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Lecture No.11

Chapter 4

Contemporary Engineering Economics

Copyright 2010

Contemporary Engineering Economics, 5th edition, 2010

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Equivalence Calculations using Effective

Interest RatesStep 1: Identify the payment period (e.g., annual,

quarter, month, week, etc)

Step 2: Identify the interest period (e.g., annually,

quarterly, monthly, etc)

Step 3: Find the effective interest rate that covers thepayment period.

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Case I: When Payment Period is Equal to

Compounding Period

Step 1: Identify the number of compounding periods (M) per

year

Step 2: Compute the effective interest rate per payment

period (i)

Step 3: Determine the total number of payment periods (N)

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Example 4.4:

Calculating Auto

Loan PaymentsGiven:

MSRP = $20,870

Discounts & Rebates =

$2,443

Net sale price = $18,427

Down payment = $3,427

Dealers interest rate =

6.25% APR

Length of financing = 72

months

Find: the monthly payment

(A)

Solution:

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Dollars Down

in the DrainSuppose you drink a cupof coffee ($3.00 a cup) on

the way to work every

morning for 30 years. If you

put the money in the bank

for the same period, howmuch would you have,

assuming your accounts

earns a 5% interest

compounded daily.

NOTE: Assume you drink a

cup of coffee every day

including weekends.

Solution:

Payment period = daily

Compounding period = daily

Contemporary Engineering Economics, 5th edition, 2010

5%0.0137% per day

365

30 365 10,950 days

$3( / ,0.0137%,10950)

$76,246

i

N

F F A

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Case II: When Payment Periods Differ from

Compounding Periods Step 1: Identify the following parameters.

M = No. of compounding periods

K= No. of payment periods per year

C= No. of interest periods per payment period Step 2: Compute the effective interest rate per payment

period.

For discrete compounding

For continuous compounding

Step 3: Find the total no. of payment periods.

N= K(no. of years)

Step 4: Use iand Nin the appropriate equivalence formula.

Contemporary Engineering Economics, 5th edition, 2010

[1 / ] 1Ci r CK

/ 1r Ki e

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Example 4.5 Compounding

Occurs More Frequently than

Payments are Made (Discrete

Case)Given:A = $1,500 per quarter, r= 6% per year, M = 12compounding periods per year,and N= 2 years

Find: F

Step 1:M = 12 compoundingperiods/year K= 4 paymentperiods/year C= 3 interest periodsper quarter

Step 2:

Step 3: N= 4(2) = 8

Solution:

F= $1,500 (F/A, 1.5075%, 8)= $14,216.24

Contemporary Engineering Economics, 5th edition, 2010

30.06

1 112

1.5075% per quarter

i

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Example 4.6

Compounding is Less

Frequent than

Payments

Given:A = $500 per month, r=10% per year, M = 4 quarterlycompounding periods per year, andN= 10 years

Find: F

Step 1:

M = 4 compoundingperiods/year K= 12 paymentperiods/year C= 1/3 interest periodper quarter

Step 2:

Step 3: N= 4(2) = 8

Solution:

F= $500 (F/A, 0.826%, 120)= $101,907.89

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0.101 14

0.826%

i

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A Decision Flow Chart on How to Compute the

Effective Interest Rate per Payment Period

Contemporary Engineering Economics, 5th edition, 2010

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Key Points

Financial institutions often quote interest rate

based on an APR.

In all financial analysis, we need to convert the APR

into an appropriate effective interest rate based ona payment period.

When payment period and interest period differ,

calculate an effective interest rate that covers the

payment period. Then use the appropriate interest

formulas to determine the equivalent values