equivalence and compound interest anastasia lidya m

Equivalence and Compound interest

Anastasia Lidya M.

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Economic Equivalence

• What do we mean by “economic equivalence?”

• Why do we need to establish an economic equivalence?

• How do we establish an economic equivalence?

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Economic Equivalence

• Economic equivalence exists between cash flows that have the same economic effect and could therefore be traded for one another.

• Even though the amounts and timing of the cash flows may differ, the appropriate interest rate makes them equal.

Single-Payment Factors(F/P and P/F)

• Objective:– Derive factors to determine the present or future

worth of a cash flow• Cash Flow Diagram – basic format

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0 1 2 3 n-1 n

P0

Fn

i% / period

P0 = Fn1/(1+i)n →(P/F,i%,n) factor: Excel: =PV(i%,n,,F)

Fn = P0(1+i)n →(F/P,i%,n) factor: Excel: =FV(i%,n,,P)

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Single-Payment Factors(F/P and P/F)

• If you deposit P dollars today for N periods at i, you will have F dollars at the end of period N.

• F dollars at the end of period N is equal to a single sum P dollars now, if your earning power is measured in terms of interest rate i.

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Equivalence Between Two Cash Flows

• Step 1: Determine the base period, say, year 5.

• Step 2: Identify the interest rate to use.

• Step 3: Calculate equivalence value.

$3,000$2,042

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i F

i F

i F

6%, 042 1 0 06 733

8%, 042 1 0 08 000

10%, 042 1 0 10

5

5

5

$2, ( . ) $2,

$2, ( . ) $3,

$2, ( . ) $3,289

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Example 2.1: Equivalence

Various dollar amounts that will be economically equivalent to $3,000 in 5 years, given an interest rate of 8%

Uniform-Series: Present Worth Factor (P/A) and

Capital Recovery Factor(A/P)

• Cash flow profile for P/A factor

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. . . . 0 1 2 3 n-2 n-1 n$A per interest period

i% per interest period

Required: To find P given A

Cash flows are equal, uninterrupted and flow at the end of each interest period

Find P

(P/A) Factor Derivation• Setup the following:

• Multiply by to obtain a second equation…

• Subtract (1) from (2) to yield…

1 2 1

1 1 1 1..

(1 ) (1 ) (1 ) (1 )n nP A

i i i i

2 3 1

1 1 1 1..

1 (1 ) (1 ) (1 ) (1 )n n

PA

i i i i i

9

1

(1+i)

(1)

(2)

1

1 1

1 (1 ) (1 )n

iP Ai i i

(3)

Example 2.2. Uniform Series: Find A, Given P, i, and N

Given: P = $250,000, N = 6 years, and i = 8% per year

Find: A Formula to use:

Capital Recovery Factor

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Example 2.3. – Deferred Loan Repayment

Given: P = $250,000, N = 6 years, and i = 8% per year, but the first payment occurs at the end of year 2 Find: A

Step 1: Find the equivalent amount of borrowing at the end of year 1:

Step 2: Use the capital recovery factor to find the size of annual installment:

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Example 2.4. Uniform Series: Find P, Given A, i, and N

Given: A = $10,576,923, N = 26 years, and i = 5% per year Find: P Formula to use:

Present Worth Factor

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Equal Payment Series

0 1 2 N

0 1 2 NA A A

F

P

0 N

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Equal-Payment Series Compound Amount Factor

Formula

0 1 2 N0 1 2 N

A A A

F

0 1 2N

A A A

F

=

0 1 2 N0 1 2 N

A A A

F

0 1 2N

A A A

F

=

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An Alternate Way of Calculating the Equivalent Future Worth, F

0 1 2 N 0 1 2 N

A A A

F

A(1+i)N-1

A(1+i)N-2

A

1 2 (1 ) 1(1 ) (1 )

NN N i

F A i A i A Ai

1 2 (1 ) 1

(1 ) (1 )N

N N iF A i A i A A

i

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Example 2.5. Uniform Series: Find F, Given i, A, and N

Given: A = $3,000, N = 10 years, and i = 7% per year

Find: F

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Example 2.6. Handling Time Shifts: Find F, Given i, A, and N

Given: A = $3,000, N = 10 years, and i = 7% per year

Find: F

Each payment has been shifted to one year earlier, thus each payment would be compounded for one extra year

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ANSI Standard Notation for Interest Factors

• Standard notation has been adopted to represent the various interest factors

• Consists of two cash flow symbols, the interest rate, and the number of time periods

• General form: (X/Y,i%,n)• X represents what is unknown• Y represents what is known• i and n represent input parameters; can be known or

unknown depending upon the problem

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Notation - continued

• Example: (F/P,6%,20) is read as:– To find F, given P when the interest rate is 6%

and the number of time periods equals 20.

• In problem formulation, the standard notation is often used in place of the closed-form equivalent relations (factor)

• Tables at the back of the text provide tabulations of common values for i% and n

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Interpolation in Interest Tables

• When using tabulated interest tables one might be forced to approximate a factor that is not tabulated

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Arithmetic Gradient Factors(P/G) and (A/G)

• Cash flow profile

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0 1 2 3 n-1 n

A1+G

A1+2G

A1+(n-2)G

A1+(n-1)G

Find P, given gradient cash flow G

CFn = A1 ± (n-1)G

Base amount = A1

Gradient Example

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0 1 2 3 4 5 6 7

$100

$200

$300

$400

$500

$600

$700

Gradients have two components:

1. The base amount and the gradient

2. The base amount (above) = $100/time period

Gradient Components

Slide Sets to accompany Blank & Tarquin, Engineering Economy, 6th Edition, 2005

© 2005 by McGraw-Hill, New York, N.Y All Rights Reserved 2-23

…….. 0 1 2 3 n-2 n-1 n

Base amount = A / period

0G

1G 2G

(n-3)G (n-2)G (n-1)G

Present worth point is 1 period to the left of the 0G cash flow

For present worth of the base amount, use the P/A factor (already known)

For present worth of the gradient series, use the P/G factor (to be derived)

Find P of gradient series

Example 2.7. – Linear Gradient: Find A, Given A1, G, i, and N

Given: A1 = $1,000, G = $300, N = 6 years, and i = 10% per year Find: A

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Example 2.8.Declining Linear Gradient Series Given: A1 = $1,200, G = -$200, N = 5 years, and i = 10% per year Find: F

Strategy: Since we have no interest formula to compute the future worth of a linear gradient series directly, we first find the equivalent present worth of the gradient series and then convert this P to its equivalent F.

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Gradient Decomposition

• As we know, arithmetic gradients are comprised of two components1. Gradient component2. Base amount

• When working with a cash flow containing a gradient, the (P/G) factor is only for the gradient component

• Apply the (P/A) factor to work on the base amount component

• P = PW(gradient) + PW(base amount)

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Use of the (A/G) Factor

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0 1 2 3 n-1 n

G

2G

(n-2)G

(n-1)G

Find A, given gradient cash flow G

CFn = (n-1)G

Equivalent A of gradient series

A A A . . . A A

A = G(A/G,i,n)

Geometric Gradient Series Factor

• Geometric Gradient– Cash flow series that starts with a base amount A1

– Increases or decreases from period to period by a constant percentage amount

– This uniform rate of change defines…• A GEOMETRIC GRADIENT• Notation:

g = the constant rate of change, in decimal form, by which future amounts increase or decrease from one time period to the next

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Typical Geometric Gradient

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Required: Find a factor (P/A,g%,i%,n) that will convert future cash flows to a single present worth value at time t = 0

A1

A1(1+g)A1(1+g)2

. . . .0 1 2 3 n-2 n-1 n

A1(1+g)n-1

Given A1, i%, and g%

Two Forms to Consider…

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1

(1 )g

nAP

i

1

11

1 g i

n

g

g

iP A

i g

Case: g = i Case: g = i

A1 is the starting cash flow

There is NO base amount associated with a geometric gradient

The remaining cash flows are generated from the A1 starting value

No tables available to tabulate this factor…too many combinations of i% and g% to support tables

To use the (P/A,g%,i%,n) factor

Example 2.9. Retirement Plan – Saving $1 Million

Given: F = $1,000,000, g = 6%, i = 8%, and N = 20

Find: A1

Solution:

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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

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• Effective interest rate per year, ia, is the annual interest rate taking into account the effect of any compounding during the year.

• In Example we saw that $100 left in the savings account for one year increased to $105.06, so the interest paid was $5.06. The effective interest rate per year, ia, is $5.06/$100.00 = 0.0506 = 5.06%.

Nominal Versus Effective Interest Rates

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Financial Jargon

Nominal interest rate

Annual percentagerate (APR)

Interest period

18% Compounded Monthly

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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?

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

F i

I

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

F i

I

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Effective Annual Interest Rate (Yield)

• Formula:

r = nominal interest rate per year

ia = 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

(1 ) 1Ma

ri

M 120.18

1 1 19.56%12ai

120.181 1 19.56%

12ai

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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:

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(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

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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

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

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.

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Effective Interest Rate per Payment Period (i)

Formula:

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

1 1Cr

iCK

1 1Cr

iCK

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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/

lim 1 1

1r K

C

C

ri

CK

e

/

lim 1 1

1r K

C

C

ri

CK

e

0.12/4 1

3.045% per quarteri e

0.12/1 112.75% per year

ai e

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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

1

[1 / ] 1

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

Ci r CK

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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

3

[1 / ] 1

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

Ci r CK

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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

i r CK C

[ / ]

[ . / ( )( )]

.

1 1

1 0 08 13 4 1

2 0186%

13

per quarter

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Case 3: 8% compounded continuously

Payment 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

/

0.02

1

12.0201% per quarter

r Ki e

e

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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

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ExerciseA loan shark lends money on the following

terms: "If I give you $50 on Monday, you owe me $60 on the following Monday."

(a) What nominal interest rate per year (r) is the loan shark charging?

(b) What effective interest rate per year (ia)is he charging?

(c) If the loan shark started with $50 and was able to keep it, as well as all the money he received, out in loans at all times, how much money would he have at the end of one year?

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answer

Effective interest rate per year (ia) =

F = P(F/P, i, n)60 = 50 (F/P, i, 1)(F/P, i, 1) = 1,2

Therefore, i= 20% per weekNominal interest rate per year = 52 weeks x 0,20 = 10,40 = 1040%

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Exercise

• On January 1, a woman deposits $5000 in a' credit union that pays 8% nominal annual interest, compounded quarterly. She wishes to withdrawall the money in five equal yearly sums ,beginning December 31 of the first year. How much should she withdraw each year?

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Continuous compounding

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Continuous compounding

• A man deposited $500 per year into a credit union that paid 5% interest, compounded annually. At the end of 5 years, he had $2763 in the credit union. How much would he have if the institution paid 5% nominal interest, compounded continuously?

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