Chapter 6 Zica

7/27/2019 Chapter 6 Zica

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Formula (6.3) is often referred to as the present value formula.

Example 1

K25 000 000 is invested for three years at an interest rate of 15%.

a) Calculate the simple interest paid in any one year.

b) Calculate the total value of savings at the end of one, two, and three years. Showthat the total value of savings at the end of successive years is an arithmetic

progression.

c) Calculate the present value (principal) when the future value (total value) is K3750 000 after three years.

a) Start by writing down any information given in the question.

P = 25 000 000;10015=r and t = 3 years.

Therefore:

00025011

310015

00000025

K

t r P I

=

=

=

b) Using formula (6.2), the total value of the savings after t years iscalculated as:

After 1 year, t = 1

00050032

)3.1(00000025

))2(15.1(000000252,2

00028750

))1(15.1(00000025

)1(

2

1

K

At years After

K

A

rt P At

==

+==

=+=

+=

After 3 years, t = 3


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76.0787732

)773078757.2(0000001

)12.1(0000001)1(

9,10012

,0000001

99

K

r P A

yearst r P

t

=

=

+=+=

===

Example 3

a) Calculate the amount owed on a loan of K5 000 000 over 4 years at aninterest rate of 12.5% compounded annually.

20.0330098)125.1(0000005)1(

4,125.100

5.12,0000005

44

K r P A

yearst r P

t

=

=+=

====

b) Musenge places K25 000 on deposit in a bank earning 5% compoundinterest per annum. Find the amount that would have accumulated:

i) After 1 year ii) After 2 yearsiii) After three years.

The final amount accumulated (terminal value), nr P S )1( +=

Where P = Principal r = Interest rate per annum

n = time.

i) 000500287)15.1(00000025 K S ==

ii) 50030623)15.1(00000025 2 K S ==

iii) .87502138)15.1(000000253

K S ==

6.3 Terminal Values

Comparison of Projects.


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If we were to be given the choice between two Projects A and B, the expected profits of which over the next four years are:

A: K25,000,000 at 5% inter per annum

B: K30,000,000 per annum

Which would we prefer (assuming both require the same initial outlay)?

On Project A, we have to compound each flow by adding on interest at 5% pa for the number of years remaining until the end of the projects, that is the year 1 cashflow of K25,000,000 earns 3 years interest and is thus worth K107 753 125 at theend of 4 years. We have compounded the flows to produce what is termed as theTerminal value of each flow.

Project A Cashflows K

Year 1 625,940,28)05.1(00000025 3 =

Year 2 25 000 000 2)05.1( = 27,562,500

Year 3 25 000 000(1.05) = 26,250,000

ear 4 25 00 000 25,000,000

K107,753,125=========

Project B

Year 120,000,000 = 120,000,000

Net Terminal Value


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a) An initial outlay of K500 000 which will generate the following cash flows.

Year Cash flow

1 30 0002 20 0003 40 0004 50 000

The annual interest rate available for deposit is 7.5%.

b) An initial outlay of K5 000 000 which will generate the following cash flows:

Year Cash flow

1 2 500 0002 2 000 0003 -4 5 000 000

The annual interest rate available for deposits is 10%.

c) An initial outlay of K3 000 000 which will generate cash flows of k120 0000 for four years. The annual interest rate available for deposits is 8.5%.

Solution

The net terminal values are calculated as follows:

a) Year Cash flow Compound factor Terminal value

00.00050000504

00.00043)075.1(000403

50.11223)075.1(000202

91.26837)075.1(000301

)57.734667()075.1()000500(0

2

3

4

(514 353.16)

So the project would be rejected. b) Year Cash flow Compound factor Terminal value


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000000500000054

3

0004202)10.1(00000022

0053273)10.1(00050021

)5003207()10.1()0000005(0

2

3

4

3 427 000

So the project would be accepted.

c) Year Cash flow Compound factor Terminal value

00.000200100012004

00.0003021)085.1(00012003

00.6704121)085.1(00020012

95.7463251)085.1(00012001

)10.5761574()085.1()0000003(0

2

3

4

(1 289 840.85)=========

So the project would be rejected.

6.4 Other Applications of the Compound Interest Formula

In the Compound Interest Formula, there are four variables, .,, t and r P At If any three of these variables are given, the fourth may be determined. In some


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cases, you will require the rules for indices and logs. For example, a generalexpression for r may be derived as:

t t r P A )1( +=

dividing throughout by P, we get

t t r P A

)1( +=

then take the t th root on both sides

r P A t t +=

1

1

making r the subject of the formula gives the required formula for r .

)5.6(1

1

=

t

P

Ar t

Example 4

a) Find the compound interest rate required for K15 000 000 to grow toK25 000 000 in 5 years.

b) A bank pays 13.5% interest compounded annually. How long will it takefor K15 000 000 to grow to K20 000 000?

a) Calculate r when given

5,00000015,000000255 === t and P A

Direct from the formula, (6.5)


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%.8.10108.01108.1

1000001500000025 5

1

===

=

r or

r

An interest rate of 10.8% is required if this investment is to be doubled invalue over a period of 5 years.

b) When %.5.1300000015,00000025 === r and P At Direct fromcompound interest formula (6.4),

t

t

)135.1(667.1

)135.1(0000001500000025

=+=

taking natural logarithms or common logarithms on both sides gives us

0355.4

12663265.0511025603.0

)135.1ln()667.1ln(

)135.1ln()667.1ln(

=

=

=

=

t

t

So, at 13.5% interest, it will take over 4 years for the investment to doublein value.

6.5 Present Value At Compound Interest

At present value of a future sum, t A , is the amount which, when put on depositnow (i.e. P t ,0= ), at (r%) rate of interest, will grow to the value of t KA after t


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years. The present value, P, is calculated by arranging the compound interest(6.4).

)6.6()1(

)1(

+=

+=

t

t

t

t

r

A P

r P A

Example 5

If the future value of an investment is K25 000 000 invested at 12.5% compoundinterest per annum for five years. Compute the present value.

yearst r K A 5,125.,000000255 === . Using formula (6.6)

93.22387313

)125.1(00000025

5

K P

P

=

=

6.6 Nominal and Effective Rates

When interest is compounded several times per year, for example it may becompounded daily, weekly, monthly, quarterly, semi-annually or continuously.Each period is called a conversion period or interest period. Then the amount of

the future value is given by the formula

where t mn = = total number of conversion periodsm = conversion periods per year t = number of years

Example 6

K25 000 000 is invested for five years at 12.5%. Calculate the total value of theinvestment when compounded

i) Monthly

)7.6(1

+=

mt

t m

r P A


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ii) Daily

Assume a year has 365 days.

i) P = 25 000 000, r = 12.5% = .125, t = 5 and m = 12. Usingformula (6.7) the total value after three years with 60512 === t mn conversion periods is calculated as:

11.40255546

)01041666.1(00000025

12125.

100000025

1

60

60

5

K

P

mr

P P mt

t

==

+=

+=

ii) 18255365,365,5%,5.12,00000025 ======= t mnmt r P

00070046)868.1(00000025

365125.

100000025

1

1825

5

K

mr

P P t m

==

+=

+=

When working with problems involving interest we use the term payment period as follows:

Annually Once per year Semi-annually Twice per year Quarterly 4 times per year Monthly 12 times per year Daily 365 timer per year (some basics use 360 times per

year)

6.7 Effective Rate Of Interest

Interest rates are usually cited as nominal rates of interest expressed as per annumfigures. However, as compounding may occur several times during the year withthe nominal rate, the amount owed or accumulated will be different from thatcalculated by compounding once a year. So a standard measure used to comparethe amount earned (owed) at quoted nominal rates of interest when compounding


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is done several time per year is called the annual percentage rate (APR) oreffective annual rate or effective rate of interest.

Let us consider formula (6.7)

mt

t mr P A +=

1 nominal rate compounded m times per year.

t t APR P A )1( += APR rate compounded annually. Note that m = 1 i.e, once

per year.

Equating the two amounts since they are the same, we have

t mt

APR P mr

P )1(1 +=

+

making APR subject of the formula, we have

)8.6(11

+=

m

m

r APR

Example 7

Interest on a savings account is payable semi-annually at a (nominal) rate of 12.5% per annum. What is the effective rate of interest?

R = 0.125 and m = 2 because interest is payable twice yearly, so

1289.0

12

125.01

2

=

+= APR

The effective rate of interest is 12.89%

Example 8

a) A finance company advertises money at 25% nominal interest, but compoundedquarterly. Find the effective interest rate (APR)

b) Two banks in a local town quote the following nominal interest rates. Bank X pays interest on a saving account at 12.5% compounded monthly and bank Y pays12.5% on a savings account compounded quarterly, which pays its savers themost interest?


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a) r = .25, and m = 4 because interest is payable 4 times a year, so

274.0

14

25.01

4

=

+= APR

The effective rate of interest is 27.4%.

b) Bank X pays interest of 12.5% compounded monthly

%2.13

132.0112125.

112

=

=

+= APR

Bank Y offers the greater effective interest rate and thus pays its saversmore interest.

6.8 Investment Appraisal

If an initial investment will bring in payments at future times, the payments arecalled cash flows . The net present value , denoted NPV, of cash flows is definedto be the sum of the present values of the cash flows (revenue), minus the initialinvestment (cost). If NPV > 0, then the investment is profitable, if NPV < 0 theinvestment is not profitable.

Example 9

Suppose that you can invest K100 000 000 in a business that guarantees you cashflows at the end of years 1, 2 and 3, as indicated in the table. Assume an interestrate of 12.5% compounded annually and find the net present value of the cashflows.

Year Cash Flow

1 5 000 000


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2 400 0003 200 000

NPV = cash inflows cash outflows

(Revenue) (cost)

79.03909995

00000010039.46614038.04931644.4444444

000000100)125.1(000200)125.1(000400)125.1(0000005 321

K =++=

++=

Since NPV < 0, the business venture is not profitable. If one considers the timevalue of money, it would be better to invest the K100 000 000 in a bank paying12.5%.

6.9 Internal Rate Of Return

The discount rate at which a project has a net present value of zero is called theInternal rate of return (IRR). There is no precise formula for calculating theIRR of a given project. However, it can be estimated (using linear interpolationtechnique) with:

a) graphically, or

b) by formula given by IRR = )9.6(12

1221

NPV NPV NPV r NPV r

Both techniques need that the NPV is calculated using two different discountrates. We explain the two methods in the following examples.

Example 10

A project involves an initial outlay of K100 000 000. The expected cash flow at the endof the next four years is given as given as follows (the amounts are in million of kwachas).

Year 1 2 3 4Cash flows 50 125 167 182


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a) Determine IRR graphically by plotting the NPV against r. For r = 0.05, 0.08, .10, .20.

b) By formula, show that the value of the IRR is slightly different when calculated

from pairs of points.

a) Graphically. Excel is ideal at this point. Since calculation of NPVsrequires the repeated use of the formula t t r A NPV

+= )1( where is the symbol for the sum of several NPVs. Then we plot the curve of

NPV against r .

Table 6.1 Excel Sheet for Calculating NPVs at Different Interest Rates

t Cashflow

r = 0.05 r = 0.08 r = 0.10 r = 0.20 r =0.24

01234

30050

125167182

30047.61905113.3787144.2609149.7319154.9905

30046.2963

107.1674132.5700133.7754119.8091

30045.45455103.3058125.4696124.3084

98.53835

30041.66667

80.000085.504074.5472

-18.2821

-30040.32258

81.295387.5893776.98119-13.8113

NPV

200

150

100


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50

0 -50 0.1 0.2 0.3 r

IRR

Figure 6.1

The NPV for each discount rate is plotted in Figure 6.1. The IRR is the value of r at which this graph crosses the horizontal axis. In Figure 6.1 this point is

between r = 10% and 20%, but considerably closer to 20%.

b) In Table 6.1 are several positive and negative NPVs. Therefore, the IRR given byformula (6.9) is calculated from any such pair.

Summarising points already calculated.

Points A B C D E

r 0.05 24.020.010.008.0

NPV 154.9905 119.8091 98.53835 18.2821 13.8113

i) From points C and D


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1843.0

82045.11653586.21

53835.982821.18)53835.98(20.0)2821.18(10.0

12

1221

=

=

=

=

IRR

NPV NPV NPV r NPV r

IRR

Therefore, IRR = 18.43%.

ii) From points B and E,

22346.0

6204.133

859088.29

8091.1198113.13)8091.119(24.0)8113.13(08.0

12

1221

=

=

=

=

NPV NPV NPV r NPV r

IRR

Therefore, IRR = 22.346%

The two results (i) and (ii) demonstrate that slightly different estimates arecalculated from different pairs of points.

6.10 Comparison of Appraisal Techniques: NPV and IRR


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When comparing the profitability of two or more projects, the most profitable project would be (a) the project with the largest NPV, (b) the project with thelargest IRR.

The advantages of NPV method is that :

i) It gives results in cash termsii) It is practical as it discounts net cash flows.

The disadvantage is that it relies on the choice of one discount rate which meansthat a change in the discount rate could lead to a change in the choice of project.

The advantage of the IRR is that it does not depend on external rates of interest.A major weakness is that the method does not differentiate between the scale of

projects; for example, one project might involve a cash flow in units of K5 000 000 while another involve units of K5. Not that in most cases where two

or more similar project are being ranked in order of preference, the methods of NPV and IRR will generally agree on the best project but this is not a hard andfast rule.

Example 11

The net cash flow for two projects, A and B, is as follows:

K000, 000

Year 0 1 2 3 4

Project A -45.0 -13.5 18 27 36Project B -22.5 -9 4.5 13.5 22.5

a) Use the net present value criterion to decide which project is the most profitable if a discount rate of

i) 8%, andii) 12% is used.

b) Calculate the IRR of each project. Which project would now beconsidered more profitable?


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Project A

Year Net Flow DiscountFactor at

8%

PresentValue

Discountfactor at

12%

PresentValue

0

1

2

3

4

-45

-13.5

18

27

36

1.0000

0.9259

0.8573

0.7938

0.7350

-45.0

-12.50

15.43

21.43

26.465.82

1.0000

0.8929

0.7972

0.7118

0.6355

-45.00

-12.05

14.35

19.22

22.88-0.68

116.0

42.67464.0

42.6048.06984.0

82.56.0)6.0)(08.,10()82.5(12.

12

2112

+=

=

=

=+=

+=

NPV NPV NPV r NPV r

IRR

Project B

Year Net Flow DiscountFactor at8%

PresentValue

Discountfactor at12%

PresentValue


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0

1

2

3

4

-22.5

-9

4.5

13.5

22.5

1.0000

0.9259

0.8573

0.7938

0.7350

-22.5

-8.33

3.86

10.72

16.540.29

1.0000

0.8929

0.7972

0.7118

0.6355

22.50

8.04

3.59

9.61

14.30 3.04

083.0

33.3278.0

42.60348.02432.0

29.004.3)29.0(12.)04.3(08.0

12

1221

=

=

=

=

=

NPV NPV NPV r NPV r

IRR

From (a) at rate 8%, project A has the highest NPV (9.42) and thus would be chosen as best. (Note also that project B has the highest NPV (0.29)). And from (b) project A hasthe highest rate of return at 11.6% and thus would be chosen as best agreeing with thechoice in (a). Overall clearly project A is the best choice.

Exercise 1

1) K5 400 000 is invested at 9.5% simple interest. How much will have to beaccrued after 5 years?

2) Find the amount of:


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a) K 2 160 000 compounded at 13.5% for 3 years.

b) K5 580 000 compounded at 8.5% for 10 years.

3) Calculate the present value of K56 million that is expected to be received in fiveyears time when simple interest is 6.5%.

4) Calculate the compound interest rate required for K250 000 to grow to K450 000in 3 years time.

5) Calculate the number of years it will take a sum of K450 000 to grow to K1 800000 when invested at 4.5% interest compounded annually.

6) Calculate the APR for a 6.5% nominal rate of interest which is compounded

a) four times per year b) 12 times per year, andc) 3 times per year.

7) Two banks in a Geal town quote the following nominal interest rates: Bank Xcharges interest on a loan at 9.5% compounded semi-annually and bank Y charges9.0% on a loan compounded quarterly. Which bank charges the most interest on aloan?

8) You have a choice of two savings schemes. Scheme A offers 7.0% interest payable semi-annually and scheme B offer 6.5% interest payable quarterly.Which bank charges the most interest on a loan?

9) Find the present value of K25 million in 6 years time if the discount rate is 13.5%compounded semi-annually.

10) A Financial group can make investment of K240 million now and receive K264million in two years time. What is the internal rate of return?

11) An investment project has the following NPV calculated for a range of discountrates. Give an approximate IRR for the project.

Discount rate (%) NPV (Kmillion)5

5.56

6.5

11.257.023.421.2


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given time interval is the term of the annuity. An example of an annuity isdepositing of K100,000 in a savings account every 6 months for a year.

In this Section, we consider the amount accrued from a series of such payments and also the present values of a series of such payments which

are to be made.

The amount of an annuity is found using the following formula:

( ))10.6(

11 +=r

r R A

n

while the present value of an annuity is found using the following formula.

)11.6()1(1

)(Pr +=

r r

Rannuityof valueesent P n

where A is the amount of an ordinary annuityr is the interest rate per periodn number of period P is the present value of an ordinary annuity.

Example 11

Find the amount of an annuity consisting of payments of K225 000 at theend of every 3 months for 4 years at the rate of 6.5% compoundedquarterly.

To find the amount of the annuity we use equation (6.10) with R =22 000,

n = 4(4) = 16, and r = 01625.04065.0 =


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67.8490734

)10599851.18(000225

01625.01)01625.1(

00022516

K

A

==

=

Example 12

What is the terminal or future value of an annuity of K4 500 000 for five years at 10% of interest rate per annum?

R = 4 500 000, r = 0.10, n = 5 and n 6

00.95047227

)1051.6(0005004

10.01)10.1(

50045

K

A

=

=

=

Alternatively, we reason as follows:

The first payment will be made at the end of year 1, and so at the end of 5 years, it willhave been invested for 4 years and will have a value of 4)10.1(0005004 . The next

payment is made at the end of year 2, and so at the end of year 5 it will have beeninvested for 3 years and will have a value of .)10.1(0005004 3 It is easy to work usinga table as follows:

End of Year

Amount (K) Value of the end of Year 5


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5

4

3

2

1

00050040005004000500400050040005004

0005004)10.1(0005004)10.1(0005004

)10.1(0005004

)10.000915004

2

3

4

Notice the pattern

At the end of year 5 the total value of all the payments will be the total of the thirdcolumn. In reverse order, we can see that this is a geometric series with the first term a = 4 500 000, common ratio r = 1.10 and n = 5.

That is 432 )10.1(000450)10.1(00450)10.1(0004500)10.1(00045000004500 ++++

=

11

r r

aS n

n using formula (5.6)

=110.1

1)10.1( 55 aS

.95047227)1051.6(0004500 K ==

The present value of an annuity is the sum of the present values of all the payments. Itrepresents the amount that must be invested now to purchase the payments due in thefuture. Unless otherwise specified, we assume that each payment is made at the end of a

payment period, that is called an ordinary annuity.

Example 13

Find the present value of an annuity of K450 000 per month for 3 years at an interest rateof 6.5% compounded monthly.

Using equation (6.11), R = 450 000, .36)3(1212065.0 === nand r

Thus


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77.73168614

6371817.32(450

0054.0)0054.1(1

00045036

K

P

=

=

=

Alternatively, there will be altogether 12(3) = 36 monthly payments. The interest

rate is .0054.012065.0 = The present value of the payment is therefore:

362 )0054.1(000450

...)0054.1(

0004500054.1

000450 +++

This is a geometric series with

.360054.1

1,

0054.1

000450 === nand r a

So the sum is =

11

r r

aS n

n using formula (5.6)

77.731686141

0054.11

10054.11

0054.1000450

36

K =

=

6.12 Equation of Value

Example 14

Suppose that Mr Chilufya owes Mr Banda two sums of money: K500 000 due in2 years and K300 000 due in 5 years. If Mr Chilufya wishes to pay off the totaldebt now by a single payment, how much should the payment be? Assume aninterest of 5% compounded semiannually.

The single payment x due now must be such that it would grow and eventually pay off the debts when they are due. That is, it must equal the sum of the presentvalues of the future payments. As shown in the figure below, we have

104 )025.1(000300)025.1(000500 += x

This equation is called an equation of value. We find that


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84.33468752.23435932.452975 K x =+=

Year

0 1 2 3 4 5

SinglePayment x 500 000 300 000

Present 500 000 (1.025) -4 4 periodsValue of Dents

300 000(1.025) -10 10 periods

Replacing two future payments by a single payment now for Example 13

In general, an equation of value illustrates that when one is considering two methods of paying a debt (or other transaction), at any time the value of all payments under onemethod must equal the value of all payments under the other method.

Example 15

A debt of K15 000 000, which is due 7 years from now is instead to be paid off by three payments: K2 500 000 now, K7500 000 in 4 years, and a final payment in the 6 th year.What would this payment be if an interest rate of 5% compounded annually is assumed?

Let x be the final payment due in 6 years. Setting up the equation of value, we have

1

26

)05.1(00000015

)05.1(0007500)05.1(0002500=

++ x

Year

0 1 2 3 4 5 6 7


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2500 000 7500 000 x 15 000 000

7500 000 (1.05) 2

2500 000(1.06) 6

15 000 000 (1.06) -1

Time values of payment for Example 15

3546297.78 + 8268 750 + x = 14285714.29

x = K2470666.51

6.13 Perpetuities

When the present value of an annuity continue for an indefinitely long period of time, we have what we call Perpetuities . The present value of annuity sayreceivable in arrears in perpetuity at a given discount rate r is given by thefollowing formula:

PV of perpetuity = )12.6( = ratediscount cashflows Annual

r A

Example 15

Find the present value of K25 000 000 receivable annually in arrears at a discountrate of 7.5% . A = 25 000 000, r = 0.075

30.333333333075.

00000025 K

r A

PV ===

Example 16

The T Company is expected to pay K11 250 every 6 months indefinitely on ashare of its preferred stock. If money is worth 6.5% compounded semi annuallyto X, what should he be willing to pay for a share of the stock?

A = 11 250, r = 0.0325; then


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85.1533460325.0

11250 K

r A

PV ===

6.14 Amortization of Loans

When borrowing a sum of money from a bank or building Society for house purchase (a mortgage), it is usual to repay it by a series of regular equalinstallments. The present value of the series will be the same as the amount

borrowed.

Example 17

Chisha wishes to borrow a sum of money to buy a house. She wishes to repayexactly K2 500 000 a month for 15 years at the rate of 13.5% compoundedmonthly,. Find how much she can borrow.

Using formula (6.11), 1801215,01125.012135.

,0005002 ===== nr R .

80.750556192)02270031.77(0005002

01125.0)01125.1(1

0005002180

K ==

=

Alternatively, there will be 1801215 = . Monthly repayments of K2 500 000.The interest rate is 1.125%. The present value of repayments is therefore

( )

80.750556192

101125.1

1

101125.1

1

01125.1

0005002

180,01125.1

1,

01125.10005002

01125.10005002

...)01125.1(

0005002)01125.1(

000500201125.1

0005002

180

18032

K

nr a

=

=

===

++++

Example 18


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Lungowe borrows K11250 000 to be paid in 4 years to buy a car. How muchmust she repay per month assuming an interest rate of 7.5% a year compoundedmonthly.

Using formula (6.11)

00625.012075.0

,48)12(4,00025011

)1(1

=====

+=

r n P where

r r

R P n

Substituting these values in the above formula

=

00625.0)00625.1(1

0002501148

R

and making R the subject of the formula

70312.5 = R(0.258489819)

R = K272 012.65

6.15 Amortization ScheduleAn analysis of how each payment in the loan is handed can be given in a tablecalled an Amortization Schedule . The amortization schedule contains

i) Principal outstanding at the beginning of the period

ii) Interest for period

ii) Payment at the end of the period

iv) Principal repaid at end of period

A loan is amortized when part of each payment is used to pay interest and theremaining part is used to reduce the outstanding principal. Since each paymentreduces the outstanding principal, the interest portion of a payment decreases astime goes on. For examination purposes, a schedule would only be asked for where the period was relatively short; for example, up to 5 or 6 time periods.


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Example 19

A debt of K22 500 000 with interest at 6% compounded semi-annually isamortized by equal payments for the next 2 years.

a) Find monthly payment b) Construct an amortization schedule.

a) Using formula (6. 11)

52.1080536

)717098403.3(00050022

03.0)03.1(1

00050022

422,206.0

,00050022

4

K R

R

R

nr P

=

=

=

====

The monthly repayments must be K6 053 108.52

b) The Amortization Schedule is shown in Table 1.0

Period PrincipalOutstanding atBeginning of period

Interest forperiod

Payment atEnd of Period

Principal Repaidat End of Period

1234Total

22 500 00017 121 891.4811 582 439.70

5 876 804.37

675 000513 656.74347 473.19176 304.13

1712 434.06

6 053 180.526.053 108.526 053 108.526 053 108.50

24 212 434.06

5 378 108.525 539 457.7857 05 635.335 876 804.37

22 500 000.00

The final payment is adjusted to offset rounding errors. Note that:

1) Principal repaid = payment made + interest paid

2) Outstanding principal at beginning of period = principal outstanding(current)

At beginning of period (previous) principal repaid at end of period (previous)


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6.16 Sinking Fund

A sinking fund is a fund into which periodic payments are made in order to satisfya future obligation. This is the amount of an annuity as opposed to the present

value of an annuity in the case of a loan. Here we use formula (6.10).

+=r

r R

annuityanof Amount An 1)1(

)(

Sinking funds are commonly used for the following purposes:

a) repayments of debts

b) to provide funds to purchase a new asset when the existing asset is fullydepreciated.

c) To pay for future school fees or a pension.

Example 20

The Board of Education received permission to issue K200 000 000 I bonds to build a new block of classrooms. The board is required to make payments, every6 months into a sinking fund paying 12.5% compounded semi-annually. At theend of 15 years the bond obligation will be retired. What should each payment

be?

The payment R required twice a year to accumulate to K200 000 in 15 years

30215( = payments at a rate of interest 0625.02

125. = per payment period).

Using formula (6.10),

38.5674202

)62525618.82(

0625.01)0625.1(

000000200

1)1(

30

K R

R

R

r r R A

n

=

=

=

+=


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Example 21

A woman borrows K13 500 000, which will be paid back to the lender in one payment at the end of 4 years. She agrees to pay interest semi annually at 15%.At the same time she sets up a sinking fund in order to repay the loan at the end of 4 years. She decides to make equal deposits into her sinking fund, which earns6.5% interest compounded semi-annually.

a) What is the monthly sinking fund deposit?

b) Construct a table that shows how the sinking fund grows over time.

c) How much does she need each month to be able to pay the interest on the

loan and make the sinking fund deposit?

a) The sinking fund deposit is the value of R in the formula.

+=r r

R An 1)1(

where A equal the amount to be accumulated namely,

.2

065.0,824,0050013 ==== r and n K A The sinking fund

deposit is therefore:

55.7455041

)97161647.8(00050013

2065.

12065.0

100050013

8

K R

R

R

=

=

+

=

b) The table below shows the growth of the sinking fund over time. Theentries for payment number 8 are obtained by using the amount of an


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annuity formula for a monthly payment of K1 504 745.55 made for 8months at 6.5% compounded monthly.

79.8163186

)199259328.4(55.1504745

2065.0

12065.0

155.1504745

8

K

Total

=

=

+

=

The deposit for payment number 8, the final payment, is only K1 504745.58 because a deposit of K1 504 745.58 results in a total payment of

K134 99999 97.

Period InterestAdded

Deposit Increase inFund

Amount inFund at End of Period

1

2

3

4

5

6

7

8

0

48 904.23

99.397.85

151 532.51

205 361.55

260 940.03

318 324.81

377 574.59

1 504 745.55

1 504 745.55

1 504 745.55

1 504 745.55

1 504 745.55

1 504 745.55

1 504 745.55

1 504 745.55

1504 745.55

1 553 649.78

1 604 143.40

1 656 278.06

1 710 107.10

1 765 685.58

1 823 070.36

1 882 320.17

1 504 745.55

3 058 395.33

4 662 538.73

6 318 816.79

8 028 923.89

9 794 609.47

11 617 679.83

13 500 000.00

c) The monthly interest payment due on the loan of K13 500 000 at 15% interest isfound using the simple interest formula.

500,012,121

)15.0(00050013 K I =

=


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Thus the woman needs to be able to pay K1 504 745.55 + K1 012 500

= K2 517 245 55 each month.

Example 22

A copper mine is expected to yield an annual net return of 900 million for the next 15years, after which it will be worthless. An investor wants an annual return on theinvestment of 18%. If she can establish a sinking fund earning 12% annually, how muchshould he be willing to pay for the mine?

Let x be the purchase price. Then 0.185 x represents an 18.5% annual return oninvestment. The annual sinking fund contribution needed to obtain the amount x in 15

years is found by solving 11)1( +

r r

xn

where n = 15 and r = 0.12. The investor

should be willing to pay an amount x so that

Annual return + Annual sinking = Annual returnon investment fund requirement

00.5268052484

000000900211824239.0

000000900026824239.0185.0

00000090012.

1)12.1(185.0

115

K x

x

x x

x x

=

=

=+

=+

A purchase price of K4 248 805 526.00 will achieve the investors goals.

Example 23

A machine costing K18 million new is estimated to have after 6 years of use a scrapvalue of K1.8million. if the depreciation fund earns 3%, use the sinking fund method to:

a) Find the annual deposit into the fund,

b) Find the amount in the fund at the end of 4 years,

c) Prepare a depreciation schedule.


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a) Original cost Scrap value = 18 1.8 = 17.2

Therefore:

51.4795042

194052296.0486.0

1)03.1()03.0(2.17

03.01)03.01(2.17

6

6

=

==

+=

R

R

Age DepreciationCharge

Interestfund

Increase fund Amount infund

Book value

0

1

2

3

4

5

6

0

2 504 479.51

2 504 479.51

2 504 479.51

2 504 479.51

2 504 479.51

2 504 479.51

0

0

152522.80

232.232.87

314334.24

398898.66

486 00.00

0

2 504 479.51

2 657 002.31

2 889 235.18

3 203 569.42

3 602 468.08

4 088 468.08

0

2 504 479.51

5 084 093.41

7 741 095.72

10 477 808.1

13 296 621.85

16 200 000.02

18 000 000

15 495 520.49

12 915 906.59

10 258 904.28

7 522 191.90

4 703 378.15

1 799 999.98

The error of K0.02 in the final book value is due to rounding off all entries to 2decimal places.

Exercise 2

1) Find the present value of the given future payment at the specified interest rate.

a) K24 million due in 15 years at 5% compounded annually.


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8) A debt of K500 000 000 compounded at 18.5% is to be discharged over 4 yearsusing a sinking fund method. Find the annual payment (based on an ordinaryannuity) if the fund earns 12.5%. Draw up a schedule showing both the positionof the debt and the fund each year.

9) A debt of K2 000 000 due in 4 years and K2 000 000 due in 5 years is to be repaid by a single payment now. Find how much the payment is if an interest rate of 10% compounded quarterly is assumed.

10) A debt of K5 000 000 due in 5 years is to be repaid by a payment of K3 000 000now and a second payment at the ends of 6 years. How much should the second

payment be if the interest rate is 5% compounded quarterly?

EXAMINATION QUESTIONS WITH ANSWERS

MULTIPLE CHOICE QUESTIONS

(SECTION A)

1.1) Today Chisenga purchase an annuity of K6 750 000 per year for 15 years from aninsurance company which was 3% compounded annually. If the first payment isdue in one year, what did the annuity cost Chisenga?

a) K80 581 061.25 b) K3 037 500 c) K125 542 668.70d) K9 787 500

(Natech, 1.2 Mathematics & Statistics, November/December 2000 )

1.2) A certain machine costs K5 400 000. The depreciation for a month at the end of any month is estimated to be 5% of the value of the beginning of the month. Atwhat value is the machine cared after 24 months of use?

a) K1 659 727.08 b) K1 576 740.73 c) K11 610 000d) K90 990 000

(Natech, 1.2 Mathematics & Statistics, November/December 2000)

1.3) Find the Net present value of a project which requires an initial outlay of K50 000and guarantees you a cash flow of K30 000 per annum for the next three yearswith an interest sale of 10%.


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a) K25 608 b) K24 606 c) K50 000d) K25 000

(Natech, 1.2 Mathematics & Statistics, June 2002)

1.4) In how many years will K1 000 000 amount to K3 207 000 at 6% per annumcompound interest? (Give your answer to the nearest whole number).

a) 17 b) 18 c) 19d) 20

(Natech, 1.2 Mathematics & Statistics, June 2002 )

1.5) Calculate the annual effective rate of interest of 5% compounded monthly to twodecimal places.

a) 6% b) 5.12% c) 5% d) 4.91%


1.6) What is the present value of an annuity that pays K400 000 a month for the nextfive years if money is worth 12% compounded monthly.

a) K24 000 000 b) K17 982 015 c) K28 800 000d) K2 808 000


1.7) What is the amount for an ordinary annuity of K10 000 a year for 4 years at 8%compounded annually?

a) K45 061.12 b) K48 500.50 c) K13 300d) K13 604.89

(Natech, 1.2 Mathematics & Statistics, December 1998)

1.8) If a boy undertakes to deposit K100 on September 1, K200 on September 2, K400on September 3, K800 on September 4, and so, how much will be deposit fromSeptember 1 to September 15, inclusive?

a) K32 767.00 b) K327 670.00 c) K3 276 700.00d) K3 276.70(Natech, 1.2 Mathematics & Statistics, December 1998)

1.9) Find the compound interest on K800 000 for 2 years at 6% per annum, interest to be added half yearly.


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a) K96 200.00 b) K192 400.50 c) K100 407.05d) None of these


1.10) If K40 000 000 invested for 5 years yields a simple interest of K3 800 000, what

will be the interest on K24 000 000 invested at the same rate for 7years?

a) K380 000 b) K240 000 c) K342 000d) K760 000


MULTIPLE CHOICE QUESTIONS(SECTION B)

1.1) What sum will earn K15 750.00 simple interest in 146 days at 4.5% per annum?

a) K15 778.35 b) K284 112.30 c) K157 500.00d) K875 000.00


1.2) If 1 500 000 is deposited at simple interest of 3% per year, what amount of moneywould be in the account at the end of 12 years?

a) K180 000 b) K518 000 c) K699 000d) K18 000


1.3) What is the discountable value of a bill for K475 000 drawn on 4 th march at 3months and discounted on 10 th may at 6%?

a) K472 813.70 b) K28 500.00 c) K427 500.00d) 2 186.30



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1.4) The terms for a five year lease agreement are that, K10 million must be paid at the beginning of the first year, to be followed by four-equal payments at the beginning of years two, three, four and five at a discount rate of 8%. If the present value of the four equal payment is K26 496 000, the total amount to be paid during the lease period is close to?

a) K32 million b) K40 million c) K42 milliond) K44 million

1.5) A machine assumed to depreciate at a fixed rate of 12% per annum, will have a book value of K9 288 080 in six years time. Its purchase value to the nearest tenis:

a) K15 million b) 21.5 million c) K25 milliond) K20 million.

1.6) A government bond of K1 million is advertised to become K1.57 million after 5years. The effective annual rate of interest to the decimal place is:

a) 7.5% b) 9.0% c) 10.8%d) 11.4%

1.7) The net present value of an investment at 20% is K12 400 000 at 12% is a loss of K8 000 000. What is the internal rate of return of this investment?

a) 18% b) 17% c) 16% d) 15% (Natech, 1.2 Mathematics & Statistics, December 2003)

1.8) How long will it take for K4 275 000 to amount to K4 446 000 at 8% simpleinterest rate giving your anser in months?

a) 5 months b) 6 months c) 5.8 monthsd) 3.0 months


1.9) A bank accounts pays 12% annual interest compounded monthly. How muchmust be deposited now so that the account contains exactly K45 000 000 at theend of the year?


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i) Find the net present value of the cash flows.ii) Is the investment profitable?

b) Chipasha Mulenga recently purchased a computer for K5 400 000 and agreed to

pay it off by monthly payments of K270 000. if the store charges interest at therate of 12% compounded monthly, how many months will it take to pay off thedebt?

c) i) Find the amount of an annuity consisting of payments of K180,000 at theend of every 3 months for 3 years at the rate of 6% compounded quarterly.

iii) Also find the compounded interest.iv)


d) Miss Mwalilino has an obligation of K1 350 000 due five years from now. If interest is assumed to be 7 percent and is compounded yearly, what is the presentvalue of the obligation?

(Natech, 1.2 Mathematics & Statistics, Nov/Dec 2000)QUESTION THREE

a) Moonga borrows K500 000 now at an interest rate of 5% per annum. The loanhas the be repaid through five equal installments at the end of each year for thenext five years. Calculate the annual repayment.

b) Mr Sokonjo has just received his gratuity amount to K60 million. He wishes toinvest K50 million of the gratuity. He is now faced with a choice between twoinvestment opportunities, A and B. capital outlay for each is K50 million.

A is estimated to yield an annuity of K20 million at the end of each year receivable every 5 years.

B yields K11 million receivable at the end in perpetuity. If the discounting rate isestimated at 20% for Mr Sokonjo,

i) Evaluate the two (2) investment opportunities using Net Presentvalue (NPV) method.

ii) Recommend with a reason which one of the two (2) investmentopportunities Mr Sokonjo should choose.


QUESTION FOUR


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a) Supremo Stores advertises goods at K700 00 deposit and three further equalannual payments of K500 000 for the next three years. If the discount rate is7.5%, calculate the present value of the goods.

b) i) For how many years must I invest K20 000 if I want it to have a value of

at least K500 000 and the interest rate is 6%, payable annually?

ii) How does your answer change if interest is payable every 6 months? (Natech, 1.2 Mathematics & Statistics, June 2003)

c) Mr Musole, a sole proprietor, is paying K500 000 each quarter into a fund which pay 12% per year interest, compounded quarterly. How much will haveaccumulated in the fund by the end of the fifth year?


d) M-net offers a decorder for K540 000 and K67 500 per month for the next 12months. If interest is charged at 9% compounded monthly, find the equivalentcash value to be paid now.

(Natech, 1.2 Mathematics & Statistics, Nov/Dec 2000)

Chapter 6 Zica

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