Nominal and Effective Interest rates

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Engineering Economics

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 ?

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

Financial Jargon

18% Compounded Monthly

Nominal interest rate

Annual percentage rate (APR)

Interest period

Contemporary Engineering Economics, 5th edition, © 2010

18%

Compounded Monthly

What 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?

F

I

$1(1

 i ) 12   12

$1.1956

$1.1956 $1.00

$0.1956

Contemporary Engineering Economics, 5th edition, © 2010

Effective Annual Interest Rate (Yield)

 Formula: i a

  r

M

) M 

1

r = nominal interest rate per year i a

= effective annual interest rate

M = number of interest periods per year

 Example :

 18% compounded monthly i a

1

0.18

12

12

 What It really Means

 1.5% per month for 12 months or

 19.56% compounded once per year

Contemporary Engineering Economics, 5th edition, © 2010

Practice

Problem

Suppose your savings account pays 9% interest compounded quarterly .

(a)

(b)

(c)

Interest rate per quarter

Annual effective interest rate (i a

)

If you deposit

$10,000 for one year, how much would you have?

 Solution:

(a) Interest rate per quarter: i

9%

4

2.25%

(b) Annual effective interest rate: i a

  4  

(c) Balance at the end of one year (after 4 quarters)

F

F P 25%,4)

Contemporary Engineering Economics, 5th edition, © 2010

Nominal and Effective Interest Rates with

Different Compounding Periods

Nominal

Rate

4%

5

6

7

8

9

10

11

12

Compounding

Annually

4.00%

5.00

6.00

7.00

8.00

9.00

10.00

11.00

12.00

Effective Rates

Compounding

Semi-annually

Compounding

Quarterly

4.04% 4.06%

5.06

6.09

7.12

8.16

9.20

5.09

6.14

7.19

8.24

9.31

10.25

11.30

12.36

10.38

11.46

12.55

Compounding

Monthly

4.07%

5.12

6.17

7.23

8.30

9.38

10.47

11.57

12.68

Compounding

Daily

4.08%

5.13

6.18

7.25

8.33

9.42

10.52

11.62

12.74

Contemporary Engineering Economics, 5th edition, © 2010

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

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

Payment period

Interest period

Payment period

Interest period

Contemporary Engineering Economics, 5th edition, © 2010

Effective Interest

Rate per Payment

Period (i)

 Formula: i

1

 r

CK

C

1

C = number of interest periods per payment period

K = 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 i a

, where interest is calculated based on 9% compounded monthly and payments occur quarterly

Contemporary Engineering Economics, 5th edition, © 2010

Effective Interest Rate per

Payment Period with

Continuous Compounding

 Formula: With continuous compounding C

  i

C lim 1



 e

1 r

CK

C

1

 Example : 12% compounded continuously

 (a) effective interest rate per quarter

0.12/4 

1

3.045% per quarter

 (b) effective annual interest rate i a

 e 0.12/1

1

12.75% per year

Contemporary Engineering Economics, 5th edition, © 2010

Case 0

:

8% compounded quarterly

Payment Period = Quarter

Interest Period = Quarterly

1 st Q

2 nd Q 3 rd Q 5th Q

1 interest period

Given r = 8%,

K = 4 payments per year

C = 1 interest period per quarter

M = 4 interest periods per year i

  r CK ] C 

1

  1 

1

2.000% per quarter

Contemporary Engineering Economics, 5th edition, © 2010

Case 1

:

8% compounded monthly

Payment Period = Quarter

Interest Period = Monthly

1 st Q

2 nd Q 3 rd Q 5th Q

3 interest periods Given r = 8%,

K = 4 payments per year

C = 3 interest periods per quarter

M = 12 interest periods per year i

  r CK ] C 

1

  3 

1

2.013% per quarter

Contemporary Engineering Economics, 5th edition, © 2010

Case 2

:

8% compounded weekly

Payment Period = Quarter

Interest Period = Weekly

1 st Q

2 nd Q 3 rd Q 5th Q

13 interest periods Given r = 8%,

K = 4 payments per year

C = 13 interest periods per quarter

M = 52 interest periods per year i

[ 1

 r CK ]

C 

1

[ 1

.

/ ( )( )]

13 

1

2 0186% per quarter

Contemporary Engineering Economics, 5th edition, © 2010

Case 3

:

8% compounded continuously

Payment Period = Quarter

Interest Period = Continuously

1 st Q

2 nd Q 3 rd Q 5th Q

 interest periods Given r = 8%,

K = 4 payments per year i

 e

1

 e 0.02

1

2.0201% per quarter

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

Payments occur quarterly

8% compounded monthly

Payments occur quarterly

8% compounded weekly

Payments occur quarterly

8% compounded continuously

Payments occur quarterly

2.000% per quarter

2.013% per quarter

2.0186% per quarter

2.0201% per quarter

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

 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)

 Solution :

Cont emporary 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.

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

 Solution:

 Payment period = daily

 Compounding period = daily

N i

5%

0.0137% per day

365

30 365 10,950 days

F

$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

  r CK ] C 

1

 For continuous compounding i e

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: i

1

0.06

12

3

1

 1.5075% per quarter

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

= $14,216.24

 Step 3: N = 4(2) = 8

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: i

1

0.10

4

1/3

1

0.826%

 Step 3: N = 4(2) = 8

 Solution:

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

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