chapter 3 compound interest - الصفحات...
TRANSCRIPT
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)
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
19
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
$2000 at 12% converted semiannually. Mr . Hill
gave Mr Kan a not promising to repay the loan
with accumulated interest in 6 years. On
February 5, 1989, Mr Kan sold the note to a
buyer, who charge an interest rate of 16%
22
converted semiannually for discounting. How
much did Mr. Man get?
Solution :
Step 1:
39.4024$
)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
24
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 $20000
due in 2 years. The lender agrees to the
settlement of both obligations with a cash
29
payment. assume that the rate used equal 10%
compounded semiannually. determine the size of
the cash payments.
Solution:
Equation of value used as:
66.42821$
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)