Lecture No.11
Chapter 4
Contemporary Engineering Economics
Copyright © 2010
Contemporary Engineering Economics, 5th edition, © 2010
Equivalence Calculations using Effective
Interest Rates
Step 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 the
payment period.
Contemporary Engineering Economics, 5th edition, © 2010
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)
Contemporary Engineering Economics, 5th edition, © 2010
Example 4.4: Calculating Auto Loan Payments
 Solution:
Given:
MSRP = $20,870
Discounts & Rebates = $2,443
Net sale price = $18,427
Down payment = $3,427
Dealer’s interest rate = 6.25%
APR
Length of financing = 72
months
Find: the monthly payment (A)
Contemporary Engineering Economics, 5th edition, © 2010
Dollars Down in the Drain
Suppose you drink a cup of
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, how much
would you have, assuming
your accounts earns a 5%
interest compounded daily.
 Solution:
 Payment period = daily
 Compounding period = daily
NOTE: Assume you drink a
5%
i
 0.0137% per day
365
N  30  365  10,950 days
cup of coffee every day
including weekends.
F  $3(F / A,0.0137%,10950)
 $76,246
Contemporary Engineering Economics, 5th edition, © 2010
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
i  [1  r / CK ]C  1
 For continuous compounding
i  er / K  1
 Step 3: Find the total no. of payment periods.
 N = K (no. of years)
 Step 4: Use i and N in the appropriate equivalence formula.

Contemporary Engineering Economics, 5th edition, © 2010
Example 4.5 Compounding Occurs More Frequently than
Payments are Made (Discrete Case)
 Given: A = $1,500 per quarter, r =
6% per year, M = 12 compounding
periods per year, and N = 2 years
 Solution:
 Find: F
Step 1:
M = 12 compounding
periods/year
 K = 4 payment
periods/year
 C = 3 interest periods
per quarter
Step 2:

3
 0.06 
i  1 
 1
12 

 1.5075% per quarter
Step 3:
N = 4(2) = 8
 F = $1,500 (F/A, 1.5075%, 8)
= $12,652.61
Contemporary Engineering Economics, 5th edition, © 2010
Example 4.7 Compounding is Less Frequent than Payments
 Given: A = $500 per month, r = 10%
per year, M = 4 quarterly compounding
periods per year, and N = 10 years
 Find: F
Step 1:
 M = 4 compounding
periods/year
 K = 12 payment
periods/year
 C = 1/3 interest period per
quarter
Step 2:
 Solution:
1/3
0.10 

i  1 

4 

 0.826%
Step 3:
1
N = 4(2) = 8
F = $500 (F/A, 0.826%, 120)
= $101,907.89
Contemporary Engineering Economics, 5th edition, © 2010
A Decision Flow Chart on How to Compute the
Effective Interest Rate per Payment Period
Contemporary Engineering Economics, 5th edition, © 2010
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 on
a 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
Contemporary Engineering Economics, 5th edition, © 2010