1

Applications of Money – Time

Relationships

Chapter 5

Topics

• Evaluating the economic profitability of a single proposed

problem solution (alternative)

– Multiple alternatives in Chapter 6

• Minimum attractive rate of return (MARR) concept

• Five methods for evaluation

– Present worth

– Future worth

– Annual worth

– Internal rate of return

– External rate of return

• Payback period method as a measure of the speed with which

an investment is recovered by the cash inflows it produces

2

3

Project Evalution Example:

Federal Express

Nature of Project:

• Equip 40,000 couriers

with PowerPads

• Save 10 seconds per

pickup stop

• Investment cost: $150

million

• Expected savings: $20

million per year Federal Express

4

Ultimate Questions

• Is it worth investing $150 million to save

$20 million per year, say over 10 years?

• How long does it take to recover the initial

investment?

• What kind of interest rate should be used in

evaluating business investment

opportunities?

3

5

Project Evalution Example:

Mr. Bracewell’s Investment Problem

• Built a hydroelectric plant using his personal savings of $800,000

• Power generating capacity of 6 million kWh

• Estimated annual power sales after taxes: $120,000

• Expected service life of 50 years

�Was Bracewell's $800,000 investment a wise one?

�How long does he have to wait to recover his initial investment,

and will he ever make a profit?

6

Mr. Bracewell’s Hydro Project

4

7

Bank Loan vs. Investment Project

Bank Customer

Loan

Repayment

Company Project

Investment

Return

�Bank Loan: Loan cash flow

�Investment Project: Project cash flow

8

Describing Project Cash Flows

Year

(n)

Cash Inflows

(Benefits)

Cash Outflows

(Costs)

Net

Cash Flows

0 0 $650,000 -$650,000

1 215,500 53,000 162,500

2 215,500 53,000 162,500

… … … …

8 215,500 53,000 162,500

5

The Present Worth Method

• The Present Worth (PW) method is based

on the concept of equivalent worth of all

cash flows relative to some beginning point

in time called the present.

• All cash inflows and outflows are

discounted to the present point in time at an

interest rate that is generally the MARR.

9

10

MARR: Minimum Attractive Rate of Return

• The rate at which the firm can always invest

the money in its investment pool.

• ―hurdle rate‖

• Possibly change over the life of project

6

Determining the Minimum Rate of Return

(MARR)

Example 5.1: Estimation of the MARR,

Using an Opportunity Cost Viewpoint

Expected

Annual Rate

of Profit

Investment

Requirements

(Thousands of

Dollars)

Cumulative

Investment

40% and over $2,200 $2,200

30-39.9% 3,400 5,600

20.29.9% 6,800 12,400

10-19.9% 14,200 26,600

Below 10% 22,800 49,400

Note: All projects with a rate of profit of 10%

or greater are acceptable.

• Consider the following schedule,

which shows prospective annual

rates of profit for a company’s

portfolio of capital investment

projects (this is the demand for

capital).

• If the supply of capital obtained

from internal and external

sources has a cost of 15% per

year for the first $5,000,000

invested and then increases 1%

for every $5,000,000 thereafter,

what is the company’s MARR

when using an opportunity cost

viewpoint?

7

Example 5.1: Estimation of the MARR,

Using an Opportunity Cost Viewpoint

The Present Worth Method

• The PW of an investment alternative is a

measure of how much money an individual

or a firm could afford to pay for the

investment in excess of its cost.

• A positive PW for an investment project is a

dollar amount of profit over the minimum

amount required by investors.

• It is assumed that cash generated by the

alternative is available for other uses that

earn interest at a rate equal to the MARR.14

8

15

The Present Worth Method

(Net Present Worth Measure)

�Principle: Compute the equivalent net surplus at n = 0 for a

given interest rate of i.

�Decision Rule: Accept the project if the net surplus is

positive.

2 3 4 5

0 1

Inflow

Outflow

0

PW(i)inflow

PW(i)outflow

Net surplus

PW(i) > 0

16

Example: Tiger Machine Tool Company

PW P F P F

P F

PW

PW

( $24, ( / , ) $27, ( / , )

$55, ( / , )

$78,

( $75,

( $78, $75,

$3, ,

15%) 400 15%,1 340 15%,2

760 15%,3

553

15%) 000

15%) 553 000

553 0

inflow

outflow

Accept

$75,000

$24,400 $27,340$55,760

01 2 3

outflow

inflow

9

17

Present Worth Amounts at Varying Interest Rates:

Choice of MARR is critical

i (%) PW(i) i(%) PW(i)

0 $32,500 20 -$3,412

2 27,743 22 -5,924

4 23,309 24 -8,296

6 19,169 26 -10,539

8 15,296 28 -12,662

10 11,670 30 -14,673

12 8,270 32 -16,580

14 5,077 34 -18,360

16 2,076 36 -20,110

17.45* 0 38 -21,745

18 -751 40 -23,302

*Break even interest rate

i = MARR (%)

18

-30

-20

-10

0

10

20

30

40

0 5 10 15 20 25 30 35 40

PW

(i)

($

th

ou

san

ds)

$3553 17.45%

Break even interest rate

(or rate of return)

Accept Reject

Present Worth Profile

10

The Present

Worth Method

The higher the

interest rate and

the farther into

the future a cash

flow occurs, the

lower its PW is.

20

Meaning of Present Worth

Two possible perspectives

• Investment pool: All funds in the firm’s treasury

can be placed in investments that yield a return

equal to the MARR.

• Borrowed funds (project balance): If no funds

are available for investment, the firm can borrow

them at MARR from the capital market.

11

21

Investment Pool Concept

• The company has $75,000 available for

investment. Two choices:

– Do not invest, keep the money in the

investment pool and earn at MARR.

– Invest in the project

22

$75,000

0 1 2 3

$24,400

$27,340

$55,760

Investment pool

How much would you have if the

Investment is made?

$24,400(F/P,15%,2) = $32,269

$27,340(F/P,15%,1) = $31,441

$55,760(F/P,15%,0) = $55,760

$119,470

How much would you have if the

investment was not made?

$75,000(F/P,15%,3) = $114,066

What is the net gain from the

investment?

$119,470 - $114,066 = $5,404

Project

N = 3

Meaning of Net Present Worth

PW(15%) = $5,404(P/F,15%,3) = $3,553 Net future worth

of the project

12

23

Project Balance Concept (Borrowed funds)

N 0 1 2 3

Beginning

Balance

Interest

Payment

Project

Balance

-$75,000

-$75,000

-$75,000

-$11,250

+$24,400

-$61,850

-$61,850

-$9,278

+$27,340

-$43,788

-$43,788

-$6,568

+$55,760

+$5,404

Net future worth, FW(15%)

PW(15%) = $5,404 (P/F, 15%, 3) = $3,553

24

Project Balance Diagram60,000

40,000

20,000

0

-20,000

-40,000

-60,000

-80,000

-100,000

-120,000

0 1 2 3

-$75,000-$61,850

-$43,788

$5,404

Year(n)

Terminal project balance

(net future worth, or

project surplus)

Discounted

payback period

Pro

ject

bal

ance

($

)

13

Example 5.2: Economic Desirability of a

Project, Using Present Worth

• An investment of $10,000 can be made in a project

that will produce a uniform annual revenue of

$5,311 for five years and then have a market

(salvage) value of $2,000. Annual expenses will be

$3,000 each year. The company is willing to accept

any project that will earn 10% per year or more on

all invested capital. Show whether this is a desirable

investment by using the PW method.

Example 5.3: Evaluation of New

Equipment Purchase, Using Present Worth

• A piece of new equipment has been proposed by

engineers to increase productivity of a certain

manual welding operation. The investment cost is

$25,000, and the equipment will have a market

(salvage) value of $5,000 at the end of its expected

life of five years. Increased productivity attributable

to the equipment will amount to $8,000 per year

after extra operating costs have been subtracted from

the value of the additional production. If the firm’s

MARR is 20% per year, is this proposal a sound

one? Use the PW method.

14

Example 5.3: Evaluation of New

Equipment Purchase, Using Present Worth

Example 5.3: Evaluation of New

Equipment Purchase, Using Present Worth

15

Example 5.3: Evaluation of New

Equipment Purchase, Using Present Worth

Bond Value

• Excellent example of commercial value as being the

PW of the future net cash flows that are expected to

be received through ownership of an interest-bearing

certificate.

– The value of a bond, at any time, is the PW of future cash

receipts

16

Bond Value

Z: face, or par, value

C: redemption or disposal price (usually equal to Z)

r: bond rate (nominal interest) per interest period

N: number of periods before redemption

i: bond yield rate per period

VN: value (price) of the bond N interest periods prior

to redemption – this is a PW measure of merit

Bond Value

• The owner of a bond is paid two types of payments

by the borrower:

– Series of periodic interest payments (each equal to rZ)

until bond is retired – these constitute an annuity

– When retired, a single payment of C

• The present worth of the bond is:

NiAPrZNiFPCVN %,,|%,,|

17

Example 5.4: Finding the Current Price

(PW) of a Bond

• Find the current price (PW) of a 10-year

bond paying 6% per year (payable

semiannually) that is redeemable at par

value, if bought by a purchaser to yield 10%

per year. The face value of the bond is

$1,000.

Example 5.5: Current Price and Annual

Yield of Bond Calculations

• A bond with a face value of $5,000 pays interest

of 8% per year. This bond will be redeemed at par

value at the end of its 20-year life, and the first

interest payment is due 1 year from now.

a) How much should be paid now for this bond in

order to receive a yield of 10% per year on the

investment?

b) If this bond is purchased now for $4,600, what

annual yield would the buyer receive?

18

Example 5.6: Stan Moneymaker Wants to

Buy a Bond

• Stan Moneymaker has the opportunity to purchase a

certain U.S. Treasury bond that matures in eight years and

has a face value of $10,000. This means that Stan will

receive $10,000 cash when the bond’s maturity date is

reached. The bond stipulates a fixed nominal interest rate

of 8% per year, but interest payments are made to the

bondholder every three months; therefore, each payment

amounts to 2% of the face value.

• Stan would like to earn 10% nominal interest

(compounded quarterly) per year on his investment,

because interest rates in the economy have risen since the

bond was issued. How much should Stan be willing to pay

for the bond?

The Capitalized Worth (CW) Method

(The Capitalized Equivalent (CE) Method)

• Special variation of the PW method involving the

determination of the PW of all revenues or

expenses over an infinite length of time

• If only expenses are considered, sometimes referred

to as capitalized cost

19

37

Capitalized Equivalent Worth

�Principle: PW for a project with an annual

receipt of A over infinite (or extremely long)

service life

�Equation:

� CW(i) or CE(i) = A(P/A, i, ) = A/i

A

0

P = CE(i)

N∞

Computing PW: capitalization of project cost

Cost is called capitalized cost (the amount of money that must be invested

today in order to yield a certain return A at the end of each and every

period forever, assuming an interest rate of i

38

Given: i = 10%, N =

Find: P or CE (10%)

10

$1,000

$2,000

P = CE (10%) = ?

0

CE P F($1,

.

$1,

.( / , )

$10, ( . )

$13,

10%)000

0 10

000

0 1010%,10

000 1 0 3855

855

20

39

Comparion of Present Worth for

Long Life and Infinite Life

• Built a hydroelectric plant using his personal savings of $800,000

• Power generating capacity of 6 million kwhs

• Estimated annual power sales after taxes: $120,000

• Expected service life of 50 years

�Was Bracewell's $800,000 investment a wise one?

�How long does he have to wait to recover his initial investment,

and will he ever make a profit?

40

Mr. Bracewell’s Hydro Project

21

41

Long Service Life: 50 years

• Equivalent lump sum investment

V1 = $50K(F/P, 8%, 9) + $50K(F/P, 8%, 8) +

. . . + $100K(F/P, 8%, 1) + $60K

= $1,101K

• Equivalent lump sum benefits

V2 = $120K(P/A, 8%, 50)

= $1,468K

• Equivalent net worth

V2 - V1 = $367K > 0, Good Investment

42

Infinite Project Life

• Equivalent lump sum investment

V1 = $50K(F/P, 8%, 9) + $50K(F/P, 8%, 8) +

. . . + $100K(F/P, 8%, 1) + $60K

= $1,101K

• Equivalent lump sum benefits assuming N =

V2 = $120(P/A, 8%, )

= $120/0.08

= $1,500K

• Capitalized equivalent worth

CE(8%) = V2 - V1

= $399K > 0

Difference (50 years vs infinite) = $32,000

22

43

Long life vs infinite life at 12%

Long life: -302K

Infinite life: -299K

The difference is $3,000 (smaller than for 8%)

1. Selection of i very critical

2. We can approximate PW of long cash flows by capitalized worth (approximation improves as i increases)

3. 12% not profitable

8% profitable

44

Example: Bridge Construction

�Construction cost = $2,000,000

�Annual Maintenance cost = $50,000

�Renovation cost = $500,000 every 15 years

�Planning horizon = infinite period

�Interest rate = 5%

23

45

$500,000 $500,000 $500,000 $500,000

$2,000,000

$50,000

0 15 30 45 60

46

Solution:

• Construction Cost

P1 = $2,000,000

• Maintenance Costs

P2 = $50,000/0.05 = $1,000,000

• Renovation Costs

P3 = $500,000(P/F, 5%, 15)

+ $500,000(P/F, 5%, 30)

+ $500,000(P/F, 5%, 45)

+ $500,000(P/F, 5%, 60)

...

= {$500,000(A/F, 5%, 15)}/0.05

= $463,423

• Total Present Worth

P = P1 + P2 + P3 = $3,463,423

24

47

Alternate way to calculate P3

• Concept: Find the effective interest rate per payment period

• Effective interest rate for a 15-year cycle

i = (1 + 0.05)15 - 1 = 107.893%

• Capitalized equivalent worth

P3 = $500,000/1.07893

= $463,423

15 30 45 600

$500,000 $500,000 $500,000 $500,000

Example 5.7: Determining the Capitalized

Worth (CW) of an Endowment

• Suppose that a firm wishes to endow an advanced biological processes

laboratory at a university. The endowment principal will earn interest

that averages 8% per year, which will be sufficient to cover all

expenditures incurred in the establishment and maintenance of the

laboratory for an indefinitely long period of time (forever). Cash

requirements of the laboratory are estimated to be $100,000 now (to

establish it), $30,000 per year indefinitely, and $20,000 at the end of

every fourth year (forever) for equipment replacement.

a) For this type of problem, what analysis period is, practically speaking,

defined to be ―forever‖?

b) What amount of endowment principal is required to establish the

laboratory and then earn enough interest to support the remaining cash

requirements of this laboratory forever?

25

49

The Future Worth Method

(Future Worth Criterion)

• Given: Cash flows

and MARR (i)

• Find: The net

equivalent worth at

a time period other

than 0 (typically at

the end of project

life) $75,000

$24,400 $27,340$55,760

01 2 3

Project life

50

FW F P F P

F P

FW F P

FW

( $24, ( / , ) $27, ( / , )

$55, ( / , )

$119,

( $75, ( / , )

$114,

( $119, $114,

$5, ,

15%) 400 15%,2 340 15%,1

760 15%,0

470

15%) 000 15%,3

066

15%) 470 066

404 0

inflow

outflow

Accept

Future Worth Criterion

26

Example 5.8: The Relationship between

Future Worth and Present Worth

• (From Example 5.3) A piece of new equipment has been

proposed by engineers to increase productivity of a certain

manual welding operation. The investment cost is $25,000,

and the equipment will have a market (salvage) value of

$5,000 at the end of its expected life of five years. Increased

productivity attributable to the equipment will amount to

$8,000 per year after extra operating costs have been

subtracted from the value of the additional production. The

firm’s MARR is 20% per year. Evaluate the FW of the

potential improvement project. Show the relationship

between FW and PW for this example.

52

The Annual Worth (AW) Method

(Annual Equivalence (AE) Analysis)

• The Annual Worth

(AW) of a project is

an equal annual series

of dollar amounts, for

a stated study period,

that is equivalent to

the cash inflows and

outflows at an interest

rate that is generally

the MARR

27

53

Annual Worth Analysis

Principle: Measure investment worth on annual basis

The AW of a project is annual equivalent revenues (R)

or savings minus annual equivalent expenses (E)

less its annual equivalent Capital Recovery (CR)

amount

Benefit:

• Annual reports, yearly budgets

• Seek consistency of report format

• Determine unit cost (or unit profit)

• Facilitate unequal project life comparison

54

Annual Equivalent Worth

AE(i)=PW(i)*(A|P,i,N)

If AE(i) > 0, accept

AE(i) < 0, reject

AE(i)=0, remain indifferent

Since (A|P,i,N) > 0 for –1<i<

AE(i) > 0 if and only if PW(i) > 0

28

55

Computing Equivalent Annual Worth

$100

$50

$80$120

$70

0

2 3 4 5 6

1

A = $46.07

2 3 4 5 61

AE(12%) = $189.43(A/P, 12%, 6)

= $46.07

$189.43

00

PW(12%) = $189.43

56

Annual Equivalent Worth - Repeating Cash

Flow Cycles

$500

$700$800

$400 $400 $500

$700$800

$400 $400

$1,000 $1,000

Repeating cycle

29

57

• First Cycle:

PW(10%) = -$1,000 + $500 (P/F, 10%, 1)

+ . . . + $400 (P/F, 10%, 5)

= $1,155.68

AE(10%) = $1,155.68 (A/P, 10%, 5) = $304.87

• Both Cycles:

PW(10%) = $1,155.68 + $1,155.68 (P/F, 10%, 5)

= $1,873.27

AE(10%) = $1,873.27 (A/P, 10%,10) = $304.87

58

Annual Equivalent Cost

• When only costs are

involved, the AE

method is called the

annual equivalent cost.

• Revenues must cover

two kinds of costs:

Operating costs and

capital costs.

Capital

costs

Operating

costs

+

Annual

Equiv

alen

t C

ost

s

30

59

Capital and Operating Costs

• Capital costs are incurred by purchasing assets to be used in production and service. Normally, they are nonrecurring (one-time) costs.

• Operating costs are incurred by the operation of physical plant or equipment needed to provide service (e.g. labor and raw materials). Normally, they recur for as long as an asset is owned.

• Operating costs are on annual basis anyway. Annual equivalent of a capital cost is called capital recovery cost, CR(i).

Remember: (A|P, i, N) is called the capital recovery factor.

60

Capital (Ownership) Costs

Def: The cost of owning an equipment

is associated with two

transactions—(1) its initial cost (I)

and (2) its salvage value (S).

• Capital costs: Taking into these

sums, we calculate the capital costs

as:

since (A|F,i,N)=(A|P,i,N)-i

0 1 2 3 N

0N

I

S

CR(i)

CR i I A P i N S A F i N

I S A P i N iS

( ) ( / , , ) ( / , , )

( )( / , , )

31

61

Example - Capital Cost Calculation

• Given:

I = $200,000

N = 5 years

S = $50,000

i = 20%

• Find: CR(20%) $200,000

$50,000

50

CR i I S A P i N iS

CR A P

( ) = ( - ) ( / , , ) +

( = ($200, - $50, ) ( / , )

+ (0.20)$5 ,

= $60,

20%) 000 000 20%, 5

0 000

157

Annual cost of owning the asset at 20%

Calculation of Equivalent Annual CR Amount

• Consider a device that will cost $10,000, last five

years, and have a salvage (market) value of

$2,000. Thus, the loss in value of this asset over

five years is $8,000. The MARR is 10% per year.

32

63

Justifying an investment based on AE

Method

Given: I = $20,000, S = $4,000, N = 5 years, i = 10%

Find: see if an annual revenue of $4,400 is enough to cover the capital costs.

Solution:

CR(10%) = $4,620.76

Conclusion: Need an additional annual revenue in the amount of $220.76.

64

Applying Annual Worth Analysis

•Unit Cost (Profit) Calculation

• Unequal Service Life Comparison

• Minimum Cost Analysis

33

65

Equivalent Worth per Unit of Time

0

1 2 3

$24,400

$55,760

$27,340

$75,000Operating Hours per Year

2,000 hrs. 2,000 hrs. 2,000 hrs.

• PW (15%) = $3553

• AE (15%) = $3,553 (A/P, 15%, 3)

= $1,556• Savings per Machine Hour

= $1,556/2,000

= $0.78/hr.

Note: 3553/6000=0.59/hour:

instant savings in present

worth for each hourly use;

does not consider the time

over which the savings occur

66

Equivalent Worth per Unit of Time

(cont’d)

0

1 2 3

$24,400

$55,760

$27,340

$75,000Operating Hours per Year

1,500 hrs. 2,500 hrs. 2,000 hrs.

•Let C denote the equivalent annual savings per machine hour

• $1,556=[(C)(1500)(P|F,15%,1)

+(C)(2500)(P|F,15%,2)

+(C)(2000) (P|F,15%,3)] (A|P,15%,3)

C=$0.79/hr

34

67

Breakeven Analysis

Problem:

At i = 6%, what

should be the

reimbursement rate

per mile so that Sam

can break even?

Year

(n)

Miles

Driven

Total costs(Ownership &

Operating)

1

2

3

14,500

13,000

11,500

$4,680

$3,624

$3,421

Total 39,000 $11,725

68

First Year Second Year Third Year

Depreciation $2,879 $1,776 $1,545

Scheduled maintenance 100 153 220

Insurance 635 635 635

Registration and taxes 78 57 50

Total ownership cost $3,693 $2,621 $2,450

Nonscheduled repairs 35 85 200

Replacement tires 35 30 27

Accessories 15 13 12

Gasoline and taxes 688 650 522

Oil 80 100 100

Parking and tolls 135 125 110

Total operating cost $988 $1,003 $971

Total of all costs $4,680 $3,624 $3,421

Expected miles driven 14,500 miles 13,000 miles 11,500 miles

35

69

• Equivalent annual cost of owning and operating the car

[$4,680 (P/F, 6%, 1) + $3,624 (P/F, 6%, 2) +

$3,421 (P/F, 6%, 3)] (A/P, 6%,3)

= $3,933 per year

• Equivalent annual reimbursement

Let X = reimbursement rate per mile

[14,500X(P/F, 6%, 1) + 13,000X(P/F, 6%, 2)

+ 11,500 X (P/F, 6%, 3)] (A/P, 6%,3)

= 13.058X

• Break-even value

13.058X = 3,933

X = 30.12 cents per mile

70

Annual equivalent reimbursement as a

function of cost per mile

4000

3000

2000

1000

0 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40

Annual equivalent

cost of owning and

operating ($3,933)

Loss

Annual reimbursement

amount

Minimum

reimbursement

requirement ($0.3012)

Gain

Reimbursement rate ($) per mile (X)

An

nu

al e

qu

ival

ent

($)

36

Example 5.9: Using Annual Worth to Evaluate

the Purchase of New Equipment

• (From Example 5.3) A piece of new equipment has been

proposed by engineers to increase productivity of a certain

manual welding operation. The investment cost is $25,000,

and the equipment will have a market (salvage) value of

$5,000 at the end of its expected life of five years. Increased

productivity attributable to the equipment will amount to

$8,000 per year after extra operating costs have been

subtracted from the value of the additional production. The

firm’s MARR is 20% per year. By using the AW method,

determine whether the equipment should be recommended.

Example 5.10: Determination of Monthly

Rent by Using the Annual Worth Method

• An investment company is

considering building a 25-unit

apartment complex in a growing

town. Because of the long-term

growth potential of the town, it is

felt that the company could

average 90% of full occupancy for

the complex each year. If the

following items are reasonably

accurate estimates, what is the

minimun monthly rent that should

be charged if a 12% MARR (per

year) is desired. Use the AW

method.

Land investment

cost

$50,000

Building

investment cost

$225,000

Study period 20 years

Rent per unit per

month

?

Upkeep expense

per unit per month

$35

Property taxes and

insurance per year

10% of total initial

investment

37

Rate of Return

• Definition: A relative percentage method which

measures the yield as a percentage of investment

over the life of a project

• Example: Vincent Van Gogh’s painting ―Irises‖

• John Whitney Payson bought the art at $80,000.

• John sold the art at $53.9 million in 40 years.

• What is the rate of return on John’s investment?

Rate of Return

• Given: P =$80,000, F

= $53.9M, and N = 40

years

• Find: i

• Solution:

$80,000

$53.9M

$53. $80, ( )

.

9 000 1

17 68%

40M i

i

0

40

38

Why is the ROR measure so popular?

• This project will bring in a 15% rate of return on

investment.

• This project will result in a net surplus of $10,000

in NPW.

Which statement is easier to understand?

In 1970, when Wal-Mart Stores, Inc. went

public, an investment of 100 shares cost

$1,650. That investment would have

been worth $13,312,000 on January 31,

2000.

What is the rate of return on that

investment?

Meaning of Rate of Return

39

Solution:

0

30

$13,312,000

$1,650

Given: P = $1,650

F = $13,312,000

N = 30

Find i:

$13,312,000 = $1,650 (1 + i )30

i = 34.97%

NiPF )1(

Rate of Return

Suppose that you invested that amount ($1,650) in

a savings account at 6% per year. Then, you could

have only $9,477 on January, 2000.

What is the meaning of this 6% interest here?

This is your opportunity cost if putting money in

savings account was the best you can do at that

time!

40

So, in 1970, as long as you earn more than 6%

interest in another investment, you will take that

investment.

Therefore, that 6% is viewed as a minimum

attractive rate of return (or required rate of return).

So, you can apply the following decision rule, to see

if the proposed investment is a good one.

ROR > MARR

Return on Investment

(ROR Defn 1)

• Definition 1: Rate of return (ROR) is defined

as the interest rate earned on the unpaid balance

of an installment loan.

• Example: A bank lends $10,000 and receives

annual payment of $4,021 over 3 years. The

bank is said to earn a return of 10% on its loan

of $10,000.

41

Loan Balance Calculation

A = $10,000 (A/P, 10%, 3)

= $4,021

Unpaid Return on Unpaid

balance unpaid balance

at beg. balance Payment at the end

Year of year (10%) received of year

0

1

2

3

-$10,000

-$10,000

-$6,979

-$366

-$1,000

-$698

-$366

+$4,021

+$4,021

+$4,021

-$10,000

-$6,979

-$3,656

0

A return of 10% on the amount still outstanding at the

beginning of each year

Return on Investment

(ROR Defn 2)

• Definition 2: Rate of return (ROR) is the

break-even interest rate, i*, which equates the

present worth of a project’s cash outflows to the

present worth of its cash inflows.

• Mathematical Relation:

PW i PW i PW i( ) ( ) ( )* * *

cash inflows cash outflows

0

42

Return on Invested Capital

(ROR Defn 3)

• Definition 3: Return on invested capital is

defined as the interest rate earned on the

unrecovered project balance of an investment

project such that, when the project terminates,

the uncovered project balance is zero. It is

commonly known as internal rate of return

(IRR).

• Example: A company invests $10,000 in a

computer and results in equivalent annual labor

savings of $4,021 over 3 years. The company is

said to earn a return of 10% on its investment of

$10,000.

Project Balance Calculation

0 1 2 3

Beginning

project balance

Return on

invested capital

Cash generated

from project

Ending project

balance

-$10,000 -$6,979 -$3,656

-$1,000 -$697 -$365

-$10,000 +$4,021 +$4,021 +$4,021

-$10,000 -$6,979 -$3,656 0

The firm earns a 10% rate of return on funds that remain internally invested

in the project. Since the return is internal to the project, we call it internal

rate of return. Investment firm lender, project borrower

43

The Internal Rate of Return Method

• The Internal Rate of Return method is the most

widely used rate of return method for performing

engineering economic analyses.

• This method solves for the interest rate that equates

the equivalent worth of an alternative’s cash inflows

(receipts or savings) to the equivalent worth of cash

outflows (expenditures, including investment costs).

– Equivalent worth may be computed with any of the three

methods (PW, FW, AW).

• The resultant interest rate is termed the Internal Rate

of Return (IRR).

The Internal Rate of Return Method

• For a single alternative, from the lender’s

viewpoint, the IRR is not positive unless

1) Both receipts and expenses are present in the cash

flow pattern

2) The sum of receipts exceeds the sum of all cash

outflows

44

The Internal Rate of Return Method

General form of PW

versus interest rate

for an alternative

with a single

investment cost at

the present time

followed by a series

of positive cash

inflows over N

Plot of PW versus Interest Rate

Investment Balance Diagram

Investment Balance Diagram Showing IRR

45

Investment Balance Diagram

• Investment balance diagram: How much of the original

investment in an alternative is still to be recovered as a

function of time.

• Downward arrows represent annual returns, Rk-Ek

against the unrecovered investment

• Dashed lines indicate the opportunity cost of interest, or

profit, on the beginning-of-year investment balance

• The IRR is the value of i that causes the unrecovered

investment balance to exactly equal 0 at the end of the

study period and thus represents the internal earning

rate of a project

Methods for Finding Rate of Return

• Investment Classification

– Simple Investment

– Nonsimple Investment

• Computational Methods

– Direct Solution Method

– Trial-and-Error Method

– Computer Solution Method

46

Investment Classification

Simple Investment

• Def: Initial cash flows

are negative, and only

one sign change

occurs in the net cash

flows series.

• Example: -$100, $250,

$300 (-, +, +)

• ROR: A unique ROR

Nonsimple Investment

• Def: Initial cash flows are negative, but more than one sign changes in the remaining cash flow series.

• Example: -$100, $300, -$120 (-, +, -)

• ROR: A possibility of multiple RORs

Period

(N)

Project A Project B Project C

0 -$1,000 -$1,000 +$1,000

1 -500 3,900 -450

2 800 -5,030 -450

3 1,500 2,145 -450

4 2,000

Project A is a simple investment.

Project B is a nonsimple investment.

Project C is a simple borrowing.

47

Computational Methods

Direct

Solution

Direct

Solution

Trial &

Error

Method

Computer

Solution

MethodLog Quadratic

n Project A Project B Project C Project D

0 -$1,000 -$2,000 -$75,000 -$10,000

1 0 1,300 24,400 20,000

2 0 1,500 27,340 20,000

3 0 55,760 25,000

4 1,500

Direct Solution Methods

• Project A • Project B

PW ii i

xi

PW i x x

x

x

i

ii

ii

i i

( ) $2,$1,

( )

$1,

( )

,

( ) , , ,

:

.

. .

, *

000300

1

500

10

1

1

2 000 1 300 1 500

0 8

0 81

125%, 1667

1

1160%

100% 25%.

2

2

Let then

Solve for

or -1.667

Solving for yields

Since the project' s

$1, $1, ( / , , )

$1, $1, ( )

. ( )

ln .ln( )

. ln( )

.

.

000 500 4

000 500 1

0 6667 1

0 6667

41

0101365 1

1

1

4

4

0 101365

0 101365

P F i

i

i

i

i

e i

i e

10.67%

48

Trial and Error Method – Project C

• Step 1: Guess an interest

rate, say, i = 15%

• Step 2: Compute PW(i)

at the guessed i value.

PW (15%) = $3,553

• Step 3: If PW(i) > 0, then

increase i. If PW(i) < 0,

then decrease i.

PW(18%) = -$749

• Step 4: If you bracket the

solution, you use a linear

interpolation to approximate

the solution

3,553

0

-749

15% i 18%

749553,3

553,3%3%15i

%45.17

Graphical Solution – Project D

• Step 1: Create a NPW plot using Excel.

• Step 2: Identify the point at which the curve crosses the horizontal axis closely approximates the i*.

• Note: This method is particularly useful for projects with multiple rates of return, as most financial softwares would fail to find all the multiple i*s.

49

Basic Decision Rule

(single project evaluation):

If ROR > MARR, Accept

This rule does not work for a situation where

an investment has multiple rates of return

Multiple Rates of Return Problem

• Find the rate(s) of return:

PW ii i

( ) $1,$2, $1,

( )

000300

1

320

1

0

2

$1,000

$2,300

$1,320

50

Let Then,

Solving for yields,

or

Solving for yields

or 20%

xi

PW ii i

x x

x

x x

i

i

1

1

000300

1

320

1

000 300 320

0

10 11 10 12

10%

2

2

.

( ) $1,$2,

( )

$1,

( )

$1, $2, $1,

/ /

NPW Plot for a Nonsimple Investment

with Multiple Rates of Return

51

Project Balance Calculation

n = 0 n = 1 n = 2

Beg. Balance

Interest

Payment -$1,000

-$1,000

-$200

+$2,300

+$1,100

+$220

-$1,320

Ending Balance -$1,000 +$1,100 $0

i* =20%

Cash borrowed (released) from the project is assumed to

earn the same interest rate through external investment

as money that remains internally invested.

Critical Issue: Can the company be able to invest the money

released from the project at 20% externally in

Period 1?

If your MARR is exactly 20%, the answer is ―yes‖, because it

represents the rate at which the firm can always invest

the money in its investment pool. Then, the 20% is also true

IRR for the project.

Suppose your MARR is 15% instead of 20%. The assumption

used in calculating i* is no longer valid.

Therefore, neither 10% nor 20% is a true IRR.

52

• If NPW criterion is used at MARR = 15%

PW(15%) = -$1,000

+ $2,300 (P/F, 15%, 1)

- $1,320 (P/F, 15%, 2 )

= $1.89 > 0

Accept the investment

How to Proceed: If you encounter multiple rates of

return, abandon the IRR analysis and use the NPW

criterion (or use alternative procedures such as the

external rate of return method).

Decision Rules for Nonsimple Investment

• A possibility of multiple RORs.• If PW (i) plot looks like this,

then, IRR = ROR.

If IRR > MARR, Accept

• If PW(i) plot looks like this, Then, IRR ROR (i*).• Find the true IRR by using

other procedures or,• Abandon the IRR method

and use the PW method.

i*i

i

i*

i*

PW

(i)

53

Example 5.11

• A capital investment of $10,000 can be

made in a project that will produce a

uniform annual revenue of $5,311 for five

years and then have a salvage value of

$2,000. Annual expenses will be $3,000.

The company is willing to accept any

project that will earn at least 10% per year

on all invested capital. Determine whether it

is acceptable, by using the IRR method.

54

Example 5.11

Example 5.12

• A piece of new equipment has been proposed by

engineers to increase productivity of a certain manual

welding operation. The investment cost is $25,000,

and the equipment will have a market (salvage) value

of $5,000 at the end of its expected life of five years.

Increased productivity attributable to the equipment

will amount to $8,000 per year after extra operating

costs have been subtracted from the value of the

additional production. Evaluate the IRR of the

proposed equipment. Is the investment a good one?

The MARR is 20% per year.

55

Example 5.12

Example 5.12: Present Worth Profile

IRR approximately 22%

56

Example 5.12

Investment (or Project) Balance Diagram

Example 5.13

• In 1915, Albert Epstein borrowed $7,000 from a large New

York bank on the condition that he would repay 7% of the

loan every three months, until a total of 50 payments had

been made. At the time of the 50th payment, the $7,000

loan would be completely repaid.

• Albert computed his annual interest rate to be

[0.07(7000)*4]/7000=0.28 (28%).

1. What true effective annual interest rate did Albert pay?

2. What, if anything, was wrong with his calculation?

57

Example 5.13

Example 5.13:

What was wrong with his calculation?

20 quarterly

payments of

$490

50 quarterly

payments of

$490

70 quarterly

payments of

$490

True annual

effective rate

14.5% 30% 31%

His answer is the same (28% ) whether he

makes 20, 50, 70 payments (that is, insensitive

to how long the payments are made).

58

Example 5.14

• The Fly-by-Night finance company advertises a ―bargain 6%

plan‖ for financing the purchase of automobiles.

• To the amount of loan being financed, 6% is added for each

year money is owed. This total is then divided by the number

of months over which the payments are to be made, and the

result is the amount of the monthly payments. For example, a

woman purchases a $10,000 automobile under this plan and

makes an initial cash payment of $2,500. She wishes to pay

the $7,500 balance in 24 monthly payments.

• What effective annual rate of interest does she actually pay?

Example 5.14

59

Example 5.15

• A small airline executive charter company needs to borrow

$160,000 to purchase a prototype synthetic vision system

(SVS) for one of its business jets. The SVS is intended to

improve the pilots’ situational awareness when visibility is

impaired. The local (and only) banker makes this statement:

• ―We can loan you $160,000 at a very favorable rate of 12%

per year for a five-year loan. However, to secure this loan, you

must agree to establish a checking account (with no interest) in

which the minimum average balance is $32,000. In addition,

your interest payments are due at the end of each year, and the

principal will be repaid in a lump-sum amount at the end of

year five. ‖

• What is the true effective annual interest rate being charged?

Example 5.15

60

Difficulties with the IRR Method

• The reinvestment assumption of the IRR method

may not be valid.

– Suppose MARR=20% and, IRR for a project is 42.4%: it

may not be possible to reinvest net cash proceeds from the

project at much more than 20%.

– PW, AW, FW method assume that positive recovered

funds are reinvested at the MARR

• Computational difficulty

• Possibility of multiple rates of return in some

problems

The External Rate of Return

(ERR) Method

• Directly takes into account the interest rate

() external to a project at which net cash

flow generated (or required) by the project

over its life can be reinvested (or borrowed).

• If this external reinvestment rate, which is

usually the firm’s MARR, happens to equal

the project’s IRR, then the ERR method

produces results identical to those of the IRR

method.

61

The External Rate of Return

(ERR) Method

Step 1: Discount all net cash outflows to 0.

Step 2: Compound all net cash inflows to 0.

Step 3: ERR is the rate that establishes equivalence

between the two quantities.

Project acceptable when the ERR is greater than or

equal to the firm’s MARR.

The External Rate of Return

(ERR) Method

• Two advantages over IRR:

– Easier to obtain

– Not subject to the possibility of multiple rates of return

62

Example 5.16

• Referring to Example 5.12, suppose that

= MARR = 20% per year.

• What is the project’s ERR, and is the

project acceptable?

Example 5.16

63

Example 5.17

• When =15% and MARR = 20% per year,

determine whether the project (with the

cash flow given next) is acceptable. Notice

in this example that the use of an %

different from the MARR is illustrated. This

might occur if, for some reason, part or all

of the funds related to a project are

―handled‖ outside the firm’s normal capital

structure.

Example 5.17

64

Appendix 5-A The Multiple Rate of Return

Problem with the IRR Method:

Example 5-A-1

• Plot the present worth versus interest rate for the

following cash flows. Are there multiple IRRs? If

so, what do they mean?

Year k Net Cash Flow i% PW(i%)

0 500 0 250

1 -1000 10 150

2 0 20 32

3 250 30 0

4 250 40 -11

5 250 62 0

80 24

Example 5-A-1

65

Example 5-A-2

• Use the ERR method to

analyze the cash flow

pattern in the accompanying

table.

• The IRR is indeterminant

(none exists), so the IRR is

not a workable procedure.

The external reinvestment

rate is 12% per year, and the

MARR equals 15%.

Year Cash Flows

0 5,000

1 -7,000

2 2,000

3 2,000

Example 5-A-2

66

Summary of Decision Rules

132

Payback (Payout) Period Method

How fast can I recover my initial investment?

• Conventional payback method (time value of money

ignored)

• Discounted payback method

• Payback screening - If the payback period is within

acceptable range, formal project evaluation may start (a

high-tech firm may have a short time limit since products

rapidly become obsolete).

67

133

Conventional-Payback Method

�Principle:

How fast can I recover my initial investment?

�Method:

Based on cumulative cash flow

�Screening Guideline:

If the payback period is less than or equal to

some specified payback period, the project

would be considered for further analysis.

�Weakness:

Does not consider the time value of money

134

Example: Conventional Payback Period

N Cash Flow Cum. Flow

0

1

2

3

4

5

6

-$105,000+$20,000

$15,000

$25,000

$35,000

$45,000

$45,000

$35,000

-$85,000

-$70,000

-$45,000

-$10,000

$35,000

$80,000

$115,000

Payback period should occurs somewhere

between N = 3 and N = 4.

68

135

-100,000

-50,000

0

50,000

100,000

150,000

0 1 2 3 4 5 6

Years (n)

3.2 years

Payback period

$85,000

$15,000

$25,000

$35,000$45,000 $45,000

$35,000

0

1 2 3 4 5 6

Years

An

nu

al c

ash

flo

w

Cu

mu

lati

ve

cash

flo

w (

$)

136

Payback Method Pitfall

n Project 1

Payback:3 years

Project 2

Payback:3.6 years

0 -90000 -90000

1 30000 25000

2 30000 25000

3 30000 25000

4 1000 25000

5 1000 25000

6 1000 25000

• Fails to measure profitability

• Ignores timing of cash flows

69

137

Discounted Payback Method

Principle:

How fast can I recover my initial investment

plus interest?

Method:

Based on cumulative discounted cash flow

Screening Guideline:

If the discounted payback period (DPP) is less

than or equal to some specified payback period,

the project would be considered for further

analysis.

Weakness:

Cash flows occurring after DPP are ignored

138

Discounted Payback Period Calculation

Period Cash Flow Cost of Funds

(15%)

Cumulative

Cash Flow

0 -$85,000 0 -$85,000

1 15,000 -$85,000(0.15)= -$12,750 -82,750

2 25,000 -$82,750(0.15)= -12,413 -70,163

3 35,000 -$70,163(0.15)= -10,524 -45,687

4 45,000 -$45,687(0.15)=-6,853 -7,540

5 45,000 -$7,540(0.15)= -1,131 36,329

6 35,000 $36,329(0.15)= 5,449 76,778

-85000(1.15)+15000= -82750

70

Example 5.12 revisited