”. But, in this case, those words were not mentioned. Itappears that the rate of 7% stated is intended to be an effective semi-annual rate, equiva-lently that the nominal annual rate is 14%. However, that would be a much different ratethan the others given in the problem. Because of this ambiguity, I will solve the problemwith both interpretations.

Assuming 3.5% effective per half year: (a) For both bonds we know that F = C =

1000, i = 0.0375. By the Premium/Discount Formula applied to the first bondwe obtain

907 = 1000 + (30 − 37.5)an0.0375 ,

implying that an 0.0375 = 12.4. Substitution of this value into the same formulaapplied to the second bond yields a price for that bond of

1000 + (Fr −Ci)an 0.0375 = 1000 + (35 − 37.5)an 0.0375

= 1000 − 31 = 961.

(b) In this case the Premium/Discount Formula applied to the first bond gives

960.45 = 1100 + (30 − 41.25)an 0.0375 ,

which implies that an 0.0375 = 12.404444. Substitution of this value into the sameformula applied to the second bond yields a price for that bond of

1150 + (Fr −Ci)an 0.0375 = 1150 + (35 − 43.125)an 0.035

= 1150 − 8.125(12.404444) = 1049.21

Assuming 7% effective per half year: (a) For both bonds we know that F = C =

1000, i = 0.0375. By the Premium/Discount Formula applied to the first bondwe obtain

907 = 1000 + (30 − 37.5)an0.0375 ,

implying that an 0.0375 = 12.4. Substitution of this value into the same formulaapplied to the second bond yields a price for that bond of

1000 + (Fr −Ci)an 0.0375 = 1000 + (70 − 37.5)an 0.0375

= 1000 + 403 = 1403.


(b) In this case the Premium/Discount Formula applied to the first bond gives

960.45 = 1100 + (30 − 41.25)an 0.0375 ,

which implies that an 0.0375 = 12.404444. Substitution of this value into the sameformula applied to the second bond yields a price for that bond of

1150 + (Fr −Ci)an 0.0375 = 1150 + (70 − 43.125)an 0.035

= 1150 + 26.875(12.404444) = 1483.37

3. A loan of 96,000 is being repaid by a monthly payment of interest at a nominal rateof 6.0% compounded monthly, together with a monthly payment into a sinking fundwhich earns interest at the rate of 4.0% per annum, also compounded monthly. The totalmonthly payment is constant, at 1,000, with the exception of the very last payment, whichcould be less than 1,000, in order to bring the fund up to its target level of 96,000; at thattime the entire sinking fund will be paid over to the lender and the loan will have beenrepaid. Determine when the final deposit into the sinking fund will be made, and itsamount.Solution: The monthly interest payment to the lender will be

112

(6%) × 96, 000 = 480,

so the monthly contribution to the sinking fund will be 1, 000 − 480 = 520. Regularcontributions of 520 will be made for as long as

520 · sn 13 % ≤ 96, 000

⇔ 520((1.00333333)n − 1

0.00333333

)≤ 96, 000

⇔ (1.0033333)n ≤ 1.615385⇔ n ≤ 144.1116636

i.e., until n = 144. Immediately after that payment the balance in the sinking fund willbe 520 · s144 1

3 % = 95, 906.43598. But, by the time for the next payment, this will havegrown to

1.0033333 × 95, 906.44 = 96, 226.12

so, the amount needed to bring the balance up to the target is −226.12: the lender willreceive a refund! (Note that this refund is from the sinking fund only — he still has topay interest of 480 to the lender.)

4. A 5000 par value 18-year bond has 7% semiannual coupons, and is callable at the end ofthe 12th through the 17th years at par.


(a) Find the price to yield 6% convertible semiannually.(b) Find the price to yield 8% convertible semiannually.(c) Find the price to yield 8% convertible semiannually, if the bond pays a premium of

250 if it is called.(d) Find the price to yield 6% convertible semiannually, if the bond pays a premium of

250 if it is called at the end of years ##12, 13, 14, and a premium of 150 if it is calledat the end of years 15 or 16. (It may still be called at the end of year 17 withoutpremium, and otherwise will mature at the end of year 18.)

Solution:

(a) F = C = 5, 000, n = 36, r = 3.5%, i = 3%. Let m be the coupon number at whosedate the bond is called or matures. Then, by the Premium/Discount Formula

P = 5000 + (175 − 150)am 3% , (n = 24, 26, 28, 30, 32, 34, 36).

Without knowing which will be the date of call — if any — we take the worstpossible date in order to minimize the price; since am 3% is an increasing function ofm, and is multiplied by a positive number, 25, we minimize by making m as small aspossible, i.e., 2 × 12 = 24:

P = 5000 + (175 − 150)a24 3% = 5000.00 + 25 × 16.9355 = 5, 423.39 .

(b) When the semi-annual yield rate is 4%, the multiplier is negative, 175 − 200 = −25,and we must choose the largest value of m, i.e., m = 36, for a price of

P = 5000 + (175 − 200)a36 4% = 5000.00 − 25 × 18.9083 = 4, 527.29 .

(c) If the bond should be called at the time of any of coupons ##24, . . . , 34, C = 5, 250and

P = 5250 + (175 − 210)am 4%

which is minimized when m = 34: P = 4, 605.61. We must compare this with theprice if not called,

5000 + (175 − 200)a36 4% = 4527.293 ,

so the price will be the latter, 4,527.29.(d) We will have to determine the minimum of

5250 + (175 − 157.50)am 3% if m = 24, 26, 28 (127)5150 + (175 − 154.50)am 3% if m = 30, 32 (128)

5000 + (175 − 150)am 4% if m = 34, 36 (129)


i.e., of

5, 250 + (175 − 157.51)a24 3% = 5, 250 + 17.50 × 16.9355 = 5546.375, 150 + (175 − 154.50)a30 3% = 5, 150 + 20.50 × 19.6004 = 5551.81

5, 000 + (175 − 150)a34 3% = 5, 000 + 25 × 21.1318 = 5528.30

so the price should be 5528.30.

5. (a) Construct a bond amortization schedule for a 4 year bond of face amount 5000,redeemable at 5250 with semiannual coupons, if the coupon rate is 6% and the yieldrate is 7% — both converted semiannually. Use the format


0...

(b) Construct a bond amortization schedule for a 4 year bond of face amount 5000,redeemable at 5250 with semiannual coupons, if the coupon rate is 7% and the yieldrate is 6% — both converted semiannually.

Solution:

(a) By the Basic Formula the purchase price of the bond will be

5, 250(1.035)−8 + 150a83.5% = 3, 986.910670 + 150(6.8740) = 5, 018.01


0 5,018.011 150.00 175.63 -25.63 5,043.64 .2 150.00 176.53 -26.53 5,070.173 150.00 177.46 -27.46 5,097.634 150.00 178.42 -28.42 5,126.055 150.00 179.41 -29.41 5,155.466 150.00 180.44 -30.44 5,185.907 150.00 181.51 -31.51 5,217.418 150.00 182.61 -32.61 5,250.02

(b) By the Basic Formula the purchase price of the bond will now be

5, 250(1.03)−8 + 175a83% = 4, 144.398480 + 175(7.019692190) = 5, 372.84 .



0 5,372.841 175.00 161.19 13.81 5,359.032 175.00 160.77 14.23 5,344.803 175.00 160.34 14.66 5,330.144 175.00 159.90 15.10 5,315.045 175.00 159.45 15.55 5,299.496 175.00 158.98 16.02 5,283.477 175.00 158.50 16.50 5,266.978 175.00 158.01 16.99 5,249.98

6. An 8-year bond of face value 10,000 with semiannual coupons, redeemable at par, ispurchased at a premium to yield 6% convertible semiannually.

(a) If the book value (just after the payment of the coupon) six months before the re-demption date is 10024.27, find the total amount of premium or discount in theoriginal purchase price.

(b) Determine the nominal annual coupon rate of the bond, compounded semiannually.(c) Give the amortization table for the last 2 years.

Solution:

(a) The book value just after the pænultimate41 coupon is

10, 024.27 = 10, 000v + Fr · a10.03

= 10, 000v + Fr · v =10000 + Fr

1.03

soFr = 1.03 × 10, 024.27 − 10000 = 325 .

Knowing the amount of each coupon we can now evaluate the purchase price of thebond to have been

10, 000(1.03)−16 + 325a160.03 = 6, 231.67 + 4, 082.36= 10, 314.03 .

The bond was purchased at a premium of 10, 314.03 − 10, 000 = 314.03.(b) The rate per period was 325

10,000 = 3.25%; hence the nominal rate compounded semi-annually, is 2 × 3.25% = 6.5%.

41preceding the last, i.e., 2nd last


(c) For convenience we could compile this table backwards, beginning with Time=16.We were given that B15 = 10, 024.27. Hence the Principal Adjustment contained inthe 16th coupon is

10, 024.27 − 10, 000 = 24.27.

The book value at Time=15 will be

(10, 000)(1.03)−2 + 325((1.03)−1 + (1.03)−2

)= 10, 047.84 .

The book value at Time=14 will be

(10, 000)(1.03)−3 + 325(1.03−1 + (1.03)−2 + (1.03)−3) = 10, 070.72 .

etc. Alternatively, we could observe by the Basic Formula that the book value justafter the payment of the 12th coupon is

10, 000(1.03)−4 + 325a4 0.03 = 10, 092.93

and so we may construct the table:Time Coupon Interest Principal Book

Value Adjustment Value12 325.00 . . . . . . 10,092.9313 325.00 302.79 22.21 10,070.7214 325.00 302.12 22.88 10,047.8415 325.00 301.44 23.56 10,024.2816 325.00 300.73 24.27 10,000.01

7.3.6 Solutions to Problems on the Class Test

Distribution Date: Mounted on the Web on Saturday, March 12th, 2005Distributed in hard copy on Wednesday, March 16th, 2005

(Subject to correction)

There were 4 versions of this test. The final grades of all were scaled upwards slightly, byslightly different factors, in an attempt to equalize the difficulty of the versions.

Problems on rates of interest and discount1. Problem 1 on Version 1 Showing your work in detail, determine each of the follow-

ing; the rates you determine should be accurate to 4 decimal places, or as a percentageaccurate to 2 decimal places:


(a) [4 MARKS] the nominal annual interest rate, compounded monthly, correspondingto an effective annual interest rate of i = 6%

(b) [4 MARKS] the effective annual interest rate corresponding to a nominal discountrate, compounded quarterly, of d = 3.6%

(c) [4 MARKS] the effective monthly interest rate corresponding to a force of interestof δ = 0.06.

Solution:

(a) The effective interest rate per month is (1.06)1

12 − 1 = 0.4867551%; the nominalannual interest rate, compounded monthly, i(12), is 12 times this rate, 5.8410612%.

(b) 1−d =(1 − d(4)

4

)4= (1−0.009)4 = 0.9644830906. 1+i = 1

1−d = (0.9644830906)−1 =

1.036824813, so the equivalent effective annual interest rate is 3.6824813%.(c) Since δ = 0.05 = ln(1 + i), 1 + i = e0.06.

i(12) = 12(e

0.0612 − 1

)= 12 × 0.5012521% = 6.0150252%

The effective monthly interest rate isi(12)

12= 0.5012521.

2. Problem 2 on Version 2 Showing your work in detail, determine each of the follow-ing; the rates you determine should be accurate to 4 decimal places, or as a percentageaccurate to 2 decimal places:

(a) [4 MARKS] the effective annual discount rate corresponding to a nominal interestrate, compounded quarterly, of i = 2.4%

(b) [4 MARKS] the nominal annual interest rate, compounded quarterly, equivalent toan effective semi-annual discount rate of d = 4%

(c) [4 MARKS] the effective semi-annual interest rate corresponding to a force of inter-est of δ = 0.04.

Solution:

(a) The effective interest rate per quarter is 14 × 2.4% = 0.006. If i and d be the effective

annual interest and discount rates, then

1 − d =1

1 + i=

1(1.006)4 ,

sod = 1 − 1

(1.006)4 = 2.36442751%

(b) The annual discount factor corresponding to an effective semi-annual discount rateof 4% is (1− 0.04)2; the annual accumulation factor will, therefore, be (0.96)−2. The


accumulation factor corresponding to a quarter of a year will be (0.96)−24 , and so the

equivalent effective quarterly interest rate is (0.96)−24 −1; the nominal annual interest

rate, compounded quarterly, is 4 times this, i.e.

4((0.96)

−24 − 1

)= 8.2482904%.

(c) If i be the equivalent effective annual interest rate, then ln(1 + i) = 0.04, so 1 + i =

e0.04. The effective semi-annual interest rate corresponding is√

1 + i − 1 = e0.02 − 1 = 2.0201340% .


(a) [4 MARKS] the effective annual interest rate, corresponding to an nominal annualinterest rate of i = 6% compounded every 4 months.

(b) [4 MARKS] the effective monthly interest rate corresponding to a force of interestof δ = 0.12.

(c) [4 MARKS] the effective annual interest rate corresponding to d(4) = 3.2%

Solution:

(a) A nominal interest rate of 6% compounded every 4 months is equivalent to aneffective interest rate of 2% for a 4-month period. The accumulation factor for1 year will be (1.02)3, and the equivalent annual interest rate will, therefore, be(1.02)3 − 1 = 6.1208%.

(b) If i be the equivalent annual interest rate, ln(1 + i) = 0.12, so 1 + i = e0.12. Theaccumulation factor for 1 month will be (1 + i)

112 = e0.01, so the effective interest rate

for 1 month will be e0.01 − 1 = 1.0050167%.(c) Let i and d denote the effective annual interest and discount rates. Then

1 + i =1

1 − d=

1(1 − d(4)

4

)4 =1

(1 − 0.008)4 = 1.032650385

so the effective annual interest rate is 3.2650385%.


(a) [4 MARKS] the nominal annual interest rate, compounded quarterly, equivalent toan effective semi-annual discount rate of d = 4%


(b) [4 MARKS] the effective semi-annual interest rate corresponding to a force of inter-est of δ = 0.04.

(c) [4 MARKS] the effective annual discount rate corresponding to a nominal interestrate, compounded quarterly, of i = 2.5%

Solution:

(a) The discount factor for a year is (1 − 0.04)2 = (0.96)2. If i is the effective annualinterest rate, then 1 + i = (0.96)−2, and the accumulation factor for one quarter of ayear is (1 + i)

14 = (0.96)−

24 . The effective interest rate for one quarter of a year is,

therefore, (0.96)−24 − 1, and the nominal annual interest rate, compounded quarterly,

is4((0.96)−

24 − 1

)= 4 × 2.0620726% = 8.2482904%.

(b) If i be the equivalent effective annual interest rate, then ln(1 + i) = 0.04, so 1 + i =

e0.04, and (1 + i)12 = e0.02 is the accumulation factor for half a year. The effective

semi-annual interest rate is, therefore

e0.02 − 1 = 2.0201340%.

(c) The accumulation factor for a year is (1 + 0.0254 )4 = 1.025235353, so the effective

annual interest rate is 2.5235353%. If d is the effective annual discount rate, then

1 + d =

(1 +

0.0254

)−4

= 0.9753857951

so d = 2.46142049%.

Problems on the values of annuities and perpetuities with constant payments1. Problem 2 on Version 1 For each of the following sequences of payments, determine,

as of the given time, and for the given interest or discount rate, the value, showing all ofyour work. Before determining the numeric value you are expected to express the valueusing standard symbols.

(a) [6 MARKS] the value now of 20 payments of 1 at the end of every year starting 1year from now, at an interest rate of 4%

(b) [6 MARKS] the value 1 year ago of 10 payments of 1 at the end of every half-year,the first to be paid 5 years from now, at a nominal interest rate of 8% compoundedsemi-annually

(c) [6 MARKS] 300 payments of 1 at the end of every month, as of the date of the 100thpayment, which has just been made; the interest rate is 12% compounded monthly

Solution:


(a) This question was ambiguous. One reading was that the payments would start oneyear from now, in which case the value is

a20 4% =1 − (1.04)−20

0.04= 13.59032635 .

Another reading was that the first payment would be at the end of the year thatstarts one year from now, in which case the preceding value should be multiplied by

v =1

1.04, yielding

v · a20 4% =1 − (1.04)−20

(0.04)(1.04)= 13.06762149 .

(b) The effective half-yearly interest rate is 12 · 8% = 4%. The value of the payments 4.5

years from now is a10 4%. The value 1 year ago — i.e. 5.5 years prior to the valuejust given, is

v11 · a10 4% = (1.04)−11 · 1 − (1.04)−10

0.04= 5.268683237 .

(c) The effective monthly interest rate is 112 · 12% = 1%. The value of the payments just

after the last of them is s300 1%. Precisely 200 months earlier the value is

v−200 · s300 1% =(1.01)100 − (1.01)−200

0.01= 256.8127449 .

2. Problem 3 on Version 2 For each of the following sequences of payments, determine,as of the given time, and for the given interest or discount rate, the value, showing all ofyour work. Before determining the numeric value you are expected to express the valueusing standard symbols.

(a) [6 MARKS] the value of 25 payments of 1 at the end of every year, the last onehaving just been made, at an interest rate of 6%

(b) [6 MARKS] the value 4 months from now of 30 payments of 1 at the end of every4 months, the first to be paid 2 years from now, at a nominal interest rate of 9%compounded 3 times a year

(c) [6 MARKS] 120 payments of 1 at the end of every 2 months, as of the date of the90th payment, which has just been made; the interest rate is 6% compounded every2 months.

Solution:

(a) s25 6% =(1.06)25 − 1

0.06= 54.86451200 .


(b) The effective interest rate per 4 months is 13 (9%) = 3%. The first payment is due

24 months from now; if we wish to interpret the payments as an annuity-immediate,then the value 20 months from now will be a30 3%. The value 4 months from now,i.e., 16 months, or 4

3 of a year before “the clock starts” will be

v4a30 3% = (1.03)−4 · 1 − (1.03)−30

0.03

=(1.03)−4 − (1.03)−34

0.03= 17.41473827 .

(c) The effective interest rate for a 2-month period is 212 · 6% = 1%. At the time of the

120th payment the value of all the payments is s120 1%. 30 payments prior to thatdate, the value is

v30s120 1% = (1.01)−30 · (1.01)120 − 10.01

= 170.6709757 .


(a) [6 MARKS] the value one half-year ago of 12 payments of 1 at the end of everyhalf-year, the first to be paid 4 years from now, at a nominal interest rate of 6%compounded semi-annually

(b) [6 MARKS] 180 payments of 1 at the end of every month, as of the date of the 100thpayment, which has just been made; the interest rate is 18% compounded monthly

(c) [6 MARKS] the value now of 16 payments of 1 at the end of every year starting 1year from now, at an interest rate of 4%

Solution:

(a) The effective semi-annual interest rate is 12 (6%) = 3%. Three and one-half years

from now the payments will be worth a12 3%. One half-year ago, they were worth

v8 · a12 3% =v8 − v20

i=

(1.03)−8 − (1.03)−20

0.03= 7.857782670 .

(b) The effective monthly interest rate is 112(18%) = 1.5%. As of the last payment the

payments are worth s180 1.5%; 80 payments earlier the value is

v80 · s180 1.5% =(1.015)100 − (1.015)−80

0.015= 275.2103668 .


(c) This problem is ambiguous. Does the word “starting” refer to the payments or to theyears? If it refers to the payments, the value is

a16 4% =1 − (1.04)−16

0.04= 11.65229561 .

If the word refers to the years, then one has an annuity-immediate which has beendeferred one year, and the value is

v · a16 4% =1 − (1.04)−16

(0.04)(1.04)= 11.20413039 .


(a) [6 MARKS] 90 payments of 1 at the end of every 2 months, as of the date of the18th payment, which has just been made; the interest rate is 9% compounded every2 months.

(b) [6 MARKS] the value of 20 payments of 1 at the end of every year, the last onehaving just been made, at an interest rate of 5%

(c) [6 MARKS] the value 4 months from now of 45 payments of 1 at the end of every4 months, the first to be paid 3 years from now, at a nominal interest rate of 6%compounded 3 times a year

Solution:

(a) The effective interest rate for a 2-month period is 212(9%) = 1.5%. Two months

before the first payment the value of all the payments is a90 1.5%. Immediately afterthe 18th payment the value of all the payments is

(1.015)18 · a90 1.5% =(1.015)18 − (1.015)−72

0.015= 64.33404251 .

(b) s20 5% =(1.05)20−1

0.05 = 33.06595410 .

(c) The effective interest rate per 4-month period is 412(6%) = 2%. Two and two-thirds

years from now the value of the payments is a45 2%. Hence one-third year from nowthe value will be

(1.02)−7 · a45 2% =(1.02)−7 − (1.02)−52

0.02= 25.67295884 .


Problems on increasing and decreasing annuities and perpetuities1. Problem 3(a) on Version 1 [6 MARKS] An annuity at interest rate i consists of payments

of 10 now, 12 at the end of 1 year, 14 at the end of 2 years, increasing by a constantamount until the last payment in the amount of 40, is to be evaluated as of 1 year ago.Express its value in terms of symbols (Ia)n, (Is)n, an, sn, i, v, but do not evaluate.Solution: The total increase of 40−10 = 30 will be spread over 40−10

2 = 15 years: the lastpayment will be 15 years from now. The value 1 year ago was

8 · a16 + 2(Ia)16

2. Problem 4(a) on Version 2 [6 MARKS] An annuity at interest rate i consists of paymentsof 100 now, 95 at the end of 1 year, 90 at the end of 2 years, decreasing by a constantamount until the last payment in the amount of 25, is to be evaluated as of the time of thepayment of 95. Express its value then in terms of symbols (Ia)n, (Is)n, an, sn, i, v, but donot evaluate.Solution: The total decrease of the annuity is 75, at 5 per year; so the last payment willbe 75

5 = 15 years from now. The value is

100(1 + i) + 95(1 + a14

)− 5(Ia)14

3. Problem 1(a) on Version 3 [6 MARKS] An annuity at interest rate i consists of paymentsof 10 now, 13 at the end of 1 year, 16 at the end of 2 years, increasing by a constantamount until the last payment in the amount of 40, and is to be evaluated as of 2 yearsago. Express its value in terms of symbols (Ia)n, (Is)n, an, sn, i, v, but do not evaluate.Solution: The increase of 40 − 10 = 30 will be spread over 30

3 = 10 years. The value ofthe annuity today is 10a11 + 3(Ia)10. As of 2 years ago the value was

10v · a11 + 3v2 · (Ia)10

4. Problem 2(a) on Version 4 [6 MARKS] An annuity at interest rate i consists of paymentsof 118 now, 112 at the end of 1 year, 106 at the end of 2 years, decreasing until the lastpayment in the amount of 34, the totality to be evaluated as of the time of the payment of112. Express its value in terms of symbols (Ia)n, (Is)n, an, sn, i, v, but do not evaluate.Solution: The total decrease of 118 − 34 = 84 is to be spread over 84

6 = 14 years. Thevalue of the payments at time 1 is

118(1 + i) + 112 + 112a14 − 6(Ia)13


Problems on combinations of annuities and perpetuities1. Problem 3(b) on Version 1 [12 MARKS] The accumulated value just after the last pay-

ment under a 15-year annuity of 1000 per year, paying interest at the rate of 8% perannum effective, is to be used to purchase a perpetuity, first payment to be made 2 yearsafter the last payment under the annuity. Showing all your work, determine the size ofthe payments under the perpetuity, assuming that the interest rate from the time of thelast payment under the 15-year annuity is 5%.Solution: Let the payment under the perpetuity be X. Then the equation of value at time15 is

1000 · s15 8% =X

0.05− X

1.05=

X(0.05)(1.05)

since the first payment of the perpetuity-immediate “whose clock begins to tick at time15” is not made. Solving the equation yields X = 1425.49.

2. Problem 4(b) on Version 2 [12 MARKS] The accumulated value just after the last pay-ment under a 12-year annuity of 1000 per year, paying interest at the rate of 6% perannum effective, is to be used to purchase a 10-year annuity at an interest rate of 7%, firstpayment to be made 4 years after the last payment under the annuity. Showing all yourwork, determine the size of the payments under the 10-year annuity. Assume that the 7%rate begins from the time of the last payment under the 12-year annuity.Solution: Let X denote the payment under the 10-year annuity. The value of the paymentsunder the 12-year annuity at the time of the last of them is 1000 · s12 6% = (1.07)−3a10 7%.The value of the payments under the deferred 10-year annuity, as of that same date, isX

(a13 7% − a3 7%

). Equating these amounts and solving, we obtain

X =1000 · s12 6%

a13 7% − a3 7%

=100(1.07)3s12 6%

a10 7%

= 2942.43

3. Problem 1(b) on Version 3 [12 MARKS] The accumulated value just after the last pay-ment under a 15-year annuity of 1000 per year, paying interest at the rate of 9% perannum effective, is to be used to purchase a perpetuity at an interest rate of 8%, firstpayment to be made at the same time as the last payment under the annuity. Showing allyour work, determine the size of the payments under the perpetuity.Solution: Denote the amount of the payments under the perpetuity by X. The equation ofvalue at time 15 is

1000 · s15 9% = X · a∞ 8% = X · 1.080.08

implying that X =0.08×1000×((1.09)15−1)

1.08×0.09 = 2174.88


4. Problem 2(b) on Version 4 [12 MARKS] The accumulated value just after the last pay-ment under a 14-year annuity of 1000 per year, paying interest at the rate of 10% perannum effective, is to be used to purchase a 12-year annuity-immediate at an interest rateof 7%, first payment to be made 1 year after the last payment under the 14-year annu-ity. Showing all your work, determine the size of the payments under the 12-year annuity.Assume that the 7% rate applies from the time of the last payment under the 10% annuity.Solution: If we denote the amount of the payments under the 12-year annuity by X, theequation of value at time 14 is

1000 · s14 10% = X · a12 7%

implying that X = 3522.11.

Problems on drop and balloon payments1. Problem 4 on Version 1 [18 MARKS] A loan of 6000 is to be repaid by annual payments

of 350 to commence at the end of the 1st year, and to continue thereafter for as long asnecessary. Find the time and amount of the final payment if the final payment is to belarger than the regular payments. Assume i = 4%.Solution: The balloon payment will take place at the time — call it n — which is thelargest solution of the inequality

350 · an 4% < 6000

i.e., of the inequalities

1 − (1.04)−n

0.04<

6000350

(1.04)−n >110350

n <ln 350 − ln 110

ln 1.04= 29.51126321 .

It follows that the last payment is at the end of the 29th year. The amount of the paymentis

6000(1.04)29 − 350(1.04)((1.04)28 − 1

0.04

)= 523.70851

2. Problem 1 on Version 2 [18 MARKS] A loan of 7000 is to be repaid by annual paymentsof 450 to commence immediately, and to continue at the beginning of each year for aslong as necessary. Find the time and amount of the final payment if the final payment isto be no larger than the regular payments. Assume i = 6%.


Solution: The drop payment will take place at the time — call it n — which is the smallestsolution of the inequality

450 · an+1 6% ≥ 7000

i.e., of the inequalities

(1.06) · 1 − (1.06)−(n+1)

0.06)≥ 7000

450

(1.06)−(n+1) ≤ 1 − 7000450

· 6106

=19

159

n + 1 ≥ − ln 19159

ln 1.06= 36.45967106

implying that n = 36. The amount of the final drop payment will be

7000(1.06)36 − 450s36 6% = 7000(1.06)36 − (450) · 1.060.06

((1.06)36 − 1

)

= 210.11059

so the final payment is in the amount of 210.11.

3. Problem 2 on Version 3 [18 MARKS] A loan of 8000 is to be repaid by annual paymentsof 500 to commence at the end of the 1st year, and to continue thereafter for as long asnecessary. Find the time and amount of the final payment if the final payment is to besmaller than the regular payments. Assume i = 5%.Solution: Let the final payment be made at time n, which will be the smallest integer suchthat

500 · an 5% > 8000⇔ 1 − (1.05)−n > 16(0.05) = 0.8

⇔ n > − ln 0.2ln 1.05

= 32.98693373 ,

implying that n = 33. If the present value of the deficiency of the last payment from 500be denoted by X, then

X = 500 · an 5% − 8000

= 500(1 − (1.05)−33

0.05− 16

)= 1.27460500

so the final payment is

500.00 − (1.27460500)(1.05)33 = 493.6229109

or 493.62.


4. Problem 3 on Version 4 [18 MARKS] A loan of 9000 is to be repaid by annual paymentsof 800 to commence immediately, and to continue at the beginning of each year for aslong as necessary. Find the time and amount of the final payment if the final payment isto be no smaller than the regular payments. Assume i = 8%.Solution: Let the final payment be at time n, i.e., be the (n + 1)st payment. Then n willbe the largest integer such that

800 · an+1 8% ≤ 9000

⇔ an+1 8% ≤908

⇔ 1 − 1.08−(n+1) ≤ 908· 0.08

1.08

⇔ n + 1 ≤ ln 6ln 1.08

= 23.28138292

from which it follows that n = 22. The excess of the final payment over 800 will be

9000(1.08)22 − 800 · s23 = 9000(1.08)22 − 800 · (1.08)23 − 10.08

= 214.22726 ,

The final payment will, therefore, be 800 + 214.22726 = 1014.23.

7.3.7 Final Examination 2004/2005

McGILL UNIVERSITY — FACULTY OF SCIENCE— FINAL EXAMINATION

MATH 329 2005 01 — THEORY OF INTEREST

Instructions

1. Fill in the above clearly.

2. Do not tear pages from this book; all your writing — even rough work — must be handedin.

3. Calculators. While you are permitted to use a calculator to perform arithmetic and/orexponential calculations, you must not use the calculator to calculate such actuarial func-tions as ani, sni, (Ia)ni, (Is)ni, (Da)ni, (Ds)ni, etc. without first stating a formula for the


value of the function in terms of exponentials and/or polynomials involving n and theinterest rate. You must not use your calculator in any programmed calculations. If yourcalculator has memories, you are expected to have cleared them before the examination.

4. This examination booklet consists of this cover, Pages 1 through 7 containing questions;and Pages 8 and 9, which are blank. For all problems you are expected to show all yourwork, and to simplify algebraic and numerical answers as much as you can. All solutionsare to be written in the space provided on the page where the question is printed. Whenthat space is exhausted, you may write on the facing page. Any solution may be continuedon the last pages, or the back cover of the booklet, but you must indicate any continuationclearly at the bottom of the page where the question is printed! You may do rough workanywhere in the booklet.

5. You are advised to spend the first few minutes scanning the problems. (Please inform theinvigilator if you find that your booklet is defective.)

6. Several useful formulas are printed on page 3143. You should not assume that any ofthese formulas is/are required in the solution of any of the problems on this examination.

1. Showing your work in detail, determine each of the following; the rates you determineshould be accurate to 4 decimal places, or as a percentage accurate to 2 decimal places.

(a) [2 MARKS] The nominal annual interest rate, compounded quarterly, correspondingto an effective annual interest rate of 8%.

(b) [2 MARKS] The effective annual interest rate corresponding to a nominal discountrate, compounded monthly, of 6%.

(c) [3 MARKS] The effective semi-annual interest rate corresponding to a force of in-terest of δ = 0.04.

(d) [3 MARKS] d( 12 ), corresponding to a nominal annual interest rate, compounded ev-

ery 2 months, of 12%.

2. Showing all your work, determine for each of the following sequences of payments, thevalue as of the given time and for the given interest or discount rate.

(a) [3 MARKS] The value as of one year ago of 30 payments of 1 at the end of everyhalf-year, the first to be paid 18 months from now, at a nominal interest rate of 9%compounded semi-annually.

(b) [3 MARKS] The value now of 24 payments of 1 at the end of every 3 months, thefirst to be paid one year from now, at an effective annual interest rate of 9%.

(c) [4 MARKS] The value exactly 5 years from now of a perpetuity consisting of anunending sequence of payments of 1 at intervals of one year, the first to be paid in6 months from now, at a nominal annual interest rate of 4% compounded every 4months.


3. [5 MARKS] Give a formula in terms of i alone, for the value of the following, showingall your work:. An annuity at an effective semiannual interest rate of i, of 100 now, 97 sixmonths from now, 94 one year from now, the payments decreasing by a constant amountuntil a final payment of 34.

Table 6: Several Useful Formulæ that you were not expected to memorize

You may wish to make use of some of the following formulæ, but remember that your final

answers for this problem must be expressed in terms of i alone.

(Ia)n i =an i − nvn

i (Is)n i =sn i − n

i

(Is)n i =sn+1 i

− (n+1)

i (Ia)∞ i =a∞ i

i

(Da)n i =n − an i

i (Ds)n i =n(1+i)n − sn i

i

4. The purchase of a new condominium is partially financed by a mortgage of 120,000payable to the vendor; the mortgage is amortized over 25 years, with a level payment atthe end of each month, at a nominal annual rate of 6% compounded monthly.

(a) [2 MARKS] Determine the level monthly payments under this mortgage.(b) [4 MARKS] Divide the 120th payment into principal and interest.(c) [9 MARKS] When the payments from the mortgagor start to arrive, the vendor de-

posits them into an account earning 4% per annum compounded monthly. He willcontinue to do so until the account contains exactly 120,000, with the last depositbeing possibly smaller than the others. Determine the time and amount of that lastdeposit.

5. The Smiths have sold their house for 500,000. They wish to purchase a 20-year annuity-certain with the money, but are concerned that living costs will be rising, and should beplanned for by having the amount of the payment in any month 1.005 times the amountof the payment for the preceding month. The first payment is to be 1 month after the sale.The interest rate is 4% per annum, compounded monthly.

(a) [4 MARKS] Showing all your work, determine the amount of the first payment.


(b) [2 MARKS] Determine the amount of the last payment, to be made 20 years fromnow.

(c) [4 MARKS] Suppose that, with the same first payment, the annuity payments willincrease by a constant amount and not by a constant factor. Showing all your work,determine the amount of the last payment. (You may wish to use the formulæ inTable 3 on page 3143 of this examination.)


You may wish to make use of some of the following formulæ, but remember that your final

answers for this problem must be expressed in terms of i alone.



i

(Is)n i =sn+1 i

− (n+1)

i (Ia)∞ i =a∞ i

i

(Da)n i =n − an i


i

6. X sold Y his home for 500,000. In addition to a down payment of 100,000 cash, Ymortgaged the property to X for 400,000. The mortgage provides for level quarterlypayments over 25 years, at a nominal rate of 4% compounded every 3 months.

(a) [5 MARKS] Construct an amortization table showing the first 6 payments, under thefollowing headings:

Payment Payment Interest Principal OutstandingNumber Amount Paid Repaid Loan Balance

01 . . . . . . . . . . . .

. . . . . . . . . . . . . . .6 . . . . . . . . . . . .

(b) [5 MARKS] After 10 years — just after receiving the 40th payment — X sells the(remaining payments of the) mortgage to Z at a price to yield 6% convertible quar-terly. Determine the price P that Z pays, and construct the first 3 and last 3 lines of


an amortization table for the 60 payments showing, under the following headings,how Z is recovering her investment, under the following headings:

Payment Payment Interest Principal OutstandingNumber Amount at new rate Repaid Loan Balance

0 P1 . . . . . . . . . . . .

. . . . . . . . . . . . . . .3 . . . . . . . . . . . .

58. . . . . . . . . . . . . . .59. . . . . . . . . . . . . . .60. . . . . . . . . . . . . . .

7. Consider a 10,000 par-value 10-year bond with semi-annual coupons paying interest at5% compounded semiannually. Suppose that the bond can be redeemed for 11,000 atthe time of either of the 16th or 17th coupons, at 10,500 at the time of the 18th or 19thcoupons, or can be held until maturity, at the time of the 20th coupon, where it matureswithout premium.

(a) [5 MARKS] What price should an investor pay in order to be certain of a yield rateof 6% compounded semiannually?

(b) [5 MARKS] What price should an investor pay in order to be certain of a yield rateof 4% compounded semiannually?

7.3.8 Supplemental/Deferred Examination 2004/2005

Instructions



3. Calculators. While you are permitted to use a calculator to perform arithmetic and/orexponential calculations, you must not use the calculator to calculate such actuarial func-tions as ani, sni, (Ia)ni, (Is)ni, (Da)ni, (Ds)ni, etc. without first stating a formula for thevalue of the function in terms of exponentials and/or polynomials involving n and theinterest rate. You must not use your calculator in any programmed calculations. If yourcalculator has memories, you are expected to have cleared them before the examination.

4. This examination booklet consists of this cover, Pages 1 through 7 containing questions;and Pages 8 and 9, which are blank. For all problems you are expected to show all yourwork, and to simplify algebraic and numerical answers as much as you can. All solutionsare to be written in the space provided on the page where the question is printed. When


that space is exhausted, you may write on the facing page. Any solution may be continuedon the last pages, or the back cover of the booklet, but you must indicate any continuationclearly at the bottom of the page where the question is printed! You may do rough workanywhere in the booklet.




(a) [2 MARKS] The effective annual discount rate corresponding to a nominal interestrate, compounded monthly, of 9%.

(b) [2 MARKS] The nominal annual interest rate, compounded quarterly, correspondingto an effective annual interest rate of 6%.

(c) [3 MARKS] d( 13 ), corresponding to a nominal annual interest rate, compounded ev-

ery 2 months, of 6%.(d) [3 MARKS] The effective quarterly interest rate corresponding to a force of interest

of δ = 0.06.


(a) [3 MARKS] The value now of 36 payments, each of 1, at the end of every 4 months,the first to be paid one year from now, at an effective annual interest rate of 9%.

(b) [3 MARKS] The value as of 6 months ago of 30 payments of 1 at the end of everyhalf-year, the first to be paid 18 months from now, at a nominal interest rate of 9%compounded semi-annually.

(c) [4 MARKS] The value exactly 7 years from now of a perpetuity consisting of anunending sequence of payments of 1 at intervals of one year, the first to be paid in12 months from now, at a nominal annual interest rate of 12% compounded every 4months.

3. [5 MARKS] Give a formula for the value of the following, showing all your work:. Anannuity at an effective semiannual interest rate of i, of 100 now, 103 six months fromnow, 106 one year from now, the payments increasing by a constant amount until a finalpayment of 340. Express your formula in terms of any of the functions i, v, ani, sni, (Ia)ni,(Is)ni, (Da)ni, (Ds)ni.

4. The purchase of a new home is partially financed by a mortgage of 150,000 payable tothe vendor; the mortgage is amortized over 25 years, with a level payment at the end ofeach month, at a nominal annual rate of 9% compounded monthly.



If you wish, you may make use of the following formulæ.



i

(Is)n i =sn+1 i

− (n+1)

i (Ia)∞ i =a∞ i

i

(Da)n i =n − an i


i

(a) [2 MARKS] Determine the level monthly payments under this mortgage.(b) [4 MARKS] Divide the 100th payment into principal and interest.(c) [9 MARKS] When the payments from the mortgagor arrive, the vendor deposits

them into an account earning 6% per annum compounded monthly. He will continueto do so until the account contains exactly 150,000, with the last deposit being atleast as large as the others. Determine the time and amount of that last deposit.

5. The Trudeaus have sold their house for 500,000. They wish to purchase a 24-yearannuity-certain with the money, but are concerned that living costs will be rising, andshould be planned for by having the amount of the payment in any month 1.005 times theamount of the payment for the preceding month. The first payment is to be 1 month afterthe sale. The interest rate is 6% per annum, compounded monthly.

(a) [4 MARKS] Showing all your work, determine the amount of the first payment.(b) [2 MARKS] Determine the amount of the last payment, to be made 24 years from

now.(c) [4 MARKS] Suppose that, with the same first payment, the annuity payments will

increase by a constant amount and not by a constant factor. Showing all your work,determine the amount of the last payment. (You may wish to use the formulæ inTable 3 on page 3147 of this examination.)

6. X sold Y his home for 500,000. In addition to a down payment of 100,000 cash, Ymortgaged the property to X for 400,000. The mortgage provides for level quarterlypayments over 20 years, at a nominal rate of 4% compounded every 3 months.

(a) [5 MARKS] Construct an amortization table showing the first 6 payments, under the


following headings:Payment Payment Interest Principal OutstandingNumber Amount Paid Repaid Loan Balance

01 . . . . . . . . . . . .

. . . . . . . . . . . . . . .6 . . . . . . . . . . . .

(b) [5 MARKS] After 5 years — just after receiving the 20th payment — X sells the(remaining payments of the) mortgage to Z at a price to yield 6% convertible quar-terly. Determine the price P that Z pays, and construct the first 3 and last 3 lines ofan amortization table for the remaining 60 payments showing, under the followingheadings, how Z is recovering her investment, under the following headings:

Payment Payment Interest Principal OutstandingNumber Amount at new rate Repaid Loan Balance

0 P1 . . . . . . . . . . . .

. . . . . . . . . . . . . . .3 . . . . . . . . . . . .

58. . . . . . . . . . . . . . .59. . . . . . . . . . . . . . .60. . . . . . . . . . . . . . .

7. Consider a 10,000 par-value 20-year bond with semi-annual coupons paying interest at5% compounded semiannually. Suppose that the bond can be redeemed for 11,000 atthe time of any of coupons ##14, 15, 16, 17, 18, 19, 20; at 10,500 at the time of anyof coupons ##22, 24, 26, 28, 30, 32, 34, 36, 38; or can be held until maturity, where itmatures without premium.

(a) [4 MARKS] What price should an investor pay in order to be certain of a yield rateof 6% compounded semiannually?

(b) [6 MARKS] What price should an investor pay in order to be certain of a yield rateof 4% compounded semiannually?

7.4 2008/20097.4.1 First Problem Assignment, with Solutions

These problems were to be solved with full solutions, modelled either on the solutions toproblems in the textbook, presented in class, or in the notes on the Web for this or previousyears. The essence is that the reader should be able to reconstruct every step of the prooffrom what you have written: getting the right answer is never enough. Students were advised


that they were “not being graded for elegance, but simply for the proof being logical, withoutserious gaps. While the data given may sometimes not justify a large number of decimalplaces, you should show your intermediate calculations in sufficient detail that the grader candetermine that your calculations are correct.”

1. Find the accumulated value of 1000 at the end of 4 years

(a) if the nominal annual rate of discount is 8%, convertible quarterly;(b) if the nominal annual rate of interest is 8%, convertible quarterly;(c) if the nominal annual rate of interest is 100%, convertible annually;(d) if the nominal annual rate of discount is 50%, convertible annually.

Solution:

(a) The accumulated value is 1000(1 − 0.02)−4×4 = 13861.60248 or 1,381.60.(b) The accumulated value is 1000(1.02)4×4 = 1372.7857 or 1,372.79.(c) The accumulated value is 1000(1 + 1)4 = 16, 000.(d) The accumulated value is 1000(0.5)−4 = 16, 000.

2. Find the nominal rate of discount convertible every 3 months which is equivalent to anominal rate of interest of 8% converted every 2 years.Solution: An equation of value is (cf. [5, equation (1.23a)])

(1 − d(4)

4

)−4

=

1 +i(

12 )12

12

which implies that

1 − d(4)

4=

1 +0.08

12

− 1

8

= 1.16−18 ,

implying thatd(4) = 4

(1 − 1.16−

18)

= 0.0735258496

or 7.35%.

3. A treasury bill which costs 99.00 matures after 6 months at 100.00

(a) What is the effective annual rate of discount?(b) What is the effective annual rate of interest?(c) What is the nominal annual rate of discount compounded half-yearly?(d) What is the nominal annual rate of interest compounded every 4 months?(e) If interest is interpreted as being simple, what is the annual rate of simple interest?

Solution:


(a) If d is the effective annual rate of discount, then an equation of value is

99 = (1 − d)12 100 ,

hence d = 1 − (0.99)2 = 1.99%.(b) If i is the effective annual rate of interest, then an equation of value is

100 = (1 + i)12 99 ,

hence i =

(10099

)2

− 1 = 0.020304050 or 2.03%.

(c) An equation of value is

99 =

(1 − d(2)

2

)100 ,

hence d(2) = 2.00%.(d) An equation of value is

100 =

(1 +

i(3)

3

) 64

99 ,

hence i(3) = 3

(10099

) 23

− 1

= 0.020168163 or 2.017%.

(e) If i is the rate of simple interest, then 1+i2

=10099⇒ i = 2

(10099− 1

)= 0.020202020

or 2.02%.

4. (a) Find the length of time for 10,000 to accumulate at compound interest to 25,000, ifinvested at 6.00% per annum, accumulated monthly.

(b) Find the length of time for 10,000 to accumulate to 25,000, if interest is compoundedmonthly for full months, but is simple 6% annual interest for proper fractions of amonth. (In this part of the problem it is important that you show all your work care-fully, as the grader may be unable to reconstruct the reasoning from the computationsalone.)

Solution:

(a) If n denotes the number of years, then an equation of value is

(1 +

0.0612

)12n

10000 = 25000 ,

so (1.005)12n = 2.5000, and n = ln 2.5ln 1.005 = 15.30965924 or 15.31 years.


(b) We have proved in the preceding part that the number of months required is approxi-mately 15.30965924×12 = 183.7, so the time will be 183 full months plus a fractionof a month. In 183 full months 10000 accumulates at compound interest to

(1 +

0.0612

)183

10000 = 24910.89 .

Let t denote the fraction of a month needed to this amount to accumulate at simpleinterest to 25000. Then an equation of value is

(1 +

0.0612

)183 (1 + 0.06 · t

12

)10000 ,

yielding

t =

2500010000(1.005)183 − 1

0.005= 0.7154036000

or 0.72 months. Thus the total time required for the accumulation is 183.72 months,

or 15+3.7212

= 15.31 years. The condition that interest be simple has not been strongenough to affect the final answer (because of the low rate of interest, and the shorttime involved).

5. (cf. [5, Example 1.16, p. 35]) Suppose that money accumulates at an increasing force ofinterest δt = 0.04 + 0.005t for 0 ≤ t ≤ 20.

(a) Find the value at time t = 2 of three payments of 50 to be paid at times t = 0, 1, 2.(b) Find the constant rate of interest compounded annually which would produce the

same accumulation at time t = 2.

(Note: Since the force of interest is not constant, this is not compound interest, and youmust not use results like δ = ln(1 + i), which was proved on the assumption of compoundinterest.)Solution:

(a) Using formula [5, (1.27), p. 29]), which states that a(t) = e

t∫0δr dr

, we find the accu-mulation at time t = 2 to be

50

e2∫

0δr dr

+ e

2∫1δr dr

+ e

2∫2δr dr

= 50

e2∫

0(0.04+0.005r) dr

+ e

2∫1

(0.04+0.005r) dr+ e

2∫2

(0.04+0.005r) dr


= 50(e[0.04r+0.0025r2]2

0 + e[0.04r+0.0025r2]21 + e[0.04r+0.0025r2]2

2

)

= 50(e0.09 + e0.0475 + e0

)

= 50(1.094174284 + 1.048646201 + 1) = 157.1410242 .

(b) Suppose the annual rate of compound interest is i. Then an equation of value is

50((1 + i)2 + (1 + i)1 + 1

)= 156.8795170 .

This is a quadratic equation, equivalent to (1 + i)2 + (1 + i) − 2.13759034 = 0 .Solving by completion of the square yields i+ 3

2 = ±√2.38759034 = ±1.545182947,implying that 0 = 0.045182947 or i = −3.045182947. The second root is absurd,and extraneous. Hence the equivalent constant rate of compound interest for thisparticular configuration of payments is 4.52%.

7.4.2 Second Problem Assignment, with Solutions

These problems are to be solved with full solutions, modelled either on the solutionsto problems in the textbook, presented in class, or in the notes on the Web for thisor previous years. The essence is that the reader should be able to reconstruct everystep of the proof from what you have written: getting the right answer is neverenough. You are not being graded for elegance, but simply for the proof beinglogical, without serious gaps. While the data given may sometimes not justify alarge number of decimal places, you should show your intermediate calculations insufficient detail that the grader can determine that your calculations are correct.

1. K has 3,000 in a bank account that pays interest monthly at the nominal annual ratei(12) = 1.8%. She needs to withdraw 5,000 a year from now to pay her tuition, but plansto make equal deposits 3 months from now and 9 months from now.

(a) Showing all your work, and using an equation of value, determine what the size ofthese equal deposits should be in order that the balance in the account 13 monthsfrom now will be 1,000.

(b) Repeat the calculations if the interest rate for the account is now an effective annualrate of 1.8%; there is no change in the timing of the payments or the compoundingof interest.

(Since the interest rate is very low here, it is important that you not round your computa-tions too early.)Solution:(a) Denote the payment that must be made 3 months from now, and again 9 months from

now by X. The effective interest rate per month isi(12)

12= 0.15%. An equation of


value as of a comparison date 13 months from now is

3000(1 +

i(12)

12

)13

+ X(1 +

i(12)

12

)10

+ X(1 +

i(12)

12

)4

− 5000(1 +

i(12)

12

)1

= 1000

⇔ 3000(1.0015)13 + X(1.0015)10 + X(1.0015)4 − 5000(1.0015) = 1000

⇔ X =1000 − 3000(1.0015)13 + 5000(1.0015)

(1.0015)10 + (1.0015)4 = 1458.833537

so the amount to be deposited at each of the two given times is 1,458.83.

(b) Since(1 + i(12)

12

)12= 1 + i = 1.018,

1 +i(12)

12= 1.018

112 = 1.001487765 ,

so the amount of each of the equal payments will be

X =1000 − 3000(1.001487765)13 + 5000(1.001487765)

1.00148776510 + 1.0014877654 = 1459.168645

or 1,459.17.2. Funds A and B have not been receiving contributions for the past 10 years, and will

receive no future contributions. Fund A has been accumulating, and will accumulate atan effective annual rate of 7%; Fund B has been accumulating and will accumulate at anominal annual rate of 8% compounded semi-annually. At the end of 30 years the totalof the two funds will be 50,000. At the end of 12 years Fund A will be 90% the size ofFund B.(a) Determine the values of each of the funds as of 4 years ago.(b) Determine when the two funds will be equal.Solution: Denote the respective values of Funds A and B now by A and B. Then the twohypotheses give rise to the following equations:

A(1.07)30 + B(1.04)60 = 50000A(1.07)12 = 0.9B(1.04)24

We can solve these two linear equations in two unknowns by eliminating one variablebetween them and thereby determine first one variable, then the other. The solutionsthereby obtained by these (pre-college) operations are

A =0.9(50000)

0.9(1.07)30 + (1.04)36 · (1.07)12 = 2796.095323

B =50000

0.9(1.04)24 · (1.07)18 + (1.04)60 = 2729.698321


so we may take A = 2, 796.10, B = 2, 729.70.

(a) As of 4 years ago Fund A had value 2, 796.10(1.07)−4 = 2, 133.13, while Fund Bhad value 2, 729.70(1.04)−8 = 1994.57.

(b) The funds will be equal at time t, when A · (1.07)t = B · (1.04)2t , equivalently, when(

1.07(1.04)2

)t

=BA,

which we can solve by taking logarithms:

t =ln B − ln A

ln 1.07 − 2 ln 1.04= 2.228815096

i.e., 2.229 years from now (i.e., approximately 2 years and 84 days from now).

Some students read the phrase “At the end of 30 years” differently from what was in-tended, and counted the 30 years from 10 years ago. While this reading could be justified,some of these students appear to have read “as of 4 years ago” as meaning 6 years in thefuture, and that reading is perhaps not so obvious. In any case, the grader accepted thesereadings, and advises that “the answers (were) A =4370.33andB =4196.18.”

3. Dr. X’s daughter has given birth to his first granddaughter today, and the doctor wishes toestablish a fund to pay the new baby’s tuition expenses in medical school 20 years fromnow. If he assumes that the fund will need to contain 1,000,000 in 20 years, how muchshould he invest semi-annually now? He plans to deposit the same fixed amount at theend of every half-year for 7 years, and wishes to base his computations on an effectiveannual interest rate of 6%; the first deposit will be 6 months from now.Solution: Let i denote the effective semi-annual interest rate, and X denote the amount ofDr. X’s semi-annual contribution. Then (1+ i)2 = 1.06, so i =

√1.06−1 = 0.029563014.

An equation of value as of today is X · a20 i = 1000000(1.06)−20, implying that

X =1000000(1.06)−20(0.029563014)

1 − (1.029563014)14 =1000000(1.06)−20(0.029563014)

1 − (1.06)7 = 27520.77408 ,

so Dr. X’s semi-annual contribution for seven years should be 27,520.77.

4. Mary purchases a high definition flat-screen TV worth 500 by paying an immediate“down” payment and monthly payments of 25 for 15 months.

(a) If the monthly payments begin one month after the purchase, and the nominal an-nual interest rate compounded monthly is 20%, determine the amount of the downpayment.


(b) The store offers customers the option of delaying the first payment until 1 year fromthe date of purchase. If Mary elects to accept this option, and wishes to pay the samedown payment now as previously, what should be the amount of each of her 15 equalregular payments? (Hint: One way to approach this problem — not the only way —is to express the [deferred] payment annuity as the difference of two annuities, bothwith first payment 1 month from now.)

Solution:

(a) Let D denote the down payment. Then an equation of value is

D + 25a15 2012 % = 500 ⇒ D = 500 − 25 · 1 − (1.0166666667)−15

0.0166666667⇒ D = 500 − 25(13.17557219) = 170.6106952 ,

so the down payment will be 170.61.(b) Let the amount of each of the regular payments be X. We can think of the payment

annuity as part of an annuity beginning 1 month from now, but extending for 26months; we will have to subtract the present value of the first 11 payments. Anequation of value now is

170.61 + X ·(a26 20

12 % − a11 2012 %

)= 500

⇒ X =329.39(

1−(1.066666667)−26

0.0166666667

)−

(1−(1.066666667)−11

0.0166666667

)

⇒ X =329.39(0.0166666667)

(1.066666667)−11 − (1.066666667)−26 = 29.98509009 ,

so the regular monthly payments should be in the amount 29.99. We can also think ofeach of the previous payments now being delayed by 11 months; thus the paymentswill each be

25(1.0166666667)11 = 29.98502678

or 29.99.

7.4.3 Third Problem Assignment, with Solutions

1. X borrows 20,000 and starts repaying it with semi-annual payments of 1,200, at an effec-tive annual rate of 6%. After he makes 10 payments he is laid off from his job, and asksthe lenders to permit him to delay repayment. The lenders permit him to skip just onepayment, but tell him that interest rates have risen, and that they will agree to the delayonly if X agrees to accept the new rates as of the last payment made. X counters thathe can’t pay more than 1,200 per half-year, and agrees that, when he resumes payments


after the one missed payment, the payments will be at a constant level of not more than1,200 per half year at a nominal annual rate of 8%, compounded semi-annually.

(a) Determine, as of the date of the last payment made at the effective annual rate of 6%,the amount of loan still outstanding.

(b) Determine, as of the date of the missed payment, the amount L of loan still outstand-ing.

(c) By solving an inequality for the smallest value of n such that 1200·an ≥ L, determinethe minimum number n of payments that X will still have to make.

(d) For the minimum number of payments n that you have determined above, find theexact amount of each of the equal payments.

(e) Suppose that X had had the foresight to know, at the time he was borrowing 20,000,that he would not be able to make the projected 11th payment. Determine, as of theoriginal date of the loan, the number m of payments he would have to make after themissed payment, if all except the last of the 10 + m payments are to be equal 1,200,and the last “drop” payment could be smaller.

(f) Continuing the discussion in the previous item, suppose that, knowing that he wasgoing to have to make 10 + 0 + m payments not to exceed 1200, X contracted atthe time of the loan to make all of these payments equal (except for the paymentof 0 five and one-half years into the loan). What would be the amount of theseequal payments? (Hint: There are different ways of approaching this problem. Onemethod would be to proceed as above:• Assume the first 10 payments are at the full amount of 1,200, and determine the

amount L outstanding at the time of the missed payment.• Determine the smallest m such that 1200 · am ≥ L.• Then determine the exact level amount of the 10+0+m payments (remembering

that there is a hiatus between the 10th payment and the next).)

Solution:

(a) The effective annual interest rate is 6%. The corresponding nominal annual interest

rate, compounded semi-annually, is i(2), given by(1 +

i(2)

2

)2

= 1.06; the effective

semi-annual interest rate is i(2)

2 = (1.06)12 −1, which I will denote by j: computations

of the value of the annuity of payments must be based on this interest rate; (1 + j)2 =

1.06. Immediately after the 10th payment the value of the payments already made is

1200 · s10 j = 1200 · (1 + j)10 − 1j

= 1200 · (1.06)5 − 1

(1.06)12 − 1


= 13729.00252

The amount of the loan outstanding just after the 10th payment is

20000(1.06)5 − 13729.00252 = 13035.50904

or 13,035.51. This amount will be repaid by an annuity-immediate of level paymentsof at most 1,200, one payment deferred, at an effective interest rate per period of8%2 = 4% per period.

(b) At the time of the missed payment the outstanding loan has grown to13, 035.51(1.04) = 13556.9304 or L = 13, 556.93.

(c)

1200 · an 4% ≥ 13, 556.93 ⇔ 1200 · 1 − (1.04)−n

0.04≥ 13, 556.93

⇔ (1.04)−n ≤ 1 − 13556.93 · 0.041200

= .5481023333

⇔ −n ≤ ln .5481023333ln 1.04

= −15.33101317

⇔ n ≥ 15.33101317

Thus it is going to take at least 16 more payments to repay the loan at the new interestrate.

(d) The remainder of the loan can be repaid in 16 equal payments of13, 556.93

a16 4%

=

13, 556.93 · (0.04)1 − (1.04)−16 = 1163.455722, i.e., a new level payment of 1,163.46.

(e) Under the original effective semi-annual interest rate of j =√

1.06 − 1, the out-standing loan at the time of the missed payment would be L = 13, 035.51

√1.06 =

13420.87896 or 13, 420.88.

1200 · am j ≥ 13, 420.88 ⇔ 1200 · 1 − (1.06)−m2

√1.06 − 1

≥ 13, 420.88

⇔ (1.06)−m2 ≤ 1 − 13420.88 · (√1.06 − 1)

1200= .6693652806

⇔ m ≥ −2 ln .6693652806ln 1.06

= 13.77837241 ,

so 14 payments are needed after the missed payment.(f) Suppose the amount of each payment will be Y . We can think of a 25-payment

annuity-immediate with the projected 11th payment missing.

Y(a25 j − 1.06−

112)

= 20000


soY =

200001−(1.06)−

252√

1.06−1− (1.06)−

112

= 1192.432517 ,

and the level payment will be 1,192.43.

2. Suppose that you know that, for some n and for some interest rate i, an i = X, and sn+1 i =

Y . Use your (algebraic) knowledge of the values of the annuity functions to determine aformula which expresses i in terms of X,Y . Verify your work by selecting values of i andn and computing X and Y , and checking that the formula is correct in that special case.Solution: There will be many ways of solving this problem, but the solutions must beequivalent. To begin with, it would be useful to have the annuity information all forthe same numbers of payments. This can be done by, for example, recalling that sn+1 iincludes a last payment of 1 which has not had time to accumulate. Hence sn i = sn+1 i −1 = Y − 1. Since

X = an i = vn · sn i = vn(Y − 1) ,

we now know that(1 + i)n = v−n =

Y − 1X

.

Henced =

an

1 − vn =X

1 − XY−1

=X(Y − 1)Y − 1 − X

.

Since (1 − d)(1 + i) = 1, it follows that 1 + i =X(Y−1)

Y(X−1)+1 and

i =Y − X − 1

Y(X − 1) + 1.

To verify, I am selecting the value i = 4%, and the number of years as n = 10. 42 Then X =

an i =1 + i

i· (1 − vn) =

1.040.04

·(1 − 1.04−10

)= 8.435331611. Y = sn+1 i =

(1 + i)n+1 − 1i

=

(1.04)11 − 10.04

= 13.48635140. ThenY − X − 1

Y(X − 1) + 1=

13.48635140 − 8.435331611 − 113.48635140 × 7.435331611 + 1

=

0.03999999993, showing a small rounding error from 4%.

3. George has just begun working at the type of job he has dreamed of, where he wouldwork for 20 years and then retire for life. He plans to invest 1,000 at the beginning ofevery month that he works. When he retires he will spend 150,000 to purchase a chalet inthe mountains, and will begin to draw fixed monthly payments from the balance. The firstpayment he receives will be one month after he retires — i.e., 1 month after he draws the

42It might be tempting to verify with a trivial case, e.g., n = 0, where X = 0 and Y = 1; but that leads to anindeterminate ratio — clearly the possibility that Y−X = 1 should have been excluded somewhere in the problem.


capital for his chalet. The fixed payments he receives will begin a perpetuity, which willcontinue to his estate, forever (unless his executors negotiate a termination with the trustcompany that holds the account). Suppose that the interest rate during the years whenhe is working is 6%, compounded monthly, and that the interest rate for the perpetuity-immediate is 7% annual, effective. Determine, to the nearest unit, the constant amount ofhis perpetuity payments.Solution: At the time of his retirement, George’s monthly investments have accumulatedto 1000 · s240 1

2 %, so the amount available to purchase his perpetuity is 1000 · s240 12 % −

150000 . Denote the amount of the level monthly payments of his perpetuity by X, andobserve that the effective monthly interest rate is (1.07)

112 − 1. An equation of value just

after the removal of the funds for the chalet is

1000 · s240 12 % − 150000 =

X

(1.07)112 − 1

hence

X =((1.07)

112 − 1

)·(1000(1.005) · (1.005)240 − 1

0.005− 150000

)= 1777.39 ,

so the monthly payments under the perpetuity are in the amount of 1764.

4. Juliette was born on February 29th, 1980, and has always felt sorry for people born on thatday, because of the paucity of their birthdays. After making her first billion, she decidedto endow the Leap Year Foundation, with the purpose of making awards to persons with29 February as their birthday. Awards totalling 20,000 will be made on New Year’sDay of every leap year — assume that a leap year is one divisible by 4, excluding yearsdivisible by 100 except those divisible by 400, which are leap years. If the Foundationmakes its first awards in 2012, determine how much money Juliette needs to contributeon April 1st, 2009 to fund the Foundation, if the effective annual rate of interest is takento be i = 5%.Solution:

(a) As of the year 2012, the cost of awards in every year divisible by 4 would be 20000 ·a∞ j, where j1 is the effective rate of interest for a 4-year period, i.e., it is 4i(

14 ) =

(1.05)4 − 1 = 0.21550625 or 21.550625%; as of April 1st, 2009, this cost is

(1.05)−2 34 · 20000 ·

(1 +

10.21550625

)= 98, 640.84 .

(b) Funding not required for the years divisible by 100 (including those divisible by400) would be worth 20000 · a∞ j2 as of the year 2100, where j2 is the effective rate


of interest for a 100-year period, i.e., j2 = (1.05)100 − 1 = 130.5012578; as of April1st, 2009, this cost is

(1.05)−90 34 · 20000 ·

(1 +

1130.5012578

)= 240.67 .

(c) Fund that is required for the years divisible by 400 would be worth 20000 · a∞ j3 asof the year 2400, where j3 is the effective rate of interest for a 400-year period, i.e.,j3 = (1.05)400 − 1 = 299, 033, 350.2488392; as of April 1st, 2009, this cost is

(1.05)−390 34 · 20000 ·

(1 +

1299, 033, 350.2488392

)= 0.0001050 ,

which is negligible.

Thus the cost to Juliette on April 1st, 2009, is 98, 640.84 − 240.67 + 0 = 98, 400.17.

5. The owner of a perpetuity-immediate which pays 10,000 per year wishes to adjust thepayments so that he receives 12 equal monthly payments through the year. If the monthlypayments work out to 825 each, what is the effective annual interest rate? (It is intendedthat you determine this rate by first “trapping” the correct rate in an interval between twointerest rates, and then repeatedly cut the interval in half by testing the midpoint. Findingthe ends of the first interval you work with is easy, particularly if you don’t demand thatit be small. For example, see what happens if you take an interest rate of 0%. Applythis “bisection” method 10 times, and use the result to obtain an approximation to theeffective annual interest rate the question requests. The validity of this procedure derivesfrom the Intermediate Value Theorem, based on the continuity of the inverse function ofsn i, i.e., of i as a function of sn i for fixed n.)Solution: Let j denote the effective monthly interest rate. Then 825 · s12 j = 10000, so

s12 j = 12.12121212... Clearly j > 0, since s12 0 = 12 exactly. s12 0.01 =1.0112 − 1

0.01=

12.6825, so we know the correct interest rate will lie between 0% and 1%.

(a) Next we test the midpoint of the interval [0, 0.01]: s12 0.005 =1.00512 − 1

0.005= 12.33556;

since this exceeds 12.1212, the correct value will be in the interval [0, 0.005].(b) We test the midpoint of this half-interval, i.e., i2 = 0.0025, and find s12 0.0025 =

1.002512 − 10.0025

= 12.1664, which is still larger than 12.1212.... The correct value willlie in the interval [0, 0.0025].

(c) We test the midpoint of this interval, i.e., i3 = 0.00125, and find s12 0.00125 =1.0012512 − 1

0.00125=

12.0828 < 12.121212..., implying that the solution lies in the interval [0.00125, 0.0025].


(d) We test the midpoint 0.001875 of [0.00125, 0.0025]: s12 0.001875 =1.00187512 − 1

0.001875=

12.12453 > 12.121212, showing that the correct value is in the interval [0.00125, 0.001875].

(e) We test the mid-point 0.0015625: s12 0.0015625 =1.001562512 − 1

0.0015625= 12.10366 <

12.121212, showing that the correct value is in the interval [0.0015625, 0.001875].

(f) We test the mid-point 0.00171875: s12 0.00171875 =1.0017187512 − 1

0.00171875= 12.11409 <

12.121212, showing that the correct value is in the interval [0.00171875, 0.001875].

(g) We test the mid-point 0.001796875: s12 0.001796875 =1.00179687512 − 1

0.001796875= 12.11931.

The correct interest rate is in the interval [0.001796875, 0.001875].

(h) We test the mid-point 0.001835937: s12 0.001835937 =1.00183593712 − 1

0.001835937= 12.12192 >

12.121212, so the correct value is in the interval [0.001796875, 0.001835937].

(i) We test the mid-point 0.00181641: s12 0.00181641 =1.0018164112 − 1

0.00181641= 12.12058 <

12.121212 so the solution will be in the interval [0.00181641, 0.001835937], withmid-point 0.00182617.

(j) We test 0.00182617: s12 0.00182617 =1.0018261712 − 1

0.00182617= 12.12126.

The effective annual interest rate is approximately (1.00182617)12 − 1 = 0.022135489,i.e., approximately 2.2.

7.4.4 Fourth Problem Assignment, with Solutions

1. It is New Years Day, and Jacques and Jacqueline have just taken out a mortgage on theirnew condo, in the amount of 100,000. They have agreed that, in any calendar year exceptpossibly the last year of the mortgage, Jacques’ contribution will be exactly twice that ofJacqueline’s. The annual interest rate, compounded semi-annually, is 6%,

(a) Suppose that Jacques will always pay 2X on the last day of April; and Jacquelinewill pay X at the end of December. Determine the value of X if the mortgage is to bepaid off in 20 years.

(b) Suppose that Jacques will always pay 6000 on the last day of April; and Jacquelinewill pay 3000 on the last day of December, until a final year wherein the paymentsof Jacques and Jacqueline — still in the ratio of 2:1 — would not be as large asin previous years. Determine when that final year occurs, and what the respectivepayments of Jacques and Jacqueline will be that year.

(c) Suppose that Jacques will always pay 6000 on the last day of April; and Jacquelinewill pay 3000 on the last day of December, until a time when the payment which


is due exceeds the amount owing on the mortgage, in which case the payment isreduced to simply pay off the mortgage. Determine when this final year occurs,which of Jacques and Jacqueline has a reduced payment, and what that reducedpayment is in that final year.

Solution: The effective annual interest rate is (1.03)2 − 1 = 6.09%; the accumulationfactor for a 4-month period is, therefore, (1.03)

23 .

(a) Jacques’ contributions may be interpreted as an annuity-immediate which has beenadvanced through two-thirds of a year, i.e., through 4

3 half-years. Accordingly thepresent value of Jacques’ contributions is (1.03)

43 · 2X · a20 (1.03)2−1; Jacqueline’s con-

tributions also may be interpreted as an annuity-immediate which is today worthX · a20 (1.03)2−1. Equating the sum of the present values of these contributions to thepresent value of the loan, we obtain the equation of value

(1.03)43 · 2X · a20 (1.03)2−1 + X · a20 (1.03)2−1 = 100, 000

⇔ X(2(1.03)

43 + 1

)· 1 − (1.03)−40

(1.03)2 − 1= 100, 000

⇔ X =100, 000

((1.03)2 − 1

)(2(1.03)

43 + 1

)· (1 − (1.03)−40) = 2851.016174

so Jacques’ annual payment on each April 30th is 5702.03, while Jacqueline’s pay-ment on 31st December of each year is 2851.02. These payments will continue for20 years.

(b) Let n denote the number of full years in which Jacques and Jacqueline both makefull payments respectively of 6000 and 3000. Then n is the largest integer such that

(6000(1.03)

43 + 3000

)· an (1.03)2−1 ≤ 100, 000

⇔ 1 − (1.03)−2n ≤100000

((1.03)2 − 1

)(6000(1.03)

43 + 3000

) = 0.6590058875

⇔ (1.03)−2n ≥ 0.3409941125

⇔ n ≤ − ln 0.3409941125ln 1.03

= 18.19914857

At the end of the 18th year, just after Jacqueline’s payment, the amount outstandingon the mortgage is

(1.03)36100000 − (1.03)43 6000 · s18 (1.03)2−1 − 3000 · s18 (1.03)2−1

= (1.03)36100000 −3000

(2 · (1.03)

43 + 1

)·((1.03)36 − 1

)

(1.03)2 − 1= 1776.0318.


If Jacques pays 2Y at the end of April, and Jacqueline Y at the end of 1 year, anequation of value at the beginning of the 19th year is

1776.03 = (1.03)−23 · 2Y + (1.03)−2Y ,

implying that Y = 611.67, so Jacques will pay 1223.34 at the end of April, andJacqueline will pay 611.67 at the end of the year.

(c) The computations in the previous case show that the mortgage will definitely be paidoff in the 19th year. The balance of 1776.03 which is outstanding at the beginning ofthe year grows to (1776.03)(1.03)

23 = 1811.38, which becomes Jacques’ payment:

the mortgage is paid off, and Jacqueline will pay nothing at the end of the 19th year.

2. Mary is contributing to an RSSP for her retirement in 3 years by purchasing semi-annually a Certificate of Deposit which expires on the date of her retirement. She invests10,000 every half-year, with the last investment just one-half year before retirement. Sup-pose that the interest rate on the certificate she is purchasing now is 3%, and that the ratesfall by 1

2% each half-year, until the last half-year before retirement, when the rate reachesjust 1

2%; these rates, each applying to a half-year period, are all nominal annual rates,compounded semi-annually. Determine the magnitude of Mary’s account on the day sheretires.

(a) Compute the value of the account at retirement, using the Yield Curve Method [5, p.96].

(b) Compute the value of the account at retirement, using the Portfolio Rate Method [5,p. 96].

(Of course, Mary doesn’t have the choice of which method will be used: in either the fineprint of her contract or in the established practices of the banking institution the methodwill be made more explicit.)Solution:

(a) Using the Yield Curve method, we accumulate each deposit of 10,000 at the rate ineffect when it was deposited; remember that the rate in effect for each half-year isone-half of the quoted, nominal rate. The value of her account will be

10000((1.0025)1 + (1.0050)2 + (1.0075)3 + (1.0100)4 + (1.0125)5 + (1.0150)6

)

= 62, 333.24 .

(b) Using the Portfolio Rate Method the first payment of 10,000 accumulates to 10000(1.015)just before the 2nd deposit. That deposit increases the capital in the account to10000(1.0150+1), which accumulates to 10000((1.0150+1)1.0125+1) just after the3rd deposit. Just prior to the 4th deposit the account has grown to 10000((1.0150 +

1)1.0125 + 1)1.0100, becoming 10000(((1.0150 + 1)1.0125 + 1)1.0100 + 1) just after


the 4th deposit; to 10000((((1.0150 + 1)1.0125 + 1)1.0100 + 1)1.0075 + 1) after the5th deposit; and to

10000(((((1.0150 + 1)1.0125 + 1)1.0100 + 1)1.0075 + 1)1.0050 + 1) = 61, 266.24389

after the last deposit; between this deposit and retirement the account accumulatesby an accumulation factor of 1.0025, to a final balance of 61,419.41.

3. Jack’s mortgage, in the amount of 80,000, requires regular payments of 600 every month,at an interest rate of 8%, compounded semiannually.

(a) Suppose that Jack has agreed that the last payment will be a “balloon” payment ofat least 600. Determine when that payment is due, and its amount.

(b) Suppose that Jack has agreed that the last payment will be a ”drop” payment of atmost 600. Determine when that payment is due, and its amount.

Solution:

(a) Let n be the time of the balloon payment. Then n will be the largest integer suchthat 600 · sn < 80, 000(1 + i)n, and the last payment will be in the amount of 600increased by 80, 000− 600 · sn, where the interest rate per month is i = (1.04)

16 − 1 =

0.006558197 or 0.6558197%. We have

600 · sn < 80, 000(1 + i)n

⇔ 600 · (1 + i)n − 1i

< 80, 000(1 + i)n

⇔ (1 + i)n <600

600 − 80, 000i=

11 − 400i

3

⇔ n <− ln(1 − 400i

3 )ln(1 + i)

=−6 ln

(1 − 400

3 (0.006558197))

ln(1.04)= 317.4132252

The largest integer solution is, therefore n = 317. The balloon payment will occurin month 317 of the repayment. The payment will be equal to

600 + 80, 000(1.04)3176 − 600 · (1.006558197)317 − 1

0.006558197= 846.78

(b) In the case of a drop payment we need to find the smallest n such that the reversal ofthe preceding inequality be satisfied. The preceding calculations show that this willbe n = 318. In that case the excess of 600 of the amount of the (drop) payment willbe

−80, 000(1.04)3186 + 600 · (1.006558197)318 − 1

0.006558197= 351.6041 ,

so the amount of the drop payment will be 248.40.


4. Under a nominal interest rate of 6% compounded continuously an annuity pays 10,000per year for 5 years. Determine

(a) The value of the annuity 2 years before the payment begins.(b) The total annual payment under a continuous perpetuity at the same rate of interest,

whose value at time 0 is the same as this annuity.(c) The time t such that the value at time 0 of the portion of the annuity up to time t is

equal to the value at time 0 of the portion of the annuity from time t to time 5.

Solution: The intention was that this annuity should be considered as being continuous,i.e., paying infinitesimal amounts throughout its 5-year term. Some students interpretedthe problem as referring to a discrete annuity; in that case it wasn’t clear when the pay-ments would be made, but some assumed them to be annual; one problem with thisdiscrete interpretation is that the second part of the problem no longer has a clearly de-fined solution. The interpretation as a continuous annuity was also indicated by the use ofthe word “payment”, rather than the plural: rather than a sequence of discrete payments,there was to be one continuous payment.

Continuous Annuity(a) The annual accumulation factor corresponding to a nominal annual rate of 6%

compounded instantaneously is 1 + i = e0.06 = 1.061836547; hence v = e−0.06,and ln v = −0.06. The value at time t = 0 of all payments under the annuity is

10, 000a5 = 10, 000∫ 5

0vt dt = 10, 000

[vt

ln v

]5

0=

10, 000−0.06

(e−0.3 − 1

)= 43, 196.96 .

Two years before payment begins the annuity is worth only v2 = e−0.12 times thisamount, i.e., 38, 312.27.

(b) Let the total annual payment under the continuous perpetuity be X. Then anequation of value at time t = 0 is

10, 000a5 = 10, 000∫ 5

0vt dt = X

∫ ∞

0vt dt = Xa∞ ,

implying that

X =43, 196.96

limb→∞

1−0.06

(e−0.06b − 1

) =43, 196.96

1−0.06 · (−1)

= (0.06)(43, 196.96) = 2591.82.

(c) The value t we seek satisfies the equation

10, 000at = (0.5)(43, 196.96)


⇔ 10, 000−0.06

(e−0.06t − 1

)= 21, 598.48

⇔ e−0.06t = 1 − (0.000006)(21, 598.48) = 0.87040912⇔ t = 2.313198750 .

Discrete Annuity with Annual PaymentsIn this case one can interpret the annuity aseither an annuity-immediate or an annuity-due. I will interpret it as an annuity-immediate.

(a) The effective annual accumulation factor is limm→∞

(1 +

0.06m

)= e0.06 = 1.061836547,

so the effective annual interest rate is j = 6.183655%. The value 2 years beforethe first payment is that of an annuity-immediate deferred one year, i.e.,

10, 000 ·1∣∣∣∣a5 i = 10, 000 · 1 − (1 + j)−5

j(1 + j)= 39, 473.13 .

(b) The value of this annuity at time t = 0 will depend on whether it’s interpreted asan annuity-immediate or an annuity-due. If it is an annuity-immediate, its valueis 1 + j times the preceding, i.e., 41,914.01; if it is interpreted as an annuity-due,then the value is increased by multiplication by another factor 1 + j, i.e., it is44,505.83.

(c) It’s not clear how to interpret this problem in the discrete case; we can solve anequation and obtain a fractional number of years; but how do we interpret thatfraction, in view of the fact that payments are only at the end or beginning of ayear?

5. Gasoline costs 1.00 per litre now, and the price will increase by 0.05 per week for thenext year. If my tank holds 65 litres, and money is worth 4.5% per annum effective, whatwill be the value 52 weeks from now, just after the tank has been filled, of the gasolinethat I have purchased, if I fill the tank completely every 2 weeks?Solution: The cost of filling the tank is increasing by the amount of 65×0.10 = 6.50 fromone filling to the next. I am assuming that the tank completely every 2 weeks, but thattoday’s filling is not included in the cost for a full year, and my computations will startwith the value one year from now of the tank filled two weeks from now. To the amountof 6.50(Is)26, whose starting payment is 6.50, we must add (65 × 1.00) − 6.50 + 6.50 =

65.00 times s26. The effective interest rate for a two-week period is i = (1.045)126 − 1 =

0.001694391. The terminal value will be

6.50(Is)26 + 65.00s26

= 6.50( s26 − 26

0.001694391

)+ 65.00

((1.001694391)26 − 1

0.001694391

)


= 6.50

(1.001694391) · (1.001694391)26−1

0.001694391 − 260.001694391

+ 65.00((1.001694391)26 − 1

0.001694391

)

= 6.50

(1.001694391) · 1.045−1

0.001694391 − 260.001694391

+ 65.00(

1.045 − 10.001694391

)

= 6.50

(1.001694391) · 0.045

0.001694391 − 260.001694391

+ 65.00(

0.0450.001694391

)= 4040.328165

or 4,040.33.

7.4.5 Fifth Problem Assignment, with Solutions

1. George’s older brother has sold his company for a large cash payment. He wishes to placehis capital in secure investments which will provide him a stream of income to replacethe salary he was previously drawing from the company. As George is purchasing acondo, and needs immediate capital for a large down payment, they come to an agreementwhereby George will borrow 100,000: he plans to repay the loan at the end of 10 years,and will pay annual interest monthly at the rate of 5% per annum effective. In additionto this down payment, George will assume from the vendor of his condo a mortgage for400,000 at a nominal annual rate of 4%, compounded semi-annually, under which hemakes equal monthly payments at the end of each month to amortize the mortgage loanin 20 years. (For the purpose of this problem assume that a year consists of exactly 12months.)(a) After 3 years George realizes that he needs to prepare to repay the loan to his brother,

and decides to accumulate a sinking fund for the purpose, a fund which will attain abalance of 100,000 when his loan to his brother comes due. The best interest rate hecan now secure is 3% per annum, compounded monthly. Determine the amount ofhis equal monthly payments into the sinking fund.

(b) Set up portions of a table which will show the amortization of George’s mortgagewith the following columns: Mortgage Payment Number, Interest Component, Prin-cipal Component, Outstanding Principal After Payment, as of the following dates:t = 0 (when mortgage is signed), t = 12, t = 24, t = 36, t = 239.

(c) Determine George’s total monthly cost for these loans during years 0-3, years 3-10,and years 10-20.

Solution:(a) The regular monthly payment into the sinking fund, beginning at the end of the first

month of the 4th year, is

100, 000s84 1

4 %

=(100, 000)(0.0025)

(1.0025)84 − 1= 1071.33 .


(b) The mortgage interest is at an effective monthly rate of (1.02)16 − 1 = 0.003305890;

the regular monthly payment under the mortgage will be

400, 000a240 0.330589%

=(400, 000)(0.003305890)

1 − (1.003305890)−240 = 2416.985777 .

The principal outstanding just after the 11th payment is

400000(1.003305890)11 − 2416.985777 · s11 0.003305890

= 400000(1.003305890)11 −(2416.985777)

((1.003305890)11 − 1

)

0.003305890= 387, 758.06 ,

so the interest component of the 12th payment is (0.003305890) · (387, 758.06) =

1, 281.89, and the principal component will be 2, 416.99 − 1, 281.89 = 1, 135.10;the outstanding principal after the 12th payment will be 387, 758.06 − 1, 135.10 =

386, 622.96.The principal outstanding just after the 23rd payment is

400000(1.003305890)23 − 2416.985777 · s23 0.003305890

= 400000(1.003305890)23 −(2416.985777)

((1.003305890)23 − 1

)

0.003305890= 373, 886.44 ,



372, 705.48.The principal outstanding just after the 35th payment is

400000(1.003305890)35 − 2416.985777 · s35 0.003305890

= 400, 000(1.003305890)35 −(2, 416.985777)

((1.003305890)35 − 1

)

0.003305890= 359, 454.40 ,



358, 225.73.


The principal outstanding just after the 238th payment is

400000(1.003305890)238 − 2416.985777 · s238 0.003305890

= 400000(1.003305890)238 −(2416.985777)

((1.003305890)238 − 1

)

0.003305890= 4, 810.11 ,


15.90, and the principal component will be 2, 416.99 − 15.90 = 2, 401.09; the out-standing principal after the 239th payment will be 4, 810.11− 2, 401.09 = 2, 409.02.

(c) The effective monthly rate of interest on his loan from his brother is0.0512

= (1.05)112−

1 = 0.004074124, so his monthly interest cost on this loan is 407.41; this cost ispresent up to the end of the 10th year.

Month Mortgage Interest to Brother Sinking Fund Total1–36 2,416.99 407.41 2,824.40

37–120 2,416.99 407.41 1,071.33 3,895.7321-240 2,416.99 2,416.99

2. A borrower is repaying a loan at a quarterly cost of 5,000 for 10 years. One-quarter of theloan is amortized by equal quarterly payments at 5% effective, included in each quarterlyexpenditure of 5,000. The other three-quarters of the loan is repaid by the sinking fundmethod, where that portion of the principal is repaid when the loan matures by the pro-ceeds of a sinking fund which accumulates at a nominal annual rate of 4%, compoundedquarterly: until maturity the lender pays only quarterly interest at 5% (annual effective)on the non-amortized part of the loan, and settles that part of the loan at maturity by trans-ferring to the lender the balance in the sinking fund. Showing all your work, determinethe amount of the loan.Solution: For the amortized portion of the loan the effective quarterly interest rate isi = (1.05)

14 − 1 = 0.012272234 = 1.2272234%. Let the total principal of the loan be

denoted by L. For the portion of the loan amortized by quarterly payments, the quar-

terly cost isL

4 · a40 1.2272234%

=L(0.012272234)

4(1 − (1.012272234)−40) = 0.007946552452L . The quar-

terly interest due on the non-amortized portion of the loan is 0.012272234 × 34× L =

0.00920417550L. For that portion of the loan the quarterly contribution to the sinking

fund is34 · L

s40 0.01

=0.0075L

(1.01)40 − 1= 0.01534169847L. We sum the three components of the

quarterly contribution and equate to 5,000:

0.007946552452L + 0.00920417550L + 0.01534169847L = 5, 000 ,


which implies that L = 153, 882.01.

3. An investor is considering the purchase of a 15-year bond with par value 100. bearingsemi-annual coupons at a nominal annual interest rate of 4%, convertible semi-annually.The bond is callable at 109.00 on any coupon date from 5 years to 10 years, inclusive; at104.50 from on any coupon date from 10.5 years to 14.5 years; and it matures at 100.00in 15 years. Showing details of your work, determine what price an investor should payin order that he be guaranteed:

(a) an effective annual yield rate of 3%; or(b) a nominal annual yield rate of 5%, convertible every 6 months.

Solution: Let t denote the time in half-years, so 0 ≤ t ≤ 30.

(a) I am using the “Premium/Discount Formula”, P = C + (Fr − Ci)an. The effectivesemi-annual interest rate is 1.03

12 − 1 = 1.4889156%. As a function of the half-year

t at the end of which the bond is redeemed, the value of the bond is as follows:

t ≥ t ≤ Price10 20 109.00 + (2 − 1.622918)at 1.4889156%21 29 104.50 + (2 − 1.5559168)at 1.4889156%30 30 100.00 + (2 − 1.4889156)a30 1.4889156%

.

Because the parenthesized quantity Fr − Ci is positive, these prices are minimizedwhen t is minimal in each given range. Thus we need to compare

t Formula Price10 109.00 + (2 − 1.622918)a10 1.4889156% 112.47962921 104.50 + (2 − 1.5559168)a21 1.4889156% 112.458233530 100.00 + (2 − 1.4889156)a30 1.4889156% 112.29970329

.

Thus the lowest price will occur if the bond is not called: the purchaser should notpay more than 112.30 to guarantee an effective annual yield of 3%.

(b) I am again using the “Premium/Discount Formula”, P = C + (Fr − Ci)an. Theeffective semi-annual interest rate is 2.5%. As a function of the half-year t at the endof which the bond is redeemed, the value of the bond is as follows:

t ≥ t ≤ Price10 20 109.00 + (2 − 2.7250)at21 29 104.50 + (2 − 2.6125)at30 30 100.00 + (2 − 2.5000)a30

.

As the parenthesized quantity Fr − Ci is always negative, the values are minimizedwhen t is maximal in the given range. Thus we need only consider the following


values:t Formula Price

20 109.00 + (2 − 2.7250)a20 97.7029 104.50 + (2 − 2.6125)a29 91.9730 100.00 + (2 − 2.5000)a30 89.53

.

It follows that the worst price will occur if the bond is not called: the purchasershould not pay more than 89.53 to guarantee a yield of 5% converted semi-annually.

7.4.6 Draft Solutions to Problems on the 2009 Class Tests

The tests were administered on 13 March, 2009.

Version 11. Showing your work in detail, determine each of the following; the rates you determine

should be accurate to 4 decimal places, or as a percentage accurate to 2 decimal places:



(c) [4 MARKS] the effective monthly interest rate corresponding to nominal annual rateof interest of 4%, compounded continuously.

Solution:

(a) d(4) = 3%⇒ effective discount rate for 3 months is 34%. As a 2-year period contains

214

= 8 disjoint 3-month periods, the accumulation factor for a 2-year period is (1 −0.0075)−8 = 1.062076688, so the effective interest rate for a 2-year period is 6.21%.

(b)(1 +

i(6)

6

)6

= 1 + i = 1.06 ⇒ i(6) = 6((1.06)

16 − 1

)= 0.058552764, so the nominal

annual interest rate, compounded every two months is 5.86%.(c) When a nominal interest rate of 4% is compounded continuously, the annual accu-

mulation factor is limm→∞

(1 +

0.04m

)m

= e0.04; hence the effective accumulation factor

for 1 month is(e0.04

) 112

= e0.0412 = 1.003338895. The effective monthly interest rate is

0.334%.

2. For each of the following sequences of payments, determine, as of the given time, andfor the given interest or discount rate, the value, showing all of your work. Before deter-mining the numeric value you are expected


• to express the value using standard symbols, and• to state the formula that you have used to evaluate the standard symbol.

(a) [6 MARKS] the value now of 25 payments of 5 each at the end of every year — thefirst payment being exactly 2 years from now — at an interest rate of 3%;

(b) [6 MARKS] the value four years after the last payment of 12 payments of 3 at theend of every half-year, the first to be paid 2 years from now, at a nominal interestrate of 7% compounded every 3 months;

(c) [6 MARKS] 200 payments of 1 at the end of every month, as of the date of the 120thpayment, which has just been made; the interest rate is 6% compounded monthly.

Solution:

(a) 5 ·1∣∣∣∣a25 3% =

51.03

· 1 − (1.03)−25

0.03= 84.52984317 or 84.53.

(b) It’s irrelevant when the payments start, since we are evaluating as of 4 years afterthe last payment. The effective interest rate per half-year is j =

(1 + 0.07

4

)2 − 1 =

0.035306250. The value just after the 12th payment is 3 · s12 j = 3 · (1 + j)12 − 1j

,

and we must accumulate this through 4× 4 = 8 quarter-years. The value at that timewill be (

1 +0.07

4

)16

· 3 ·(1 + 0.07

4

)24 − 1

0.035306250= 57.92186862 .

(c) The effective interest rate per month is 612 = 0.5%. The value of the 200 payments 1

month before the first of them is a200 0.5%, so the value just after the 120th payment is

(1.005)120 · a200 0.5% = (1.005)120 · 1 − (1.005)−200

0.005= 229.6816522 .

3. (a) [12 MARKS] The accumulated value just after the last payment under a 20-yearannuity of 10,000 per year, paying interest at the rate of 5% per annum effective,is to be used to purchase a perpetuity, first payment to be made 2 years after thelast payment under the annuity. Showing all your work, determine the size of thepayments under the perpetuity, assuming that the interest rate from the time of thelast payment under the 20-year annuity is 4%.

(b) [6 MARKS] An annuity at interest rate i consists of payments of 10 now, 13 at theend of one year, 16 at the end of two years, increasing by a constant amount untilthe last payment in the amount of 70, is to be evaluated as of 6 years ago. Expressits value in terms of symbols (Ia)n, (Is)n, an, sn, i but do not evaluate.

Solution:


(a) If we interpret the perpetuity as a perpetuity-immediate, then the amount availablefor its purchase is

(1.04) × 10000 × s20 5% = 10400 · (1.05)20 − 10.05

= 343, 885.92 .

If X is the payment size under the perpetuity, then an equation of value is 343, 885.92 =X

0.04, so the payment size is 0.04(343, 885.92) = 13, 755.44.

(b) The total number of payments under this increasing annuity is70 − 10

3+ 1 = 21.

The value of this increasing annuity one year ago was 3(Ia)21 + 7a21. To obtain itsvalue as of 6 years ago we have to discount back through 5 more years, obtainingthe value (1 + i)−5

(3(Ia)21 + 7a21

).

4. [18 MARKS] A loan of 8000 is to be repaid by annual payments of 430 to commenceat the end of the 1st half-year, and to continue thereafter for as long as necessary. Findthe time and amount of the final payment if the final payment is to be no smaller than theregular payments. Assume the interest rate per half-year is i = 5%.Solution: Suppose that the final payment is the nth. Then n is the largest integer suchthat43

8000 > 430 · an ⇔ 1 − (1.05)−n <8000430

· (0.05)

⇔ (1.05)−n > 1 − 0.9302325582

⇔ n < − ln(0.06976)ln(1.05)

= 54.57222523.

Hence n = 54. The last payment is equal to

430 + (1.05)54(8000 − 430 · a54

)= 8000(1.05)54 + 430

(1 − (1.05)54 − 1

0.05

)= 666.78 .

Version 21. [18 MARKS] A loan of 12,000 is to be repaid by semi-annual payments of 800 to com-

mence immediately, and to continue at the beginning of each half-year for as long asnecessary. Find the time and amount of the final payment if the final payment is to beno larger than the regular payments. Assume an interest rate of 14% compounded semi-annually.

43It’s not enough to state an inequality for n — you need to make it clear which n you will choose once youhave solved the inequality.


Solution: Let n be the number of the last payment, which is to be a drop payment. Thenan inequality of value as of time 0 is 800an 7% ≥ 12, 000, where n is to be the smallest44

integer with this property. Solving this inequality we obtain

800an 7% ≥ 12, 000 ⇔ (1.07)−n ≤ 1 − 15(0.07)1.07

= 0.0186915888

⇔ n ≥ − ln(0.0186915888)ln(1.07)

= 58.81999926 .

Thus the last payment will be the 59th. Since the 1st payment is at time 0, the 59thpayment is at the beginning of the 59th year, i.e.: it is at time 58. The amount of thatpayment will be

12, 000(1.07)58 − 800s58 = 12, 000(1.07)58 − 800(1.07)((1.07)58 − 1

0.07

)= 659.96

2. Showing your work in detail, determine each of the following; the rates you determineshould be accurate to 4 decimal places, or as a percentage accurate to 2 decimal places:

(a) [4 MARKS] the effective annual discount rate corresponding to a nominal interestrate, compounded quarterly, of i = 2.4%

(b) [4 MARKS] the nominal annual discount rate, compounded quarterly, equivalent toan effective semi-annual interest rate of i = 4%

(c) [4 MARKS] the effective semi-annual interest rate corresponding to a force of inter-est of δ = 0.04.

Solution:

(a) The effective annual interest rate is (1.006)4−1 = 0.024216865. The effective annualdiscount rate d is given by (1−d)(1.024216865) = 1, so d = 0.0236442748 or 2.36%to the desired accuracy.

(b) An effective semi-annual interest rate of 4% produces an annual accumulation factorof (1.04)2 = (1 − d)−1, so d = 1 − (1.04)−2 = 0.0754 is the effective annual discountrate. Since

(1 − d(4)

4

)4= 1 − d = (1.04)−

24 , d(4) = 0.0776773 or 7.77%.

(c) The accumulation factor for one-half year is e∫ 0.5

0 δ dt = e[0.04t]0.50 = e0.02 = 1.020201340,

so the effective semi-annual interest rate is 2.02%.





(a) [6 MARKS] the value of 20 payments of 6 at the end of every year, the last one made3 years ago, at an interest rate of 4%;

(b) [6 MARKS] the value 6 months from now of 34 payments of 10 at the end of every3 months, the first to be paid 6 years from now, at a nominal interest rate of 12%compounded 4 times a year;

(c) [6 MARKS] 150 payments of 2 at the end of every 4 months, as of the date of the100th payment, which has just been made; the interest rate is 8% compounded every4 months.

Solution:

(a) 6(1 + i)3 · s20 4% = 6 · (1.04)23 − (1.04)3

0.04= 200.98.

(b) The effective interest rate per 3 months is 124 = 3%. As of 5 years and 9 months

from now, the payments are worth 10 · a34 3%; as of 6 months from now the value is

1024−1−2

∣∣∣∣·a34 3% =10

0.03

((1.03)−21 − (1.03)−55

)= 113.59.

(c) When the interest rate is stated as being “compounded every 4 months”, the clearintention is that the rate is a nominal one: the effective interest rate per 4 monthsis .08

3 %. Four months before the first payment the annuity is worth 2a150 83 % = 2 ·

1 −(1 + 0.08

3

)−150

0.083

= 37.5

1 −(1 +

0.083

)−150 = 73.55242200. Just after the 100th

payment the annuity is worth

(1 +

0.083

)100

(73.55242200) = 1022.22 .

(Alternatively, the value could be viewed as

2(s100 8

3 % + a50 83 %

)

= 75

(1 +

0.083

)100

−(1 +

0.083

)−50 = 1022.217167 .

4. (a) [6 MARKS] An annuity at interest rate i consists of payments of 260 now, 235 at theend of 1 year, 210 at the end of 2 years, decreases by a constant amount until the lastpayment in the amount of 35, is to be evaluated as of just after the third payment.Express its value then in terms of symbols (Da)n, (Ds)n, an, sn, i, but do not evaluate.


(b) [12 MARKS] The accumulated value just after the last payment under a 14-yearannuity of 9,000 per year, paying interest at the rate of 8% per annum effective, isto be used to purchase a 20-year annuity at an interest rate of 6%, first payment tobe made 3 years after the last payment under the annuity. Showing all your work,determine the size of the payments under the 20-year annuity. Assume that the 6%rate is in effect from the time of the last payment under the 14-year annuity.

Solution:

(a) The number of payments is 260−3525 + 1 = 10. The value one year before the first

payment is 25(Da)10 + 10a10. Just after the 3rd payment this decreasing annuity is

worth (1 + i)3(25(Da)10 + 10a10

).

(b) Let X be the amount of each payment under the 20-year annuity. Then an equationof value just after the last payment under the 14-year annuity is

9000s14 8% = X2

∣∣∣∣a206%

= 9000 ·(1.08)14−1

0.08

(1.06)−2 · 1−(1.06)−20

0.06

=9000(0.06)(1.06)2

((1.08)14 − 1

)

0.08(1 − (1.06)−20)

= 21348.97 .

Version 31. (a) [6 MARKS] An annuity at interest rate i consists of payments of 24 now, 29 at the

end of one year, 34 at the end of two years, increasing by a constant amount until thelast payment in the amount of 224, and is to be evaluated as of 4 years ago. Expressits value then in terms of symbols (Ia)n, (Is)n, an, sn, i, but do not evaluate.

(b) [12 MARKS] The accumulated value just after the last payment under a 22-yearannuity of 11,000 per year, paying interest at the rate of 8% per annum effective,is to be used to purchase a perpetuity at an interest rate of 6%, first payment to bemade at the same time as the last payment under the annuity. Showing all your work,determine the size of the payments under the perpetuity.

Solution:

(a) The payments are differing by 5, so the number of payments is 224−245 + 1 = 41.

One year before the first payment the annuity is worth 5(Ia)41 + 19a41. As of 4years ago the annuity must be discounted an additional 3 years, so its value then was(1 + i)−3

(5(Ia)41 + 19a41

).



(b) If we denote the amount of each payment under the perpetuity by X, then an equationof value at the time of the first payment of the perpetuity-due is 11, 000s22 8% =

X · a∞ 6%, which can be solved to yield

X =11000

((1.08)22−1

0.08

)

1.060.06

= 34529.68 .

2. [18 MARKS] A loan of 9,000 is to be repaid by annual payments of 1,000 to commenceat the end of the 1st year, and to continue thereafter for as long as necessary. Find thetime and amount of the final payment if the final payment is to be smaller than the regularpayments. Assume i = 6%.Solution: Suppose that the last payment, which is to be a drop payment, is the nth. Thenn is the smallest45 integer such that

9000 < 1000an ⇔ (1.06)−n < 1 − 9(0.06) = 0.46⇔ n > 13.32664048

Hence the drop payment is the 14th. Just after the 13th payment the amount still owingis 9000(1.06)13 − 1000s13, so the amount of the 14th payment is

1.06(9000(1.06)13 − 1000 · (1.06)13 − 1

0.06

)= 333.07 .


(a) [4 MARKS] the effective annual interest rate, corresponding to an nominal annualinterest rate of i = 6% compounded every 4 months.

(b) [4 MARKS] the effective monthly interest rate corresponding to a force of interestof δ = 0.3.

(c) [4 MARKS] the effective annual interest rate corresponding to d(4) = 3.20%

Solution:

(a) Since 1+i =(1 + 0.06

3

)3, the effective annual rate is (1.02)3−1 = 0.061208, or 6.12%.

(b) The accumulation factor for one month is

e∫ 1

120 0.3 dt = e0.025 = 1.025315121 ,

so the equivalent effective monthly interest rate is 2.53%.45It’s not enough to state an inequality for n — you need to make it clear which n you will choose once you

have solved the inequality.


(c) The effective annual accumulation factor is(1 − 0.032

4

)−4= 1.032650385, so the

equivalent effective annual interest rate is 3.27%.



(a) [6 MARKS] the value 3 months ago of 24 payments of 5 at the end of every half-year,the first to be paid 7 years from now, at a nominal interest rate of 9% compoundedsemi-annually;

(b) [6 MARKS] 240 payments of 10 at the end of every month, as of the date of the150th payment, which has just been made; the interest rate is 12% compoundedmonthly

(c) [6 MARKS] the value now of 18 payments of 500 at the end of every year, the firstpayment 8 years from now, at an interest rate of 5%

Solution:

(a) As of one half-year before the first payment, the annuity is worth 5a24 4.5%. Since thefirst payment will be in 7 years, the annuity, interpreted as an annuity-immediate,is being deferred through 13 half-years, and is worth today 5(1.045)−13a24 4.5%. Thevalue a quarter-year ago, i.e., half of a half-year ago, was

5(1.045)−13.5a24 4.5% = 5(1.045)−13.5 · 1 − (1.045)−24

0.045= 40.01 .

(b) The effective monthly interest rate is 1%. The value of the payments which have beenmade is 10s150 1%, and the value of the future payments is 10a90 1%; so the present

value is 10s150 1%+10a90 1% =10

0.01

((1.01)150 − 1 + 1 − (1.01)−90

)= 1000

((1.01)150 − (1.01)−90

)=

4, 040.03 .

(c) 5007

∣∣∣∣a18 5% = 500(a25 5% − a7 5%

)=

500((1.05)−7 − (1.05)−25

)

0.05= 4153.79.

Version 41. For each of the following sequences of payments, determine, as of the given time, and

for the given interest or discount rate, the value, showing all of your work. Before deter-mining the numeric value you are expected

• to express the value using standard symbols, and


• to state the formula that you have used to evaluate the standard symbol.

(a) [6 MARKS] 72 payments of 1 at the end of every 3 months, as of the date of the24th payment, which has just been made; interest is at a nominal annual rate of 12%compounded every 4 months;

(b) [6 MARKS] the value of 24 payments of 7 at the end of every half-year, the lastone having just been made, at a nominal annual interest rate of 6% compoundedquarterly;

(c) [6 MARKS] the value now of 50 payments of 2 at the end of every 8 months, thefirst to be paid 2 years from now, at a nominal interest rate of 4.5% compounded 3times every 2 years.

Solution:

(a) The effective interest rate for 4 months is 123 % = 4%. But the compounding/payment

interval for the annuity is 3 months, for which the accumulation factor will be(1.04)

34 = 1.029852445, and the interest rate will be j = 2.9852445%. The value of

the annuity at the indicated date is

s24 j + a48 j =

((1.04)18 − 1

)+

(1 − (1.04)−36

)

0.029852445= 59.69855344 .

(b) The effective interest rate per quarter-year is 1.5%; the effective interest rate perhalf-year is (1.015)2 − 1 = 0.030225. The value of the annuity is, therefore,

7s24 0.030225 = 7 · (1.030225)24 − 10.030225

= 241.67 .

(c) Interest is compounded every 2/3 of a year, i.e., 32 times a year. Hence the annual

accumulation factor is1 +

0.04532

32

, and the accumulation factor for 8 months is

1 +

0.04532

32

23

= 1 +0.045

32

= 1.03 ,

i.e., the effective interest rate for an 8-month period is 3%. Since the first paymentis after 2 years — i.e., at the end of the 3rd interest compounding interval from now,the value of the annuity is

2 ·2∣∣∣∣a50 3% = 2 · (1.03)−2 − (1.03)−52

0.03= 48.51.


2. (a) [6 MARKS] An annuity at interest rate i consists of payments of 1,118 now, 1,112at the end of 1 year, 1,106 at the end of 2 years, decreasing until the last paymentin the amount of 1,058, the totality to be evaluated as of the time of the payment of1,106. Express its value then in terms of symbols (Da)n, (Ds)n, an, sn, i, but do notevaluate.

(b) [12 MARKS] The accumulated value just after the last payment under a 19-yearannuity of 1000 per year, paying interest at the rate of 8% per annum effective, isto be used to purchase a 10-year annuity-immediate at an interest rate of 6%, firstpayment to be made 1 year after the last payment under the 19-year annuity. Showingall your work, determine the size of the payments under the 10-year annuity. Assumethat the 6% rate applies from the time of the last payment under the 8% annuity.

Solution:

(a) The number of payments is 1118−10586 + 1 = 11. The value of this decreasing an-

nuity as of the last payment is 6(Ds)11 + 1052s11. The payment of 1,106 occurs1106−1058

6 = 8 years earlier, so the value of the decreasing annuity at that time is(1 + i)−8

(6(Ds)11 + 1052s11

)= (1 + i)3

(6(Da)11 + 1052a11

).

(b) If X denotes the size of payments under the 10-payment annuity, an equation of valuejust after the last payment under the first annuity is

1000 · s19 8% = X · a10 6% ⇔ X = 1000 ·s19 8%

a10 6%

⇔ X =1000 ·

((1.08)19 − 1

)· (0.06)

(1 − (1.06)−10)

) · (0.08)= 5631.219160

so the second annuity has 10 equal payments of 5,631.22.

3. [18 MARKS] A loan of 18,000 is to be repaid by annual payments of 2,500 to commenceimmediately, and to continue at the beginning of each year for as long as necessary. Findthe time and amount of the final payment if the final payment is to be no smaller than theregular payments. Assume i = 13%.Solution: Since the 1st payment is at time t = 0, we may assume that it is the (n + 1)stpayment (made at time t = n) which is the balloon payment. Then n is the largest46

integer such that

18000 − 2500an+1 > 0 ⇔ 15500 − 2500an > 0

⇔ 1 − (1.13)−n <0.13(15500)

2500= 0.806




⇔ −n >ln(0.194)ln(1.13)

= −13.4

⇔ n < 13.4 ,

so n = 13, the balloon payment is the 14th (at time t = 13). The amount outstanding onthe loan just after the 13th payment (at time t = 12) is 15500(1.13)12−2500s123059.66382;so the amount of the balloon payment is

1.13(15500(1.13)12 − 2500s12

)= (1.13)

(15500(1.13)12 − 2500 · (1.13)12 − 1

0.13

)

= 3457.42 .


(a) [4 MARKS] the nominal annual interest rate, compounded quarterly, equivalent toan effective semi-annual discount rate of d = 4%

(b) [4 MARKS] the effective semi-annual interest rate corresponding to a force of inter-est of δ = 0.14.

(c) [4 MARKS] the effective annual discount rate corresponding to a nominal interestrate, compounded quarterly, of i = 2.5%

Solution:

(a) The effective annual accumulation factor is (1 − 0.04)−2. If i(4) denotes the nominalannual interest rate, compounded quarterly, then we have

(1 + i(4)

4

)4= (1 − 0.04)−2,

so i(4) = 4((0.96)−

12 − 1

)= 0.082482904 or 8.25%.

(b) The accumulation factor for half a year is

e∫ 0.05

0 δt dt = e[0.14t]0.50 = e0.07 = 1.072508181 ,

so the effective semi-annual interest rate is 7.25%.

(c) In one year 1 accumulates to(1 +

0.0254

)4

= 1.025235353, whose reciprocal is

0.9753857951, so the effective annual discount rate is 1−0.9753857951 = 0.0246142049or 2.46%.

7.4.7 Final Examination 2008/2009

Instructions









(a) [2 MARKS] The nominal annual interest rate, compounded every 5 months, corre-sponding to an effective annual interest rate of 12%.

(b) [2 MARKS] The effective semi-annual interest rate corresponding to a nominal dis-count rate, compounded every 4 months, of 12%.

(c) [3 MARKS] The value to which 1000 will grow from time t = 0 to time t = 1 undera force of interest of δt = 0.02t3.

(d) [3 MARKS] The effective annual discount rate corresponding to a nominal discountrate, compounded every 3 months, of 12%.


(a) [3 MARKS] The value as of one year ago of 36 payments of 1 at the end of everyhalf-year, the first to be paid 12 months from now, at a nominal discount rate of 9%compounded semi-annually.

(b) [3 MARKS] The value now of 24 payments of 1 at the end of every 3 months, the firstto be paid one year from now, at a rate of interest of 3% compounded instantaneously.


(c) [4 MARKS] The value exactly 10 years after the first payment, and just after thepayment due on that day, of a perpetuity consisting of an unending sequence ofpayments of 1 at intervals of one year, at a nominal annual interest rate of 5% com-pounded every 9 months.

3. [10 MARKS] A monthly annuity to a pensioner begins with a payment of 6,000, and in-creases by the amount of 100 from each payment to the next. Suppose that the pensionersurvives until the day when he has just received a payment of 8,000. If the interest rateis 4% compounded monthly, determine the value, as of that day of the payment of 8,000,of all the payments received under this annuity.




i

(Is)n i =sn+1 i

− (n+1)

i (Ia)∞ i =a∞ i

i

(Da)n i =n − an i


i

4. X has sold Y his home for 450,000. In addition to a down payment of 100,000 cash,Y mortgaged the property to X for 350,000. The mortgage provides for level monthlypayments in four 5-year periods, so that the loan will be amortized in 20 years. Ini-tially interest is charged at a nominal rate of 4.8% compounded monthly. However, it isunderstood that this rate will be renegotiated at the end of each 5-year period.



01 . . . . . . . . . . . .

. . . . . . . . . . . . . . .. . . . . . . . . . . .

(b) [5 MARKS] After 5 years — just after the 60th payment — the interest rate changesto a nominal annual rate of 6% compounded monthly. Based on the amount outstand-


ing at the loan at that time the level monthly payment is recalculated. Determine theamount of that level payment, and calculate the interest and principal portions of the70th payment.

5. [5 MARKS] Select one of the following identities

1 = ian + vn (130)

1 = (1 + i)n − isn (131)

1an

=1sn

+ i (132)


i(133)

and prove it “verbally”.

6. The terms of a loan of 20,000 require that a payment of 1,000 be made every 3 monthsfor as long as possible. The next payment after the last payment of 1,000 — called the“drop” payment — will be less than 1,000, and will complete the repayment of the loan.Interest is earned by the lender at an effective semi-annual rate of 4%.

(a) [5 MARKS] Determine when that “drop” payment is made.(b) [5 MARKS] Determine the amount of the “drop” payment.

7. [10 MARKS] A borrower is repaying a loan at a semi-annual cost of 1,000 over a periodof 15 years. Two-thirds of the loan is amortized by equal payments at 6% effective,included in each semi-annual expenditure of 1,000. The other one-third of the loan isrepaid by the sinking fund method, where the principal is repaid when the loan matures,and the sinking fund accumulates at a nominal annual rate of 5%, compounded semi-annually: until maturity the lender pays only semi-annual interest at 6% on the non-amortized part of the loan, and settles that part of the loan at maturity by transferring tothe lender the balance in the sinking fund. Determine the amount of the loan.

8. [10 MARKS] A 10,000 par value 4% bond with semi-annual coupons matures at the endof 10 years. The bond is callable at 10,500 at the ends of years 4 through 6, at 10,250at the ends of years 7 through 9, and at 10,000 at the end of year 10. Showing detailedwork, find the maximum price that an investor can pay and still be certain of a yield rateof 5% convertible semiannually.

7.4.8 Supplemental/Deferred Examination 2008/2009

Instructions






4. Accuracy. The final answers in all questions should be accurate to the nearest 100th of aunit, or better. There is no penalty for stating answers accurate to more than two decimalplaces.




8. Regular and translation dictionaries are permitted.


(a) [3 MARKS] The value as of 9 months ago of 36 payments of 1 at the end of everyhalf-year, the first to be paid 3 months from now, at a nominal annual discount rateof 8% compounded semi-annually.

(b) [3 MARKS] The value now of 12 payments of 1 at the end of every 4 months, the firstto be paid one year from now, at a nominal annual rate of interest of 5% compoundedinstantaneously.

(c) [4 MARKS] The value exactly 6 years after the first payment, and just after the pay-ment due on that day, of all payments under a perpetuity consisting of an unendingsequence of payments of 100 at intervals of one year, at a nominal annual annualinterest rate of 4.75% compounded every 8 months.



(a) [2 MARKS] The nominal annual interest rate, compounded every 7 months, corre-sponding to an effective annual interest rate of 14%.

(b) [2 MARKS] The effective semi-annual discount rate corresponding to a nominalinterest rate, compounded every 4 months, of 10%.

(c) [3 MARKS] The value to which 10,000 will grow from time t = 0 to time t = 10under a force of interest of δt = 0.01t2.

(d) [3 MARKS] The effective annual discount rate corresponding to a nominal annualdiscount rate, compounded every 4 months, of 12%.

3. [10 MARKS] A 10,000 par value 5% bond with semi-annual coupons matures at the endof 10 years. The bond is callable at 11,000 at the ends of years 4 through 7, at 10,400 atthe ends of years 8 through 9, and at par at the end of 9 1

2 years. Showing detailed work,find the maximum price that an investor can pay and still be certain of a yield rate of 3%convertible semiannually.

4. [10 MARKS] A monthly annuity to a pensioner begins with a payment of 5,000, and in-creases by the amount of 250 from each payment to the next. Suppose that the pensionersurvives until the day when she has just received a payment of 9,000. If the interest rateis 6% compounded monthly, determine the value, as of that day of the payment of 9,000,of all the payments received under this annuity.




i

(Is)n i =sn+1 i

− (n+1)

i (Ia)∞ i =a∞ i

i

(Da)n i =n − an i


i

5. X has sold Y his home for 500,000. In addition to a down payment of 140,000 cash,Y mortgaged the property to X for 360,000. The mortgage provides for level monthlypayments in five 4-year periods, so that the loan will be amortized in 20 years. Initially


interest is charged at a nominal rate of 4.2% compounded monthly. However, it is under-stood that this rate will be renegotiated at the end of each 4-year period.



01 . . . . . . . . . . . .

. . . . . . . . . . . . . . .. . . . . . . . . . . .

(b) [5 MARKS] After 4 years — just after the 48th payment — the interest rate changesto a nominal annual rate of 6.72% compounded monthly. Based on the amount out-standing at the loan at that time the level monthly payment is recalculated. Determinethe amount of that level payment, and calculate the interest and principal portions ofthe 100th payment (after the signing of the mortgage).

6. [5 MARKS] Select one of the following identities

1 = ian + vn (134)

1 = (1 + i)n − isn (135)

1an

=1sn

+ i (136)


i(137)

and prove it “verbally”. (It is not sufficient to simply describe each of the terms in thegiven identity in words.)

7. The terms of a loan of 20,000 require that a payment of 900 be made every 3 monthsfor as long as possible, but that the next payment after the last payment of 900 — calledthe “balloon” payment — will be more than 900, and will complete the repayment of theloan. Interest is earned by the lender at an effective semi-annual rate of 3%.

(a) [5 MARKS] Determine when that “balloon” payment is made.(b) [5 MARKS] Determine the amount of the “balloon” payment.

8. [10 MARKS] A borrower is repaying a loan at a semi-annual cost of 5,000 over a periodof 12 years. Three-fifths of the loan is amortized by equal payments at an effective annual


interest rate of 4%, included in each semi-annual expenditure of 5,000. The other two-fifths of the loan is repaid by the sinking fund method, where the principal is repaidwhen the loan matures, and the sinking fund accumulates at a nominal annual rate of 4%,compounded semi-annually: until maturity the borrower pays only semi-annual interestat an effective annual rate of 4% on the non-amortized part of the loan, and settles thatpart of the loan at maturity by transferring to the lender the balance in the sinking fund.Determine the amount of the loan, and the amounts that the borrower pays semi-annuallyin amortization payment to the lender, interest on the non-amortized portion of the loan,and contribution to the sinking fund.

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