5. more interest formula (part ii)

5: More Interest Formulas

(continued…)

Dr. Mohsin Siddique

Assistant Professor

[email protected]

Ext: 29431

Date: 28/10/2014

Engineering Economics

University of SharjahDept. of Civil and Env. Engg.

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Part I

Outcome of Today’s Lecture

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� After completing this lecture…

� The students should be able to:

� Understand geometric series compound interest formulas

More interest Formulas

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� Uniform Series

� Arithmetic Gradient

� Geometric Gradient

� Nominal and Effective Interest

� Continuous Compounding

Geometric Gradient Series

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� Instead of constant amount of increase, sometimes cash flows increase by a uniform rate of increase g (constant percentage amount) every subsequent period.

� For example: If the maintenance costs of car are $100 for the first year and they are increasing at a uniform rate, g, of 10% per year

100

110

121

133.1

146.41

1A

( )gAA += 112

( ) ( )2123 11 gAgAA +=+=

10 2 3 4 5

( ) ( )4145 11 gAgAA +=+=

( ) 11 1 −+= n

n gAA

...


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� Let’s write a present worth value for each period individually, and add them up

� Recall:

( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) nnnn igAigA

igAigAiAP−−+−−

−−−

++++++

++++++++=

1111

...111111

112

1

321

211

11

Eq. (1)

( )( ) n

n

iF

iP−+=

+=

1P

1F

Multiply Eq. (1) by (1+g)/(1+i) to obtain

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) 1

11

1

431

321

211

11

1111

...11111111−−−−

−−−−

++++++

+++++++++=++nnnn igAigA

igAigAigAigP

Eq. (2)


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� Subtract Equation (2) from Equation (1) to yield

( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( )( ) ( ) ( )[ ]

( ) ( )( )

−−+++−=

++−=−−+

++−=+−+

++−+=++−

−

−

−

−−−−

gi

igAP

igAgiP

igAAgPiP

igAiAigPP

nn

nn

nn

nn

11

111

11111

1111

11111

1

1

111

11

11

11

-

Eq. (3)

Eq. (4)Where gi ≠

Where is called geometric series present worth factor and has notation

( ) ( )( )

−−+++− −

gi

ig nn

11

111

( )nigAP ,,,/

( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) nnnn igAigA

igAigAiAP−−+−−

−−−

++++++

++++++++=

1111

...111111

112

1

321

211

11

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) 1

11

1

431

321

211

11

1111

...11111111−−−−

−−−−

++++++

+++++++++=++nnnn igAigA

igAigAigAigP


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� Example : Suppose you have a vehicle. The first year maintenance cost is estimated to be $100. The rate of increase in each subsequent year is 10%. You want to know the present worth of the cost of the first five years of maintenance, given i = 8%.

� Solution:

� 1. Repeated Present-Worth (Step-by-Step) Approach:


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� 2. Using geometric series present worth factor formula

Multiple Compounding

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� The time standard for interest computations is One Year.

� Many banks compound interest multiple times during the year.

� e.g.: 12% per year, compounded monthly (1% interest is paid monthly)

� e.g.: 8% per year, compounded semi-annually (4% interest is paid twice a year or once every 6 months)

� Compounding is not less important than interest. You have to know all the info to make a good decision.

Multiple Compounding

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� You need to pay attention to the following terms:

� Time Period –The period over which the interest is expressed (always stated).

� e.g.: “6% per year”

� Compounding Period (sub-period) – The shortest time unit over which interest is charged or earned.

� e.g.: If interest is “6% per year compounded monthly”, compounding period is one month

� Compounding Frequency –The number of times (m) that compounding occurs within time period.

� Compounding semi-annually: m = 2; Compounding quarterly: m = 4

� Compounding monthly: m = 12; Compounding weekly: m = 52; compounding daily: m = 365

Nominal and Effective Interest Rate

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� Two types of interest are typically quoted:

� 1. Nominal interest rate, r, is an annual interest rate without considering the effect of (sub-period) compounding.

� 2. Effective interest rate, i or ia , is the actual rate that applies for a stated period of time which takes into account the effect of (sub-period) compounding.

� Sometimes one interest rate is quoted, sometimes another is quoted. If you confuse the two you can make a bad decision.

� Effective interest is the “real” interest rate over a period of time; Nominal rate is just given for simplicity (per year)

� All interest formulas use the effective interest rate


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� Let r=nominal interest rate per interest period (usually one year)

� i=effective interest rate per interest period (sub-period compounding)

� m=Compounding frequency (No. of compounding sub period per time period)

� r/m=interest rate per compounding sub-period

1=P

( )11 /11 mrF +=

( ) ( )2112 /11/1 mrmrFF +=+=

1

0

2 3 4

( ) ( )4134 /11/1 mrmrFF +=+=

...

interest period (one year) sub period (quarter)

( )mmrF /11 +=

( ) ( )11 1111 aiiF +=+=

( ) 1/1 −+= ma mri

( ) ( )ma mri /11 +=+

Thus in general form we can write

Moreover, we also know that

Thus


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� Example:

� Given an interest rate of 12% per year, compounded quarterly:

� Nominal rate=r = 12%

� Compounding frequency=m=4

� Effective (Actual) rate =r/m= 12%/4 = 3% per quarter

� Effective rate per year = [1+(0.12/4)]4-1= 0.1255=12.55%

� Investing $1 at 3% per quarter is equivalent to investing $1 at 12.55% annually


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� Example: A bank pays 1.5% interest every three months. What are the nominal and effective interest rates per year?

� Solution:

Effective interest rate per three months=1.5%

Nominal interest rate per year = r = 4 x 1.5% = 6% a year

Effective interest rate per year= ia= (1 + r/m)m–1 = (1.015)4–1 =

=0.06136

=6.14% a year


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� Example: $10K is borrowed for 2 years at an interest rate of 24% per year compounded quarterly. If the same sum of money could be borrowed for the same period at the same interest rate of 24% per year compounded annually, how much could be saved in interest charges?

� Solution

Interest=F-P

� Interest charges for quarterly compounding:

� 10,000(1+24%/4)2x4-10,000 = $5938.48

� Interest charges for annually compounding:

� 10,000(1+24%)2-10,000 = $5376.00

� Savings: $5938.48 -$5376 = $562.48


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� Example 4-15: A loan shark lends money on the following terms. “If I give you $50 on Monday, then you give back $60 the following Monday.”

� Solution

� 1.What is the nominal rate, r ?

� The loan shark charges i= 20% a week:

� 60 = 50 (1+i) [Note we have solved 60 = 50(F/P,i,1) for i]

� i= 0.2

� We know m = 52, so r = 52 x i= 10.4, or 1,040% a year

� 2. What is the effective rate, ia?

� ia= (1 + r/m)m–1 = (1+10.4/52)52–1 =13,104

� This means about 1,310,400 % a year !!!!


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� Example 4-16: You deposit $5,000 in a bank paying 8% nominal interest, compounded quarterly. You want to withdraw the money in five equal yearly sums, beginning Dec. 31 of the first year. How much should you withdraw each year ?

08.0quarterly compounded %8 == yearlyr

4

0

8 12 16

(year)

w w w w w

20 20 months

1 2 3 4 5 5 years

( ) ( ) yearlymri ma %24.814/08.011/1 4 =−+=−+=

$5,000

Effective annual interest:


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� This diagram may be solved directly to determine the annual withdrawal W with the capital recovery factor

� The depositor should withdraw $1260 per year

4

0

8 12 16

(year)

w w w w w

20 20 months

1 2 3 4 5 5 years$5,000

( ) ( )( ) 1260$

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1%,,/ =

−++==

n

n

i

iiPniPAPW

Continuous Compounding

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� Continuous compounding is when m is infinite

� i.e. m�∞: r/m�0:

( )( )mn

ma

mrPF

mri

/1

1/1

+=

−+=

( )( ) rnmn

rma

PemrPF

emri

=+=

−=−+=

/1

11/1


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� Continuous compounding can sometimes be used to simplify computations, and for theoretical purposes.

� The previous equation illustrates that (er – 1) is a good approximation of (1 + r/m)m -1 for large m (i.e., ∞).

� This means there are continuous compounding versions of the formulas we have seen earlier.

� F = P ern is analogous to F = P (F/P,r,n): (F/P,r,n)∞= ern

� P = F e-rn is analogous to P = F (P/F,r,n): (P/F,r,n)∞= e-rn


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Thank You

Feel Free to Contact

[email protected]

Tel. +971 6 5050943 (Ext. 2943)