INDR 202 - Chapter 7

INDR  202

ENGINEERING  ECONOMICS

CHAPTER  7

RATE-­OF-­RETURN  ANALYSIS

SPRING  2015

INSTRUCTOR:  BORA  ÇEKYAY

COURSE  CONTENT

Engineering Economic Decisions Time Value of Money

Money Management Inflation

Present-­Worth Analysis

Annual-­Equivalence Analysis Rate-­of-­Return Analysis

Benefit-­Cost Analysis

Depreciation & Income Taxes Project Cash-­Flow Analysis Project Uncertainty

RATE-­OF-­RETURN  ANALYSIS

3

Rate  of  Return

Internal  Rate  of  Return

Internal-­Rate-­of-­Return  Analysis

RATE  OF  RETURN

RATE  OF  RETURN

Rate  of  Return

q What  it  is:  Interest  earned  on  your  invested  capital,  or   commonly  known  as  internal  rate  of  return  (IRR)

Why  ROR  measure  is  so  popular?

q

_{This  project  will  bring  in  a}  

_{15%  rate  of  return}  

on  investment.

q

This  project  will  result  in  a  

net  surplus  of  

$10,000  

in  NPW.

EXAMPLE  1:  VAN  GOGH  PAINTING

7

Bought for $80,000

Sold for $53.9M after 40 years

$80,000

$53.9M

0

40

$80,000 1 + 𝑖

= $53.9𝑀

Investing  in  Wal-­‐

Mart  Stock

In  October  1,1970,   when  Wal-­‐Mart   Stores,  Inc.  went   public,  an  

investment  of  100   shares  cost  $1,650.     That  investment   would  have  been   worth  $10,053,632   on  September  30,   2009.  What  is  the  

Wal-­‐Mart  Investment  Problem

—

Given

:

—

P

=  $1,650

—

F

=  $10,053,632

—

N

=  29  years

—

Find

:  

i

—

Formula  to  Use

:

—

F

=  

P(1  +  i)

—

$10,053,632  =  $1,650(1  +  i)

—

i

=  25.04%

—

Cash  Flow  Diagram

$10,053,632

$1,650

1970

How  Good  the  Wal-­‐Mart  Investment  Was  and  

What  to  Compare  with?

If  you  took  out  $1,650  from   your  savings  account  and   invested  in  Wal-­‐Mart  stock,   you  could  have

If  you  did  not  invest  $1,650  in   Wal-­‐Mart  stock,  what  could   use  your  money  for?

— $10,053,632

— Or  equivalent  to  earning  

25.04%  interest  on  your   savings  account.

— If  the  best  you  could  do  was  

to  leave  the  money  in  the   savings  account  to  earn  6%   interest  over  29  years,  you   will  have  $16,010.

— What  is  the  meaning  of  6%  

interest?    This  will  be  your  

opportunity  cost  rate  or  

Is  This  a  Good  investment?

q

In  1970,  as  long  as  you  could  earn  more  than  a  6%  

interest  in  another  investment  opportunity,  you  would  

take  that  investment.

q

Therefore,  that  6%  is  viewed  as  a  

minimum  

attractive  rate  of  return

(or  required  rate  of  return).  

This  is  the  interest  rate  used  in  NPW  analysis.

q

So,  to  see  if  the  proposed  investment  is  a  good  one,  

you  adopt  the  following  decision  rule.

RATE  OF  RETURN  ON  INVESTMENT

YIELD

INTERNAL  RATE  OF  RETURN

MARGINAL  EFFICIENCY  OF  CAPITAL

DEFINITION  1

interest  rate  earned  on  unpaid  balance  of  a  loan

DEFINITION  2

EXAMPLE  3:  BANK  LOAN

13

A  bank  lends  $10,000  &  receives  3  annual  payments  of  $4,021.

$10,000 𝐴|𝑃, 10%, 3 = $4,021

Year _{Unpaid Balance}Beginning _{Unpaid  Balance}Return  on   _ReceivedPayment   _{Unpaid Balance}Ending

0 -­$10,000

1 -­$10,000 -­$1,000 $4,021 -­$6,979 2 -­$6,979 -­$698 $4,021 -­$3,656

3 -­$3,656 -­$366 $4,021 $0

DEFINITION  1

RATE  OF  RETURN  ON  INVESTMENT

DEFINITION  2

break-­even  interest  rate  at  which  the  project’s  

present  worth  is  zero

RATE  OF  RETURN  ON  INVESTMENT

15

DEFINITION  3

EXAMPLE  4:  CAPITAL  INVESTMENT

A  company  invests  $10,000  in  computers  &  saves  $4,021  in   equivalent  annual  labor  costs  over  3  years.

Year _{Unpaid Balance}Beginning _{Unpaid  Balance}Return  on   _ReceivedPayment   _{Unpaid Balance}Ending

0 -­$10,000

1 -­$10,000 -­$1,000 $4,021 -­$6,979 2 -­$6,979 -­$698 $4,021 -­$3,656

3 -­$3,656 -­$366 $4,021 $0

DEFINITION  3

CALCULATING  RATE  OF  RETURN

17

STEP  1:  INVESTMENT  CLASSIFICATION

§

SIMPLE  INVESTMENT

§

NONSIMPLE  INVESTMENT

STEP  2:  COMPUTATIONAL  METHODS

INVESTMENT  CLASSIFICATION

number  of  sign  changes  in  net  cash  flow  series

(ignoring  zero  cash  flows)

Simple  Investment

exactly  one  sign  change

e.g.,  (-­$100,  $250,  $300)      

à

(-­,  +,  +)

UNIQUE  ROR

Nonsimple Investment

more  than  one  sign  change

e.g.,  (-­$100,  $250,  -­$300)      

à

(-­,  +,  -­)

Simple  versus  Nonsimple  Investments

Simple  investment

Nonsimple investment

— Definition:  An  investment  

with  only  one  sign  change   in  the  net  cash  flow  series.

EXAMPLE  5:  INVESTMENT  CLASSIFICATION

Cash  Flow  Sign  at  Period _Changes#  Sign   Investment  _Type

0 1 2 3 4 5

-­ + + + + + 1 Simple

-­ -­ + + 0 + 1 Simple

-­ + -­ + + -­ 4 Nonsimple

EXAMPLE  6:  RATE  OF  RETURN

21

Period 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:  simple  investment [1  sign  change  &  initial  <  0]

Project  B:  nonsimple investment [3  sign  changes]

EXAMPLE  6:  RATE  OF  RETURN

𝑃𝑊_< 𝑖 = −$1,000 − $500 1 + 𝑖 +

$800

1 + 𝑖 > +

$1,500 1 + 𝑖 ? +

$2,000 1 + 𝑖 (

-­1000 0 1000 2000 3000

0 0.2 0.4 0.6 0.8 1 1.2 𝑃𝑊_< 𝑖

𝑖

𝑃𝑊_< 0 = $2,800

𝑃𝑊_< ∞ = −$1,000

EXAMPLE  6:  RATE  OF  RETURN

23

𝑃𝑊_A 𝑖 = −$1,000 + $3,900 1 + 𝑖 −

$5,030 1 + 𝑖 > +

$2,145 1 + 𝑖 ?

-­20 -­10 0 10 20

0 0.2 0.4 0.6 0.8

𝑃𝑊_A 𝑖

𝑖

10% 30% 50%

1 2 3 4 𝑃𝑊A 0 = $15

EXAMPLE  6:  RATE  OF  RETURN

𝑃𝑊_B 𝑖 = $1,000 − $450

1 + 𝑖 −

$450

1 + 𝑖 > −

$450 1 + 𝑖 ?

𝑃𝑊_B 0 = −$350

𝑃𝑊_B ∞ = $1,000

-­200 0 200 400 600 800

0 0.2 0.4 0.6 0.8 1 1.2

𝑃𝑊_B 𝑖

𝑖

Predicting  Multiple  i*s

—

Net  Cash  Flow  Rule  of  Signs

The  number  of  real  i*s  that  are  greater  than  -­‐100%  for  a   project  with  N periods  is  never  greater  than  the  number  of   sign  changes  in  the  sequence  of  the  cash  flows.  A  zero  cash   flow  is  ignored.

—

Accumulated  Cash  Flow  Sign  Test

Predicting  the  Number  of  

i

*s

Net  Cash-­‐Flow  Rule  of  Signs

Accumulated  Cash-­‐Flow  Sign  

Test

• No.  of  real  i*s  ≤ 3

• This  implies  that  the  project  could  

EXAMPLE  7:  RATE  OF  RETURN

27

Period Project  A Project  B Project  C Project  D 0 -­$3,000 -­$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

EXAMPLE  7:  RATE  OF  RETURN

Period Project  A 0 -­$3,000

1 $0

2 $0

3 $0

4 $4,500

$3,000 1 + 𝑖

= $4,500

4 ln 1 + 𝑖 = ln 1.5

1 + 𝑖 = 𝑒

𝑖 = 10.67%

EXAMPLE  7:  RATE  OF  RETURN

29

Period Project  B 0 -­$2,000 1 $1,300 2 $1,500 3

4

−$2,000 +

$1,300

1 + 𝑖

+

$1, 500

1 + 𝑖

= 0

−2 + 1.3𝑥 + 1.5𝑥

= 0

𝑥 =

−1.3 ± 1.3

>

_{− 4 1.5 −2}

2 1.5

𝑥

= 0.8, 𝑥

= −1.667

𝑖

= 25%, 𝑖

= −160%

𝑖 = 25%

(since  

−100% < 𝑖 < ∞

)

TRIAL-­AND-­ERROR  METHOD

(FOR  SIMPLE  INVESTMENTS)

STEP 1.

Guess an interest rate

𝑖

(e.g., MARR).

STEP 2.

Compute

𝑃𝑊 𝑖

.

STEP 3.

If

𝑃𝑊 𝑖 < 0

, decrease

𝑖

& go to STEP 2.

If

𝑃𝑊 𝑖 > 0

, increase

𝑖

& go to STEP 2.

STEP 4.

Given

𝑖

with

𝑃𝑊 𝑖

< 0

&

𝑖

with

𝑃𝑊 𝑖

> 0

,

use linear interpolation to approximate

𝑖

.

𝑖

≅ 𝑖

+ 𝑖

− 𝑖

𝑃𝑊 𝑖

EXAMPLE  7:  RATE  OF  RETURN

31

Period Project  C 0 -­$75,000 1 $24,400 2 $27,340 3 $55,760 4

STEP  1. 𝑖 = 15%

STEP  2. 𝑃𝑊 15% = $3,553

STEP  3. 𝑃𝑊 15% > 0,  let  𝑖 = 18%.

STEP  2. 𝑃𝑊 18% = −$749

STEP  4:  𝑖_G = 18% & 𝑖_> = 15%

𝑖∗ ≅ 0.15 + 0.03 3,553

3,553 + 749

𝑖∗ ≅ 17.45%

EXAMPLE  7:  RATE  OF  RETURN

3,553

15% -­749

18% 0

𝑖∗

Remark. The smaller 𝑖_> − 𝑖_G is, the smaller will be the linear-­

33

STEP 1.

Plot

𝑃𝑊 𝑖

in Excel.

STEP 2.

Find the point where the curve intersects the

horizontal axis. This interest rate is an approximation

to the rate of return.

This method is useful especially for projects with

multiple rates of return, since most financial software

will fail to identify all possible rates.

EXAMPLE  7:  RATE  OF  RETURN

Period Project  D 0 -­$10,000 1 $20,000 2 $20,000 3 -­$25,000

4 -­2000

0 2000 4000 6000 8000

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8

𝑃𝑊_M 𝑖

𝑖

𝑖∗ = 140%

INTERNAL-­RATE-­OF-­RETURN  ANALYSIS

35

Simple-­investment  projects  have  a  unique  rate  of  return.

Evaluating  a  single  

simple-­investment

project:

EXAMPLE  8:  MULTIPLE  RATES  OF  RETURN

𝑃𝑊 𝑖 = −$1𝑀 + $2.3𝑀

1 + 𝑖 −

$1. 32𝑀

1 + 𝑖 > = 0

−1 + 2.3𝑥 − 1.32𝑥> = 0

𝑥_G = 10

11 , 𝑥> = 10 12

𝑖_G = 10%, 𝑖_> = 20%

$1M

$2.3M

$1.32M

EXAMPLE  8:  MULTIPLE  RATES  OF  RETURN

37

Neither 10% nor 20% represents the true IRR of this project.

-­20000 -­15000 -­10000 -­5000 0 5000

0 0.05 0.1 0.15 0.2 0.25 0.3 𝑃𝑊 𝑖

𝑖

𝑃𝑊 𝑖 > 0

EXAMPLE  8:  MULTIPLE  RATES  OF  RETURN

Period 0 1 2

Beginning  Balance Interest

Cash  Flow -­$1.00M

-­$1.00M

-­$0.20M

+$2.30M

+$1.10M

+$0.22M

-­$1.32M

Ending  Balance -­$1.00M +$1.10M $0

Cash released from the project is assumed to earn the same interest rate through external investment as money that remains internally invested.

EXAMPLE  8:  MULTIPLE  RATES  OF  RETURN

39

Will  the  company  be  able  to  invest  the  money  released  from   the  project  at  20%  externally  in  Period  1?

Yes  if MARR  is  exactly  20%,  since  MARR  is the  rate  at  which   the  firm  can  always  invest  the  money  in  its  investment  pool.  

In  this  case,  20%  is  also  the  true  IRR  for  the  project.

If  MARR  is  15%  instead  of  20%, the  assumption  used  in   calculating  𝑖∗ is  no  longer  valid,  so  neither  10%  nor  20%  is  

EXAMPLE  8:  MULTIPLE  RATES  OF  RETURN

Period 0 1 2

Beginning  Balance Interest

Cash  Flow −$1𝑀

−$1𝑀

𝑖 −$1𝑀

$2.3𝑀

$1.3𝑀 − $𝑖𝑀

15% $1.3𝑀 − $𝑖𝑀

−$1.32𝑀

Ending  Balance −$1𝑀 $1.3𝑀 − $𝑖𝑀 $0

CASE  1:  

$1.3𝑀 − $𝑖𝑀 > 0

invested  at  MARR  15%

EXAMPLE  8:  MULTIPLE  RATES  OF  RETURN

41

Period 0 1 2

Beginning  Balance Interest

Cash  Flow −$1𝑀

−$1𝑀

𝑖 −$1𝑀

$2.3𝑀

$1.3𝑀 − $𝑖𝑀

𝑖 $1.3𝑀 − $𝑖𝑀

−$1.32𝑀

Ending  Balance −$1𝑀 $1.3𝑀 − $𝑖𝑀 $0

CASE  2:  

$1.3𝑀 − $𝑖𝑀 < 0

invested  at  

𝑖

$1.3𝑀 − 𝑖 +

𝑖 $1.3𝑀 − $𝑖𝑀

− $1.32𝑀 = 0

𝑖 = 10% < 1.3

or  

𝑖 = 20% < 1.3

EXAMPLE  8:  MULTIPLE  RATES  OF  RETURN

CASE  2: No  solution such  that

1.3 − 𝑖 < 0

CASE  1: (Correct  Situation)  

1.3 − 𝑖 > 0

IRR = 15.22% > MARR

The  project  is  acceptable.

𝑃𝑊 15% = $1,890 > 0

Accept  the  project.

MULTIPLE  RATES  OF  RETURN

43

NET-­CASH-­FLOW  RULE  OF  SIGNS

The  number  of  real  RORs  is  less  than  or  equal  to  the  

number  of  sign  changes  in  net  cash  flow  series.

EXAMPLE  9:  MULTIPLE  RATES  OF  RETURN

Period Net  Cash  Flow

0 -­$100

1 -­$20

2 $50

3 0

4 $60

5 -­$30

6 $100

3  sign  changes

MULTIPLE  RATES  OF  RETURN

45

Period  𝑛 Cash  Flow  𝐴_S Sum  𝑆_S

0 𝐴₎ 𝑆₎ = 𝐴₎

1 𝐴_G 𝑆_G = 𝑆₎ + 𝐴_G

2 𝐴_> 𝑆_> = 𝑆_G + 𝐴_>

⋮ ⋮ ⋮

N 𝐴_V 𝑆_V = 𝑆_VWG + 𝐴_V

CUMULATIVE-­CASH-­FLOW  RULE  OF  SIGNS

If  

𝑆

< 0

,  

𝑆

≠ 0

,  and  

𝑆

changes  sign  ONLY  ONCE,  

EXAMPLE  9:  MULTIPLE  RATES  OF  RETURN

Period Net  Cash  _Flow Cumulative_{Cash  Flow}

0 -­$100 -­$100

1 -­$20 -­$120

2 $50 -­$70

3 0 -­$70

4 $60 -­$10

5 -­$30 -­$40

6 $100 $60

NONSIMPLE  INVESTMENTS

Unique  ROR:

IRR  =  ROR

Accept  if  IRR  >  MARR.

Multiple  RORs:

IRR  ≠  ROR

Find  the  true  IRR.

DECISION  RULES

number  of  sign  changes  in  project  cash  flow  series

ONE

MULTIPLE

simple  investment

nonsimple  investment

IRR  =  

𝑖

NONE

Accept  if  IRR  >  MARR.

no  ROR

Use  PW  or  AE  criterion.

EXAMPLE  10:  INCREMENTAL  ROR  ANALYSIS

49

Period Project  A1 Project  A2

0 -­$1,000 -­$5,000

1 $2,000 $7,000

IRR 100% 40%

EXAMPLE  10:  INCREMENTAL  ROR  ANALYSIS

If MARR = 10%, investing the additional $4,000 in A2 yields additional $5,000, which is equivalent to earning at rate 25%. Hence, incremental investment in A2 is justified.

Period Project  A1 Project  A2

0 -­$1,000 -­$5,000

1 $2,000 $7,000

IRR 100% 40%

𝑃𝑊 10% $818 $1,364

Incremental   Investment  (A2-­A1)

-­$4,000

$5,000

25%

INCREMENTAL  ROR  ANALYSIS

51

STEP 1.

Compute incremental cash flows by subtracting

cash flows of the lower investment cost project (A) from

those of the higher investment project (B). [IRR for each

project must exceed the MARR when do-­nothing option

is available]

STEP 2.

Compute

IRR

for incremental investment.

EXAMPLE  11:  INCREMENTAL  ROR  ANALYSIS

IRR

> 10%

&  

IRR

> 10%  

Select B2.

Period Project  _B1 Project  _B2

0 -­$3,000 -­$12,000

1 $1,350 $4,200

2 $1,800 $6,225

3 $1,500 $6,330

IRR 25% 17.43%

EXAMPLE  11:  INCREMENTAL  ROR  ANALYSIS

EXAMPLE  12:  INCREMENTAL  ROR  ANALYSIS

Period Project  C1 Project  C2

0 -­$9,000 -­$9,000

1 $480 $5,800

2 $3,700 $3,250

3 $6,550 $2,000

4 $3,780 $1,561

IRR 18% 20%

MARR  =  12%    C1-­C2  OR  C2-­C1

C1-­C2

$0 -­$5,320

$450

$4,550 $2,219

EXAMPLE  13:  INCREMENTAL  ROR  ANALYSIS

55

MARR  =  15%

Step  1. Eliminate  projects with  IRR < MARR.

Step  2. Compare  D1  &  D2,  IRR_MGWM> = 27.61% > 15%,  select  D1.

Step  3. Compare  D1  &  D3,  IRR_M?WMG = 8.8% < 15%,  select  D1.

Select  D1.

Period Project  D1 Project  D2 Project  D3

0 -­$2,000 -­$1,000 -­$3,000

1 $1,500 $800 $1,500

2 $1,000 $500 $2,000

3 $800 $500 $1,000

EXAMPLE  14:  INCREMENTAL  ROR  ANALYSIS

Items CMS  Option FMS  Option

Annual  O&M  costs

Annual  labor  cost $1,169,600 $707,200

Annual  material  cost 832,320 598,400

Annual  overhead  cost 3,150,000 1,950,000

Annual  tooling  cost 470,000 300,000

Annual  inventory  cost 141,000 31,500

Annual  income  taxes 1,650,000 1,917,000

Total  annual  costs $7,412,920 $5,504,100

Investment $4,500,000 $12,500,000

EXAMPLE  14:  INCREMENTAL  ROR  ANALYSIS

57

Year CMS  Option FMS  Option

0 -­$4,500,000 -­$12,500,000

1 -­$7,412,920 -­$5,504,100

2 -­$7,412,920 -­$5,504,100

3 -­$7,412,920 -­$5,504,100

4 -­$7,412,920 -­$5,504,100

5 -­$7,412,920 -­$5,504,100

6 -­$7,412,920 -­$5,504,100

Salvage $500,000 $1,000,000

Incremental (FMS-­CMS) -­$8,000,000 $1,908,820 $1,908,820 $1,908,820 $1,908,820 $1,908,820 $2,408,820

IRR

= 12.43% < 15%

EXAMPLE  15:  UNEQUAL  PROJECT  LIVES

Year Model  A Model  B

0 -­$12,500 -­$15,000

1 -­$5,000 -­$4,000

2 -­$5,500 -­$4,500

3 -­$6,000+$2,000 -­$5,000

4 -­$5,500+$1,500

5

Required  service  life: 5  years

Lease  Model  A  for  remaining  years  at  annual  cost $11,500

EXAMPLE  15:  UNEQUAL  PROJECT  LIVES

59

Year Model  A Model  B

0 -­$12,500 -­$15,000

1 -­$5,000 -­$4,000

2 -­$5,500 -­$4,500

3 -­$6,000+$2,000 -­$5,000

4 -­$6,500-­$5,000 -­$5,500+$1,500

5 -­$6,500-­$5,000 -­$6,500-­$5,000

IRR

= 45.67% > 15%

(Although  nonsimple,  B-­A  has  a unique ROR, so  ROR  =  IRR.)

SUMMARY

Rate of return

is the interest rate earned on unrecovered

project balances such that an investment’s cash receipts

make the terminal project balance equal to zero.

Since it provides an intuitive & understandable measure

of project profitability, managers typically prefer ROR

over other equivalence measures.

A project’s rate of return can be determined by finding a

break-­even interest rate that equates the present worth

of its cash flows to zero.

SUMMARY

61

The first step in rate-­of-­return analysis is to classify the

investment as either simple or nonsimple.

A

simple investment

involves a negative initial cash flow

& its net cash flow series exhibits only one sign change.

The net cash flow series of a

nonsimple investment

exhibit more than one sign change.

Multiple RORs

occur only in nonsimple investments, but

not all nonsimple investments have multiple RORs.

SUMMARY

For a

nonsimple investment

, because of the possibility of

having multiple rates of return, it is recommended the

IRR analysis be abandoned & either PW or AW analysis

be used to make an accept/reject decision.

Incremental rate-­of-­return analysis

is used to select