1

Chapter 3

Compound interest

2

Simple interest and compound amount formula

Formula for compound amount interest is:

niPS )1(

Where :

S: the amount at compound interest

P: the principal

i: the rate per conversion period

n: the number of conversion periods

the factor ni)1( called the accumulation

factor.

Note:

1) In many business transactions , the interest is

computed annually, semiannually, quarterly,

monthly, daily, or at some other time interval.

2)The important rate is the interest rate per

conversion period, which is designated by the

symbol i.

3

3) The quoted annual rate is called the nominal

rate, and is indicated by the symbol m. the

equation relating j, m, i is i=j/m , or j= i. m

("jim"), the symbol )(mj means a nominal rate

converted m times a year.

4) When no conversion period is stated in

problem assume that the interest is compounded

annually .

Finding n

Example 1

How many semiannual conversion periods are

there from June 1, 1989, to December 1, 1994?

Solution:

Year Month day

1994 12 1

-1989 -6 -1

5 years 6 months 0 days

4

n = (5 x 2) + 6/6 = 10 + 1 = 11

after subtraction we multiply the number of

years by the period per year and divide the

number of months by the months in a period,

and add the results to get the number of periods.

Example 2

How many quarterly conversion period are

from November 15, 1990, to August 15, 1999.

Solution:

n = ( 8 X 4 ) + 9/3 = 32 + 3 = 35.

Example 3

Find the computed a mount of $25 invested at

6% converted quarterly for 5 years.

5

Solution :

67.33$3469.125

)4

06.01(25)1( 45

niPS

Example 4

A principal of $1000 is deposited at 6% for 10

years. What will be the computed amount and

the compound interest if the interest is

compounded annually, semiannually, quarterly,

and monthly?

Solution :

- interest computed annually:

85.1790$

)06.01(1000)1( 10

niPS

Compound interest= S - P

=$1790.85 - $1000=$790.85.

- interest computed semiannually:

6

11.1806$

)2

06.01(1000)1( 210

niPS

Compound interest= S - P

=$1806.11 - $1000=$806.11.

- interest computed quarterly :

02.1814$

)4

06.01(1000)1( 410

niPS

Compound interest= S - P

=$1814.02 - $1000=$814.02.

- interest computed monthly :

40.1819$

)12

06.01(1000)1( 1210

niPS

Compound interest= S - P

=$1819.40 - $1000=$819.40

7

Example 5

A person aged 60 put $10000 in a deferred a

account paying 8% converted quarterly. The a

account is to mature in 5 years. Find the amount

at that time.

Solution:

Substituting P=10000, i=0.08/4=0.02, n=5*4=20

47.14859$

)4

08.01(10000)1( 45

niPS

Example 6

A bank pays 7.8% compounded quarterly on

savings accounts. A woman puts $5000 into such

an account on July 1, 1990. find amount in the

account on January 1, 1995.

Solution :

8

N = 4 * 4 + 6/3 = 18 periods

48.7078$

)4

078.01(5000)1( 18

niPS

Example 7

A depositor planned to leave $2000 in saving

and loan association paying 5% compounded

semiannually for a period of 5 years. At the end

of 2.5 years the depositor had to withdraw

$1000. what amount will be in the account at the

end of original 5-year period?

Solution:

First we compute the a mount after 2.5 years

9

82.2262$

)2

05.01(2000)1( 25.2

niPS

After withdrawal of $1000 the final balance is $

2262.82 - $ 1000 = $1262.82

The amount at the next 2.5 years

76.1428$

)2

05.01(82.1262)1( 25.2

niPS

Alternate solution

Using the equation of value

11

x 510 )2

05.01(1000)

2

05.01(2000

x = $ 1428.76

low of organic growth

example 1

during the period 1970 – 1980 , the population

of a city increased at rate of a bout 3% a year. If

the population in 1980 was 300 000 , what is the

predicted population in 1990?

Solution:

Substituting P=300 000 , i=0.03 , n=10

403175

)03.01(300000)1( 10

niPS

Example 2:

11

During the period 1970- 1975 the earning per

share of Oklahoma natural Gas Company

common stock increased at about 9% a year

compounded. The earnings per share for 1975

was $2.92. assuming that the same rate of

increase continues, predict the earnings per

share for 1990.

Solution:

Substituting P=2.92 , i=0.09 , n=15

63.10$

)09.01(92.2)1( 15

niPS

Example 3:

During the period 1970-1980 the population of a

city increased 8% . if the population was 500

000 in 1980, what is the estimated population for

12

2000 assuming that the same rate of growth

continues?

Substituting P=500 000 , i=0.08 , n=2

583200

)08.01(500000)1( 2

niPS

H.W :page 119 Exercise 3a (all)

Example ( daily compounding)

Compute the amount of 1-factor for 5%

converted daily for 2 days for 360- and – 365

years.

Solution :

For 360 day year, we find that

13

0002777.1$

)360

05.01()

3601( 22

j

S

For 365 day year, we find that

00027399.1$

)365

05.01()

3651( 22

j

S

H.W :page 126 Exercise 3b (all)

14

15

Interest for a part of a period.

when deriving the compound interest formula,

we assume that the time would be an integer

number of conversion periods, when there is a part

of period, the usual practice is to allow simple

16

interest for this time on the computed amount at the

end of the last whole period.

Example

At 7% compounded semiannually, $2000 will

amount to how much in 3 years and 5 months?

The total time in this case is 6 period and 5

months left over.

Solution :

The amount after 6 period is

17

51.2458$

)2

07.01(2000)1( 6

niPS

The simple interest for the remaining 5 months is

I = P. r t = 2458.51 * 0.07 *(5/12)= $71.71

So the amount at the end if 3 years and 5 months is

2458.51 + 71.71 = $2530.22

Amount at changing rates

Example :

A Principal of $900 earns 6% converted

quarterly for 4 years and then 7% converted

semiannually for 2 more years. Find the final

amount.

Solution :

18

First we find the amount at the end of 4 years (

16 period)

09.1142$

)4

06.01(900)1( 16

niPS

Second we find the amount at the next of 2 yeas

years ( 4 period)

57.1310$

)2

07.01(09.1142)1( 4

niPS

H.W :page 130 Exercise 3c (all)

Present value at compound interest

ni

SP

)1(

Where :

P: the principal or present value

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S: the amount due in the future

i: the rate per period

n: the number of periods

Example :1

Find the present value of $5000 due in 4 years if

money is worth 8% compounded semiannually.

Solution:

Substituting S =5000, i=0.08 , n=8

45.3653$

)2

08.01(

5000

)1( 8

ni

SP

Example :2

Find the present value of $7500 due in 4 years if

money is worth 14% compounded monthly.

Solution:

Substituting S =7500 , i=0.14/12 , n=4*12=48

21

98.4297$

)12

14.01(

7500

)1( 48

ni

SP

Example :3

How much must be invested in an account

paying 8.4% compounded monthly in order to

accumulate to $15000 in 5 years.

Solution:

Substituting S =$15000 , i=0.084/12 = 0.007 ,

n=5*12=60

13.9870$

)12

084.01(

15000

)1( 60

ni

SP

Example :4

A note with a matuary value of $1000 is due in3

years and 8 months. What is its present value at

6% compounded semiannually?

Solution:

21

Number of integer period = 7 , and the part of

last period = 2 months

09.999$

)12

206.01(

1000

).1(

tr

SP

04.805$

)2

06.01(

09.999

)1( 7

ni

SP

Example 5 :

On August 5, 1985, Mr. Kane loaded Ms. Hill

$2000 at 12% converted semiannually. Mr . Hill

gave Mr Kan a not promising to repay the loan

with accumulated interest in 6 years. On

February 5, 1989, Mr Kan sold the note to a

buyer, who charge an interest rate of 16%

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converted semiannually for discounting. How

much did Mr. Man get?

Solution :

Step 1:

39.4024$

)2

12.01(2000)1( 12

niPS

Step 2:

23

95.2738$

)2

16.01(

39.4024

)1( 5

ni

SP

Example 6:

A person can buy a piece of property for $4500

cash or $2000 down and $3000 in 3 years. If the

person has money earning 6% converted

semiannually, which is the better purchase plan

and how much now?

Solution:

We get the present value of $3000 due in 3 years

at 6% compounded semiannually.

45.2512$

)2

06.01(

3000

)1( 6

ni

SP

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Adding this amount to the $2000 down payment

makes the present value of this payment plan

$4512.45.

By paying $4500 cash, the buyer saves $12.45

now.

Example 7:

A piece of property can be purchased for $ 2850

cash or for $3000 in 12 months. Which is the

better plan for the buyer if money is worth 7%

compounded quarterly? Find the cash ( present

value) equivalent of the saving made by

adopting the better plan.

Solution :

Put the focal date now:

88.2798$

)4

07.01(

3000

)1( 4

ni

SP

25

It is better to pay later. And the cash equivalent

of the saving is = 2850 – 2798.88 = $ 51.12

H.W :page 138 Exercise 3d, and 3e page 143

(all)

Finding the rate

Example

If $500 amounts to $700 in 5 years with interest

compounded quarterly, what is the rate of interest?

Solution :

26

%79.60679.0

01697.04.

01697.0

01697.14.11

4.1)1(

)1(500700

)1(

20

20

20

imj

i

i

i

i

iPS n

Economic analysis:

Example:

Per capita personal income in the united state

increased from $8421 in 1980 to $13157 in 1987.

what was the annual compounded rate of return?

Solution:

27

%58.60658.0

0658.1562.11

562.1)1(

)1(842113157

)1(

7

7

7

i

i

i

i

iPS n

H.W :page 151 Exercise 3f (all)

Finding the time :

Example 1:

How long will it take $200 to amount to $350 at

7% compounded semiannually?

Solution:

28

years 8.13362

16.26723t

16.26723)035.1log(

)75.1log(

)75.1log()035.1log(

)75.1log()035.01log(

75.1)035.01(

)2

07.01(200350

)1(

n

n

iPS

n

n

n

n

H.W :page 175 Exercise 3g (all)

Equation of value

Example 1:

A person owes $20000 due in 1 year and $20000

due in 2 years. The lender agrees to the

settlement of both obligations with a cash

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payment. assume that the rate used equal 10%

compounded semiannually. determine the size of

the cash payments.

Solution:

Equation of value used as:

66.42821$

07.2468156.18140

)2

10.01(

30000

)2

10.01(

20000

42

x

Example 2:

31

A person owes $50000 due now. The lender

agrees to settle this obligation with 2 equal

payments in 1 and 2 years, respectively, find the

size of the payments if the settlement is based on

9%.

Solution:

44.28423$

59405$09.2

)09.01(50000)9.01( 21

x

x

xx

Example 3:

31

A piece of property is sold for $50000. the buyer

pays $20000 cash, and signs a non-interest-

bearing not for $10000 due in 1 year, and second

a non-interest-bearing not for $10000 due in 2

years. If the seller charges 10% compounded

annually, what a non-interest-bearing not due in

3 years will pay off the debt?

Solution:

Equation of value as follows:

16830$

110001210039930

)10.01(10000)10.01(10000)10.01(30000 123

x

x

x

32

Example 4:

A person owes $20000 due in 3 years with

interest at 10% compounded quarterly, and

$10000 due in 5 years with interest at 8%. If

money is worth 9% what single payment 6 years

hence will be equivalent to the original

obligations?.

Solution:

First we obtain the maturity values of the debts:

33

The amount of %20000 after 3 years =

78.26897$)4

10.01(20000 12

The amount of $10000 after 5 years =

28.14693$)08.01(10000 5

The equation of value as:

09.50849$

68.1601541.34833

)09.01(28.14693)09.01(78.26897 13

x

Example 5:

On June 1, 1991, a person obtained a $5000 loan

for which payment of $1000 on the principal

plus 6% interest on the unpaid balance will be

made every 6 months. The payment schedule for

this loan is given in the following chart:

34

Pay

men

t

nu

mb

er

Date Total

payment

Payment

on

interest

Payment

on

principal

Balance

of loan

June 1, 1991 5000

1 Dec.1,1991 1300 300*

1000 4000

2 June 1,1992 1240 240 1000 3000

3 Dec 1, 1992 1180 180 1000 2000

4 June 1, 1993 1120 120 1000 1000

5 Dec 1,1993 1060 60 1000 0

* interest at Dec. 1,1991 = 5000*0.06*1

On June 1, 1992 the lender sells this contract to

a buyer how wants a yield of 16% converted

semiannually. Find the sale price.

Solution :

35

Put the focal date at June 1, 1992, the equation

value is:

27.2894$

46.84122.96059.1092

)2

16.01(

1060

)2

16.01(

1120

)2

16.01(

1180

321

x

36

EXAMPLE 6.

A head of household stipulates in a will that

$30000 from the bequeathed estate is to be

placed in a fund from which each of the three

children in a family is to receive an equal

amount upon reached age 21. when the head of

the household dies, the children are ages 19, 16,

and 14. if the fund is invested at 85 converted

semiannually, how much does each receive?

Solution :

37

The equation value as:

56.14232$

300005774754.06755642.08548042.0

)2

08.01()

2

08.01()

2

08.01(

30000014104

x

xxx

xxx

H.W :page 168, 177 Exercise 3h, 3i (all)

38

Coefficient.

39

Solution: