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Engineering Economics Contemporary Engineering Economics, 5th edition, © 2010.

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Nominal and Effective Interest Rates Engineering Economics Contemporary Engineering Economics, 5th edition, © 2010
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Page 1: Engineering Economics Contemporary Engineering Economics, 5th edition, © 2010.

Nominal and Effective Interest Rates

Engineering Economics

Contemporary Engineering Economics, 5th edition, © 2010

Page 2: Engineering Economics Contemporary Engineering Economics, 5th edition, © 2010.

Contemporary Engineering Economics, 5th edition, © 2010

Understanding Money and Its Management – Main Focus

1. If payments occur more frequently than annual, how do you calculate economic equivalence?

2. If interest period is other than annual, how do you calculate economic equivalence?

3. How are commercial loans structured?4. How would you manage your debt?

Page 3: Engineering Economics Contemporary Engineering Economics, 5th edition, © 2010.

Nominal Versus Effective Interest Rates

Nominal Interest Rate:

Interest rate quoted based on an annual period

Effective Interest Rate:Actual interest earned or paid in a year or some other time period

Contemporary Engineering Economics, 5th edition, © 2010

Page 4: Engineering Economics Contemporary Engineering Economics, 5th edition, © 2010.

Financial Jargon

Contemporary Engineering Economics, 5th edition, © 2010

Nominal interest rate

Annual percentagerate (APR)

Interest period

18% Compounded Monthly

Page 5: Engineering Economics Contemporary Engineering Economics, 5th edition, © 2010.

18% Compounded MonthlyWhat It Really Means?

Interest rate per month (i) = 18%/12 = 1.5%

Number of interest periods per year (N) = 12

In words,Bank will charge 1.5% interest

each month on your unpaid balance, if you borrowed money.

You will earn 1.5% interest each month on your remaining balance, if you deposited money.

Question: Suppose that you invest $1 for 1 year at 18% compounded monthly. How much interest would you earn?

Contemporary Engineering Economics, 5th edition, © 2010

12 12$1(1 ) $1(1 0.015)$1.1956$1.1956 $1.00 $0.1956

F i

I

Page 6: Engineering Economics Contemporary Engineering Economics, 5th edition, © 2010.

Effective Annual Interest Rate (Yield)Formula:

r = nominal interest rate per yearia = effective annual interest rate

M = number of interest periods per year

Example: 18% compounded

monthly

What It really Means

1.5% per month for 12 months or

19.56% compounded once per year

Contemporary Engineering Economics, 5th edition, © 2010

(1 ) 1Ma

ri

M 120.18

1 1 19.56%12ai

Page 7: Engineering Economics Contemporary Engineering Economics, 5th edition, © 2010.

Practice ProblemSuppose your savings account pays 9% interest compounded quarterly. (a) Interest rate per

quarter(b) Annual effective

interest rate (ia)

(c) If you deposit $10,000 for one year, how much would you have?

Solution:

Contemporary Engineering Economics, 5th edition, © 2010

4

(a) Interest rate per quarter:9%

2.25%4

(b) Annual effective interest rate:

(1 0.0225) 1 9.31%(c) Balance at the end of one year (after 4 quarters) $10,000( / ,2.

a

i

i

F F P

25%,4) $10,000( / ,9.31%,1) $10,931

F P

Page 8: Engineering Economics Contemporary Engineering Economics, 5th edition, © 2010.

Contemporary Engineering Economics, 5th edition, © 2010

Nominal and Effective Interest Rates with Different Compounding Periods

Effective Rates

Nominal Rate

Compounding Annually

Compounding Semi-annually

Compounding Quarterly

Compounding Monthly

Compounding Daily

4% 4.00% 4.04% 4.06% 4.07% 4.08%

5 5.00 5.06 5.09 5.12 5.13

6 6.00 6.09 6.14 6.17 6.18

7 7.00 7.12 7.19 7.23 7.25

8 8.00 8.16 8.24 8.30 8.33

9 9.00 9.20 9.31 9.38 9.42

10 10.00 10.25 10.38 10.47 10.52

11 11.00 11.30 11.46 11.57 11.62

12 12.00 12.36 12.55 12.68 12.74

Page 9: Engineering Economics Contemporary Engineering Economics, 5th edition, © 2010.

Why Do We Need an Effective Interest Rate per Payment Period?

Contemporary Engineering Economics, 5th edition, © 2010

Payment period

Interest period

Payment period

Interest period

Whenever payment and compounding periods differ from each other, one or the other must be transformed so thatboth conform to the same unit of time.

Page 10: Engineering Economics Contemporary Engineering Economics, 5th edition, © 2010.

Effective Interest Rate per Payment Period (i) Formula:

C = number of interest periods per payment periodK = number of payment periods per year CK = total number of interest periods per year, or M r/K = nominal interest rate per payment period

Functional Relationships among r, i, and ia, where interest is calculated based on 9% compounded monthly and payments occur quarterly

Contemporary Engineering Economics, 5th edition, © 2010

1 1Cr

iCK

Page 11: Engineering Economics Contemporary Engineering Economics, 5th edition, © 2010.

Effective Interest Rate per Payment Period with Continuous Compounding

Formula: With continuous compounding

Example: 12% compounded continuously(a) effective interest rate per quarter

(b) effective annual interest rate

Contemporary Engineering Economics, 5th edition, © 2010

C

/

lim 1 1

1r K

C

C

ri

CK

e

0.12/4 13.045% per quarter

i e

0.12/1 112.75% per year

ai e

Page 12: Engineering Economics Contemporary Engineering Economics, 5th edition, © 2010.

Contemporary Engineering Economics, 5th edition, © 2010

Case 0: 8% compounded quarterlyPayment Period = QuarterInterest Period = Quarterly

1 interest period

Given r = 8%,K = 4 payments per yearC = 1 interest period per quarterM = 4 interest periods per year

2nd Q 3rd Q 5th Q1st Q

1

[1 / ] 1

[1 0.08 / (1)(4)] 12.000% per quarter

Ci r CK

Page 13: Engineering Economics Contemporary Engineering Economics, 5th edition, © 2010.

Contemporary Engineering Economics, 5th edition, © 2010

Case 1: 8% compounded monthlyPayment Period = QuarterInterest Period = Monthly

3 interest periods Given r = 8%,

K = 4 payments per yearC = 3 interest periods per quarterM = 12 interest periods per year

2nd Q 3rd Q 5th Q1st Q

3

[1 / ] 1

[1 0.08 / (3)(4)] 12.013% per quarter

Ci r CK

Page 14: Engineering Economics Contemporary Engineering Economics, 5th edition, © 2010.

Contemporary Engineering Economics, 5th edition, © 2010

Case 2: 8% compounded weeklyPayment Period = QuarterInterest Period = Weekly

13 interest periods Given r = 8%,

K = 4 payments per yearC = 13 interest periods per quarterM = 52 interest periods per year

2nd Q 3rd Q 5th Q1st Q

i r CK C

[ / ]

[ . / ( )( )]

.

1 1

1 0 08 13 4 1

2 0186%

13

per quarter

Page 15: Engineering Economics Contemporary Engineering Economics, 5th edition, © 2010.

Contemporary Engineering Economics, 5th edition, © 2010

Case 3: 8% compounded continuouslyPayment Period = QuarterInterest Period = Continuously

interest periods Given r = 8%,

K = 4 payments per year

2nd Q 3rd Q 5th Q1st Q

/

0.02

1

12.0201% per quarter

r Ki e

e

Page 16: Engineering Economics Contemporary Engineering Economics, 5th edition, © 2010.

Contemporary Engineering Economics, 5th edition, © 2010

Summary: Effective Interest Rates per Quarter at Varying Compounding Frequencies

Case 0 Case 1 Case 2 Case 3

8% compounded quarterly

8% compounded monthly

8% compounded weekly

8% compounded continuously

Payments occur quarterly

Payments occur quarterly

Payments occur quarterly

Payments occur quarterly

2.000% per quarter

2.013% per quarter

2.0186% per quarter

2.0201% per quarter

Page 17: Engineering Economics Contemporary Engineering Economics, 5th edition, © 2010.

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 the payment period.

Contemporary Engineering Economics, 5th edition, © 2010

Page 18: Engineering Economics 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

Page 19: Engineering Economics Contemporary Engineering Economics, 5th edition, © 2010.

Example 4.4: Calculating Auto Loan PaymentsGiven:

MSRP = $20,870Discounts & Rebates = $2,443Net sale price = $18,427Down payment = $3,427Dealer’s interest rate = 6.25% APRLength of financing = 72 months

Find: the monthly payment (A)

Solution:

Contemporary Engineering Economics, 5th edition, © 2010

Page 20: Engineering Economics Contemporary Engineering Economics, 5th edition, © 2010.

Dollars Down in the DrainSuppose 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.

NOTE: Assume you drink a cup of coffee every day including weekends.

Solution:Payment period = dailyCompounding period = daily

Contemporary Engineering Economics, 5th edition, © 2010

5%0.0137% per day

36530 365 10,950 days$3( / ,0.0137%,10950)$76,246

i

NF F A

Page 21: Engineering Economics 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

For continuous compounding

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

[1 / ] 1Ci r CK

/ 1r Ki e

Page 22: Engineering Economics 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

Find: F

Step 1: M = 12 compounding periods/year K = 4 payment periods/year C = 3 interest periods per 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

30.061 1

121.5075% per quarter

i

Page 23: Engineering Economics Contemporary Engineering Economics, 5th edition, © 2010.

Example 4.6 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:

Step 3: N = 4(2) = 8

Solution:

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

Contemporary Engineering Economics, 5th edition, © 2010

1/30.101 1

40.826%

i

Page 24: Engineering Economics 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

Page 25: Engineering Economics Contemporary Engineering Economics, 5th edition, © 2010.

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


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