Introductory maths analysis chapter 05 official

INTRODUCTORY MATHEMATICAL INTRODUCTORY MATHEMATICAL ANALYSISANALYSISFor Business, Economics, and the Life and Social Sciences

2007 Pearson Education Asia

Chapter 5 Chapter 5 Mathematics of Finance Mathematics of Finance


INTRODUCTORY MATHEMATICAL ANALYSIS

0. Review of Algebra

1. Applications and More Algebra

2. Functions and Graphs

3. Lines, Parabolas, and Systems

4. Exponential and Logarithmic Functions

5. Mathematics of Finance

6. Matrix Algebra

7. Linear Programming

8. Introduction to Probability and Statistics


9. Additional Topics in Probability

10. Limits and Continuity

11. Differentiation

12. Additional Differentiation Topics

13. Curve Sketching

14. Integration

15. Methods and Applications of Integration

16. Continuous Random Variables

17. Multivariable Calculus

INTRODUCTORY MATHEMATICAL ANALYSIS


• To solve interest problems which require logarithms.

• To solve problems involving the time value of money.

• To solve problems with interest is compounded continuously.

• To introduce the notions of ordinary annuities and annuities due.

• To learn how to amortize a loan and set up an amortization schedule.

Chapter 5: Mathematics of Finance

Chapter ObjectivesChapter Objectives


Compound Interest

Present Value

Interest Compounded Continuously

Annuities

Amortization of Loans

5.1)

5.2)

5.3)

5.4)


Chapter OutlineChapter Outline

5.5)



5.1 Compound Interest5.1 Compound Interest

Example 1 – Compound Interest

• Compound amount S at the end of n interest periods at the periodic rate of r is as

nrPS 1

Suppose that $500 amounted to $588.38 in a savings account after three years. If interest was compounded semiannually, find the nominal rate of interest, compounded semiannually, that was earned by the money.


Solution:

There are 2 × 3 = 6 interest periods.

The semiannual rate was 2.75%, so the nominal rate was 5.5 % compounded semiannually.


5.1 Compound Interest


0275.01500

38.588

500

38.5881

500

38.5881

38.5881500

6

6

6

6

r

r

r

r


How long will it take for $600 to amount to $900 at an annual rate of 6% compounded quarterly?

Solution:

The periodic rate is r = 0.06/4 = 0.015.

It will take .




233.27015.1ln

5.1ln

5.1ln015.1ln

5.1ln015.1ln

5.1015.1

015.1600900

n

n

n

n

n

months 9 years,68083.6 21

4233.27




Example 5 – Effective Rate

Effective Rate

• The effective rate re for a year is given by

11

n

e n

rr

To what amount will $12,000 accumulate in 15 years if it is invested at an effective rate of 5%?

Solution: 14.947,24$05.1000,12 15 S




Example 7 – Comparing Interest RatesIf an investor has a choice of investing money at 6% compounded daily or % compounded quarterly, which is the better choice?

Solution:

Respective effective rates of interest are

The 2nd choice gives a higher effective rate.

%27.614

06125.01

and %18.61365

06.01

4

365

e

e

r

r

8

16



5.2 Present Value5.2 Present Value

Example 1 – Present Value

• P that must be invested at r for n interest periods so that the present value, S is given by

Find the present value of $1000 due after three years if the interest rate is 9% compounded monthly.

Solution:

For interest rate, .

Principle value is .

nrSP 1

15.764$0075.11000 36 P

0075.012/09.0 r



5.2 Present Value

Example 3 – Equation of Value

A debt of $3000 due six years from now is instead to be paid off by three payments: • $500 now, • $1500 in three years, and • a final payment at the end of five years.

What would this payment be if an interest rate of 6% compounded annually is assumed?



5.2 Present Value

Solution:

The equation of value is

27.1257$

02.160002.11000 208

x



5.2 Present Value

Example 5 – Net Present Value

You can invest $20,000 in a business that guarantees you cash flows at the end of years 2, 3, and 5 as indicated in the table.

Assume an interest rate of 7% compounded annually and find the net present value of the cash flows.

Net Present Value

investment Initial - values present of Sum NPV ValuePresent Net

Year Cash Flow

2 $10,000

3 8000

5 6000



5.2 Present Value

Example 5 – Net Present Value

Solution:

31.457$

000,2007.1600007.1800007.1000,10NPV 532



5.3 Interest Compounded Continuously5.3 Interest Compounded Continuously

Example 1 – Compound Amount

Compound Amount under Continuous Interest

• The compound amount S is defined as

If $100 is invested at an annual rate of 5% compounded continuously, find the compound amount at the end ofa. 1 year.

b. 5 years.

kt

k

rPS

1

13.105$100 105.0 ePeS rt

40.128$100100 25.0505.0 eeS



5.3 Interest Compounded Continuously

Effective Rate under Continuous Interest

• Effective rate with annual r compounded continuously is .1 r

e er

Present Value under Continuous Interest

• Present value P at the end of t years at an annual r compounded continuously is .rtSeP



5.3 Interest Compounded Continuously

Example 3 – Trust Fund

A trust fund is being set up by a single payment so that at the end of 20 years there will be $25,000 in the fund. If interest compounded continuously at an annual rate of 7%, how much money should be paid into the fund initially?

Solution:

We want the present value of $25,000 due in 20 years.

6165$000,25

000,254.1

2007.0

e

eSeP rt



5.4 Annuities5.4 Annuities

Example 1 – Geometric Sequences

Sequences and Geometric Series

• A geometric sequence with first term a and common ratio r is defined as

a. The geometric sequence with a = 3, common ratio 1/2 , and n = 5 is

0 where,...,,,, 132 aarararara n

432

2

13 ,

2

13 ,

2

13 ,

2

13 ,3



5.4 Annuities

Example 1 – Geometric Sequences

b. Geometric sequence with a = 1, r = 0.1, and

n = 4.

c. Geometric sequence with a = Pe−kI , r = e−kI ,

n = d.

001.0 ,01.0 ,1.0 ,1

dkIkIkI PePePe ,...,, 2

Sum of Geometric Series

• The sum of a geometric series of n terms, with first term a, is given by

1r for

1

11

0

r

raars

nn

i

i



5.4 Annuities

Example 3 – Sum of Geometric Series

Find the sum of the geometric series:

Solution: For a = 1, r = 1/2, and n = 7

62

2

1...

2

1

2

11

64

127

21

1

21

11

1

1

21

128127

7

r

ras

n

Present Value of an Annuity

• The present value of an annuity (A) is the sum of the present values of all the payments.

nrRrRrRA 1...11 21



5.4 Annuities

Example 5 – Present Value of Annuity

Find the present value of an annuity of $100 per month for years at an interest rate of 6% compounded monthly.

Solution: For R = 100, r = 0.06/12 = 0.005, n = ( )(12) = 42

From Appendix B, .

Hence,

005.042__100aA

798300.37005.042

__ a

83.3779$798300.37100 A

2

13

2

13



5.4 Annuities

Example 7 – Periodic Payment of Annuity

If $10,000 is used to purchase an annuity consisting of equal payments at the end of each year for the next four years and the interest rate is 6% compounded annually, find the amount of each payment.

Solution: For A= $10,000, n = 4, r = 0.06,

91.2885$465106.3

000,10000,10

000,10

06.04

06.04

____

__

aa

AR

Ra

rn



5.4 Annuities

Example 9 – Amount of Annuity

Amount of an Annuity

• The amount S of ordinary annuity of R for n periods at r per period is

Find S consisting of payments of $50 at the end of every 3 months for 3 years at 6% compounded quarterly. Also, find the compound interest.

Solution: For R=50, n=4(3)=12, r=0.06/4=0.015,

r

rRS

n 11

06.652$041211.135050015.012

__ S

06.52$501206.652 Interest Compund



5.4 Annuities

Example 11 – Sinking Fund

A sinking fund is a fund into which periodic payments are made in order to satisfy a future obligation. A machine costing $7000 is replaced at the end of 8 years, at which time it will have a salvage value of $700. A sinking fund is set up. The amount in the fund at the end of 8 years is to be the difference between the replacement cost and the salvage value. If equal payments are placed in the fund at the end of each quarter and the fund earns 8% compounded quarterly, what should each payment be?



5.4 Annuities

Example 11 – Sinking Fund

Solution:

Amount needed after 8 years = 7000 − 700 = $6300.

For n = 4(8) = 32, r = 0.08/4 = 0.02, and S = 6300,the periodic payment R of an annuity is

45.142$6300

6300

02.032

02.032

________

____

ss

SR

Rs

rn



5.5 Amortization of Loans5.5 Amortization of LoansAmortization Formulas



5.5 Amortization of Loans

Example 1 – Amortizing a Loan

A person amortizes a loan of $170,000 by obtaining a 20-year mortgage at 7.5% compounded monthly. Find

a.monthly payment,

b.total interest charges, and

c.principal remaining after five years.



5.5 Amortization of Loans

Example 1 – Amortizing a Loan

Solution:a. Monthly payment:

b. Total interest charge:

c. Principal value:

51.1369$

00625.01

00625.0000,170

240

R

40.682,158$000,17051.1369240

74.733,147$

00625.0

00625.0151.1369

180

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