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Chapter one

means that interest calculated not only on the initial investment but Compounding

also on the interest of previous period.

In the compound interest the money growth faster, then investment in simple

interest.

The compound interest may be compounded either:

1.Annually(One a year)

2.Half yearly(semiannually)(Twice a year)

3.Quartenly (4 times a year)

4.Monthly (12 times a year)

5.Weekly (52 times a year)

6. Daily (365 times a year)

When computing compound interest , the first thing to do is find what is called the

period interest rate(PIR),

𝑃𝐼𝑅 = 𝒀𝒆𝒂𝒓𝒍𝒚 𝒊𝒏𝒕𝒆𝒓𝒔𝒕 𝒓𝒂𝒕𝒆

𝑵𝒖𝒎𝒃𝒆𝒓 𝒐𝒇 𝒊𝒏𝒕𝒆𝒓𝒔𝒕 𝒑𝒆𝒓 𝒚𝒆𝒂𝒓

or r=R/n

To find the PIR when interest is compounded

Yearly : Divide the yearly interest rate by 1

Half yearly: Divide the yearly interest rate by 2

Quarterly: Divide the yearly interest rate by 4

Monthly: Divide the yearly interest rate by 12

Weekly: Divide the yearly interest rate by 52

Daily: Divide the yearly interest rate by 365

Example 1

Find the Period Interest Rate(PIR) , If the annual interest rate is 8% and the interst is

compounded Quarterly.

Solution:

If the interest is compounded Quarterly, then the number of interest payment per

year is 4, hence

PIR=𝟖%

𝟒 = 𝟎.𝟎𝟖

𝟒

0.02=2% =

The next example shows the difference between simple and compound interest.

Example 2

Find the amount of the

a.Simple interest

b.The Compound interest on a 2000$ investment on amount rate of 4% for 3 years.

Solution:

a.For simple interest, use the formula

I=PRT when P=2000, R=4% & T=3

I=PRT=2000$x0.04x3=240$.

b.For compound interest, the calculations need to be done three times, one for each

year:

For year 1:

I=PRT=2000x0.04x1=80$

During the second year, the 80$ interest also generates interest, and so the principle

is

2000+80=2080$

For year 2:

I=PRT=2080*0.04x1=83.20$

For year 3:

The principle is

2080+83.20=2163.20 $ , Hence:

I=PRT=2163.20x0.04x1=86.53$

Hence the total compound interest is

80+83.20+86.53=249.73$

The total a mount of the investment is

2000+249.73=2249.73$

If interest is compound half yearly, the same problem would require six calculations.

If interest is compounded quarterly , the problem would require 12 calculations and

so on.

The following formula can be used:

FV=S=P(1+r)N

Where

FV(or S)= Future value( a mount)

P=Principle

r=Period interest rate r=PIR

N=number of periods per year times( The number of years)

Example 3

Find the future value and compound interest on 2000$ investment at a rate 4%

compounded yearly for 3 years.

Solution

In this case, the period interest rate is 4% since the interest is compound yearly and

N=1x3=3

FV=(1+r)N=2000(1+(0.04/1))3x1 =2249.73$

The amount of the compound interest is:

I=FV-P= 2249.73-2000=249.73$

4Example

Find the future value and compound interest on a 6000$ on a investments at a rate

10% compounded half yearly for 6 years.

Solution

In this case, the period interest rate is (10%)/2=0.05 or 5%

The number of period is 2x6=12, then

FV=(1+r)N=6000(1+0.05)12 =10,775.14$

The amount of the compound interest is:

10775.14-6000=4775.14

Example 5

Find the future value and compound interest on a 500$ on a investments at a rate 6%

compounded daily for 5 years.

:Solution

In this case, the period interest rate(PTR)=(6%)/365=1.64x10-4

The number of period is 365x5=1825

FV=P(1+r)N=500(1+1.64x10-4 ) 1825 =500(1.000164) 1825 =500X1.3498=674.9127

The compound interest is 674.9127-500=174. 9127$.

Effective Rate of Interst

When the interest compound more than once a year it is sometimes necessary to

determine on equivalent simple interest rate.

For example, suppose a person invested 800$ I a saving account for 1 year at a rate of

8% compounded quarterly.

The future value would be

FV=P(1+r)N=800(1+0.08/4)1x4 =865.95$

The total interest is (865.95-800=65.95$).

is rate interestsimple able ar compthe Now,

)=(65.95/(800x1))=0.0824 or 8.24%*T)p(R=(I/

In other words, if a person invested 800$ in saving account paying 8.24% simple

interest for 1 year, he/she would receive 65.95$ in interest, which is the same amount

as the person investing 800$ for 1 year at 8% compounded quarterly .

The actual interest or simple interest rate that is equivalent to compound interest is

. Efficient Ratecalled the

this rate is called the annual percentage rate APR. an lo For the

The following formula can be used for calculating the effective rate.

, where 1-

N

E=(1+r)

E=effective rate,

r=Period interest rate (R/n) and n= the number of period per year.

Example 6

Find the effective rate of interest is equivalent to a 5% rate compounded

semiannually (half yearly).

Solution:

Since the interest is compounded half yearly

r=(R/n)=(0.05/2)=0.025$

E=(1+r)n-1=(1+0.025)2 -1=0.0506 or 5.06%.

eciationrDep

Business most often need to pu rchase equipment , furniture buildings and often

.assetsnecessary items in order to operate. These items are called

The worth of an item most often decrease over the year. This decrease in worth is

Depreciation due to what is called

.

The straight Line Method

Depreciation D= 𝐎𝐫𝐢𝐠𝐢𝐧𝐚𝐥 𝐜𝐨𝐬𝐭 𝐨𝐫 𝐑𝐞𝐩𝐥𝐚𝐜𝐞𝐦𝐞𝐧𝐭 𝐕𝐚𝐥𝐮𝐞−𝐒𝐜𝐫𝐚𝐩 𝐯𝐚𝐥𝐮𝐞 𝐨𝐫 𝐒𝐚𝐥𝐯𝐚𝐠𝐞 𝐯𝐚𝐥𝐮𝐞

𝐥𝐢𝐟𝐞 𝐨𝐟 𝐭𝐡𝐞 𝐩𝐫𝐨𝐩𝐞𝐫𝐭𝐲 𝐢𝐧 𝐲𝐞𝐚𝐫𝐬 = 𝑪−𝑺

𝒏

Where,

C – Original cost or Replacement Value

S – Scrap value or Salvage value

n - life of the property in years

Example 7

A factory machine was purchased for 15000$ and has an estimated lifetime of (5)

year. If the scrap value at the end of( 5) years is estimated at 3000$ , find the

amount of the depreciation for each year.

Solution:

Find the depreciation amount (DA),

DA=Original price-scrap value=15000-3000=12000$.

Divide the depreciation amount by the estimated lifetime

12000/5=2400$ per year.

In other words, the value of the machine depr eciates 2400$ per year.

Note: The book value(the value of an item after depreciation is calculated)

t

LS-P

t-PlineStraight for Value Book

8Example

Find the book value(the value of an item after depreciation is calculated) at the end

of 4 years of a digital video camera costing 24000$ with an expected life time of 6

years. The scrap value at the end of lifetime is 3ooo$.

Solution:

e Constant Percentage Method Th

Note: The book value for the Constant Percentage Method

BV=p(1-r) n

Example 9

A car costing 2400$ depreciates 25% of its value each year. Find the book value at

the end of 5 years.

Solution:

BV=p(1-r) n

B=24000(1-0.25)5=24000(0.75)5 =5698.31 $

t

LS-P

t-PlineStraight for Value Book

$ 10000

63000 - 24000

4 24000 (BV) Value

Book

Example 10

Determine the rate of depreciation for the machine costing 40000$ which is

estimated to have a useful lifetime of 5 years and scrap value of 5000$.

Solution:

BV=p(1-r) n , 5000=40000(1-r) 5

(1-r)5 =5000/40000=0.125

1-r=(0.125)1/5=(0.125)0.2 =0.65975

-r=0.65975-1=-0.340246 r=0.340246=0.34 r=34%

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