fundamentals of engineering economics instructor’s manual

Instructor’s Manual

FUNDAMENTALS OF ENGINEERING

ECONOMICS, Second Edition

Chan S. Park Auburn University

Instructor Solutions Manual to accompany Fundamentals of Engineering Economics, Second Edition, by Chan S. Park.ISBN-13: 9780132209618. © 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.

This material is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storagein a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise.

Contents PREFACE Chapter 1 Engineering Economic Decisions Chapter 2 Time Value of Money Chapter 3 Understanding Money Management Chapter 4 Equivalence Calculations under Inflation Chapter 5 Present-Worth Analysis Chapter 6 Annual Equivalence Analysis Chapter 7 Rate-of-Return Analysis Chapter 8 Accounting for Depreciation and Income Taxes Chapter 9 Project Cash Flow Analysis Chapter 10 Handling Project Uncertainty Chapter 11 Replacement Decisions Chapter 12 Benefit-Cost Analysis Chapter 13 Understanding Financial Statements

ii



Preface This Instructor's Manual to Fundamentals of Engineering Economics, 2nd edition contains detailed solutions to all the end-of-chapter problems (except some open-ended case problems). The problem solutions follow topical headings listed in the main text to indicate the generic content of each problem. The headings are provided to aid you in selecting your preferred mix of problem types for homework assignments. Companion Book Website Two important resources are available from the book’s companion website under “Instructor Recourse Center” at http://www.prenhall.com/park.

• Instructor’ Manual – Each end-of-chapter problems can be downloaded from this website. Each chapter solutions come in two file formats: one in Microsoft WORD file format and the other in PDF format. In creating many mathematical equations the WORD files, I used the MathType™ Equation Editor (version 6.0) from Design Science. If you have not installed this MathType™ on your PC, some of the characters may not be displaced correctly. If that is the case, just open the PDF version the file.

• A comprehensive set of lecture notes in PowerPoint format is also

available to instructors who adopt the text. Solutions Provided in Excel Worksheet

• Fundamentals of Engineering Economics contains nearly 493 problems including 41 short-case assignment problems. Many of them come with multiple-part questions, bringing the total number of questions to nearly 900.

• The problems listed under ``Short Case Studies with Excel" require more involved assumptions, solution steps, and calculations.

• Many of the Excel worksheets in WORD files appear as “Microsoft

Excel Worksheet Object,” so just place the cursor over the worksheet area and double-click will convert them into live Excel worksheet. This will allow you to modify the solution and answer many what-if questions.

Some Deviations in Answers

• Your independent solutions to the text problems may yield answers slightly different from mine due to rounding differences. In some cases, the interpretations and assumptions students bring to the problems may differ from my own in creating them, again resulting in different answers.

• Some of the answers (not the procedures) to the end-of-chapter problems are posted in the book's companion website, so that students can check the answers on their own.

iii



http://www.prenhall.com/park

Having made these disclaimers, I wish to emphasize that it is my and the publisher's intention to provide the most accurate solutions possible. Thus, if you discover any typo errors, or disagree strongly with the interpretation and assumptions required of a particular solution, please do not hesitate to report your concerns to me via email at [email protected] so they may be corrected in subsequent printings on time. I will also plan on posting any errors on the book's web site under "Instructor Resource Center." Chan S. Park Auburn December 2007

iv



mailto:[email protected]

Chapter 1 Engineering Economic Decisions There are no end-of-chapter questions in this introductory chapter. However, the following questions could be added as a part of instruction:

1. Ask students to review the contents of The Wall Street Journal for the past 3 months. Then, identify and categorize the types of investment decisions appeared in the journal according to the types of strategic economics decisions discussed in the text.

2. Work in small groups and brainstorm ideas about how a common appliance,

device or tool could be redesigned to improve it in some way. Identify the steps involved and the economic factors, which you would need to consider prior to making a decision to manufacture the redesigned product. A detailed design and actual cost estimates are not required. Some items, which could be considered for this redesign exercise, are: a shopping cart, telephone, can opener, screwdriver, etc.

3. Many oil price forecasts in the early 2000 indicated that the price of oil in the year

2007 would not exceed $50 per barrel. What is the price of today? Why are these prices so difficult to predict? Imagine what the consequences would be if you used these optimistic estimates in your economic analysis in your early project undertaking. What would be some practical ways to consider this type of variation in economic analysis?

Chapter 2: Time Value of Money 2.1) I = iPN = (0.09)($3,000)(5) = $1,350 2.2)

• Simple interest:

(1 )$4,000 $2,000(1 0.08 )

12.5 years (or 13 years)

F P iNN

N

= += +=

• Compound interest:

$4,000 $2,000(1 0.07)2 1.07

log 2 log 1.0710.24 years (or 11 years)

N

N

NN

= +

===

2.3)


(0.07)($10,000)(20)$14,000

I iPN= ==


I = P (1+ i)N −1⎡⎣ ⎤⎦ = $10,000 (1.07)20 −1⎡⎣ ⎤⎦= $28,696.84

2.4)


5$1,000(1 0.06)$1,338.23

F = +=


$1,000(1 0.07(5))$1,350

F = +=

The simple interest option is better.



2

2.5) • Loan balance calculation:

End of period Principal Payment

Interest Payment

Remaining Balance

0 $0.00 $0.00 $5,000.00 1 $835.46 $450.00 $4,164.54 2 $910.65 $374.81 $3,253.89 3 $992.61 $292.85 $2,261.28 4 $1,081.94 $203.52 $1,179.33 5 $1,179.32 $106.14 $0.00

2.6) $8,000( / ,8%,5) $8,000(0.6806) $5, 444.8P P F= = = 2.7) F = $20,000(F / P,10%,2) = $20,000(1.21) = $24,200 2.8)

• Alternative 1

$100P =

• Alternative 2

P = $120(P / F ,8%,2) = $120(0.8573) = $102.88

• Alternative 2 is preferred 2.9) (a) F = $7,000(F / P,9%,8) = $7,000(1.9926) = $13,948.2

(b) F = $1,250(F / P,4%,12) = $1,250(1.6010) = $2,001.25 (c) F = $5,000(F / P,7%,31) = $5,000(8.1451) = $40,725.5 (d) F = $20,000(F / P,6%,7) = $20,000(1.5036) = $30,072

2.10) (a) P = $4,500(P / F ,7%,6) = $4,500(0.6663) = $2,998.35

(b) P = $6,000(P / F ,8%,15) = $6,000(0.3152) = $1,891.2 (c) P = $20,000(P / F ,9%,5) = $20,000(0.6499) = $12,998 (d) P = $12,000(P / F ,10%,8) = $12,000(0.4665) = $5,598

2



3

2.11) (a) P = $6,000(P / F ,8%,5) = $6,000(0.6806) = $4,083.6 (b) F = $15,000(F / P,8%,4) = $15,000(1.3605) = $20,407.5 2.12)

3 (1 0.07)

log 3 log 1.0716.24 years (or 17 years)

NF P PN

N

= = +==

2.13)

• 2 (1 0.12

log 2 log 1.126.12 years

NF P PN

N

= = +==

)

• Rule of 72:

72 /12 6 years= 2.14)

P = $35,000(P / F ,9%,4) + $10,000(P / F ,9%,2)= $35,000(0.7084) + $10,000(0.8417)= $33,211

2.15)


I = iPN = (0.1)($1,000)(3) = $300


I = P (1+ i)N −1⎡⎣ ⎤⎦ = $1,000 (1+ .095)3 −1⎡⎣ ⎤⎦= $312.93

• Susan’s balance will be greater by $12.93.

2.16) P =

$3,0001.062 +

$3,5001.063 +

$4,0001.064 +

$6,0001.065 = $13,260.58

2.17)

$1,000( / ,8%,10) $1,500( / ,8%,8) $2,000( / ,8%,6)$8,109.05

F F P F PF P

= ++=

3



4

2.18)

$3,000,000 $2, 400,000( / ,8%,5) $3,000,000( / ,8%,5)( / ,8%,5)$20,734,774.86

P PP A P F

= ++=

A

2.19)

$3,000( / ,9%, 2) $4,000( / ,9%,5) $5,000( / ,9%,7)$7,859.7

P P F P FP F

= ++=

2.20)

• Method 1:

F = $2,000(1.05)(1.1)(1.15) + $3,000(1.1)(1.15) + $5,000= $11,451.5

• Method 2:

( )$6,451.50

$5,100

$2,000(1.05) $3,000 (1.10)(1.15) $5,000

$11, 451.50

F = + +

=

2.21)

$150,000 $20,000( / ,9%,5) $10,000( / ,9%,3) ( / ,9%,6)

$134,046.98P A P F X P F

X= − +=

2.22)

F = $80,000 = $10,000(1.08)5 + $12,000(1.08)3 + X (1.08)2

X = $43,029.99 2.23)

100(1.08)4 = 8(1.08)3 + 9(1.08)2 +10(1.08) +11+ XX = $93.67

This is the minimum selling price. So if John can sell the stock for a higher price than $93.67, his return on investment will be higher than 8%.

2.24) (a) F = $3,000(F / A,7%,8) = $3,000(10.2598) = $30,779.4 (b) F = $3,000(F / A,7%,8)(1.07) = $32,933.96

4



5

2.25) (a) F = $5,000(F / A,6%,6) = $5,000(6.9753) = $34,876.5 (b) F = $9,000(F / A,7.25%,9) = $108,928.76 (c) F = $12,000(F / A,8%,25) = $12,000(73.1059) = $877,270.8 (d) F = $6,000(F / A,9.75%,10) = $94,485.71 2.26) (a) A = $15,000( A / F ,5%,13) = $15,000(0.0565) = $847.5

(b) A = $20,000( A / F ,6%,8) = $20,000(0.1010) = $2,020 (c) A = $5,000( A / F ,8%,25) = $5,000(0.0137) = $68.5 (d) A = $4,000( A / F ,6.85%,8) = $391.98

2.27)

( )

$35,000 $3,000( / ,6%, )( / ,6%, ) 11.6666

1 0.06 111.6666

0.06log(1.06) log(1.7)

9.11 years

N

F A NF A N

N

N

==

+ −=

⋅ =

=

2.28)

$10,000 = A(F / A,7%,5)A = $1,738.92

2.29)

F = $500(1.1)10 + $1,000(1.1)8 + $1,000(1.1)6

+$1,000(1.1)4 + $1,000(1.1)2 + $1,000= $8,886.12

2.30) (a) A = $15,000( A / P,8%,5) = $15,000(0.2505) = $3,757.5 (b) A = $3,500( A / P,9.5%,4) = $1,092.22 (c) A = $8,000( A / P,11%,3) = $8,000(0.4092) = $3,273.6 (d) A = $25,000( A / P,6%,20) = $25,000(0.0872) = $2,180 2.31)

• Equal annual payment amount:

A = $20,000( A / P,10%,3) = $20,000(0.4021) = $8,042

• Loan balance calculation:

5



6

End of period Principal Payment

Interest Payment

Remaining Balance

0 $0.00 $0.00 $20,000.00 1 $6,042.00 $2,000.00 $13,958.00 2 $6,646.20 $1,395.80 $7,311.80 3 $7,310.82 $731.18 $0

Interest payment for the second year = $1,395.80

2.32) (a) P = $5,000(P / A,6%,8) = $5,000(6.2098) = $31,049 (b) P = $7,500(P / A,9%,10) = $7,500(6.4177) = $48,132.75 (c) P = $1,500(P / A,7.25%,6) = $7,094.96 (d) P = $9,000(P / A,8.75%,30) = $94,551.83

2.33) (a)

( A / P,6.25%,36) =0.0625 1+ 0.0625( )36

1+ 0.0625( )36−1

= 0.07044

(b) ( )( )

125

125

1 0.0925 1( / ,9.25%,125) 10.81064

0.0925 1 0.0925P A

+ −= =

+

2.34) F = $400(F / A,9%,15)(1.09) = $400(29.3609)(1.09) = $12,801.35 2.35)

F = F1 + F2

= $5,000(F / A,8%,5) + $2,000(F / G,8%,5)= $5,000(F / A,8%,5) + $2,000( A / G,8%,5)(F / A,8%,5)= $5,000(5.8666) + $2,000(1.8465)(5.8666)= $50,998.35

2.36)

$1, 200( / ,9%,5) $200( / ,9%,5)$1, 200( / ,9%,5) $200( / ,9%,5)( / ,9%,5)$1, 200(5.9847) $200(7.1110)(1.5386)$4,993.44

F F A F GF A P G F P

= −= −= −=

6



7

2.37)

P = $100(P / F ,8%,1) + $150(P / F ,8%,3)+$200(P / F ,8%,5) + $250(P / F ,8%,7)+$300(P / F ,8%,9) + $350(P / F ,8%,11)= $793.83

2.38)

A = $15,000 − $3,000( A / G,9%,10)= $15,000 − $3,000(3.7978)= $3,606.6

2.39)

$1, 000( / , 9%,8) $250( / , 9%,8)$1, 000(5.5348) $250(16.8877)$9, 756.73

P P A P G= += +=

2.40)

C(P / G,12%,6) = $800(F / A,12%,4)+[$1,000 − $200(P / G,12%,4)](F / P,12%,4)

C(8.9302) = $800(4.7793)+[$1,000 − $200(4.1273)](1.5735)

C = $458.90 2.41) (a)

( ) ( )( )

1,

7 7

$3,000,000( / 10%,12%,7)

1 1 0.1 1 0.12$3,000,000

0.12 0.1$10,686,037.81

P P A−

= −

− − += ⋅

− −

=

(b) Note that the oil price increases at the annual rate of 5% while the oil production decreases at the annual rate of 10%. Therefore, the annual revenue can be expressed as follows:

1 1

1

1

$30(1 0.05) 100,000(1 0.10)

$3,000,000(0.945)$3,000,000(1 0.055)

n nn

n

n

A − −

−

−

= + −

=

= −

This revenue series is equivalent to a decreasing geometric gradient series with g = -5.5%.

7



8

N An

1 $3,000,000 2 $2,835,000 3 $2,679,075 4 $2,531,726 5 $2,392,481 6 $2,260,894 7 $2,136,545

( ) (( )

)1,

7 7

$3,000,000( / 5.5%,12%,7)

1 1 0.055 1 0.12$3,000,000

0.12 0.055$11,923,948.35

P P A−

= −

− − += ⋅

− −

=

(c) Computing the present worth of the remaining series 4 5 6 7( , , , )A A A A

at the end of period 3 gives

( ) (( )

)1

4 4

$2,531,730( / , 5.5%,12%,4)

1 1 0.055 1 0.12$2,531,730

0.12 0.055$7,134,825.54

P P A−

= −

− − += ⋅

− −

=

2.42)

P = An(1+ i)−n

n=1

20

∑

= (2,000,000)n(1.06)n−1(1.06)−n

n=1

20

∑

= (2,000,000 / 1.06) n(1.061.06n=1

20

∑ )n

= (2,000,000 / 1.06) nn=1

20

∑

= (2,000,000 / 1.06)20(21)

2= $396,226,415.1

2.43) (a) The withdrawal series would be:

8



9

2

3

4

Period Withdrawal11 $3,00012 $3,000(1.06)13 $3,000(1.06)14 $3,000(1.06)15 $3,000(1.06)

Equivalent worth of the withdrawal series at period 10, using i = 8%:

P = $3,000(P / A1,6%,8%,5)

ŹŹ= $3,000 ⋅1− 1+ 0.06( )5 1+ 0.08( )−5

0.08 − 0.06( )ŹŹ= $13,383.92

Assuming that each deposit is made at the end of each year, the following equivalence must be hold:

$13,384 ( / ,8%,10)

14.4866$923.88

A F AA

A

===

(b) Equivalent present worth of the withdrawal series at 6%

15$3,000( / ,6%,6%,5) $3,000 $14,150.94

1 0.06P P A= = =

+

$14,151= A(F / A,6%,10)=13.1808A

A= $1,073.60

2.44)

P = [$100(F / A,10%,8) + $50(F / A,10%,6)+$50(F / A,10%,4)](P / F ,10%,8)= [$100(11.4359) + $50(7.7156)+$50(4.6410)](0.4665)= $821.70

2.45) Select (a).

9



10

2.46)

P(1.1) + $500 = $300(P / F ,10%,2)+$300(P / F ,10%,3) + $800(P / F ,10%,4)= $300(0.8264)+$300(0.7513) + $800(0.6830)

P = $472.46

2.47)

Computing the equivalent worth at period 3 will require only two different types of interest factors.

V1,3 = $120(P / A,10%,5)(F / P,10%,3)

= $120(3.7908)(1.3310)= $605.466

V2,3 = A(P / A,10%,2)(F / P,10%,3) + A(P / A,10%,2)

= A(1.7355)(1.3310) + A(1.7355)= 4.04545A

A = $605.466 / 4.04545= $149.67

2.48)

P1,1 = $200(P / A,10%,4) −100(P / A,10%,2)

= $200(3.1699) −100(1.7355)= 460.43

2,1 ( / ,10%, 4)

(3.1699)4.1699

P X X P AX X

X

= +

= +=

P1,1 = P2,1

$460.43= 4.1699XX = $110.42

2.49)

10



11

1

2

1 2

$50( / ,10%,4) $35( / ,10%,2)( / ,10%,2)$50(3.1699) $35(1.7355)(0.8264)208.6926

( / ,10%,4) ( / ,10%,2)( / ,10%,1)(3.1699) (1.7355)(0.9091)

4.7476

$43.96

P P A P A P F

P C P A C P A P FC C

C

P PC

= += +== += +=

==

2.50)

C(F / A,9%,8) = $5,000(P / A,9%,2)C(11.0285) = $5,000(1.7591)

C = $797.52 2.51) The original cash flow series is

n An

0 $0 1 $800 2 $820 3 $840 4 $860 5 $880 6 $900 7 $920 8 $300 9 $300

10 $300 - $500 2.52)

2C + C(P / A,12%,7)(P / F ,12%,1)= $1,200(P / A,12%,8) − 400(P / A,12%,4)

2C + C(4.5638)(0.8929)= $1,200(4.9676) − 400(3.0373)

11



12

6.075C = $4,746.20C = $781.27

2.53)

200(1.06)(1.08)(1.12)(1.15) (1.08)(1.12)(1.15) $300(1.15)$1000

247.9 1.39104 345 10001.39104 360.1

$258.87

X

XX

X

++=

+ + ==

=

2.54) Computing the equivalent worth at n = 5,

X = $5,000(F / A,10%,5) + $5,000(P / A,10%,5)= $5,000(6.1051) + $5,000(3.7908)= $49,475.5

2.55)

A(F / A,8%,18) = $20,000 + $20,000(P / A,8%,3)A(37.4502) = $20,000 + $20,000(2.5771)

= $71542

A = $1910.32

2.56)

[ ][ ]

1,0

2,0

$500 $500( / ,10%,5)$500 $500(3.7908)$2,395.4

( / ,10%,1) ( / ,10%,4)

(0.9091) (0.6830)1.5921

$1,504.55

P P A

P X P F P F

XX

X

= +

= +=

= +

= +

=

=

12



13

2.57)

P1,2 = X (P / F ,8%,3)

= X (0.7938)

P2,2 = 800(P / A,8%,10)

= 800(6.7101)= 5368.08

X = 6,762.512.58)

( / ,9%,5)( / ,9%,1) $4,000

(3.8897)(0.9174) $4,000$1,120.95

C P A P FC

C

===

2.59)

(1.05)(1.08)(1.1)(1.06)$1,000(1.08)(1.1)(1.06) $1,500(1.1)(1.06)$1,000(1.06) $1000

(1.322244) $5,068.28$3,833.09

P

PP

= ++ +==

2.60)

• Exact:

5

5

2 (1 )2 (1 )

log 2 5 log(1 )14.87%

P P ii

ii

= +

= += +=

• Rule of 72:

72 / i = 5yearsi =14.4%

2.61)

• 1

51

5

$150( / , ,5) $50( / , ,1)

(1 ) 1$150 $50 (1 )(1 )

P P A i P F

i ii i

i

−

= −

⎛ ⎞+ −= −⎜ ⎟+⎝ ⎠

⋅ +

13



14

• 2 2 3 4 5

$200 $150 $50 $200 $50(1 ) (1 ) (1 ) (1 ) (1 )

Pi i i i

= + + + ++ + + + + i

• and solving i with Excel Goal Seek function, P1 = P2

14.96%i = 2.62)

5

$35,000 $10,000( / , ,5)$10,000(1 )28.47%

F P ii

i

=

= +=

2.63) The equivalent future worth of the prize payment series at the end of

Year 20 (or beginning of Year 21) is

F1 = $1,952,381(F / A,6%,20)= $1,952,381(36.7856)= $71,819,506.51

The equivalent future worth of the lottery receipts is

F2 = ($36,100,000 − $1,952,381)(F / P,6%,20)= ($36,100,000 − $1,952,381)(3.2071)= $109,514,828.9

The resulting surplus at the end of Year 20 is

F2 − F1 = $109,514,828.9 − $71,819,506.51= $37,695,322.4

2.64)

$1,000(F / P,9.4%,5) + $500(F / A,9.4%,5)

= $1,000((1+ 0.094)5) + $500((1+ 0.094)5 −1

0.094)

= $1,000(1.5671) + $500(6.0326)= $4,583.4

$4,583.4(F / P,9.4%,60)= $4,583.4((1+ 0.094)60 )= $4,583.4(219.3)= $1,005,141.21

14



15

The main question is whether or not the U.S. government will be able to invest the social security deposits at 9.4% interest over 60 years.

2.65)

Contract $3,875,000 $3,125,000( / ,6%,1) $5,525,000( / ,6%,2) $8,875,000( / ,6%,7)$3,875,000 $2,550,000(0.9434) $5,525,000(0.8900) $8,875,000(0.6651)$39,548,212.5

P PP FP F

= +

+ ++= ++ ++=

F

15



Chapter 3 Understanding Money Management 3.1) (a)

r = 1.5% ×12 = 18%

(b) 12(1 0.015) 1 19.56%ai = + − =

3.2)

• Nominal interest rate: 0.95% 12 11.40%r = × =

• Effective annual interest rate:

12(1 0.0095) 1 12.01%ai = + − =

3.3) Assume a continuous compounding:

0.0755

7.55%7.842%

1 1 0.07842a

ra

ri

i e e

==

= − = − ≈ 3.4) : $400, $26.61, 16 weeks,Given P A N= = =

$400 $26.61( / , ,16)P A i=

Solve by Excel Goal Seek for i = 0.74385% per week

(a) Nominal interest rate:

0.74385% 52 38.6802%r = × = (b) Effective annual interest rate:

52(1 0.0074385) 1 47.0159%ai = + − = 3.5) Effective interest rate per payment period:

$1080 = $1000(1+ i)i = 8% per week

(a) Nominal interest rate:

r = 8% × 52 = 416%



(b) Effective annual interest rate:

ia = (1+ 0.08)52 −1= 5,370.6% 3.6)

$15,000 = $493.93(P / A, i,36)(P / A,i,36) = 30.3686

Use Excel to calculate : i

i = 0.95% per monthr = 0.95×12 = 11.4%

3.7)

$16,000 = $517.78(P / A,i,36)(P / A,i,36) = 30.901155

i = 0.85% per monthr = 0.85×12 = 10.2%

3.8)

$20,000 $922.90( / , , 24)( / , , 24) 21.6708

0.8333%0.8333% 12 10%

P A iP A i

iAPR

==== × =

3.9)

$24,000 = $583.66(P / A,i,48)(P / A,i,48) = 41.1198

i = 0.65%ia = (1+ 0.0065)12 −1= 8.085%

3.10)

a) i = (1+0.0912

)1 −1= 0.75%

b) i = (1+0.0912

)3 −1= 2.267%

c) i = (1+0.0912

)6 −1= 4.585%

d) i = (1+0.0912

)12 −1= 9.381%

2



3.11)

i = (1+

0.0912

)3 −1= 2.267%

3.12)

i = e0.0612 −1= 0.501%

3.13) What will be the amount accumulated by each of these present investments?

(a)

F = $4,500(F / P,4.5%,20) = $10,852.71

(b) $8,500( / , 2%,60) $27,888.76F F P= =

(c)

F = $18,600(F / P,0.5%,84) = $28,278.88 3.14) (a)

$5,000( / , 4%, 20) $148,890.39F F A= =

(b) $9,000( / , 2%, 24) $273,796.76F F A= =

(c)

F = $3,000(F / A,0.75%,168) = $1,003,554.24 3.15) (a)

A = $15,000( A / F ,4%,20) = $504

(b) A = $2,000( A / F ,1.5%,60) = $20.79

(c)

A = $48,000( A / F ,0.6125%,60) = $664.4 3.16) (a)

P = $1,000(P / A,4.5%,20) = $13,007.94

(b) P = $7,000(P / A,2%,20) = $114,459.8

3



(c)

P = $6,000(P / A,0.75%,96) = $409,550.63 3.17)

• Equivalent future worth of the receipts:

FW = $1,500(F / P,2%,2) + $1,500(F / P,2%,4) + $1,500(F / P,2%,6) + $2,500= $7,373.5

• Equivalent future worth of deposits:

FD = C(F / A,2%,8) +C(F / P,2%,8)= 9.7547C

Letting and solving for C yields WF F= D

$755.89C =

3.18) (d)

3quarter

0.12(1 ) 1 3.03% per quarter12

i = + − =

Effective interest rate per payment period i = (1 + 0.01)3 – 1 = 3.03%

0 1 2 3 4 5 6 7 8 9 10 11 12

$1,000

4



3.19) (d) 3.20)

A = $50,000( A / F ,0.5%,24)= $1,966.03

3.21)

• The balance just before the transfer:

F9 = $3,000(F / P,0.5%,108) + $4,000(F / P,0.5%,72)+$6,000(F / P,0.5%,48)= $18,492.21

Therefore, the remaining balance after the transfer will be

$18,492.21× (

12

) = $9,246.1. This remaining balance will continue to grow at 6%

interest compounded monthly. Then, the balance 6years after the transfer will be

F15 = $9,246.11(F / P,0.5%,72)= $13,240.84

• The funds transferred to another account will earn 8% interest compounded

quarterly. The resulting balance 6 years after the transfer will be

F15 = $9,246.11(F / P,2%,24)= $14,871.79

3.22) Establish the cash flow equivalence at the end of 25 years. Referring A to his quarterly deposit amount, we obtain the following:

40.08(1 ) 1 8.243%4

( / , 2%,100) $53,000( / ,8.243%,10)312.2323 $351,769.13

$1,126.63

ai

A F A P AAA

= + − =

===

5



3.23) $100,000 $1,000( / ,9% /12, )

( / ,0.75%, ) 100P A N

P A N==

N = 185.53 months or 15.46 years

3.24) Given: r = 6% per year compounded quarterly, N = 60 quarterly deposits, date of last deposit = date of first withdrawal of $50,000, four withdrawals. We can calculate i =

1.5% per quarter compounded quarterly and ia = (1+0.06

4)4 −1= 6.136% . To find A,

the amount of quarterly deposit,

( / ,1.5%,60) $50,000 $50,000( / ,6.136%,3)$183,314 / 96.2147

A F A P AA= +=

= $1,905.26 3.25) Setting the equivalence relationship at the end of 15 years gives

isa = (1+0.082 ⋅2

)2 −1= 4.04%

A(F / A,2%,60) = $45,000(P / A,4.04%,10)114.0515A = $364,266

A = $364,266 / 114.0515= 3,193.87

3.26) Given i = 6%/12 = 0.5% per month,

$500,000( / ,0.5%,60)A A P= = $9,650

3.27) First compute the equivalent present worth of the energy cost during the first operating cycle:

P = $50(P / A,0.75%,3)(P / F ,0.75%,1) + $80(P / A,0.75%,3)(P / F ,0.75%,7)= $371.08

Then, compute the total present worth of the energy cost over 5 operating cycles.

P = $371.08 + $371.08(P / F ,0.75%,12) + $371.08(P / F ,0.75%,24)+$371.08(P / F ,0.75%,36) + $371.08(P / F ,0.75%,48)= $1,563.27

6



3.28)

• Option 1

1.06(1 ) 1 1.5%4

$1,000( / ,1.5%,40)( / ,1.5%,60) $132,587

i

F F A F P

= + − =

= =

• Option 2

4.06(1 ) 1 6.136%4

$6,000( / ,6.136%,15) $141,110

i

F F A

= + − =

= =

• Option 2 – Option 1 = $141,110 – 132,587 = $8,523 • Select (b)

3.29) Given: r = 7% compounded daily, N = 25 years

• Since deposits are made at year end, find the effective annual interest rate:

365(1 0.07 / 365) 1 7.25%ai = + − =

• Then, find the total amount accumulated at the end of 25 years:

F = $3,250(F / A,7.25%,25) + $150(F / G,7.25%,25)= $3,250(F / A,7.25%,25) + $150(P / G,7.25%,25)(F / P,7.25%,25)= $297,016.95

0 1 2 3 4 5 6 7 8 9 10 11 12 May June July Aug. Sept. Oct. Nov. Dec Jan. Feb. Mar. Apr.

$50 $50 $50 $80 $80 $80

7



3.30) Given:

3 (1 )log3 log(1 )

log3/ log(1 )

NiN i

N i

= += += +

(a)

i = (1+ 0.0225)4 −1= 9.31% : N = 12.34 years

(b)

i = (1+ 0.09 / 12)12 −1= 9.38% : N = 12.25 years

(c)

i = e0.09 −1= 9.42% : N = 12.21 years 3.31) (a)

iq = (1+0.09

4)1 −1= 2.25%

P = $3,000(P / A,2.25%,60) = $98,247

(b)

iq = (1+0.094 ⋅3

)3 −1= 2.2669%

P = $3,000(P / A,2.2669%,60) = $97,857.9

(c)

iq = e0.09

4 −1= 2.2755%

P = $3,000(P / A,2.2755%,60) = $97,661.1

3.32)

i = e0.07 −1= 7.251%F = A(F / A,i, N )= $2,000(F / A,7.251%,8)= $20,706

3.33) Given: A = $1,000, N = 80 quarters, r = 8% per year

(a)

F = $1,500(F / A,2%,80) = $290,658

8



(b)

F = $1,500(F / A,2.0133%,80) = $292,546.5 (c)

F = $1,500(F / A,2.020%,80) = $293,503.35

3.34)

i = e0.085/ 4 −1= 2.1477%A = $15,000( A / P,2.1477%,16)= $1,117.5

3.35) (a)

F = $5,000(F / A,0.7444%,72)= $474,014.38

(b)

F = $5,000(F / A,0.75%,72)= $475,035.14

(c)

F = $5,000(F / A,0.75282%,72)= $475,550.21

3.36) Nominal interest rate per quarter = 8%/4 = 2%

Effective interest rate per quarter 0.02 1 2.020%e= − =

A = $20,000( A / P,2.020%,20) = $1,226

3.37)

i = e0.0975/ 4 −1= 2.4675%P = $1,500(P / A,2.4675%,20) = $23,455.65

3.38) Equivalent present worth of the series of equal quarterly payments of $3,000 over 10 years at 8% compounded continuously: i = e0.02 −1= 2.02013%

$3,000(P / A,2.02013%,40) = $81,777.6

9



Equivalent future worth of $81,777.6 at the end of 15 years:

0.08

15

1 8.3287%$81,777.6( / ,8.3287%,15) $271,511

ai eV F P

= − == =

3.39)

• Effective interest rate for Bank A

i = (1+

0.184

)4 −1= 19.252%

• Effective interest rate for Bank B

i = (1+

0.175365

)365 −1= 19.119%

• Select (c) 3.40) (a)

• Bank A: 12(1 0.0155) 1 20.27% per yearai = + − =

• Bank B: ia = (1+ 0.195 / 12)12 −1= 21.34% per year

(b) Given , find the total cost of credit card usage for each bank over 24 months. We first need to find the effective interest rate per payment period (month—30 days per month):

6% / 365 0.01644% per dayi = =

i = (1+ 0.0001644)30 −1= 0.494%

• Monthly interest payment:

Bank A: $300(0.0155) = $4.65/month

Bank B: $300(0.19512

) = $4.875/month

We also assume that the $300 remaining balance will be paid off at the end of 24 months.

• Bank A:

P = $20 + $4.65(P / A,0.494%,24) + $20(P / F ,0.494%,12)= $143.85

• Bank B: P = $4.13(P / A,0.494%,24) = $93.25

Select Bank B

10



3.41) Loan repayment schedule: $20,000( / ,0.75%, 48) $497.90A A P= =

End of month

Interest Payment

Repayment of Principal

Remaining Balance

0 $0.00 $0.00 $20,000.00 1 $150.00 $347.90 $19,652.10 2 $147.39 $350.51 $19,301.59 3 $144.76 $353.14 $18,948.45 4 $142.11 $355.79 $18,592.67 5 $139.44 $358.46 $18,234.21 6 $136.76 $361.14 $17,873.07

3.42) Given: P = $150,000, N = 360 months, i = 0.75% per month

(a)

A = $150,000( A / P,0.75%,360)= $1,200

(b) If r = 9.75% APR after 5 years, we want to find new annual amount A:

i = 0.8125% per month. First, find the remaining balance at the end of 60 months:

B60 = $1,200(P / A,0.75%,300)= $142,993.92

Then, find the new monthly payments:

$142,993.92( / ,0.8125%,300)$1, 274.27

A A P==

3.43) (a) 1. $1 4,000( / ,0.75%, 24)A P

(b) 3. 12 ( / ,0.75%,12)B A P A=

3.44) Based on effective monthly compounding-Given i = 9.25%/365= 0.02534% per day,

and N = 48 months:

11



i = (1+ 0.0002534)30 −1= 0.763075%

A = $7,000( A / P,0.763075%,48)= $175 per month

I = $175× 48 − $7,000= $1,400

3.45) Given: P = $15,000, r = 9% per year compounded monthly, N= 36 months, i = 0.75% per month:

A = P( A / P,0.75%,36)= $15,000(0.0318)= $477

To find payoff balance immediately after 20th payment:

B20 = $477(P / A,0.75%,16)= $477(15.0243)= $7,166.59

3.46) Given i = 8.5%/12 per month, and N = 180 months,

$210,000( / ,0.7083%,180)$2,067.90

A A P==

3.47) Given: P = $350,000, N = 240 months, i = 0.75% per month:

A = $350,000( A / P,0.75%,240)= $3,150

• Total payment:

$3,150 × 60 = $189,000 • Remaining balance at the end of 5 years (60 months):

$3,150(P / A,0.75%,180) = $310,569.21 • Reduction in principal:

$350,000 − $310,569.21= $39,430.79

12



• Interest payment:

$189,000 − $39,430.79 = $149,569.21 3.48) Given: purchase price = $300,000, down payment = $45,000, N = 360 months, and

i = 0.75% per month:

A = $255,000( A / P,0.75%,360)= $2,051.79

To find minimum acceptable monthly salary:

Monthly salary =A

0.25

=$2,051.79

0.25= $8,207.16

3.49) Given: purchase price = $180,000, down payment (sunk equity) = $30,000, i = 0.75%

per month, and N = 360 months,

• Monthly payment:

A = $150,000( A / P,0.75%,360)= $1,200

• Balance at the end of 5 years (60 months): :

60 $1, 200( / ,0.75%,300)$142,993.92

B P A=

=

• Realized equity = sales price – balance remaining – sunk equity:

$205,000 − $142,993.92 − $30,000 = $32,006.1 The $32,006.1 represents the net gains (before tax) from the transaction.

3.50) Given: i = 0.75% per month, mortgages’ for families A, B and C have identical

remaining balances prior to the 20th payment = $100,000, find interest on 20th payment for A, B, and C. With equal balances, all will pay the same interest.

13



$100,000(0.0075) = $750 3.51) Given: loan amount = $130,000, points charged = 3%, N = 360 months, i = 0.75% per month, actual amount loaned $130,000(0.97) = $126,100:

A = $130,000( A / P,0.75%,360)= $1,040

To find the effective interest rate on this loan

$126,100 = $1,040(P / A, i,360)i = 0.7732% per monthr = 0.7732% ×12 = 9.2784%ia = (1+ 0.007732)12 −1= 9.683% per year

3.52) (a)

$44,000 = $6,600(P / A,i,5) + $2,200(P / G,i,5)i = 6.913745%

(b)

Amount borrowed = $44,000Total payment made = $6,600 + $8,800 + $11,000

+$13,200 + $15,400= $55,000

Interest payment = $55,000 − $44,000= $11,000

Period Beginning Balance Interest Payment Repayment

Ending Balance

1 $44,000.00 $3,042.05 ($6,600.00) $40,442.05 2 $40,442.05 $2,796.06 ($8,800.00) $34,438.11 3 $34,438.11 $2,380.96 ($11,000.00) $25,819.08 4 $25,819.08 $1,785.07 ($13,200.00) $14,404.14 5 $14,404.14 $995.87 ($15,400.00) $0.00 $11,000.01 ($55,000.00)

3.53) (a)

Amount of dealer financing = $18,400(0.90) = $16,560

14



$16,560( / ,1.125%, 48) $448.38A A P= = (b)

Assuming that the remaining balance will be financed over 44 months,

4 $448.38( / ,1.125%,44) $15,493.71$15,493.71( / ,1.02083%,44) $438.88

B P AA A P

= == =

(c)

• Interest payment to the dealer:

dealer $448.38 4 ($16,560 $15,493.71) $727.23I = × − − =

• Interest payment to the credit union:

credit

Total payment $438.88(44) $19,310.72$19,310.72 $15, 493.71 $3,817.01I

= == − =

• Total interest payment:

$727.23 $3,817.01 $4,544.24I = + =

3.54) • The monthly payment to the bank: Deferring the loan payment for 6 months is equivalent to borrowing

$4,800(F / P,1%,6) = $5,095.20

To pay off the bank loan over 36 months, the monthly payment would be

A = $5,095.20( A / P,1%,36) = $169.16 per month

• The remaining balance after making the 16th payment:

B16 = $169.16(P / A,1%,20) = $3,052.59

• The loan company will pay off this remaining balance and will charge $104 per month for 36 months. To find the effective interest rate for this new transaction, we set up the following equivalence relationship and solve for i:

15



$3,052.59 = $104(P / A,i,36)(P / A, i,36) = 29.3518

i = 1.1481%r = 1.1481% ×12 = 13.78% per year

ia = 1+0.1378

12⎛⎝⎜

⎞⎠⎟

12

−1= 14.68%

3.55)

$15,000 = A(P / A,0.667%,12) + A(P / A,0.75%,12)(P / F ,0.667%,12)= A(11.4958) + A(11.4349)(0.9234)= 22.05479A

A = $680.12

3.56) Given: i = 1% per month, deferred period = 6 months, N = 36 monthly payments, first payment due at end of month 7, the amount of initial loan = $12,000

(a) Find the monthly payment to the furniture store: first, find the loan adjustment for

deferred period

$12,000(F / P,1%,6) = $12,738

Find the monthly payments based on this adjusted loan amount

A = $12,738( A / P,1%,36) = $422.90

(b) Find the remaining balance after the 26th payment. Since there are 10 payments outstanding,

B26 = $422.90(P / A,1%,10) = $4,005.41

(c) Find the effective interest rate:

$4,005.41= $204(P / A, i,30)i = 2.9866% per monthr = 2.9866% ×12 = 35.84% per yearia = (1+ 0.029866)12 −1= 42.35% per year

3.57) Given: Purchase price = $18,000, down payment = $1,800, monthly payment (dealer financing) = $421.85, N = 48 end-of-month payments:

(a) Given: i = 11.75%/12 = 0.97917% per month

16



A = $16,200( A / P,0.97917%,48)= $16,200(0.0262)= $424.44

(b) Using dealer financing, find i:

$421.85 = $16,200( A / P, i,48)i = 0.95% per monthr = 0.95% ×12 = 11.4% per year

ia = 1+0.114

12⎛⎝⎜

⎞⎠⎟

12

−1= 12.015%

3.58)

• 24-month lease plan:

P = ($2,500 + $520) + $500 + $520(P / A,0.5%,23)−$500(P / F ,0.5%,24)= $13,884.13

• Up-front lease plan:

P = $12,780 + $500 − $500(P / F ,0.5%,24)= $12,836.4

Select the single up-front lease plan.

3.59) Given: purchase price = $85,000, down payment = $17,000

• Option 1: i = 10%/12= 0.8333% per month, N =360 months • Option 2: For the assumed mortgage,

1 1$35,394, 8.5% /12 0.70833% per month,P i= = =

1 1300 months, $285 per month;N A= = For the second mortgage,

2 2 2$32,606, 1% per month; 120 monthsP i N= = =

A2 = $32,606( A / P,1%,120) = $466.27

(a) For the second mortgage, the monthly payment will be

17



$68,000 = $285(P / A, i,300) + $466.27(P / A, i,120)i = 0.805% per monthr = 0.805% ×12 = 9.66% per yearia = (1+ 0.00805)12 −1= 10.10% per year

(b) Monthly payments:

• Option 1:

1 $68,000( / ,0.8333%,360) $596.75A A P= = • Option 2:

$285 + $466.27= $751.27 for 120 months, then $285 for remaining 180 months

(c) Total interest payment for each option:

• For Option 1: $146,826.99 • For Option 2: $50,108.14 + $23,529.22 = $73,637.36

(d) Equivalent interest rate:

$596.27( / , ,360) $285( / , ,300) $466.27( / , ,120)

0.9114% per month0.9114% 1210.9368% per year

P A i P A i P A iir

= +== ×=

12(1 0.009114) 1 11.50% per yearai = + − =3.60) No answers given, but refer to the article by Formato, Richard A., "Generalized

Formula for the Periodic Payment in a Skip Payment Loan with Arbitrary Skips," The Engineering Economist, Vol. 37, No. 4; p. 355, Summer 1992

3.61) If you left the $15,000 in your savings account, the total balance at the end of 48

months at 8% interest compounded monthly would be

$15,000( / ,8%/12,48) $20,635IF F P= =

The earned interest during this period is then

$20,635 $15,000 $5,635I = − =

Now if you borrowed $15,000 from the dealer at interest 11% compounded monthly over 48 months, the monthly payment would be

18



$15,000( / ,11%/12, 48) $388A A P= =

You can easily find the total interest payment over 48 months under this financing by

($388 48) $15,000 $3,624I = × − =

It appears that you save about $2,011 in interest ($5,635 - $3,624). However, reasoning this line neglects the time value of money for the portion of principal payments. Since your money is worth 8%/12 interest per month, you may calculate the total equivalent loan payment over the 48-month period. This is done by calculating the equivalent future worth of the loan payment series.

$388( / ,8%/12,48) $21,863.77IIF F A= =

Now compare with . The dealer financing would cost $1,229 more in future dollars at the end of the loan period.

IF IIF

3.62) (a) $60,000( / ,13% /12,360) $664A A P= =

(b)

$60,000 = $522.95(P / A,i,12)+$548.21(P / A, i,12)(P / F , i,12)+$574.62(P / A, i,12)(P / F , i,24)+$602.23(P / A,i,12)(P / F ,i,36)+$631.09(P / A, i,12)(P / F , i,48)+$661.24(P / A, i,300)(P / F ,i,60)

Solving for i by trial and error yields

i = 1.0028%ia = (1+ 0.010028)12 −1= 12.72%

Comments: With Excel, you may enter the loan payment series and use the IRR(range, guess) function to find the effective interest rate. Assuming that the loan amount (-$60,000) is entered in cell A1 and the following loan repayment series in cells A2 through A361, the effective interest rate is found with a guessed value of 11.5/12%:

( 1: 361,0.95833%) 0.010028IRR A A= =

(c) Compute the mortgage balance at the end of 5 years:

19



• Conventional mortgage:

60 $664( / ,13% /12,300) $58,873.84B P A= =

• FHA mortgage (not including the mortgage insurance):

60 $635.28( / ,11.5% /12,300) $62,498.71B P A= =

(d) Compute the total interest payment for each option:

• Conventional mortgage(using either Excel or Loan Analysis Program at the book’s website—http://www.prenhall.com/park):

$178,937.97I =

• FHA mortgage: $163,583.28I =

(e) Compute the equivalent present worth cost for each option at

per month: 6% /12 0.5%i = =

• Conventional mortgage:

$664( / ,0.5%,360) $110,749.63P P A= = • FHA mortgage including mortgage insurance:

P = $522.95(P / A,0.5%,12)+$548.21(P / A,0.5%,12)(P / F ,0.5%,12)+$574.62(P / A,0.5%,12)(P / F ,0.5%,24)+$602.23(P / A,0.5%,12)(P / F ,0.5%,36)+$631.09(P / A,0.5%,12)(P / F ,0.5%,48)+$661.24(P / A,0.5%,300)(P / F ,0.5%,60)= $105,703.95

The FHA option is more desirable (least cost).

3.63) Given: Contract amount = $4,909, A = $142.45, N = 42 months, and SUM = (42)(43)/2 = 903

(a)

$142.45 = $4,909( A / P,i,42)i = 0.9555% per monthia = (1+ 0.009555)12 −1= 12.088% per year

(b)

APR = 0.9555% ×12 = 11.466%

20



(c) Rebate factor:

42 41 40 35rebate factor 1

9031 308 / 9030.6589

+ + + += −

= −=

(d) Verify the payoff using the Rule of 78th :

B7 = $4,909 + $25− ($142.45)(7)

+$1,048.90(42 + 41+ + 35)

903= $4,934 − $997.15+ $357.76= $4,294.61

(e) Compute payoff using ( / , , ) :P A i N

B7 = $142.45(P / A,0.9555%,35)= $4,220.78

3.64)

No answers given, but refer to the website and the document http://www.studentaid.ed.gov/students/attachments/funding/PerkinsLoanInfo.pdf.

21



Chapter 4 Equivalence Calculations under Inflation 4.1)

1.1(1+ f )11 = 3.15f = 10.04%

100(1+ 0.1004)11 = 286.45

4.2) (a)

5144.5(1 ) 170.63.3766%

ff

+ ==

(b) 170.6(1+ 0.033766)2 = 182.32

4.3)

100(1+ 0.05)(1+ 0.08) = 113.40100(F / P, f ,2) = 113.40

f = 6.4894%

4.4)

f1 =538,400 − 504,000

504,000= 6.825%

f2 =577,000 − 538,400

538,400= 7.169%

f3 =629,500 − 577,000

577,000= 9.099%

f =629,500504,000

⎛⎝⎜

⎞⎠⎟

1/3

−1= 7.69%



4.5)

Given : 7%f =

1(1 0.07) 21.07 2

( ) log1.07 log 2log 2 / log1.0710.24 years

N

N

NN

+ =

====

Comments: If you use the Rule of 72, you may find 72 10.29 years7=

which is very close to the actual value. 4.6)

5

5

$70,000 $55,000(1 )(1 ) 1.27273

4.94%

ff

f

= +

+ =

=

4.7) Given: 12%, 5%,i f= = 10 annuity payments in actual dollars

P = $8,500(P / A,12%,10)= $48,026.7

Comments: Since the annuity payments are made in actual dollars, we use the market interest rate to find its equivalent lump sum amount in today’s dollars.

4.8) Given: 15%, 8%,i f= = maintenance costs are given in constant dollars,

0.15 0.08' 61 0.08

i −= =

+.48%

2



$25,000( / ,6.48%,1) $30,000( / ,6.28%, 2)$32,000( / ,6.48%,3) $35,000( / ,6.48%, 4)

$40,000( / ,6.48%,5)$132,894$132,894( / ,15%,5)$39,644.29

P P F P FP F P FP F

A A P

= ++ ++===

4.9) Given: 16%, 4%i f= =

n Actual dollars Constant Dollars 0 $2,500 $2,500(P/F,4%,0) = $2,500 4 $4,500 $4,500(P/F,4%,4) = $3,846.6 5 $3,500 $3,500(P/F,4%,5) =$ 2,876.65 7 $5,500 $5,500(P/F,4%,7) = $4,179.45

4.10) per month, Given:P = $12,000, i = 1% 0.5%f = per month

• 20th payment in actual dollars:

20

20

A = $12,000( A / P,1%,48) = $315.6

• 20th payment in constant dollars:

A ' = $315.6(P / F ,0.5%,20) = $285.65

4.11) Given: 13%, 7%i f= =

(a) Constant-dollar analysis: we need to find the inflation-free interest rate.

' 5.607%1i fi

f−

= =+

Then, find the equivalent present worth of this geometric series at ' . i

P = $15,000(P / A1,8%,5.607%,4)= $58,774.83

(b) Actual-dollar analysis

3



Period Net Cash Flowin Constant $

Conversion factor

Net Cash Flow in Actual $

1 $15,000 1(1 0.07)+ $16,050 2 $16,200 2(1 0.07)+ $18,547.4 3 $17,496 3(1 0.07)+ $21,433.35 4 $18,895.68 4(1 0.07)+ $24,768.38

$16,050( / ,13%,1) $18,547.4( / ,13%, 2)$21, 433.35( / ,13%,3) $24,768.38( / ,13%, 4)$58,774.6

P P F P FP F P F

= ++ +=

Comments: As an alternative way of finding the equivalent cash flows in actual dollars, we may use the compound growth rate (geometric growth and inflation):

g = (1+ 0.08)(1+ 0.07) −1= 15.56%

P = $15,000(1.07)(P / A1,15.56%,13%,4)= $58,774.16

4.12) Given: 9%, 3.8%i f= = , we find the inflation-free interest rate as follows:

' (0.09 0.038) /(1 0.038) 5.01%i = − + = First compute the equivalent present worth of the constant dollar series at : 'i

$1,000( / ,5.01%,3)$2,722.74

P P A==

Then, we compute the equivalent equal annual payment in actual dollars using i: $2,722.74( / ,9%,3)$1,075.63

A A P==

4.13) Given: 12%, 6%i f= = , bond interest rate = 9% compounded semiannually,

face value = $1,000

• The 16th interest payment in actual dollars:

16 $1,000(0.045) $45I = =

4



• The 16th interest payment (8th year) in constant dollars:

16' $45( / ,6%,8) $28.23I P F= =

4.14) Given: ′i = 4%, f = 5%

i = 0.04 + 0.05+ (0.04)(0.05) = 0.092

$30,000(1+ f )5(P / F ,9.2%,5)= $30,000(1+0.05)5(0.644)= $24,657.8

4.15) per month, Given: 1%i = 0.5%f = per month, P = $20,000, N = 60 months

0.01 0.005'1 0.005

0.4975%' $20,000( / ,0.4975%,60)

$386.38

i

A A P

−=

+===

4.16) Given: ' 6%, 5%, 5 years, $1.5i f N A= = = = million in constant dollars

• Market interest rate: 0.06 0.05 (0.06)(0.05) 11.3%i = + + =

• Actual dollar analysis:

Period Net Cash Flow

in Constant $

Net Cash Flow in Actual $

Equivalent Present Worth

1 $1,500,000 $1,575,000 $1,415,094 2 $1,500,000 $1,653,750 $1,334,995 3 $1,500,000 $1,736,438 $1,259,429 4 $1,500,000 $1,823,259 $1,188,140 5 $1,500,000 $1,914,422 $1,120,887

Total $6,318,545

5



$1,575,000( / ,11.3%,1)$1,914, 422( / ,11.3%,5)

$6,318,545

P P FP F

=+ +=

4.17) per month, Given: 0.75%i = 0.5%f = per month, P = $5,000, N = 24 months,

down payment = $1,000

(a) Inflation-free interest rate: 0.0075 0.005' 0.2488% per month

1 0.005i −= =

+

(b) Equal monthly payment in constant dollars:

' $5,000( / ,0.2488%,24)

$214.87A A P==

4.18) compounded monthly, Given: 6%i = 5%f = compounded annually, number

of months to deposit = 240 months, number of annual withdrawals = 15, first withdrawal = 6 months after retirement

• Effective inflation rate per half-year: Since the first withdrawal is made after 6 months from retirement, it is necessary to calculate the effective inflation rate per half-year.

f = 1+

0.052 ⋅ (1 / 2)

⎛⎝⎜

⎞⎠⎟

1/ 2

−1= 2.4695% per half-year

• Annual withdrawals in actual dollars: On a semiannual basis, the first withdrawal will be made after 41 semiannual periods. Then, we can calculate the equivalent amount of this first withdrawal in actual dollars as follows:

A = $40,000(F / P,2.4695%,41) = $108,75241

43

The second withdrawal will be made after 43 semiannual periods. The equivalent amount of this second withdrawal in actual dollars is

A = $40,000(F / P,2.4695%,43) = $114,192

6



The remaining withdrawals in actual dollars are

69

A45 = $40,000(F / P,2.4695%,45) = $119,990A47 = $40,000(F / P,2.4695%,47) = $125,895A49 = $40,000(F / P,2.4695%,49) = $132,189A51 = $40,000(F / P,2.4695%,51) = $138,799A53 = $40,000(F / P,2.4695%,53) = $145,739A55 = $40,000(F / P,2.4695%,55) = $153,026A57 = $40,000(F / P,2.4695%,57) = $160,677A59 = $40,000(F / P,2.4695%,59) = $168,711A61 = $40,000(F / P,2.4695%,61) = $177,146A63 = $40,000(F / P,2.4695%,63) = $186,003A65 = $40,000(F / P,2.4695%,65) = $195,304A67 = $40,000(F / P,2.4695%,67) = $205,069A = $40,000(F / P,2.4695%,69) = $215,322

• Equivalence calculation: To find the required equal monthly deposit amount (A), we establish the following equivalence relationship:

ia = 1+

0.061⋅ (12)

⎛⎝⎜

⎞⎠⎟

12

−1= 6.168% per year

( / ,0.5%, 240)( / ,0.5%,6) $108,752

$114,192( / ,6.168%,1)$119,900( / ,6.168%, 2)

A F A F PP FP F

=++

$215,322( / ,6.168%,14)

$1,511,533.13,174.91 per month

P F

A

+==

4.19) Gi per quarter,ven: 2%i = 6%f = per year

(a) • Actual dollar analysis:

7



A(F / A,2%,160) = $600,000(F / P,6%,40)= $6,171,431

A = $5,420.69

(b) • Effective annual interest rate:

4(1 0.08 / 4) 1 8.243%ai = + − =

• Equivalent value of $600,000 in actual dollars at the end of 63rd birthday:

$600,000( / ,6%, 40) $6,171, 431F P =

• Conversion of gradient series to equivalent uniform series:

A = G( A / G,8.243%,40)= $1,000(10.3745)= $10,374

• Amount of the first deposit : 1( )A

( A1 + $10,374)(F / A,8.243%,40) = $6,171,431

A1 = $11,968.6

4.20)

i = ′i + f + ′i f = 0.06 + 0.05+ 0.06(0.05) = 0.113

A(F / A,11.3%,8) = 40,000(P / A,11.3%,4) +1,000(P / G,11.3%,4)⎡⎣ ⎤⎦(F / P,11.3%,1)

11.9897 A = $141,930.65

A = $11,837.72

4.21) per year,Given: 8%i = 6%f = per year

(a) Freshman-year expense in actual dollars:

$40,000(F / P,6%,10) = $71,632

8



(b) Equivalent single-sum amount at 0n =

i ' =i − f1+ f

= (0.08 − 0.06) / (1+ 0.06)= 0.01887

P = [$40,000(P / A,1.887%,3)+$40,000](P / F ,1.887%,10)= $129,076.84

(c) Required annual deposit in actual dollars:

A = $129,076.84( A / P,8%,10) = $19,236.2

4.22) (a) The average annual general inflation rate:

(1+ 0.065)(1+ 0.077)(1+ 0.081) = 1.2399

(1+ f )3 = 1.2399

f = 7.4308%

(b) Constant dollars: n Actual

dollars Constant dollars

0 -$45,000 -$45,000 1 $26,000 $26,000(0.9390) = $24,414 2 $26,000 $26,000(0.8718) = $22,667 3 $26,000 $26,000(0.8065) = $20,969

Conversion factors: ( / ,6.5%,1) 0.9390

( / ,7.7%,1)( / ,6.5%,1) 0.8718( / ,8.1%,1)( / ,7.7%,1)( / ,6.5%,1) 0.8065

P FP F P F

P F P F P F

===

(c)

P = −$45,000 + $24,414(P / F ,5%,1)+$22,667(P / F ,5%,2) + $20,969(P / F ,5%,3)= $16,923.88

9



4.23) L et be the effective interest rate per month. Then,i

( )

12

12

1/12

(1 ) 1 0.0677(1 ) 1 0.0677

1 0.0677 10.5474%

ii

i

+ − =

+ = +

= + −

=

P = $10,000 +$100(P / A,0.5474%,480) = $26,938.67 4.24) To find the long-term average tuition inflation rate, we first need to find out what the tuition would be in 2005 using year 1978 as a base period. As shown in the following table, it will cost almost eight (7.9296) times higher than year 1978.

Year Tuition CPI Compound (n ) (%) Factor

2005 7.46 7.92962004 9.46 7.37912003 8.4 6.74142002 6.82 6.21902001 5.09 5.82192000 4.14 5.53991999 3.98 5.31971998 4.22 5.11611997 5.11 4.90891996 5.66 4.67031995 6 4.42011994 6.98 4.16991993 9.37 3.89781992 10.74 3.56391991 10.17 3.21831990 8.09 2.92121989 7.93 2.70251988 7.6 2.50401987 7.56 2.32711986 8.09 2.16361985 9.1 2.00161984 10.23 1.83471983 10.41 1.66441982 13.44 1.50751981 12.43 1.32891980 9.43 1.18201979 8.01 1.08011978 Base 1.0000

10



That is, 277.9296 1(1 )

7.97%f

f= +=

If a baby born in 2005 goes to college at the age of 18, the expected college tuition each school year is as follows:

Freshman: $18,000( / ,7.97%,18) $71,570

Sophomore: $18,000( / ,7.97%,19) $77, 274Junior: $18,000( / ,7.97%, 20) $83, 432Senior: $18,000( / ,7.97%, 21) $91,600

F PF PF PF P

====

There are many ways to meet the future college expenses. One of the options is to consider opening a mutual fund account and make regular contribution, say monthly, until the child reaches 18. Let’s assume that the mutual fund would grow at an 8% annual compound return. Then, we may be able to estimate the required monthly contribution (C) as follows:

18 $71,570 $77,274( / ,8%,1)$83,432( / ,8%,2) $91,600( / ,8%,3)$287,365

V P FP F P F

= +

+ +=

8%( / , , 216) $287,36512

$287,365480.0861$598.57

C F A

C

=

=

=

Since there is no way of knowing that the mutual fund will generate an 8% return over 18 years, it would be a good idea to increase the monthly contributions regularly to meet potential shortfalls in case the fund does not perform as expected.

4.25)

(a) Real after-tax yield on bond investment: • Nontaxable municipal bond:

municipal0.09 0.03' 5.825%

1 0.03i −

= =+

11



• Taxable corporate bond:

i 'corporate =

0.12(1− 0.3) − 0.031+ 0.03

= 5.243%

The municipal bond provides a greater return on investment. (b)

Given : 6%,i = and 3%f = ,

savings' 2.91i = %

Since >2.91% and >2.91%, both bond municipal'i corporate'i

investments are better than the savings account. Now to compare two mutually exclusive bond investment alternatives, we need to perform an incremental analysis.

After-tax Cash Flow n Municipal Corporate Incremental 0 -$10,000 -$10,000 0 1 $900 $840 -$60 2 $900 $840 -$60 3 $900 $840 -$60 4 $900 $840 -$60 5 $900 $840 -$60

We cannot find the rate of return on incremental investment, as returns from municipal bond dominate those from corporate bond in every year. Municipal bond is a clear choice for any value of MARR.

12



4.26) Two common approaches may be used: either (1) constant dollar analysis or (2) actual dollar analysis. In this case, it may be easier to use the constant dollar analysis, as we don’t need to project the future price increase of the subscription, assuming that the price of magazine will follow the general inflation rate. Then, we need to determine which interest rate to use in evaluating the three different subscription options. Assuming that the decision-maker’s desired inflation-free interest rate ( ) or real earnings from his or her personal investment is around 2%, we can determine the total subscription cost for life-time (say, more than 50 years) as follows:

'i

1-year subscription

2-year subscription

3-year subscription

$39$39 $1,9890.02

$72$72 $1,8540.0404

$103$103 $1,7860.06121

P

P

P

= + =

= + =

= + =

In this case, the 3-year subscription option appears to be a better choice. Note that 4.04% represents the effective interest rate for 2 years and 6.121% does for 3 years. The view taken in this calculation is that if the general inflation rate were running at 3%, the decision-maker would earn around 5% (=3% + 2%) market interest rate. Certainly the choice will change depending upon the individual decision-maker’s true earrings requirement. If we take a finite planning horizon, say 6-year, the subscription cost for each option would be as follows:

1-year subscription

2-year subscription

3-year subscription

$39 $39( / , 2%,5) $222.82

$72 $72( / , 2%, 2) $72( / , 2%, 4) $207.72

$103 $102( / , 2%,3) $200.05

P P A

P P F P F

P P F

= + =

= + + =

= + =

It still appears that the 3-year subscription is a better choice.

4.27) Four-year Full Benefits Plan for The University of Michigan at Ann Arbor: • Determine the average inflation rate for tuition and fees:

13



13$7,560 $3,191(1 )6.86%

ff= +=

• Compute the anticipated annual tuitions and fees from the perspective of a parent with a newborn in 2001, assuming that the future tuition and fees will increase at the annual rate of 6.86%:

Birthday Expected Tuition and Fees

18 Freshman 19 Sophomore 20 Junior 21 Senior

$7,560(F/P,6.86%,18) = $24,957$7,560(F/P,6.86%,19) = $26,669$7,560(F/P,6.86%,20) = $28,498$7,560(F/P,6.86%,21) = $30,453

• Determine the amount of accumulated savings versus the required college

savings in actual dollars: Sample calculations at i = 8%:

Fdeposit = $24,252(1+ 0.08)18 = $96,911

Prequired savings = $24,957 + $26,669(P / F ,8%,1)

+$28,498(P / F ,8%,2) + $30,453(P / F ,8%,3)= $98,202

Savings Rate (Market rate)

Accumulated Savings at age of

18 years

Required College Savings at age of

18 years 5% $58,365 $102,453 6% $69,223 $100,991 7% $81,970 $99,975 8% $96,911 $98,202 10% $134,839 $95,578

The breakeven interest rate is about 8.07%. In other words, if you cannot invest your money at higher than 8.07%, you are better off with the State’s Full Benefits Plan.

14



Chapter 5 Present-Worth Analysis 5.1) (a), (b), (c)

n Inflow Outflow Net Cash Flow

0 $0 $65,000 -$65,000

1 $215,500 $53,000 $162,550

2 $215,500 $53,000 $162,550

3 $215,500 $53,000 $162,550

4 $215,500 $53,000 $162,550

5 $215,500 $53,000 $162,550

6 $215,500 $53,000 $162,550

7 $215,500 $53,000 $162,550

8 $215,500 $53,000 $162,550

Annual cash inflow = $17 34,000 $15 30,000 $25 3,500 $215,500× − × + × = 5.2) Project cash flows over the project life

n Cmax Demand Revenue Expense Cost of Bldg. NCF

0 - - 1,527,776 -$1,527,776

1 6,000,000 3,000,000 16,256,976 6,462,108 - 9,794,868

2 6,000,000 3,300,000 17,882,673 7,096,319 - 10,786,354

3 6,000,000 3,630,000 19,670,941 7,793,951 - 11,876,990

4 6,000,000 3,993,000 21,638,035 8,561,346 - 13,076,689

5 6,000,000 4,392,300 23,801,838 9,405,481 - 14,396,358

6 6,000,000 4,831,530 26,182,022 10,334,029 1,545,123♠ 14,302,870

7 12,000,000 5,314,683 28,800,224 11,355,432 - 17,444,793

8 12,000,000 5,846,151 31,680,247 12,478,975 - 19,201,272

9 12,000,000 6,430,766 34,848,271 13,714,872 - 21,133,399

10 12,000,000 7,073,843 38,333,099 15,074,359 - 23,258,739

11 12,000,000 7,781,227 42,166,408 16,569,795 - 25,596,613

12 12,000,000 8,559,350 46,383,049 18,214,775 - 28,168,274

13 12,000,000 9,415,285 51,021,354 20,024,252 - 30,997,102

14 12,000,000 10,356,814 56,123,490 22,014,678 1,573,302♠ 32,535,510

15 24,000,000 11,392,495 61,735,839 24,204,145 - 37,531,693



♠: The cost of building is given as if Cmax is being built from scratch. No “credit” is given for the capacity already in place. This assumption could be rather unrealistic. In that case, what we need to do is to identify the incremental cost of adding the additional capacity above the existing capacity. 5.3)

(a) Payback period: 1 years

n Net Cash Flow Cumulative

CF

0 -$65,000 -$65,000

1 $162,500 $97,500

2 $162,500 $260,000

3 $162,500 $422,500

4 $162,500 $585,000

5 $162,500 $747,500

6 $162,500 $910,000

7 $162,500 $1,072,500

8 $162,500 $1,235,000

(b) Discounted payback period = 1 year.

n Net Cash Flow Cost of funds Cumulative CF

0 -$65,000 $0 -$65,000

1 $162,500 -$9,750 $87,750

2 $162,500 $13,163 $263,413

3 $162,500 $39,512 $465,424

4 $162,500 $69,814 $697,738

5 $162,500 $104,661 $964,899

6 $162,500 $144,735 $1,272,134

7 $162,500 $190,820 $1,625,454

8 $162,500 $243,818 $2,031,772

5.4) (a) It will take 3 years to recover the total investment.

2



n Inflow Outflow Net Cash Flow Cumulative CF

0 $0 $32,500 -$32,500 -$32,500

1 $12,000 $0 $12,000 -$20,500

2 $12,000 $0 $12,000 -$8,500

3 $12,000 $0 $12,000 $3,500

4 $12,000 $0 $12,000 $15,500

5 $17,000 $0 $17,000 $32,500

(b) It will take 4 years to recover the total investment.

n Cash Flow Cost of funds Cumulative CF

0 -$32,500 $0 -$32,500

1 $12,000 -$4,550 -$25,050

2 $12,000 -$3,507 -$16,557

3 $12,000 -$2,318 -$6,875

4 $12,000 -$962 $4,163

5 $17,000 $583 $21,745

5.5)

(a) It will take 5 years to recover the total investment.

n Cash Flow Cumulative CF

0 -$10,000 -$10,000

1 -$15,000 -$25,000

2 $8,000 -$17,000

3 $8,000 -$9,000

4 $8,000 -$1,000

5 $8,000 $7,000

(b) The total investment is recovered in year 6 (or 5.19 years)

n Cash Flow

Cost of funds (10%) Cumulative CF

3



0 -$10,000 $0 -$10,000

1 -$15,000 -$1,000 -$26,000

2 $8,000 -$2,600 -$20,600

3 $8,000 -$2,060 -$14,660

4 $8,000 -$1,466 -$8,126

5 $8,000 -$813 -$939

6 $8,000 -$94 $6,968

5.6)

(a) Payback period Project A: 5 years, Project B: 5 years, Project C: 4 years

A B C D

n CF Cum.CF CF Cum.CF CF Cum.CF CF Cum.CF

0 -$1,500 -$1,500 -$6,000 -$6,000 -$10,000 -$10,000 -$4,500 -$4,500

1 200 -1,300 2,000 -4,000 2,000 -8,000 5,000 500

2 300 -1,000 1,500 -2,500 2,000 -6,000 3,000 3,500

3 400 -600 1,500 -1,000 2,000 -4,000 -4,000 -500

4 500 -100 500 -500 5,000 1,000 1,000 500

5 300 200 500 0 5,000 6,000 1,000 1,500

6 300 500 1,500 1,500 2,000 3,500

7 300 800 3,000 6,500

8 300 1,100

(b) Project D does not have a unique payback period, as there are two payback

periods—one at year 2 and the other at period 4. However, if the project is undertaken, we would say 4 years, because that is when the project truly is financially in the clear.

(c) Discounted payback period Project A: 7 years, Project B: none, Project C: 5 years.

A B C D

n CF Cum.CF CF Cum.CF CF Cum.CF CF Cum.CF

0 -$1,500 -$1,500 -$6,000 -$6,000 -$10,000 -$10,000 -$4,500 -$4,500

4



1 200 -1,450 2,000 -4,600 2,000 -9,000 5,000 50

2 300 -1,295 1,500 -3,560 2,000 -7,900 3,000 3,055

3 400 -1,025 1,500 -2,416 2,000 -6,690 -4,000 -640

4 500 -627 500 -2,158 5,000 -2,359 1,000 297

5 300 -390 500 -1,873 5,000 2,405 1,000 1,326

6 300 -129 1,500 -561 2,000 3,459

7 300 159 3,000 6,805

8 300 474

5.7)

n Cash Flow

0 -$18,000 1 $4,800 2 $6,350 3 $7,735 4 $7,500 5 $4,300 6 $7,000 + $1,800

PW(9%) $18,000 $4,800( / ,9%,1) $6,350( / ,9%, 2)

$7,735( / ,9%,3) $7,500( / ,9%, 4)$4,300( / ,9%,5) $8,800( / ,9%,6)$11,076.22

P F P FP F P FP F P F

= − + ++ ++ +=

5.8)

(a) There is an opportunity cost of $100,000 for land, which is tied up for this project. This cost should be viewed as an investment required undertaking the project. The $25,000 license fee is considered as one time up-front cost.

n Inflow Outflow Net Cash Flow Cumulative CF

0 $0 $1,625,000 -$1,625,000 -$1,625,000

1 $500,000 $240,000 $260,,000 -$1,365,000

5



2 $500,000 $240,000 $260,000 -$1,105,000

3 $500,000 $240,000 $260,000 -$845,000

4 $500,000 $240,000 $260,000 -$585,000

5 $500,000 $240,000 $260,000 -$325,000

6 $734,010 $240,000 $494,010 $169,010

• Inflow for year 6: $500,000 + $100,000(F/P, 5%, 6) = $734,010 • Outflow for year 0: $1,500,000 + $100,000 + $25,000 = $1,625,000 • Outflow for years 1 - 6: (0.30 + 0.15 + 0.03)($500,000) = $240,000

PW(15%) $1,625,000 $260,000( / ,15%,5) $494,010( / ,15%,6)

$539,865 0P A P F= − + +

= − <

(b) No discounted payback period exist as the initial investment is not fully recovered at the end of the project period (or PW(15%) < 0)

5.9) (a)

PW(10%) $800 $3,000( / ,10%,3)$1, 453.9

PW(10%) $1,800 $600( / ,10%,1)$900( / ,10%, 2) $1,700( / ,10%,3)$766.49

PW(10%) $1,000 $1, 200( / ,10%,1)$900( / ,10%, 2) $3,500( / ,10%,3)$1282.3

PW(10%)

A

B

C

D

P F

P FP F P F

P FP F P F

= − +== − ++ +== − −+ +== $6,000 $1,900( / ,10%, 2)

$2,800( / ,10%,3)$598.91

P AP F

− ++= −

(b) Not provided.

6



5.10)

n Inflow Outflow Net Cash Flow Cumulative

CF

0 $0 $1,500,000 -$1,500,000 -$1,500,000

1 $227,000 $157,000 $70,000 -$1,430,000

2 $227,000 $157,000 $70,000 -$1,360,000

……

…

……

…

……

…

……

…

……

…

33 $227,000 $157,000 $70,000 $810,000

34 $227,000 $157,000 $70,000 $880,000

35 $452,000 $157,000 $295,000 $1,175,000

PW (12%) = −$1,500,000 + $70,000(P / A,12%,34) + $295,000(P / F ,12%,35)= −$923,453

5.11)

Given: Estimated remaining service life 25 years= , current rental income per year, for the first year increasing by $6,000 thereafter, salvage value

= $250,000 O&M costs = $65,000 = $200,000 , and MARR = 15%. Let A0 be the

maximum investment required to break even.

PW(15%) $250,000( / ,15%,5) $275,000( / ,15%,5)( / ,15%,5)$302,500( / ,15%,5)( / ,15%,10)$332,750( / ,15%,5)( / ,15%,15)$366,025( / ,15%,5)( / ,15%, 20)$65,000( / ,15%, 25) $6,000( / ,15%, 25) $200,0

P A P A P FP A P FP A P FP A P F

P A P G

= ++++− − + 00( / ,15%, 25)

$1,116,775P F

=

5.12)

7



P = −$4,000 + $3,400(P / F ,9%,1)+$3,400(P / F ,12%,1)(P / F ,9%,1)+$1,500(P / F ,10%,1)(P / F ,12%,1)(P / F ,9%,1)+$3,500(P / F ,13%,1)(P / F ,10%,1)(P / F ,12%,1)(P / F ,9%,1)+$4,300(P / F ,12%,1)(P / F ,13%,1)(P / F ,10%,1)(P / F ,12%,1)(P / F ,9%,1)= $7,858.34

5.13)

n Inflow Outflow Net Cash Flow Cumulative

CF

0 $0 $250,000 -$250,000 -$250,000

1 $160,000 $50,000 $110,000 -$140,000

2 $160,000 $50,000 $110,000 -$30,000

3 $160,000 $50,000 $110,000 $80,000

4 $160,000 $50,000 $110,000 $190,000

5 $160,000 $50,000 $110,000 $300,000

6 $160,000 $50,000 $110,000 $410,000

7 $160,000 $50,000 $110,000 $520,000

8 $160,000 $50,000 $110,000 $630,000

PW(12%) $250,000 $110,000( / ,12%,8)

$296, 440P A= − +

=

5.14) Given: Initial cost , annual savings$3,000,000= $1, 200,000= ,

Annual O&M costs $250,000= , annual income taxes $150,000= , Salvage value $2 , useful life00,000= = 10years, MARR 18%=

PW(18%) $3,000,000

[$1, 200,000 $250,000$150,000]( / ,18%,10)$200,000( / ,18%,10)$633, 482

P AP F

= −+ −−+=

The project is a profitable one.

8



5.15)

PW(13%) $5,000 $5,800( / ,13%,1)$12,400( / ,13%,2) $8,200( / ,13%,3)$15,526.86

FW(13%) $15,526.86( / ,13%,3)$22,403.88

PW(13%) $2,000 $4,400( / ,13%,1)$7,000( / ,13%,2) $3,000( / ,13%,3)$1,667

A

A

B

P FP F P F

F P

P FP F P F

= − ++ +==== − −+ +=

FW(13%) $1,667( / ,13%,3) $2,405.85PW(13%) $4,500 $6,000( / ,13%,1)

$2,000( / ,13%,2) $4,000( / ,13%,3)$3,528.6

FW(13%) $3528.6( / ,13%,3) 5,091.42PW(13%) $3,500 $1,000( / ,13%,1)

$5,000( / ,13%,

B

C

C

D

F PP F

P F P F

F PP F

P F

= == −

+ +== =

= − ++ 2) $6,000( / ,13%,3)$5,458.99

FW(13%) $5,458.99( / ,13%,3) $7,876.76D

P F

F P

+== =

5.16) (a)

Year Outflow

2008 $14,500,000

2009 $3,500,000

2010 $26,000,000

FW(15%) $14,500,000( / ,15%,2) $3,500,000( / ,15%,1)

$26,000,000$49, 201, 250

F P F P= ++=

(b)

9



( / ,15%,10) (15%)$49, 201,250

(5.0188) $49, 201,250$9,803, 450

A P A FW

AA

====

5.17)

1PB( ) $1,000(1 ) $200 $90010%

i ii= − + + = −=

5.18)

(a) In part (b), it is determined that I = 20%. Then, the original cash flows of the project is as follows:

n

nA Project Balance

0 -$1,000 -$1,0001 $100 -$1,1002 $520 -$8003 $460 -$5004 $600 $0

(b)

3PB( ) $800(1 ) $460$500

$800 $16020%

i i

ii

= − + += −

− = −=

(c) Yes

10



5.19) • For Project B:

2$650(1 ) $41625%

ii

−+ ==

Statement 3 is true.

• For Statement 1 to be true, I would have to equal 0%, since $200(1 + i) + $100 equal $300. So Statement 1 is false.

• Statement 2 is false, since FW of Project C is $150. Therefore, the correct answer is (c).

5.20)

(a) From the project balance diagram, note that 1PW(24%) 0= for project 1 and

2PW(23%) 0= for project 2.

1

2

PW(24%) $100 $40( / , 24%,1) $80( / , 24%,2)( / , 24%,3)

0PW(23%) $100 $30( / , 23%,1) ( / , 23%,2)

$80( / , 23%,3)0

P F P FX P F

P F Y P FP F

= − + ++== − + ++=

Solving for X and yields Y $29.96X = and $49.35Y = , respectively.

(b) Since , this implies thatPW(24%) 0= 3FW(24%) PB(24%) 0= = . (c)

100 30 49.35 80 $59.35100 40 80 29.96 $49.96

= 17.91%

abc

= − + + + == − + + + =

5.21)

(a) In Part (b), it is determined that i= 10%.

n nA Project Balance 0 -$1,000 -$1,0001 $200 -$9002 $490 -$5003 $550 $0

11



4 -$100 -$1005 $200 $90

(b)

2PB( ) $900(1 ) $490 $50010%

PW(10%) $90( / ,10%,5) $55.88

i ii

P F

= − + + = −== =

5.22) (a)

FW(15%) $5,000( / ,15%,5) $500( / ,15%,4)$500

$4,691FW(15%) $5,000( / ,15%,5) $2,000( / ,15%,4)

$3,500$4,741.10

FW(15%) $5,000( / ,15%,5) $3,000( / ,15%,2)$13,000

$14,960.71FW(15%) $5

A

B

C

D

F P F P

F P F P

F P F P

= − ++ −= −= − ++ +== − ++ +== − ,000( / ,15%,5) $500( / ,15%,4)

$1,250$3,676.97

FW(15%) $5,000( / ,15%,3) $1,000( / ,15%,2)$2,000

$831.87

E

F P F P

F P F P

++ +== − ++ += −

(b),(c),(d)

Project A

n Cash Flow Cost of funds Project Balance

0 -$5,000.0 $0 -$5,000

1 $500.0 -$750 -$5,250

2 $900.0 -$788 -$5,138

3 $1,000.0 -$771 -$4,908

4 $2,000.0 -$736 -$3,644

5 -$500.0 -$547 -$4,691

Discounted payback period for project A - none

12



Project B


0 -$5,000.0 $0 -$5,000

1 $2,000.0 -$750 -$3,750

2 -$3,000.0 -$563 -$7,313

3 $5,000.0 -$1,097 -$3,409

4 $5,000.0 -$511 $1,079

5 $3,500.0 $162 $4,741

Discounted payback period for project B – 4 years

Project C


0 -$5,000.0 $0 -$5,000

1 $0.0 -$750 -$5,750

2 $0.0 -$863 -$6,613

3 $3,000.0 -$992 -$4,604

4 $7,000.0 -$691 $1,705

5 $13,000.0 $256 $14,961

Discounted payback period for project C - 4years

Project D


0 -$5,000.0 $0 -$5,000

1 $500.0 -$750 -$5,250

2 $2,000.0 -$788 -$4,038

3 $3,000.0 -$606 -$1,643

4 $4,000.0 -$246 $2,110

5 $1,250.0 $317 $3,677

Discounted payback period for project D - 4years

13



Project E


0 -$5,000.0 $0 -$5,000

1 $1,000.0 -$750 -$4,750

2 $3,000.0 -$713 -$2,463

3 $2,000.0 -$369 -$832

Discounted payback period for project E - none

5.23) (a)

PW(10%) $100 $50( / ,10%,3)$100( / ,10%, 4)$400( / ,10%, 2)( / ,10%, 4)$430.20

PW(10%) $100 $40( / ,10%,3)$10( / ,10%, 2)( / ,10%,3)$12.51

PW(10%) $100 $40( / ,10%,3)$0.53

A

B

C

P AP FP A P F

P AP A P F

P A

= − +−+== − ++== −=

All projects are acceptable.

(b) FW(10%) $430.20( / ,10%,6)

$762.13FW(10%) $12.51( / ,10%,5)

$20.15FW(10%) $0.53( / ,10%,3)

$0.70

A

B

C

F P

F P

F P

======

All projects are acceptable.

(c)

14



FW( ) [ $100( / ,10%,3) $50( / ,10%,3)]( / ,15%,3)[ $100( / ,15%, 2) $400( / ,15%, 2)]$777.08

FW( ) [ $100( / ,10%,3) $40( / ,10%,3)]( / ,15%,3)$10( / ,15%, 2) $10( / ,15%,1)$23.66

FW( ) $100( / ,10

A

B

C

i F P F A F PF P F A

i F P F A F PF P F P

i F P

= − ++ − +== − ++ +== %,3)( / ,15%,3)

$40( / ,10%,3)( / ,15%,3)$1.065

F PF A F P−

=

5.24)

(a) PW(0%) 0

PW(18%) $575( / ,18%,5) $251.34PW(12%) 0

A

B

C

P F== ==

(b) Assume that 2 $500.A =

2PB(12%) $530(1.12) $500$93.60

XX= − + == −

. (c) The net cash flows for each project are as follows:

Net Cash Flow n A B C 0 -$1,000 -$1,000 -$1,000 1 $200 $500 $590 2 $200 $500 $500 3 $200 $300 -$106 4 $200 $300 $147 5 $200 $300 $100

Sample calculation for Project C:

0

1 1

PW(12%) $1,000PW(12%) $1,000(1.12) $530A

= −= − + =

Solving for A1 yields A1 = $590.

15



(d) FW(0%) 0FW(18%) $575FW(12%) 0

A

B

C

===

5.25) (a)

PW(13%) $40,000( / ,13%,5)$50,000( / ,13%,5)( / ,13%,5)($60,000 / 0.13)( / ,13%,10)$372,103.72

P AP A P F

P F

=++=

(b) PW(13%) /

$372,103.72(0.13)$48,373.48

A iA===

5.26)

PW(12%) $1,106$1,106( / ,12%,4)CE(12%) $3,034

0.12A P

=

= =

5.27) Given: compounded monthly, maintenance cost = $25,000 per year 6%r =

12(1 0.06 /12) 1

6.17%CE(6.17%) $25,000 / 0.0617

$405,186.39

ai = + −===

5.28) Given: Construction cost = $10,000,000, renovation cost = $1,000,000 every

10 years, annual costs = $100,000 and &O M 5%i = per year

(a)

16



1

2

3

1 2 3

$10,000,000$1,000,000( / ,5%,10)

0.05$1,590,000$100,000 / 0.05$2,000,000

CE(5%)$13,590,000

PA FP

P

P P P

=

=

==== + +=

(b)

1

2

3

1 2 3

$10,000,000$1,000,000( / ,5%,15)

0.05$926,000$100,000 / 0.05$2,000,000

CE(5%)$12,926,000

PA FP

P

P P P

=

=

==== + +=

(c)

• 10-year cycle with 10% of interest:

1

2

3

$10,000,000$1,000,000( / ,10%,10)

0.10$627,000$100,000 / 0.10$1,000,000

CE(10%) $11,627,000

PA FP

P

=

=

==

==

• 15-year cycle with 10% of interest:

17



1

2

3

$10,000,000$1,000,000( / ,10%,15)

0.10$315,000$100,000 / 0.10$1,000,000

CE(10%) $11,315,000

PA FP

P

=

=

==

==

As interest rate increases, CE value decreases.

5.29) Given: Cost to design and build $650,000= , rework cost= $100,000 every 10 years, new type of gear at the end of 5$50,000= th year, annual operating costs

for the first 15 years and thereafter $30,000= $35,000

$100,000( / ,8%,10)CE(8%) $650,0000.08

$50,000( / ,8%,5) $30,000( / ,8%,15)$35,000 ( / ,8%,15)

0.08$1,165,019

A F

P F P A

P F

= +

+ +

+

=

5.30)

(a) PW(0.5%) $2,160( / ,0.5%, 240)

$301, 494.47P A=

=

(b) PW(0.5%) $2,160( / ,0.5%, 480)

$392,574.78P A=

=

(c)

PW(0.5%)

$2,1600.005

$432,000

Ai

A

=

=

=

Comments: Longer life means greater total benefit, but most of the benefit is collected in the first 20 years.

18



5.31) (a)

PW(25%) $1,000 $912( / , 25%,1)$684( / ,25%,2) $456( / , 25%,3)$228( / , 25%,4) $494.22

PW(25%) $1,000 $284( / , 25%,1)$568( / , 25%,2) $852( / , 25%,3)$1,136( / , 25%,4) $492.25

A

B

P FP F P FP F

P FP F P F

P F

= − ++ ++ =

= − ++ ++ =

Select project A.

(b)

Project A


0 -$1000 $0 -$1,000

1 $912 -$250 -$338

2 $684 -$85 $262

3 $456 $65 $783

4 $228 196 $1,207

Project B


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

1 $284 -$250 -$966

2 $568 -$242 -$640

3 $852 -$160 $53

4 $1,136 $13 $1,202

Project B is exposed to higher risk of loss if either project terminates at the end of the year 2, according to the results below.

19



5.32) (a)

PW(12%) $1,000,000 $700,000( / ,12%,2)$183,070

PW(12%) $1,200,000 $700,000( / ,12%,1)$1,000,000( / ,12%,2)$222,230

A

B

P A

P FP F

= − +== − ++=

Select project B. (b)

PW(22%) $1,000,000 $700,000( / , 22%,2)$44,074

PW(22%) $1,200,000 $700,000( / , 22%,1)$1,000,000( / , 22%,2)$45,633

A

B

P A

P FP F

= − +== − ++=

Select project B. 5.33)

(a) PW(12%) $5,000 $2,610( / ,12%,1)

$2,930( / ,12%, 2) $2,300( / ,12%,3)$1,303.23

PW(12%) $3, 200 $1, 210( / ,12%,1)$1,720( / ,12%, 2) $1,500( / ,12%,3)$319.2

A

B

P FP F P F

P FP F P F

= − ++ +== − ++ +=

Select Project A. (b)

FW(12%) $1,303.23( / ,12%,3)$1,830.94

FW(12%) $319.2( / ,12%,3)$448.44

A

B

F P

F P

====

Select Project A. 5.34)

(a)

20



PW(15%) $4,000 $400( / ,15%,1)$7,000( / ,15%, 2)$1,640.83

PW(15%) $8,500 $11,500( / ,15%,1)$400( / ,15%, 2)$1,802.46

A

B

P FP F

P FP F

= − ++== − ++=

Select project B.

(b) Project B dominates Project A at any interest rate (0% to 46.7%.) as indicated in the following present worth profile. Note however that for very high interest rate (i > 46.7%), Project A is less undesirable than project B.

(X axis-interest rate, Y axis-PW(i) )

5.35) (a)

PW(15%) $15,000 $9,500( / ,15%,1)$12,500( / ,15%,2) $7,500( / ,15%,3)$7,644.03

A P FP F P F

= − ++ +=

(b) PW(15%) $25,000 ( / ,15%,2)( / ,15%,1)

$9,300$24,262.57

B X P A P F

X

= − +==

(c) Note that the net future worth of the project is equivalent to its terminal project balance.

21



3PB(15%) $7,643.7( / ,15%,3)$11,625.30

F P==

(d) Select B, which has the greater PW. 5.36)

(a) Project balances as a function of time are as follows:

Project Balances n A D 0 -$2,500 -$5,0001 -$2,100 -$6,0002 -$1,660 -$7,1003 -$1,176 -$3,8104 -$694 -$1,1915 -$163 $1,6906 $421 $3,8597 $763 $7,2458 $1,139

All figures above are rounded to nearest dollars.

(b) Knowing the relationship FW( ) PB( ) ,Ni i= FW(10%) $1,139FW(10%) $7, 245

A

D

==

(c) Assuming a required service period of 8 years

PW(10%) $7,000 $1,500( / ,10%,8)

$1,000( / ,10%,1) $500( / ,10%,2)$1,500( / ,10%,7) $1,500( / ,10%,8)

$17,794PW(10%) $5,000 $2,000( / ,10%,7)

$3,000( / ,10%,8)$16,136

B

C

P AP F P FP F P F

P AP F

= − −− −− −= −= − −

−= −

Select Project C.

22



5.37) Given: Required service period = infinite, analysis period = least common multiple service periods (6 years)

• Model A:

cycle

total

PW(12%) $11,000 $7,500( / ,12%,1)

$8,000( / ,12%,2) $5,000( / ,12%,3)$5,633.35

PW(12%) $5,633.35[1 ( / ,12%,3)]$9,643.11

P F

P F P F

P F

= − +

+ +== +=

• Model B:

cycle

total

PW(12%) $25,000 $14,500( / ,12%,1)

$18,000( / ,12%,2)$2,296.65

PW(12%) $2,296.65[1 ( / ,12%,2)( / ,12%,4)]$5,587.06

P F

P F

P FP F

= − +

+== +

+=

Model A is preferred. 5.38)

(a) Without knowing the future replacement opportunities, we may assume that both alternatives will be available in the future with the same investment and expenses. We further assume that the required service period will be indefinite.

(b) With the common service period of 24 years,

• Project A:

cycle

total

PW(10%) $900 $400( / ,10%,3)

$200( / ,10%,3)$1,744.48

PW(10%) $1,744.48[1 ( / ,33.10%,7)]$6,302.63

P A

P F

P A

= − −

+= −= − += −

Note that the effective interest rate for a 3-year cycle is 3(1.10) 1 33.10%− =

23



• Project B:

total

cyclePW(10%) $1,800 $300( / ,10%,8)

$500( / ,10%,8)$3,167.22

PW(10%) $3,167.22[1 ( / ,10%,8)

( / ,10%,16)]$5,334.03

P A

P F

P F

P F

= − −

+= −= − +

+= −

Project B is preferred.

(c)

PW(10%) $1,744.48PW(10%) $1,800 $300( / ,10%,3) ( / ,10%,3)

$2,546.06 0.7513

A

B P A S P FS

= −= − − += − +

Let PW(10%) PW(10%)A B= and solve for S.

$1,067S = 5.39)

(a) Assuming a common service period of 15 years • Project A:

cycle

total

PW(12%) $12,000 $2,000( / ,12%,5)

$2,000( / ,12%,5)$18,075

PW(12%) $18,075[1 ( / ,76.23%,2)]$34,151

P A

P F

P A

= − −

+= −= − += −

5Note : (1.12) 1 76.23%− = • Project B:

cycle

total

PW(12%) $10,000 $2,100( / ,12%,3)

$1,000( / ,12%,3)$14,332

PW(12%) $14,332[1 ( / , 40.49%,4)]$40,642

P A

P F

P A

= − −

+= −= − += −

3Note : (1.12) 1 40.49%− = Select project A.

24



(b) • Project A with 2 replacement cycles:

PW(12%) $18,074 $18,074( / ,12%,5)$28,329.67

P F= − −= −

• Project B with 4 replacement cycles where the 4th replacement cycle ends at

the end of first operating year:

PW(12%) $14,332[1 ( / ,12%,3) ( / ,12%,6)][$10,000 ($2,100 $6,000)( / ,12%,1)]( / ,12%,9)

$34,144.73

P F P FP F

P F

= − + +− + −×= −

Project A is still a better choice.

5.40) • Method A:

$10,000( / ,12%,5)CE(12%) $30,0000.12

$30,000 $13,117.50$43,117.5

AA F

= +

= +=

• Method B: $90,000( / ,12%,50)CE(12%) $75,000

0.12$75,000 $312.50$75,312.5

BA F

= +

= +=

Since values above represent cost, project A is preferred. CE(12%)

5.41) • Standard Lease Option:

SLPW(0.5%) $5,500 $1,150( / ,0.5%,24)$1,000( / ,0.5%,24)

$30,560.10

P AP F

= − −

+= −

• Single Up-Front Option:

SUPW(0.5%) $31,500 $1,000( / ,0.5%, 24)$30,612.82

P F= − += −

25



Select the standard lease option, as you will save $52.72 in present worth.

5.42) • Machine A:

PW(13%) $75, 200 ($6,800 $2, 400)( / ,13%,6)$21,000( / ,13%,6)

$101,891

P AP F

= − − ++= −

• Machine B: PW(13%) $44,000 $11,500( / ,13%,6)

$89,971P A= − −

= −

Machine B is a better choice. 5.43)

(a) • Required HP to produce 10 HP:

Motor A:

1 10 / 0.85 11.765 HPX = =

Motor B:

2 10 / 0.90 11.111 HPX = =

• Annual energy cost: Motor A:

11.765(0.7457)(1,500)(0.07) $921.18=

Motor B: 11.111(0.7457)(1,500)(0.07) $869.97=

• Equivalent cost:

(8%) $800 $921.18( / ,8%,15)$50( / ,8%,15)

$8,669(8%) $1, 200 $869.97( / ,8%,15)

$100( / ,8%,15)$8,614

A

B

PW P AP F

PW P AP F

= − −+= −= − −+= −

26



Note: Power bill is paid at years end, not monthly. Motor B is preferred.

(b) With 2,500 operating hours:

PW(8%) $800 $1,535.3( / ,8%,15)$50( / ,8%,15)

$13,925PW(8%) $1, 200 $1, 449.96( / ,8%,15)

$100( / ,8%,15)$13,579

A

B

P AP F

P AP F

= − −+= −= − −+= −

Motor B is still preferred.

5.44) Since only Model B is repeated in the future, we may have the following

sequence of replacement cycles: • Option 1: Purchase Model A now and repeat Model A forever. • Option 2: Purchase Model B now and replace it at the end of year 2 by Model

A. Then repeat Model A forever.

PW(15%) $6,600 $3,500( / ,15%,3)$1,000( / ,15%,2) $2,000( / ,15%,3)$3,462.3

AE(15%) $3,462.3( / ,15%,3)$1,516.49

PW(15%) $16,500 $11,000( / ,15%,1) $12,000( / ,15%,2)$2,138.94

AE(15%) $2,138.94

A

A

B

B

P AP F P F

A P

P F P F

= − ++ +==== − + +== ( / ,15%,2)

$1,315.7A P

=

(a)

• Option 1: $1,516.49PW(15%)

0.15$10,109.93

AAAAi

= =

=

• Option 2:

27



$1,516.49PW(15%) $2,138.94 ( / ,15%,2)0.15

$8,901.21

BAA P F= +

=

Option 1 is a better choice. (b) Let S be the salvage value of Model A at the end of year 2.

−$6,600 + $3,500(P / F ,15%,1) + ($4,500 + S )(P / F ,15%,2) = $2,138.94 Solving for S yields

S = $3,032.25 5.45)

• Since either tower will have no salvage value after 20 years, we may select the analysis period of 35 years:

Bid A

Bid B

PW(11%) $80,000 $1,000( / ,11%,35)$88,855

PW(11%) $78,000 $1,750( / ,11%,35)$93, 497

P A

P A

= − −

= −= − −

= −

Bid A is a better choice. • If we assume an infinite analysis period, the present worth of each bid will be

Bid A

Bid B

[ $80,000 $1,000( / ,11%,40)]( / ,11%,40)PW(11%)0.11

$90,341$93, 497( / ,11%,35)PW(11%)

0.11$95,985

P A A P

A P

− −=

= −−

=

= −

Bid A is still preferred. 5.46)

• Option 1: Non-deferred Plan (install remaining 7 units)

1PW(12%) $200,000 $21,000( / ,12%,8)$304,320

P A= − −= −

• Option 2: Deferred Plan

28



2PW(12%) $100,000( / ,12%, 2)$6,000( / ,12%,3)( / ,12%, 2)$160,000( / ,12%,5)$15,000( / ,12%,3)( / ,12%,5)$140,000( / ,12%,8)

$258,982

P FP A P F

P FP A P F

P F

= −−−−−= −

Option 2 is a better choice. 5.47)

• Alternative A: Once-for-all expansion

PW(15%) $30 $0.40 ( / ,15%,25)$0.85 ( / ,15%,25)

$32,559,839

A M M P AM P F

= − −+= −

• Alternative B: Incremental expansion

PW(15%) $10 $18 ( / ,15%,10)$12 ( / ,15%,15) $1.5 ( / ,15%, 25)$0.25 ( / ,15%, 25)$0.10 ( / ,15%,15)( / ,15%,10)$0.10 ( / ,15%,10)( / ,15%,15)

$17,700,745

B M M P FM P F M P F

M P AM P A P FM P A P F

= − −− +−−−= −

Select alternative B. 5.48)

• Option 1: Tank/tower installation 1(12%) $164,000PW = −

• Option 2: Tank/hill installation with the pumping equipment replaced at the end of 20 years at the same cost

PW(12%) ($120,000 $12,000)($12,000 $1,000)( / ,12%, 20)$1,000( / ,12%, 40) $1,000( / ,12%, 40)

$141,374

P FP F P A

= − +− −+ −= −


• Option 1: Process device A lasts only 4 years. You have a required service period of 6 years. If you take this option, you must consider how you will satisfy the rest of the required service period at the end of the project life. One

29



option would subcontract the remaining work for the duration of the required service period. If you select this subcontracting option along with the device A, the equivalent net present worth would be

1PW(12%) $100,000 $60,000( / ,12%, 4)$10,000( / ,12%, 4)$100,000( / ,12%,2)( / ,12%, 4)

$383, 292

P AP F

P A P F

= − −+−= −

• Option 2: This option creates no problem because its service life coincides with the required service period.

2PW(12%) $150,000 $50,000( / ,12%,6)$30,000( / ,12%,6)

$340,371

P AP F

= − −+= −

• Option 3: With the assumption that the subcontracting option would be available over the next 6 years at the same cost, the equivalent present worth would be

3PW(12%) $100,000( / ,12%,6)$411,141

P A= −= −

With the restricted assumptions above, option 2 appears to be best alternative.

30



Chapter 6 Annual Equivalence Method 6.1)

AE(9%) $20,000( / ,9%,5)$51,420

A P==

6.2)

AE(10%) ( / ,10%,3) $100,000$40,210.70A P A

A= ==

6.3)

AE(12%) $25,000( / ,12%,6)

$4,000( / ,12%,1) $13,000( / ,12%,2) $13,000( / ,12%,3)( / ,12%,6)

$13,000( / ,12%,4) $8,000( / ,12%,5) $5,500( / ,12%,6)

$3,351

A P

P F P F P FA P

P F P F P F

=−

+ +++ + +

=

⎡ ⎤⎢ ⎥⎣ ⎦

6.4)

AE(8%) $2,154( / ,8%,6)$400( / ,8%,1) ( / ,8%,2)( / ,8%,1)

( / ,8%,6)$400( / ,8%,4) ( / ,8%,2)( / ,8%,4)

$200

200 465.94 (370.36 1.65 294 1.31 )(0.2163)924.64 1,489.78 2.96

$815.68

A PP F X P A P F

A PP F X P A P F

X XX

X

=−

+⎡ ⎤+⎢ ⎥+ +⎣ ⎦=

=− + + + +=− +=



6.5)

[ ]AE(10%) $3,000( / ,10%,5)

$600( / ,10%,2) $1,000( / ,10%,3)( / ,10%,2) ( / ,10%,5)$25.43 (Accept)

AE(10%) $5,000( / ,10%,5) $500[$2,500( / ,10%,1) $1,500( / ,10%,2)$500( / ,10%,3)]( / ,10%,5)$20

A

B

A PP A P A P F A P

A PP F P F

P F A P

= −

+ +

== − ++ ++= 6.65 (Accept)

AE(10%) [ $4,000 $2,000( / ,10%,1)$2,000( / ,10%,5)]( / ,10%,5)

$781.85 (Accept)AE(10%) [ $32,000 $12,000( / ,10%,1)

$14,000( / ,10%,5)]( / ,10%,5)$6,592.33 (Accept)

C

D

P FP F A P

P FP F A P

= − −

⋅⋅⋅+== − ++ ⋅⋅⋅+=

6.6)

$1,000AE(12%) $1,000 [ ( / ,15%,6)]0.15

$1,432.3

P F= +

=

6.7)

AE(9%) $1,000( / ,9%,5) $800 $300( / ,9%, 2)( / ,9%,5)$800( / ,9%,3)( / ,9%,5)$766.65

A P P F A PP F A P

= − + ++=

6.8)

AE(13%) $4,000( / ,13%,3) $5,500( / ,13%,3)$79.75 Not Accept

AE(13%) $3,500( / ,13%,3) $1,500$300( / ,13%,3)$293.36, Accept

AE(13%) $5,000( / ,13%,3) $3,000$1,000( / ,13%,3)

$36.2, Not Accept

A

B

C

A P A F

A PA G

A PA G

= − += −= − ++== − +−= −

AE(13%) $4,500( / ,13%,3) $1,800$105.75, Not Accept

D A P= − += −

2



6.9)

2 3 4

$100 $100 $60 $60PW(14%) $240.691.14 1.14 1.14 1.14

AE(14%) $240.69( / ,14%, 4) $82.60A P

= + + + =

= =

6.10) Given: years, 1$55,000, $6,000, $5,000, $2,500, 10I S A G N= = = = = 12%i =

(a) 1AE(12%) ($55,000 $6,000)( / ,12%,10)

$6,000(0.12)$9,392

A P= −+=

(b) 2AE(12%) $5,000 $2,500( / ,12%,10)

$13,962A G= +

=

(c) AE(12%) $13,962 $9,392

$4,570= −=

This is a good investment.

6.11)

PW(15%) $15 $3.5 ( / ,15%,1) $5 ( / ,15%,2)$9 ( / ,15%,3) $12 ( / ,15%,4) $10 ( / ,15%,5)$8 ( / ,15%,6)$6.9464

AE(15%) $6.9464 ( / ,15%,6) $1.83524

M M P F M P FM P F M P F M P FM P F

MM A P M

= − − ++ + ++== =

Yes, the project is justified.

6.12)

CR(20%) ($220,000 $20,000)( / , 20%,10) $20,000(0.2)$51,700

A P= − +=

6.13)

CR(15%) ($34,000 $3,000)( / ,15%,10) $3,000(0.15)$6,628.3

A P= − +=

3



6.14)

CR(6%) ($18,500 $9,000)( / ,6%,4) (0.06)($9,000)$3,281.7

A P= − +=

6.15) Given: I = $235,000,S = $47,000, N = 5 years, 18%i =

CR(18%) ($235,000 $47,000)( / ,18%,5)

$47,000(0.18)$68,582.4

A P= −+=

6.16)

• Capital cost:

CR(15%) ($25,000 $2,000)( / ,15%,5) $2,000(0.15)

$7,160.9A P= − +

=• Annual operating costs: $15,000

AE(15%) $7,160.9 $15,000 $22,160.9= + = 6.17)

$10,000PW(8%) $10,000( / ,8%,10) ( / ,8%,10)0.06

$67,101 $77, 200$144,301

P A P F= +

= +=

The amount of additional funds should be $44,301. 6.18)

1

2

3

1 2 3

AE (10%) $100,000(0.1) $10,000AE (10%) $10,000AE (10%) $20,000( / ,10%,4) $4,310

AE(10%) AE (10%) AE (10%) AE (10%)$24,310

A F

= === =

= + +

=

6.19)

4



CR(10%) ($500,000 100,000)( / ,10%,15) 100,000(0.1)$62,600

AE(10%) $40,000 $30,000

CR(10%) AE(10%)$62,600 $10,000

6.26 (or rounds up to 7)

A P

X X

X

X

= − +== −

==

=

6.20)

(a) AE(13%) $5,000( / ,13%,4) $1,500

( $1,500)( / ,13%,2)( / ,13%,4)$181 (0.2633)( $1,500)

0$2,187

A PX P F A P

X

X

= − ++ −= − + −==

(b) AE(15%) $6,000( / ,15%, 4) $1, 200 300( / ,15%, 4)

$503.91 0A P A G= − −

= >

Accept project B.

6.21)

• Option 1: Purchase-Borrow Option: Annual repayment of loan amount of $36,000:

1

$36,000( / ,7%,5) $8,780AEC(10%) $4,000( / ,10%,5) $8,780

$9,835

A A PA P

= == +=

• Option 2: Cash Purchase Option:

5



2AEC(10%) $36,000( / ,10%,5)$9,497

A P==

Option 2 is a better choice.

6.22) The total investment consists of the sum of the initial equipment cost and the installation cost, which is . Let$195,000 R denote the break-even annual revenue.

AE(12%) $195,000( / ,12%,10) $40,000

$5,000 $15,0000

A PR

= − −− + +=

Solving for R yields $64,512R =

6.23)

• New lighting system cost:

AE(12%) $55,000( / ,12%,20) $8,000 $4,000$19,364.5

A P= +=

+

• Old lighting system cost:

AE(12%) $30,000= Annual savings from installing the new lighting system = $10,635.5

6.24)

1PW(14%) $100,000 $35,000( / , 3%,14%,5)

$14,058.92AE(14%) $14,058.92( / ,14%,5)

$4,095.13

P A

A P

= − + −===

(14%) $4,095.13 / 3,000 $1.37 per hourAE = =

6.25) Let T denote the total operating hours in full load.

• Motor I (Expensive): Annual power cost: 150 (0.746) (0.05) $6.7410.83

T T× × × =

Equivalent annual cost of operating the motor:

6



IAEC(6%) $4,500( / ,6%,10) $675 6.741$1,286.41 $6.741

A P= + + TT= +

• Motor II (Less expensive): Annual power cost:

150 (0.746) (0.05) $6.99T× × × = 380.80

T

Equivalent annual cost of operating the motor:

IIAEC(6%) $3,600( / ,6%,10) $540 $6.9938$1,029.11 $6.9938

A P TT

= + += +

Let I and solve for

I IAEC(6%) AEC(6%)= T .

$1,286.41 $6.741 $1,029.11 $6.9938T T− − = − −

1,017.8T = hours per year .26)

• Pump I:

6

I

180( )(0.746)(0.06) $9.3680.86

AEC(8%) $6,000( / ,8%,12) $500 $9.368$1, 296.2 $9.368

T T

A P TT

=

= += +

+

• Pump II:

II

180( )(0.746)(0.06) $10.0710.8

AEC(8%) $4,000( / ,8%,12) $440 $10.071$970.8 $10.071

T T

A P TT

=

= + += +

•

$1,296.2 + 9.368T = $970.8 +10.071TT = 463 hours

6.27

Capital cost

•

) •

CR(10%) ($30,000 $10,000)( / ,10%,2) (0.10)$10,000$12,524

A P= − +=

7



savings 2

$25,000 $40,000AE (10%) [ ]( / ,10%,2)1.1 1.1

$32,143.39Net annual savings =$32,143.39 $12,524

$19,619.39

A P= +

=−

=

•

hours 2

(5,000) (8,000)AE (10%) [ ]( / ,10%, 2)1.1 1.1

6, 428.68$19,619.39 6, 428.68

$3.05 per hour

C C A P

CC

C

= +

===

.28) Option 1: Pay employee per mile:

emp

Operating cost per mile

6

• $0.40• Option 2: Provide a car to loyee:

capital cost

operating cost

total cost

AE(10%) ($22,000 $5,000)( / ,10%,3)

(0.10)($5,000)$7,336

AE(10%) $1,000 ($0.22)(30,000)

$7,600AE(10%) $7,336 $7,600

$14,936

A P= −

+== +

== +

=

= $14,936 / 30,000 = $0.498 Option 1 is a better choice.

.29)

• Option 1: Purchase units from Tompkins

Unit cost

6

= $25+ ($70,000 − $35,000) / 20,000 − $3.50 = $23.25

• ption 2: Make units in house O

8



1

1

1

PW(15%) $63,000( / ,5%,15%,5)$230,241

PW(15%) $190,800( / ,6%,15%,5)$709,491

PW(15%) $139,050( / ,3%,15%,5)$490,888

AEC(15%) ($230,241 $709,491$490,888)( / ,15%,5) $70,000$496,776

dm

dl

vo

P A

P A

P A

A P

=====

== ++ +=

Unit cost $496,776 / 20,000=

$24.84=


• Capital costs:

1CR(7%) ($25,000 $2,000)( / ,7%,12)

(0.07)($2,000)$3,036

A P= −+=

• Annual battery replacement cost:

2AEC(7%) $3,000[( / ,7%,3) ( / ,7%,6)( / ,7%,9)]( / ,7%,12)$763.14

P F P FP F A P

= ++=

• Annual recharging cost:

3AEC(7%) ($0.015)(20,000) $300= = • Total annual equivalent costs:

AEC(7%) $3,036 $763.14 $300 $700$4,798.84

= + + +=

• Cost per mile:

cost/mile = $4,798.84 / 20,000 = $0.2399

6.31) • Annual total operating hours:

(0.70)(8,760) 6,132= hours per year

9



• Annual electricity generated:

50,000 6,132 306,600,000× = kilowatt-hours • Equivalent annual cost:

AEC(14%) $85,000,000( / ,14%,25) $6,000,000$18,367,364

A P= +=

• Cost per kilowatt-hour:

$18,367,364 / 306,600,000 $0.06= per kilowatt-hour .32)

• Annual equivalent revenue: 6

RevenueAE $32,000 40,000X= +

• Annual equivalent cost:

CostAEC(8%) $800,000( / ,8%, ) $133,000 $50,000( / ,8%,5)$64,000 $133,000 $8,525$205,525

A P A F= ∞ + += + +=

Revenue Cost

$32,000 40,000 $205,525

$4.34

AE AECX

X

=+ =

=

6.33)

O&M

ma

Salvage Value: $1,200,000( / ,5%,25) $4,063,680

CR(12%) ($6,000,000 $4,063,680)( / ,12%,25) (0.12)$4,063,680$734,522.4

AEC $100 12 40 $400,000 $448,000AEC(12%) CR(12%) $1,182,522.4 per year

AEC(

F P

A P

AE

=

= − +== × × + == + =

Monthly0.9489%) $1,182,522.4( / ,0.9489%,12)

$93,506 per month

A F=

=

10



6.34)

Discounted payback period at full load operation:

n

Investment

Revenue

Maintenance

cost

Net

Cash flow 0 -$30,000 -$30,000 1 $8,000 00 -$5 7,500

15 +$2,000 8,000 -500 9,500

,9%, N ) Solving fo yields

$30,000 = $7,500(P / A

r n

N = 5.179 years 6.35)

• Capital cost:

• Annual operating costs:

+ +=

• Total annual system costs:

=

• Number of rides required per year:

CR(6%) ($150,000 $3,000)( / ,6%,12)(0.06)($3,000)$17,714

A P= −+=

O&M(6%) $40,00= 0 $7,000 $2,000$49,000

AEC(6%) $17,714= +$49,000 $66,714

Number of rides $66,714 /($0.10) 667,140= = rides 6.36)

Investment cost million, plant capacity

$7= 200,000=Given: 1bs/hour, plant operating our hours per year, O&M costh s 3,600= $4= million per year, useful lif years, e 15=

salvage value=$700,000, and MARR = 15%.

11



(a)

Solving fo

PW(15%) $7,000,000 ( $4,000,000)( / ,15%,6)3.7845 $22,137,9000

R P AR

= − + −= −=

r R yields

per year

(b) Minimum processing fee per 1b (after-tax):

$5,849= ,700R

$5,849,700 $0.0081= per 1b (200,000)(3,600)

Comments: The minimum processing fee per 1b should be higher on a before-tax basis.

6. t C denote the green fee per round during the first year.

• Operating and maintenance cost:

=

• Equivalent annual revenue:

37) Le

• Capital cost:

CR(15%) ($20,000,000 $25,000,000( / ,15%,10)(0.15)($25,000,000)$2,753,740

A P= −+=

O&M(15%) $650,000 $50= + ,000( / ,15%,10)$819,160

A G

RevenueAE(15%) $15 40,00= ×

1

10 10

040,000(1.15) ( / ,5%,15%,10)( / ,15%,10)$600,000

1 (1 0.05) (1 0.15)$46,000 ( / ,15%,10)0.15 0.05

$600,000 54,752

C P A A P

C A P

C

−

+=

⎡ ⎤− + ++ ⎢ ⎥−⎣ ⎦= +

• Breakeven green fee:

= +==

$600,000 54,752+ $2,753,740 $819,160

54,752 $2,972,900$54.30

CCC

12



6.38) Let X denote the average number of round-trip passengers per year.

Capital costs:

+=

• Annual crew costs: $225,000 • Annual fuel costs for round trips:

•

CR(15%) ($12,000,000 $2,000,000)( / ,15%,15)A P= −(0.15)($2,000,000)$2,010,171

($1.10)(3,280)(2)(3)(52) $1,125,696= • Annual landing fees:

($250)(3)(52)(2) $78,000= • Annual maintenance, insurance, and catering costs:

,000 $403,500 $75$3, 400

XX

$237,500 $166,000 $75 $403,500 $75X X+ + = +

• Total equivalent annual costs:

AEC(15%) $2,010,171 $225,000 $1,125,696$78

= + ++ +

=

Solving fo

+

r X yields 1,156X = passenger round-trips per yea

or 8

r

1,156 /( )(3) 7.4152 = ≈ passengers per round trip

6.39)

• Model A:

• Model B:

Therefore, select Model B (The ROT 8).

AEC(10%) ($95,000 $12,000)( / ,10%,3) (0.1)($12,000)$3,000$37,574.3 per year

A P= − ++=

AEC(10%) ($120,000 $25,000)( / ,10%,6) (0.1)($25,000)$9,000$33,312 per year

A P= − ++=

13



6.40)

AEC(12%) ($4,500 $250)( / ,12%,10) (0.12)($250)150(0.746)$300 [ ](2,000)($0.05)

0.83$14,564$14,564unit cost

2,000$7.28 per hour

X A P= − +

+ +

=

=

=

AEC(12%) ($3,600 $100)( / ,12%,10) (0.12)($100) $500150(0.746)[ ](2,000)($0.05)

0.80$15,119$15,119unit cost2,000

$7.56 per hour

Y A P= − + +

+

=

=

=

• The difference is $ 0.28 / hour. Therefore, select Brand X.

6.41)

(a

(b) Process A:

) AE(15%) $22,000( / ,15%,4)

[$9,120 $1,280( / ,15%,4)] 2000( / ,15%,4)$684.86

AE(15%) $22,000( / ,15%,4) $7,350$356.6

A

B

A PA G A F

A P

= −+ − −= −= − += −

−$684.86 / 2,000 = −$0.3424 / hour Process B: −$356.6 / 2,000 = −$0.1783 / hour (c) Since neither option provides enough savings to recover the required investment,

the do-nothing alternative (status quo) is a better choice.

14



6.42)

Equivalent annual cost: •

AEC(13%) ($1, 200,000 $60,000)( / ,13%, 20)(0.13)($60,000) $50,000 $40,000$260,083

AEC(13%) ($750,000 $30,000)( / ,13%,10)(0.13)($30,000) $80,000 $30,000$246,596

A

B

A P

A P

= −+ + +== −+ + +=

• Processing cost per ton:

$260,083/(20)(365) $35.63AC = = per ton $246,596 /(20)(365) $33.78BC = = per ton

6.43) Assumption: jet fuel cost

Incinerator B is a better choice.

$1.80= /gallon

• System A : Equivalent annual fuel cost: A1 = ($1.80/gal)(40,000gals/1,000 hours)(2 g an end of-year convention)

[$144,0 ,6%,10%,3)]( / ,10%,3)

) ($10

(0.10)($10,000) $152, 248$189, 438

sys A

A P

,000 hours) $144,000= (assumin

1

.

AEC(10%) 00( /fuel P A=

$152, 248AEC(10% 0,000 $10,000)( / ,10%,3)A P

=−=

+ +=

• System B : Equivalent annual fuel cost: A1 = ($1.80/gal)(32,000gals/1,000 hours)(2,000 hours)

($200, 00 $20,000)( / ,10%,3)

(0.10)($20,000) $121,798$196,179

fuel

s B

P A A P

A P

=

= −

+ +=

• Equivalent operating cost ( including capital cost ) per hour:

System per hour

$115,200=

$121,798

AEC(10%) 0sy

=1

.

AEC(10%) [$115, 200( / ,6%,10%,3)]( / ,10%,3)

$189,438/ 2,000 $94.72A = =

15



System per hour System A is a better choice.

6.44) Since the required service period is 12 years and the future replacement cost for each truck remains unchanged, we can easily find the equivalent annual cost over a 12-year period by simply finding the annual equivalent cost of the first

Truck A: Four replacement cycles are required

,000) $3,000A

$196,179 / 2,000 $98.09B = =

replacement cycle for each truck.

•

AEC(12 ,12%,3)P%) ($15,000 $5,000)( /(0.12)($5

A

$7,763.50

= −+

A P

+=

• Truck B: Three replacement cycles are required

AEC(12%) ($20,0B 00 $8,000)( / ,12%,4)=(0.12)($8,000) $2,000$6,910.80

−

Truck B is a more economical choice.

6.45) (a eriod = 5 years):

+ +=

) Number of decision alternatives (required service p

Alternative Description A1 Buy Machine A and use it for 4 years.

Then lease a machine for one year.

A2 Buy Machine B and use it for 5 years.

A3 Lease a machine for 5 years. A4 Buy Machine A and use it for 4 years.

Then buy another Machine A and use it for one year.

A5 Buy Machine A and use it for 4 years. neThen buy Machine B and use it for o year.

Both A4 and A5 are feasible but m ider these alternat we ne of the machines after one-year use.

(b) With lease, the O&M c id by the leasing company:

ay be not practical alternatives. To consives, ed to know the salvage values

osts will be pa

16



• For A1:

PW(10%) $6,500 ($600 $100)( / ,10%,4)

$10,976

$2,896

P F= − + −

= −

=

Why would one change the oil filter at the end of service life? In this example, we assume that the salvage value of the asset ($600) is only feasible when the asset is maintained properly.

• For A2:

PW(10%) $8,500 $1,000( / ,10%,5)$520( / ,10%,5) $280( / ,10%,4)

$10,042

P FP A P F

= − +− −= −

• For

A2 is the best choice. 6.46)

• Option 1:

• Option 2: (18%) ($0.05 $0.215)(180,000,000)

$47,700,000cos t/lb $47,700,000 /180,000,000

$0.2650 per 1b

= +===

Option 1 is a better choice.

1

$800( / ,10%,4) $200( / ,10%,3)$100( / ,10%,2) $3,000( / ,10%,4)

P A P FP F P F

− −− −

1AEC(10%) $10,976( / ,10%,5)A P=

Note:

2

2AEC(10%) $10,042( / ,10%,5)$2,649

A P==

A3: 3AEC(10%) [$3,000 $3,000( / ,10%,4)]( / ,10%,5)

$3,300P A A P= +

=

1AEC(18%) $200,000(180)( / ,18%,20)(0.08)($200,000)(180)( / ,18%,20)($0.005 0.215)(180,000,000)$46,305,878

cos t/lb $46,305,878 /180,000,000$0.2573 per 1b

A PA F

=−+ +===

AEC 2

17



6.47) Given: Required service period= indefinite, analysis period= indefinite Plan A: Incremental investment strategy:

• Capital investment : 1CR(10%) [$1,500,000

$1,500,000( / ,10%,15)]( / ,10%, )$185,910

P F A P=+ ∞=

• Supporting equipment: P F

A P2CR(10%) [($200,000 $200,000 / 3.1772)( / ,10%,30)]

( / ,10%, )$1,507

= +× ∞

t rate for 15-year period is =

Note that the effective interes15(1 0.1) 1 3.1772+ − =

• Operating cost:

3

[$91,000( / ,10%,15)$182⎜+ ,000( / ,10%,5)( / ,10%,15)]

(10%) ( / ,10%, )$185,000[ $3,000( / ,10%, )]0.10

( / ,10%, 20)]$117,681.33

P AP A P F

A PP G

P F

⎛ ⎞⎟

⎜ ⎟OC = ∞⎜ ⎟

+ + ∞⎜ ⎟⎜ ⎟⎜ ⎟×⎝ ⎠

Note that or

=

2( / , , ) 1/P G i i∞ = ( / ,10%, ) 100P G ∞ = • Total equivalent annual worth:

AEC(10%) $185,910 $1,507 $117,681$305,098

A = + +=

Plan B: One time investment strategy: • Capital investment:

1CR(10%) $1,950,000( / ,10%, )$195,000

A P= ∞=

• Supporting equipment:

2$350,000CR(10%) ( / ,10%, )16.449

A P4

$2,128

= ∞

est rate for 30-year period is

=Note that the effective inter

18



30(1 0.1) 1 16.4494+ − =

• Operating cost:

5,000( / ,10%, )( / ,10%,15)]( / ,10%, )$80,235

P A P FA P

∞× ∞=

worth:

Plan B i

6.48)

• Installed cost per kilowatt

OC(10%) [$105,000(P A= / ,10%,15)$15+

• Total equivalent annual

AEC(10%) $195,000 $2,128 $80,235$277,363

= + +=

s a better choice.

=$84,000

60= $1,400 per kW. But if you consider the

time value of money, say 10% annual interest, the capital cost per kW without considering any salvage value at the end of its service life is as follows:

$84,000( / ,10%,10) $A P 13,671 $227.84 per kW

60 60or

$13,671 $0.026 per kWh

= =

=

• Operating cost per kilowatt-hour:

60 24 365× ×

$19,000 $0.036 (60

)(24)(365)

=

6.49)

• Make option:

19



MakeAEC(14%) $4,582,254or

$4,582,254 / (48 79,815)$1.196 / unit

=

×=

• Buy option:

6.50) Given: annual energy requirement

BuyAEC(14%) (14%) $4,331,127

($405,000 $45,000)( / ,14%,7) (0.14)($45,000) $4,331,127$90,249 $4,331,127$4,421,376

or$4,421,376 / (48 79,815)

$1.154 / unit

CR

A P

= +

= − + += +=

×=

145,000,000,000= BTUs, 1-metric

ton lbs (an approximation figure of 2,000 was mentioned in the case problem), net proceeds from demo

(a) Annual fuel costs for each alternative: • Alternative 1:

Weight of dry coal

2,204.6= lbslishing the old boiler unit $1,000=

145,000,000,000 BTUs(0.75)(14,300)

=

13,519,814= lbs 13,519,814

2,204.66,132.45

=

= tons Annual fuel cost= 6,132.45× $55.5

= $340,350.98 • Alternative 2:

Gas cost.78)(1,000,000)

= $9.5145,000,000,000(0.94)

(0

= $1,660,064.10

Oil cost= $1.45145,000,000,000(0.06)

(0.81)(139,400)

= $111,722.20 Annual fuel cost = $1,660,064.1+ $111,722.2

= $1,771,786.3 (b) Unit cost per steam pound:

20



ssuming a zero salvage value of the investment

00 $145,000 $1,000)( / , 20)

$340,350.98$659,281.25

A P= + −+=

Unit cost 5 / 145,000,000 per steam lb

• Alternat

1,000)( / ,10%,20)$1,771,786.3

A P

• Alternative 1: A

AEC(10%) ($2,570,3 ,10%

= $659,281.2 = $0.004547

ive 2: AEC(10%) ($1,289,340 $

$1,923,166.25

= −+

unit cos

=

t= $1,923,166.25 / 145,000,000 = $0.01326 per steam lb

(c) Select alternative 1.

21



Chapter 7 Rate of Return Analysis Note: Symbol convention---The symbol i* represents the breakeven interest rate that makes the PW of the project equal to zero. The symbol IRR represents the internal rate of return of the investment. For a simple (or pure) investment, IRR = i*. For a nonsimple investment, generally i* is not equal to IRR. 7.1)

$2529.24 $1000( / , %,5) $1,000( / , %,3)6%

F P i F P ii∗= +

=

7.2)

$22,000 = $547.47(P / A, i,48) i = 0.75% per month r = 0.75% ×12 = 9%

ia = (1+ 0.0075)12 −1= 9.38% per year

7.3)

$900 = $37.5(P / A,4.5%,8) + F(P / F ,4.5%,8)0.7032F = $652.6538

F = $928.12

7.4)

$53,900,000 = $80,000(F / P,i,40)i = 17.68%

7.5)

(a) Simple investment: Project A, D. Project C (Simple borrowing) (b) Non-simple investment: Project B



(c) • Project A:

*

PW( ) $22,000 $10,000( / , ,3)$10,000( / , ,3)057.81%

i PP G i

i

= − ++=

=

A i

A i

A i

P F i

• Project B:

*

PW( ) $23,000 $32,000( / , , 2)$25,000( / , ,3)082.72%

i PP F i

i

= − +−=

=

• Project C:

PW( ) $43,233 $18,000( / , ,3)0

i P= −=

i* = 12% Borrowing rate of return.

• Project D:

PW( ) $56,500 $2,500( / , ,1) $6, 459( / , , 2)$88,345( / , ,3)011.37%

i P F iP F i

i∗

= − − −+=

=

(d) The answer could project C on the grounds that that it is not in investment but a

loan. 7.6)

PW(10%) $1,500 ( / ,10%,1) $650( / ,10%,2)( / ,10%,3)

01.6604 $962.84

$580

X P F P FX P F

XX

= − + ++===

2



7.7)

PW(23%) $12,000 $2,500( / , 23%,1) $5,500( / , 23%, 2)( / , 23%, 2)( / , 23%, 2)

0$6,332 0.9743

$6, 498.93

P F P FX P A P F

XX

= − + ++===

7.8)

Use Excel or Cash Flow Analyzer to find the rate of return:

PW( ) $1,000[$50( / , ,12) $50( / , ,5) $4,000]( / , ,15)0

iF A i F A i P F i

= −+ + +=

Solving for i yields

* 12.08%i = 7.9)

(a) Classification of investment projects: • Simple projects: A, B, and E • Non-simple projects: C and D

(b)

−$150 + $60 / (1+ i) + $900 / (1+ i)2 = 0

Let 1/(1 ).X i= + Then,

2

2

$150 $60 $900 0

60 60 4(900)( 150)2(900)

X X

X

− + + =

− ± − −=

Solving for X yields

X1 = 0.37627 and X 2 = −0.44294Solving for i yields only one positive rate of return

3



i*1 = 165.78%

(c) Find by plotting the NPW as a function of interest rate: *i

Project

Number of *i

A 165.78% B 9.67% C 2.30% D 12.45% E 19.43%

7.10) Classification of investment projects

(a) o Simple projects: A, B and D o Nonsimple projects: C

(b)

o Project A: i* = 4.41%o Project B: i* = 42.46%o Project C: * 230.42%i =o Project D: i* = 57.46%

(c) Use the PW plot command provided in Cash Flow Analyzer, or you may use the

Excel’s Chart Wizard. 7.11)

(a) −$15,000 + ($9,229 − $3,000)(P / A, i,8) = 0

Solving for i yields i* = 38.45%

(b) With the geometric expense series

−$15,000 + $9,229(P / A, i,8) − $3,000(P / A1,7%, i,8) = 0

Solving for yields *i i* = 34.79%

7.12)

(a) Rate of return calculation: o Project A: i* = 25.66%o Project B: i* = 57.91%

4



(b)

PW( %) $30,000 $2,000( / , %,1) $6,000( / , %,2)$28,000( / , %,5)

PW( %) $15,000 $10,000( / , %,4) $5,000( / , %,5)PW( %) PW( %)

9.75%

A

B

A B

i P F iP F i

i P A ii i

i∗

= − + ++ ⋅⋅⋅⋅⋅ ⋅ ⋅ += − + +=

=

P F i

P F i

=

7.13)

PW(15%) $150,000 $120,000( / ,15%,5) $25,000( / ,15%,5)$264,694

P A P F= − + +=

7.14) $5,000 $4,840( / , , 2) $1,331( / , ,3) 0P F i P F i− + +

Solving for yields *i* 10%i =

Since this is a simple investment, IRR *i= . Since IRR = MARR, this project breaks even. 7.15)

(a) Since and* 10%i = PW(10%) 0= , we have

PW(10%) $2,000 $800( / ,10%,1) $900( / ,10%,2)( / ,10%,3)

0

P F P FX P F

= − + ++=

Solving for X yields $704X =

(b) Since , the project is acceptable. IRR 8%> 7.16)

PW(15%) $12,500,000 $250,000( / ,15%,5) $50,000( / ,15%,5)$14,000,000( / ,15%,5) $80,000( / ,15%,5) ( / ,15%,5)

0

$2,068,546/ 50 $41,371 per unit

P A P GP F P A A P A

AA

= − − −+ − +

=

==

5



7.17)

PW(10%) $25,000 ($3,000 )( / ,10%,6) $5,000( / ,10%,6)$25,000 4.3553 $13,065.9 $2822.37 0

$8,092.15

A P A P FA

A

= − − − += − + − + =

=

7.18)

(a)

-$20,000

-$15,000

-$10,000

-$5,000

$0

$5,000

$10,000

$15,000

$20,000

0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5

interest rat

PW(i)

(b) and *1 10%i = *

2 40%i =(c)

PW(14%) $1,000,000 $2,500,000( / ,14%,1)$1,540,000( / ,14%, 2)$8,000 0 (Accept the investment.)

P FP F

= − +−= >

6



7.19) Net cash flow (unit: million $):

n

Land

Bldg.

Equip.

Revenue

Expenses

Net Cash Flow

0 -$1.5 -$3 -$4.500001 -$4 -$4.000002 $3.50000 -$1.4000 $2.100003 $3.67500 -$1.4700 $2.205004 $3.85875 -$1.5435 $2.315255 $4.05169 -$1.6207 $2.43101

11 $5.42965 -$2.1719 $3.2577912 $5.42965 -$2.1719 $3.2577913 $5.42965 -$2.1719 $3.2577914 $2 $1.4 $0.5 $5.42965 -$2.1719 $7.15779

Rate of return calculation:

1PW( ) $4.5 $4( / , ,1) $2.1( / ,5%, ,10)( / , ,1)$3.25779( / , , 2)( / , ,11) $7.15779( / , ,14)

0

i P F i P A i P F iP A i P F i P F i

= − − ++ +

=

Solving for i yields * 24.85%i =

Since this is a simple investment, At a MARR of 15%, the project is economically attractive.

*IRR 24.85%.i= =

7.20)

(a) PW( ) $20 $8( / , ,1) $17( / , , 2) $19( / , ,3)

$18( / , , 4) $10( / , ,5) $3( / , ,6)0

i P F i P F i P F iP F i P F i P F i

= − + + ++ + +=

This is a simple investment. Therefore, Since IR the project is a acceptable.

*IRR 60.52%.i= = R 18%,>

(b) IRR 67.03%=(c) IRR 48.06%=

7



7.21)

(20%) $50,000( / , 20%,10) 0.1 ( / , 20%,10)0

0.98385 $209,625

$213,066.02

PW C P A C P F

C

C

= − + +==

=

7.22)

• Fidelity rate of return: 22$5, 245(1 ) $289,556

20%i

i+ =

=

• After 10 years, Wal-Mart investment grows to:

10$1,650(1.32) $26,499= • Reinvest in Fidelity:

22$26,499(1.20) $1,462,854= • Select (c)

7.23)

(a) Project A: IRR = 6.37% Project B: IRR = 9.18% (b) Neither project is acceptable (c)

n Project A Project B A-B 0 -$150,000.00 -$120,000.00 -$30,000.00 1 $30,000.00 $25,000.00 $5,000.00 2 $25,000.00 $15,000.00 $10,000.00 3 $120,000.00 $110,000.00 $10,000.00

Neither project. Since A-BIRR 7.87% 15% (MARR)= − < , project B is better choice if there is no “do-nothing” alternative.

7.24)

(a) Project A: IRR = 18.33% Project B: IRR = 23.77%(b) Both projects are acceptable

8



(c) n Project A Project B A-B 0 -$100,000.00 -$100,000.00 $0.00 1 $10,500.00 $70,000.00 -$59,500.00 2 $60,000.00 $50,000.00 $10,000.00 3 $80,000.00 $20,500.00 $59,500.00

IRRA-B = 8.76% < 10%( MARR) Project B is better choice. 7.25)

Option 1: Buy a certificate, Option 2: Purchase a bond, and assume that MARR 9%=

Net Cash Flow

n Option 1 Option 2 Option 1 – Option 2 0 -$10,000 -$10,0001 0 1,0002 0 1,0003 0 1,0004 0 1,0005 16,105 11,000

0 -1,000 -1,000 -1,000 -1,000 5,105

The rate of return on incremental investment is

*1 2 10% 9%i − = >

Thus, Option 1 is a better choice. 7.26)

n Project 1 Project 2 1-2 0 -$2,200.00 -$2,000.00 -$200.00 1 $1,200.00 $1,200.00 $0.00 2 $1,650.00 $1,400.00 $250.00

IRR1-2 = 11.8% < MARR and , thus select (c). IRR 2 = 18.88% > MARR 7.27) Determine the cash flow on incremental investment:

Net Cash Flow n Project A Project B B - A 0 -$2,000 -$3,000 -$1,0001 $1,400 $2,400 $1,0002 $1,640 $2,000 $360

9



* 28.11% 15%B Ai − = >


(a) IRR on the incremental investment:

Net Cash Flow n Project A1 Project A2 A2 – A1 0 -$10,000 -$12,000 -$2,0001 $5,000 $6,100 $1,1002 $5,000 $6,100 $1,1003 $5,000 $6,100 $1,100

*2 1 29.92%A Ai − =

(b) Since it is an incremental simple investment, A2-A1IRR 29.92%= 10%> .

Therefore, select project A2. 7.29)

(a) n A1 A2 A2 – A1 0 -$16,000 -$20,000 -$4,0001 $7,500 $5,000 -$2,5002 $7,500 $15,000 $7,5003 $7,500 $8,000 $500

2 1IRR 13.08%A A− =

(b) Select Project A2.

7.30) Incremental cash flows (Model A – Model B):

n A – B 0 -$2,3761 $02 $03 $04 $2,500

IRR 1.28%A B− =

If MARR Model A is preferred. 1.28%,<

10



7.31) Given: IRR for Model A: 10.49%, IRR for Model B: 12.34%; Incremental cash flows:

n B – A 0 -$2,000

1-19 $35020 $450

IRR 16.75%B A− =

Model B is a better choice. 7.32) Let A0 current practice,= A1= just-in-time system, A2 = stock less supply system.

• Comparison between A0 and A1:

n A0 A1 A1 – A0 0 0 -$2,500,000 -$2,500,000

1-8 -$5,000,000 -$2,900,000 $2,100,000

*1 0 1 0IRR 83.34% 10%A A A Ai − −= = >

A1 is a better choice.

• Comparison between A1 and A2:

n A2 A1 A2 – A1 0 -$5,000,000 -$2,500,000 -$2,500,000

1-8 -$1,400,000 -$2,900,000 $1,500,000

*2 1 2 1IRR 58.49% 10%A A A Ai − −= = >

A2 is a better choice. That means that the stockless supply system is the final choice.

7.33)

(a)

i1* = 85.08%, i2

* = 57.61%, and i 3* = 44.30%

(b) • Project 1 versus Project 2:

n Project 2 – Project 1 0 -$4,0001 $7,0002 -$1,900

This is a non-simple incremental investment. So, we may abandon the IRR

11



analysis and select the project based on the PW decision rule.

2 1PW(15%) $650− =

Select Project 2.

• Project 2 versus Project 3:

n Project 2 – Project 3 0 -$3,0001 $6,0002 -$1,400

This is another non simple incremental investment, so we may use again the PW decision rule.

2 3PW(15%) $1,158− =

Again, select Project 2.

Comments: If you want to apply the IRR decision rule to the non-simple investments, you should apply the net investment test and make the selection by calculating the return on invested capital (or true internal rate of return) as discussed in Chapter 7A. 7.34)

(a) 25.99%BIRR =(b) (15%) $10,000 $5,500( / ,15%,3) $2,558APW P A= − + =(c) Incremental analysis:

Net Cash Flow

n Project A Project B B – A 0 -$10,000 -$20,000 -$10,0001 $5,500 0 -$5,5002 $5,500 0 -$5,5003 $5,500 $40,000 $34,500

Since select project B. IRR 24.24% 15%,B A− = >

7.35) All projects would be acceptable because individual ROR exceed the MARR.

Based on the incremental analysis, we observe the following relationships: 2 1IRR 10% 15%A A− = < (Select A1)

3 1IRR 18% 15%A A− = > (Select A3)

12



3 2IRR 23% 15%A A− = > (Select A3) Therefore, A3 is the best alternative.

7.36) From the incremental rate of return table, we can deduce the following relationships:

2 1IRR 9% 15%A A− = < (Select A1)

3 2IRR 42.8% 15%A A− = > (Select A3)

4 3IRR 0% 15%A A− = < (Select A3)

5 4IRR 20.2% 15%A A− = > (Select A5)

6 5IRR 36.3% 15%A A− = > (Select A6) It is necessary to determine the preference relationship among A1, A3, and A6.

3 1IRR 16.66% 15%A A− = > (Select A3)

6 3IRR 20.18% 15%A A− = > (Select A6)

6 1IRR 18.24% 15%A A− = > (Select A6) A6 is the best alternative.

7.37)

Relationships:

IRR (D1-D2) = 27.62% > 15% (Select D1)

IRR (D1-D3) = 14.26% < 15% (Select D3)

IRR (D1-D4) = 25.24% > 15% (Select D1)

IRR (D3-D2) = 30.24% > 15% (Select D3)

IRR (D2-D4) = 17.34% > 15% (Select D2)

IRR (D3-D4) = 16.14% > 15% (Select D3)

D3 is the best alternative.

13



7.38) For each power saw model, we need to determine the incremental cash flows over the “by-hand” operation that will result over a 20-year service life.

Power Saw

Category Model A Model B Model C Investment cost $4,000 $6,000 $7,000 Salvage value $400 $600 $700 Annual labor savings $1,296 $1,725 $1,944 Annual power cost $400 $420 $480 Net annual savings $896 $1,305 $1,464

Net Cash Flow

n Model A Model B Model C 0 -$4,000 -$6,000 -$7,000 1 $896 $1,305 $1,464 2 $896 $1,305 $1,464

20 $400 + $896 $600 + $1,305 $700 + $1,464 IRR 22.03% 21.35% 20.46%

PW(10%) $3,688 $5,199 $5,568

• Model A versus Model B: ( ) $2,000 $409( / , , 20) $200( / , , 20)

0IRR 19.97% 10%

B A

B A

PW i P A i P F i−

−

= − + +== >

Select Model B. • Model B versus Model C:

( ) $1,000 $159( / , , 20) $100( / , , 20)0

IRR 15.03% 10%

C B

C B

PW i P A i P F i−

−

= − + +== >

Select Model C. The PW rule also selects Model C, as indicated in the table above.

14



7.39) With the least common multiple of 6 project years,

Net Cash Flow n Project A Project B B – A 0 -$100 -$200 -$1001 $60 $120 $602 $50 $150-$200 -$1003 $50-$100 $120 $1704 $60 $150-$200 -$1105 $50 $120 $706 $50 $150 $100

Since the incremental cash flow series is a nonsimple investment, we may abandon the IRR analysis, and use the PW decision rule.

PW(15%) $100 $60( / ,15%,1)$100( / ,15%,6)

$3.48

B A P FP F

− = − ++ +=

Since or , select PW(15%) 0,B A− > PW(15%) PW(15%)B A>project B.

Comments: Even though the incremental flow is a nonsimple, it has a unique rate of return. As shown in Chapter 7A, this incremental cash flow will pass the net investment test, indicating that the incremental cash flow is a pure investment.

IRR 15.98% 15%B A− = >


(a) Since there is not much information given regarding the future replacement options and required service period, we may assume that the required service period is indefinite and both projects can be repeated at the same cost in the future.

(b) The analysis period may be chosen as the least common multiple of project lives,

which is 3 years.

n A2 – A1 0 -$5,0001 $02 $03 $15,000

2 1IRR 44.22%A A− =

15



The MARR must be greater than 44.22% for Project A1 to be preferred. 7.41)

(a)

*

PW( ) $1,250,000 $731,500( / , ,15) $80,000( / , ,15)058.47%

i P A i

i

= − + +=

=

P F i

A i

A i

A i

(c) IRR exceeds Marco’s MARR, the project is attractive and should be accepted.

* 58.47% MARR ( 18%)i = > = 7.42)

(a) Analysis period of 40 years (unit: thousand $): • Without “mothballing” cost:

1

*

PW( ) $1,500,000 ($207,000 $69,000)( / ,0.05%, , 40)08.95%

i P

i

= − + −=

=

• With “mothballing” cost of $0.75 billion:

1

*

PW( ) $1,500,000 ($207,000 $69,000)( / ,0.05%, , 40)$750,000( / , , 40)

08.77%

i PP F i

i

= − + −−

=

=

For a 40-year analysis period, the drop of IRR with the mothballing cost is only 0.18%, which is relatively insignificant.

(b) Analysis period of 25 years (unit: thousand $): • Without “mothballing” cost:

1

*

PW( ) $1,500,000 ($207,000 $69,000)( / ,0.05%, , 25)07.84%

i P

i

= − + −=

=

• With “mothballing” cost of $0.75 billion:

16



1

*

PW( ) $1,500,000 ($207,000 $69,000)( / ,0.05%, , 25)$750,000( / , , 25)

06.80%

i PP F i

i

= − + −−

=

=

A i

For a 25-year analysis period, the drop of IRR with the mothballing cost is

about 1.04%, which is relatively significant. 7.43)

(a) Assumptions required: • There is no information regarding the expected cash flows from the current

operation if B&E Cooling decides to defer the introduction of the absorption chiller technology for 3 years. Therefore, we need to make an explicit assumption of the expected cash flow for the first three years if B&E Cooling decides to defer the decision. Assume that the annual cash flow during this period would be X.

• Another assumption we have to make is about the analysis period. Assuming that the firm will be in business for an indefinite period, we also need to make an explicit assumption regarding the future cooling technology. Since there is no information about the future cooling technology options, we may assume that the best cooling technology will be the absorption technology that will be introduced 3 years from now. Therefore, if B&E Cooling decides to select Option 1, we assume that, at the end of 8 years, Option 2 (the best cooling technology at that time) will be adopted for an indefinite period.

(b) Investment decision:

• Present worth analysis: First, we will determine the equivalent present worth for each option:

1

2

PW( ) $6M $9M( / , ,8) $1M( / , ,8)[ $5M $4M( / , ,8) $2M( / , ,8)]( / , ,8)

( / , ,8)PW( ) ( / , ,3)

[ $5M $4M( / , ,8) $2M( / , ,8)]( / , ,8)

( / , ,3)

i P A i P F iP A i P F i A P i

iP F i

i X P A iP A i P F i A P i

iP F i

= − + +− + +

+

×=

− + ++

×

Now we can determine the value X that makes the two options economically equivalent at an interest rate of 15%. In other words, if we evaluate the two present worth functions at 15%i = , we have

1

2

PW(15%) $41.32MPW(15%) 2.2832 $13.29MX

== +

17



Letting 1PW(15%) PW(15%)2= and solving for X gives

$12.28MX =

As long as the current operation continues to generate annual net revenue of at least $12.28 million for 3 years, Option 2 is a better choice.

• Rate of return analysis: The present worth analysis above indicates that, if

, the break-even rate of return on incremental investment is .

X = $12.28M*1 2 15%i − =

Therefore, the ultimate choice will depend on the level of annual revenues generated during the first 3 years when the advanced cooling technology is deferred. Clearly, if

$12.28MX < , then , Option 1 is preferred. *1 2 15%i − >

7.44)

n Current Pump(A) Larger Pump(B) B-A0 $0 -$1,600,000 -$1,600,0001 $10,000,000 $20,000,000 $10,000,0002 $10,000,000 $0 -$10,000,000

IRR = 25%400%

The incremental cash flows result in multiple rates of return (25% and 400%), so we may abandon the rate of return analysis. Using the PW analysis,

PW(20%) $1.6M $10M( / , 20%,1) $10M( / , 20%, 2)

$0.21M 0Reject the larger pump.

P F P F= − + −= − <

Comments: If we follow the procedure outlined in Appendix 7A, we will find the return on invested capital to be 4.16% at MARR of 20%, so we will reject the larger pump.

7.45) Not provided.

18



Chapter 8 Accounting for Depreciation and Income Taxes Note: For the most up-to-date depreciation and income tax information, consult the book’s website at “http://www.prenhall.com/park” and click on “Tax Information” 8.1) (a), (b), (e), (f), (h) (amortization, rather than depreciation) 8.2) The loss of value is defined as the purchase price of an asset less its market value,

also known as economic depreciation.

• Economic depreciation during 4-year ownership: $5,000 – $2,300 = $2,700• Economic depreciation during the last 3-year ownership: $2,300 – 0 = $2,300

8.3)

Total property value with the warehouse: land building

Original cost $65,000 $55,000 Adjustments to basis

add: new warehouse $50,000 demolition expense $8,000 subtract: building loss ($55,000) adjusted cost basis $65,000 $58,000

Total value = $65,000 + $58,000 = $123,000 Note that the old house that was demolished has no value. This loss may be deductible for tax purpose, but this should not be added to the cost basis of the new asset. In general, the property’s entire basis is allocated to the land only, if the company intends to demolish the building when they acquire property for business use. Then, the cost basis is increased by the net cost of demolition. (The demolition expense can be treated as a site preparation expense.)

Cost basis for depreciation: Cost basis = $8,000 + $50,000 = $58,000

8.4) Trade-in allowance:

Old molding machine (Book value) $15,000 less: trade-in allowance $20,000 Unrecognized gain $5,000 Cost of a new molding machine $105,000 less: unrecognized gain on trade-in allowance ($5,000) Cost basis of the new molding machine $100,000

Instructor Solutions Manual to accompany Fundamentals of Engineering Economics, Second Edition, by Chan S. Park.

ISBN-13: 9780132209618. © 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.This material is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage

in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise.

http://www.prenhall.com/

Comments: If the old molding machine is sold on the market (instead of traded in), there will be no unrecognizable gain. In that situation, the cost basis for the new molding machine will be just $105,000.

8.5) Trade-in allowance:

Old grinder (Book value) $12,000 less: trade-in allowance $10,000 unrecognized loss ($2,000) cost of a new grinder $45,000 plus: unrecognized loss on trade-in allowance $2,000 cost basis of new grinder $47,000

Comments: If the old grinder is sold on the market (instead of traded in), there will be no unrecognized loss. In that situation, the cost basis for the new grinder will be just $45,000. 8.6) Cost basis for flexible manufacturing cells:

flexible manufacturing cells (@$400,000×3) $1,200,000

freight charges $30,000 handling fee $15,000 site preparation costs $50,000 start up and testing costs $18,000 special wiring and material costs $2,000 cost basis $1,315,000

(Note: start-up and testing costs = $15 x 40 x 6 x 5 = $18,000) 8.7) Depreciation allowances and book values: (a) depreciation rate = 1/5 for SL,

(b) Depreciation rate = 2/5 for DDB

SL DDB

n Dn Bn Dn Bn

0 $110,000 $110,000 1 $20,000 $90,000 $44,000 $66,000 2 $20,000 $70,000 $26,400 $39,600 3 $20,000 $50,000 $15,840 $23,760 4 $20,000 $30,000 $9,504 $14,256 5 $20,000 $10,000 $4,256 $10,000

2



8.8) Given: I = $50,000, S = $5,000, N = 8 years;

DDB

n Dn Bn

0 $50,000 1 $12,500 $37,500 2 $9,375 $28,125 3 $7,031 $21,094 4 $5,273 $15,820 5 $3,955 $11,865 6 $2,966 $8,899 7 $2,225 $6,674 8 $1,674 $5,000

8.9) Given: I = $200,000, n = 3 years, N = 8 years; Assuming that N ranges from 5 years to 8 years and the salvage value is no greater than $21,000, the DDB rates are as follows:

DDB

n Dn Bn

0 $200,000 1 $50,000 $150,000 2 $37,500 $112,500 3 $28,125 $84,375 4 $21,094 $63,281 5 $15,820 $47,461 6 $11,865 $35,596 7 $8,899 $26,697 8 $6,674 $20,023

8.10) DDB switching to SL in year 5:

With switching

n Dn Bn

0 $45,000 1 $12,857 $32,143 2 $9,184 $22,959 3 $6,560 $16,399 4 $4,685 $11,714 5 $3,905 $7,809 6 $3,905 $3,905 7 $3,905 $0

3



8.11) Given: I = $90,000, S = $12,000, N = 6 years (a)

Without switching

DDB

n Dn Bn

0 $90,000 1 $30,000 $60,000 2 $20,000 $40,000 3 $13,333 $26,667 4 $8,889 $17,778 5 $5,778 $12,000 6 $0 $12,000

(b) DDB switching to SL:

with switching

From DDB to SL

n Dn Bn

0 $90,000 1 $30,000 $60,000 2 $20,000 $40,000 3 $13,333 $26,667 4 $8,889 $17,778 5 $5,778 $12,000 6 $0 $12,000

Comments: The answer is unchanged because the salvage value is relatively high.

8.12)

(a) 1 1.5 0.35

α ⎛ ⎞= =⎜ ⎟⎝ ⎠

(b) 1 (0.3)(15,000) $4,500D = = (c) 4

4 (15,000)(1 0.3) $3,601.5B = − =

8.13) Given: I = $30,000, N = 5 years, S = $3,000

4



(a) (b) DDB SL

n Dn Bn Dn Bn

0 $30,000 $30,000 1 $5,400 $24,600 $12,000 $24,600 2 $5,400 $19,200 $7,200 $19,200 3 $5,400 $13,800 $4,320 $13,800 4 $5,400 $8,400 $2,592 $8,400 5 $5,400 $3,000 $888 $3,000

8.14) Given I = $78,000, S = $8,000, N = 12 years

DDB SL

n Dn Bn n Dn Bn

0 $78,000 0 $78,000 1 $13,000 $65,000 1 $5,833 $72,167 2 $10,833 $54,167 2 $5,833 $66,333 3 $9,028 $45,139 3 $5,833 $60,500 4 $7,523 $37,616 4 $5,833 $54,667 5 $6,269 $31,346 5 $5,833 $48,833 6 $5,224 $26,122 6 $5,833 $43,000 7 $4,354 $21,768 7 $5,833 $37,167 8 $3,628 $18,140 8 $5,833 $31,333 9 $3,023 $15,117 9 $5,833 $25,500

10 $2,519 $12,597 10 $5,833 $19,667 11 $2,100 $10,498 11 $5,833 $13,833 12 $1,750 $8,748 12 $5,833 $8,000

(a)

( )$78,000 $8,000$5,833.33

12D

−= =

(b) ( ) ( )2 33 2 3 $78,000 1 (2 /12) $78,000 1 (2 /12)

$9,027.78D B B= − = − − −

=

8.15) Allowed depreciation amount

5



D =

55,000250,000

($85,000 − $5,000) = $17,600

8.16)

5,000 hrs5000 ($60,000 $8,000)

50,000$5,200

D = −

=

8.17)

Truck A:

25,000 ($50,000 $5,000) $5,625200,000

D = − =

Truck B:

12,000 ($25,000 $2,500) $2, 250120,000

D = − =

Truck C:

15,000 ($18,500 $1,500) $2,550100,000

D = − =

Truck D:

20,000 ($35,600 $3,500) $3, 210200,000

D = − =

8.18)

(a) Book depreciation:

Truck

1

2

22,000 ($25,000 $2,000) $2,530200,00025,000 ($25,000 $2,000) $2,875

200,000

D

D

= −

= −

=

=

Lathe and building:

Lathe Building

DDB SL

n Dn Bn n Dn Bn 0 $45,000 0 $800,000 1 $7,500 $37,500 1 $14,000 $786,000

6



2 $6,250 $31,250 2 $14,000 $772,000 3 $5,208 $26,042 3 $14,000 $758,000 4 $4,340 $21,701 4 $14,000 $744,000 5 $3,617 $18,084 5 $14,000 $730,000 6 $3,014 $15,070 6 $14,000 $716,000 7 $2,512 $12,559 7 $14,000 $702,000 8 $2,093 $10,466 8 $14,000 $688,000 9 $1,744 $8,721 9 $14,000 $674,000

10 $1,454 $7,268 10 $14,000 $660,000

11 $1,211 $6,056

……

.

…..

…… .

12 $1,009 $5,047 50 $14,000 $100,000

(b) Allowed annual depreciation:

With switching From DDB to SL

n Dn Bn 0 $45,000 1 $7,500 $37,500 2 $6,250 $31,250 3 $5,208 $26,042 4 $4,340 $21,701 5 $3,617 $18,084 6 $3,014 $15,070 7 $2,512 $12,559 8 $2,093 $10,466 9 $1,866 $8,599

10 $1,866 $6,733 11 $1,866 $4,866 12 $1,866 $3,000

The switching occurs at the 9th year.

8.19) (a) Straight-line

SL n Dn Bn 0 $135,000 1 $12,300 $122,700 2 $12,300 $110,400 3 $12,300 $98,100 4 $12,300 $85,800 5 $12,300 $73,500

7



6 $12,300 $61,200 7 $12,300 $48,900 8 $12,300 $36,600 9 $12,300 $24,300

10 $12,300 $12,000 (b) Units of production

23, 450 ($135,000 $12,000) $11,537.4250,000

D = − =

(c) Working hours 2, 450 ($135,000 $12,000) $10,045

30,000D = − =

(d) DDB

without switching DDB

n Dn Bn 0 $135,000 1 $27,000 $108,000 2 $21,600 $86,400 3 $17,280 $69,120 4 $13,824 $55,296 5 $11,059 $44,237 6 $8,847 $35,389 7 $7,078 $28,312 8 $5,662 $22,649 9 $4,530 $18,119

10 $3,624 $14,496 8.20) Given: I = $37,000, S = $6,000, N = 8 years, and 5-year MACRS

Book Depreciation: SL n Dn Bn

0 $37,000

1 $3,875 $33,125 2 $3,875 $29,250 3 $3,875 $25,375 4 $3,875 $21,500 5 $3,875 $17,625 6 $3,875 $13,750 7 $3,875 $9,875 8 $3,875 $6,000

8



Tax Depreciation: 5-year MACRS n Dep. Rate Dn Bn 0 $37,000.00 1 0.2 $7,400.00 $29,600.00 2 0.32 $11,840.00 $17,760.00 3 0.192 $7,104.00 $10,656.00 4 0.1152 $4,262.40 $6,393.60 5 0.1152 $4,262.40 $2,131.20 6 0.0576 $2,131.20 $0.00

8.21)

(a) Cost basis: $150,000 $5,000 $155,000+ = (b)

n Dep. Rate Dn Bn 0 $155,000 1 0.1429 $22,143 $132,857 2 0.2449 $37,959 $94,898 3 0.1749 $27,114 $67,784 4 0.1249 $19,367 $48,417 5 0.0892 $13,834 $34,584 6 0.0892 $13,834 $20,750 7 0.0892 $13,834 $6,917 8 0.0446 $6,917 ($0)

8.22) Let I denote the cost basis for the equipment.

n Dep. Rate Dn Bn 0 $100,000 1 0.1429 $14,286 $85,714 2 0.2449 $24,490 $61,224 3 0.1749 $17,493 $43,732 4 0.1249 $12,495 $31,237 5 0.0892 $8,925 $22,312 6 0.0892 $8,925 $13,387 7 0.0892 $8,925 $4,462 8 0.0446 $4,462 ($0)

9



8.23) Given: I = $58,000, S = $8,000, N = 6 years, tax depreciation method = 7 year MACRS property class

n Dep. Rate Dn Bn 0 $58,000 1 0.1429 $8,286 $49,714 2 0.2449 $14,204 $35,510 3 0.1749 $10,146 $25,364 4 0.1249 $7,247 $18,117 5 0.0892 $5,176 $12,941 6 0.0892 $5,176 $7,765 7 0.0892 $5,176 $2,588 8 0.0446 $2,588 ($0)

8.24) Given: I = $22,000 and 7-year MACRS property

n Dep. Rate Dn Bn 0 $22,000 1 0.1429 $3,143 $18,857 2 0.2449 $5,388 $13,469 3 0.1749 $3,848 $9,621 4 0.1249 $2,749 $6,872 5 0.0892 $1,963 $4,909 6 0.0892 $1,963 $2,945 7 0.0892 $1,963 $982 8 0.0446 $982 ($0)

8.25) Given: machine tool furniture warehouse$5,000, $125,000, and $335,000I I I= = =

Machine tool

n Dep. Rate Dn Bn 0 $5,000 1 0.3333 $1,667 $3,333 2 0.4444 $2,222 $1,111 3 0.1481 $741 $370 4 0.0741 $370 $0

10



CNC machine

n Dep. Rate Dn Bn 0 $125,000 1 0.1429 $17,857 $107,143 2 0.2449 $30,612 $76,531 3 0.1749 $21,866 $54,665 4 0.1249 $15,618 $39,046 5 0.0892 $11,156 $27,890 6 0.0892 $11,156 $16,734 7 0.0892 $11,156 $5,578 8 0.0446 $5,578 ($0)

Warehouse

n Dep. Rate Dn Bn 0 $335,000 1 0.0139 $4,653 $330,347 2 0.0256 $8,590 $321,757 3 0.0256 $8,590 $313,168

……

.

……

.

……

.

…..

39 0.0256 $8,590 $3,937 40 0.0118 $3,937 $0

8.26) Given: Residential real property (27.5-year), I = $120,000

n Dep. Rate Dn Bn 0 $170,000 1 0.0076 $1,288 $168,712 2 0.0364 $6,182 $162,530 3 0.0364 $6,182 $156,348 4 0.0364 $6,182 $150,167 5 0.0197 $3,348 $146,818

8.27) Given: Residential real property (27.5 year), I = $150,000

(a)

1100% 5.5 ($150,000)27.5 12

(0.016667)($150,000) $2500

D ⎛ ⎞= ⎜ ⎟⎝ ⎠

= =

11



(b) Total amount of depreciation over the 4-year ownership, assuming that the asset is sold at the end of the 4th calendar year:

n Dep. Rate Dn Bn 0 $150,000 1 0.01667 $2,500 $147,500 2 0.03636 $5,454 $142,046 3 0.03636 $5,454 $136,592 4 0.03485 $5,227 $131,365

Total amount of depreciation allowed = $18,635. Note that the 4th year depreciation reflects the mid-month convention (11.5 months).

( )4 $150,000 $2,500 2($5, 454) $5, 227$150,000 $18,635$131,365

B = − + +

= −=

8.28) Given: I = $1,000,000, 39 years-MACRS real property

n Dep. Rate Dn Bn 0 $1,000,000 1 0.007479 $7,479 $992,521 2 0.025641 $25,641 $966,880

……

.

……

.

…..

……

. 8.29) Types of depreciation method

(a). B (b). A (c). D (d). C (e). None

8.30)

(a) Book depreciation methods:

Straight-line method:

SL n Dn Bn Cum. Dn 0 $80,000 1 $14,400 $65,600 $14,400

12



2 $14,400 $51,200 $28,800 3 $14,400 $36,800 $43,200 4 $14,400 $22,400 $57,600 5 $14,400 $8,000 $72,000

DDB method:

DDB n Dn Bn Cum. Dn 0 $80,000 1 $32,000 $48,000 $32,000 2 $19,200 $28,800 $51,200 3 $11,520 $17,280 $62,720 4 $6,912 $10,368 $69,632 5 $2,368 $8,000 $72,000

(b) Tax depreciation: 7-year MACRS

n Dep. Rate Dn Bn Cum. Dn

0 $80,000 1 0.1429 $11,429 $68,571 $11,429 2 0.2449 $19,592 $48,980 $31,020 3 0.1749 $13,994 $34,985 $45,015 4 0.1249 $9,996 $24,990 $55,010 5 0.0892 $7,140 $17,850 $62,150 6 0.0892 $7,140 $10,710 $69,290 7 0.0892 $7,140 $3,570 $76,430 8 0.0446 $3,570 $0 $80,000

(c) Trade-in allowance

Book value of the old equipment (B3) $34,985 Less: Trade-in allowance $10,000 Unrecognized loss ($24,985) Cost of the new equipment $92,000 plus: unrecognized loss on trade-in $24,985 Cost basis of the new equipment $116,985

Comments: If the old equipment was sold on the market (instead of trade-in), there would be no unrecognized loss. In that situation, the cost basis for the new equipment will be just $92,000. No half-year convention is assumed in the analysis.

13



8.31) (a) and (b):

n Dep. Rate Dn Bn Bn-1 Property

taxes 0 $3,800,000 1 0.1429 $542,857 $3,257,143 $3,800,000 $45,600 2 0.2449 $930,612 $2,326,531 $3,257,143 $39,086 3 0.1749 $664,723 $1,661,808 $2,326,531 $27,918 4 0.1249 $474,802 $1,187,005 $1,661,808 $19,942 5 0.0892 $339,144 $847,861 $1,187,005 $14,244 6 0.0892 $339,144 $508,717 $847,861 $10,174 7 0.0892 $339,144 $169,572 $508,717 $6,105 8 0.0446 $169,572 ($0) $169,572 $2,035

8.32) Net income calculation:

Gross income 34,000,000$ Expenses: Sarlaries 5,000,000$ Wages 4,000,000$ Depreciation 1,000,000$ Loan interest 210,000$

Taxable income 23,790,000$ Income Taxes 8,326,499$

Net income 15,463,501$

Note: Using the tax formula in Table 8.11,

Total income taxes = $6,416,666 + 0.35($23,790,000 - $18,333,333) = $8,326,499.45 8.33) (a) Taxable income:

$2,500,000 - $1,280,000 - $128,000 = $1,092,000

(b) Income tax calculation using the tax formula from Table 8.11:

$113,900 + 0.34($1,092,000 - $335,000) = $371,280

14



8.34) (a) Depreciation expenses:

Building (39-year class, placed in service in February):

building100% 10.5$400,000

39 12$400,000(2.2436%)$8,974

D ⎛ ⎞⎛= ⎜ ⎟⎜⎝ ⎠⎝

==

⎞⎟⎠

Equipment (5-year MACRS):

( )equipment $200,000 20% $40,000D = = Total depreciation allowed in year 2008:

$8,974 $40,000 $48,974D = + =

(b) Tax liability:

Sales revenue $2,500,000 Expenses: Cost of goods sold $800,000 Bond interest $50,000 Depreciation $48,974 Taxable income $1,601,026 Income taxes $544,349 Net income $1,056,677

Note: Income taxes = $113,900 + 0.34($1,601,026 - $335,000) = $544,349 from Table 8.11. 8.35)

(a) Taxable gain:

Ordinary gains proceeds from old equipment - book value$23,000 $20,000$3,000

== −=

(b) Taxable income:

15



Gross income 2,250,000$ Interest income 6,000$ Bond interest income 4,000$ Expenses: Labor 550,000$ Materials 385,000$ Depreciation 132,500$ Interest 22,200$ Rental 45,000$ , ,$Taxable income 1,125,300$ Income taxes 382,602$ Net income 742,698$ $

Note: Income taxes = $113,900 + 0.34($1,125,300 - $335,000) = $382,602

Note: Ordinary gains are not included in this calculation, even though these gains will be treated as ordinary income. Of course, these figures can be included to find the total tax liabilities.

(c) Marginal and average tax rates:

Marginal tax rate 34%Average (effective) tax rate $382,602 / $1,125,300

34%

===

(d) Net cash flow:

Net income 742,698$ Adjustments: Add depreciation 132,500$ Proceeds from sale 23,000$ Subtract gains tax (1,020)$ Net cash flow 897,178$

8.36) (a) Income tax liability:

16



Gross revenues 1,200,000$ Expenses: Manufacturing 450,000$ Operating 120,000$ Interest 40,000$ Taxable operating income 590,000$ Adjustment: loss 15,000$ Taxable income 575,000$ Income taxes 195,500$ Net income 379,500$

Note 1: book loss = $60,000 - $75,000 = ($15,000)

Note 2: Income taxes = $113,900 + 0.34($575,000 - $335,000) = $195,500

(b) Operating income:

Taxable operating income $590,000Income taxes $200,600Net operating income $389,400 Note 1: The loss from disposal of the asset is not a part of operating activities, so it is not included in the operating income calculation. Note 2: Income taxes = $113,900 + 0.34($590,000 - $335,000)

= $200,600

(c) Net cash flow:

Net income 379,500$ Adjustments: Add depreciation 45,000$ Proceeds from sale 60,000$ Short-term loan 50,000$ Net cash flow 534,500$

8.37) (a) Disposed of in year 3:

allowed depreciation $76,000(0.20 0.32 0.192 / 2)$46,816

book value $76,000 $46,816$29,184

loss $20,000 $29,184 ($9,184)

= + +== −== − =

(b) Disposed of in year 5:

17



allowed depreciation $76,000(0.20 0.32 0.192

0.1152 0.1152 / 2)$67, 244.8

book value $76,000 $67, 244.8$8,755.2

Taxable gains $10,000 $8,755.2 $1, 224.8

= + ++ +== −== − =

(c) Disposed of in year 6:

allowed depreciation $76,000book value $0

Taxable gains $5,000

===

8.38)

allowed depreciation $300,000(0.1429 0.2449 0.17490.1249 0.0893 / 2)$219,675

book value $300,000 $219,675$80,325

= + ++ +== −=

(a) If sold at $10,000:

losses $10,000 $80,325 ($70,325)loss credit $70,325(0.34) $23,911

net loss ($70,325) $23,911 ($46,414)

= − == == + =

(b) If sold at $125,460:

gains $125,460 $80,325 $45,135gains tax $45,135(0.34) $15,346net gain $45,135 $15,346 $29,789

= − == == − =

(c) If sold at $200,000:

gains $200,000 $80,325 $119,675gains tax $119,675(0.34) $40,689.5net gain $119,675 $40,689.5 $78,985.5

= − == == − =

18



8.39) (a) Taxable operating income (Do not include ordinary gains):

Revenues: Gross income 4,250,000$ Expenses: Labor 1,550,000$ Materials 785,000$ Depreciation 332,500$ Office supplies 15,000$ Interest 42,200$ Rental 45,000$ , ,$Taxable income 1,480,300$ Income taxes 503,302$ Net income 976,998$ $

(b) Taxable gains:

$43,000 - $30,000 = $13,000 (c) Total taxes:

income taxes = $113,900 + 0.34($1,480,300 − $335,000)= $503,302

gain taxes = (0.34)($13,000)= $4,420

total taxes = $503,302 + $4,420= $507,722

8.40)

(a) Book value:

3

$4,000 $0Total depreciation (3)6

$2,000$4,000 $2,000$2,000

B

−=

== −

=

(b) Cost basis:

Depreciation base = $14,000 + $800 + $200 = $15,000 (c) Taxable gains and gains taxes

19



Taxable gain $2,500 $2,000

$500Gains tax (0.40)($500)

$200

= −===

(d) Capital gains:

3 $2,000

ordinary gain $4,000 $2,000 $2,000gain taxes $2,000(0.40) $800

capital gain $5,000 $4,000 $1,000capital gain taxes $1,000(0.40) $400

total gains taxes $800 $400 $1, 200

B =

= − == == − == =

= + =

(e) Book value at the end of year 3 under 175% DB:

3 $1,422B =

With switching From DDB to SL

n Dn Bn 0 $4,000

1 $1,167 $2,833

2 $826 $2,007

3 $585 $1,422

4 $474 $948

5 $474 $474

6 $474 $0

(f) Optimal time to switch: during the 4th year

8.41) Note: Personal income tax brackets and amount of personal exemption are

updated yearly, so you need to consult the IRS tax manual for the tax rates as well as the amount of exemption that are applicable to your tax year. In this solution, we assumed the tax rate schedule of year 2007. For 2007, the amount of personal exemption is $3,400.

20



(a) Business form: Corporate

• Corporate taxes:

Year 1 Year 2 Year 3 Gross income $180,000 $195,000 $210,000 Expenses: Salary $80,000 $90,000 $100,000 Business expenses $25,000 $30,000 $40,000

Taxable income $75,000 $75,000 $70,000 Income taxes $13,750 $13,750 $12,500

• Personal income taxes (assuming that the personal exemptions as well

as the individual tax rates remain unchanged over the 3-year period) :

Year 1 Year 2 Year 3Gross income 80,000$ 90,000$ 100,000$ Deductions: Exemptions 13,600$ 13,600$ 13,600$ Itemized deduction 16,000$ 18,000$ 20,000$ Taxable income 50,400$ 58,400$ 66,400$ Income taxes 6,778$ 7,978$ 9,448$

Note that, in Year 2007, the personal income tax rates for married filing jointly are as follows:

Schedule Y-1 — Married Filing Jointly or Qualifying Widow(er)

If taxable income is over-- But not over-- The tax is:

$0 $15,650 10% of the amount over $0 $15,650 $63,700 $1,565.00 plus 15% of the amount over 15,650 $63,700 $128,500 $8,772.50 plus 25% of the amount over 63,700 $128,500 $195,850 $24,972.50 plus 28% of the amount over 128,500$195,850 $349,700 $43,830.50 plus 33% of the amount over 195,850$349,700 no limit $94,601.00 plus 35% of the amount over 349,700

• Total taxes = corporate taxes + personal taxes:

Year 1 $13,750 $6,778 $20,528Year 2 $13,750 $7,978 $21,728Year 3 $12,500 $9,448 $21,948

= + == + == + =

21



(b) Business form: sole ownership

Year 1 Year 2 Year 3 Gross income $180,000 $195,000 $210,000 Expense: Exemptions $13,600 $13,600 $13,600 Itemized deduction $16,000 $18,000 $20,000 Business expenses $25,000 $30,000 $40,000

Taxable income $125,400 $133,400 $136,400 Income taxes $24,198 $26,345 $27,185

The corporate business form is preferred.

8.42) (a) Incremental Operating income:

Operating CostsYear 1 Year 2

Revenue $15,000,000 $15,000,000Expenses: Mfg. cost $6,000,000 $6,000,000 O&M costs $1,200,000 $1,200,000 Depreciation $714,500 $1,224,500Taxable income $7,085,500 $6,575,500Income taxes (35%)$2,479,925 $2,301,425Net income $4,605,575 $4,274,075

Year 5 Year 3 Year 4

$15,000,000 $15,000,000$15,000,000

$6,000,000 $6,000,000 $6,000,000$1,200,000 $1,200,000 $1,200,000

$223,250 $874,500 $624,500$7,576,750 $6,925,500 $7,175,500$2,651,863 $2,423,925 $2,511,425$4,924,888 $4,501,575 $4,664,075

(b) Gains or losses:

5

Total depreciation $3,661,250$5,000,000 $3,661, 250$1,338,750

Taxable gains $1,600,000 $1,338,750$261, 250

B== −== −=

22



8.43)

(a) Let denote the interest rate for a corporate bond: ci

9.5% (1 0.25)12.67%

c

c

ii= −=

(b) Let A denote the annual interest payment from the corporate bond. Since

Julie’s opportunity cost rate is 9.5%, we can establish the following equivalence relationship:

$50,000 (1 0.25) ( / ,9.5%,3) [$50,000

(1 0.25)(0.05)($50,000)]( / ,9.5%,3)1.8817 $39,510.79

A P AP F

A

= − ++ −= +

Solving for A yields

$5,574.39A = This is equivalent to receiving a bond interest rate of

$5,574.39 / $50,000 11.15%ci = =

(c)

PW(9.5%) $50,000 [$75,000($75,000 $50,000)(0.25)]( / ,9.5%,3)$2,363.70 011.20% 9.5% Better than investment in bonds.

P F

IRR

= − +− −= >= > ⇒

Investment in a tract of land is more economically desirable.

23



Chapter 9 Project Cash Flow Analysis 9.1)

• wages paid to temporary workers: Variable cost • property taxes on factory building: Fixed cost • property taxes on administrative building: Fixed cost • sales commission: Variable cost • electricity for machinery and equipment in a plant: Variable cost • heat and air-conditioning for a plant: Fixed cost • salaries paid to design engineers: Fixed cost • regular maintenance on machinery and equipment: Fixed cost • basic raw materials used in production: Variable cost • factory fire insurance: Fixed cost

9.2)

a) 6 b) 11 c) 5 (Note: It is tempting to select “1”, but the graphs are drawn on

cumulative basis) d) 4 e) 2 f) 10 g) 3 h) 7 i) 9

9.3)

a) Incremental cost

Description In-house Option

Outsourcing Option

Soldering operation $4.80 Direct materials $7.50 $6.00 Direct labor $5.00 $4.25 Mfg. Overhead $4.00 $3.40

Variable $3.80 $3.23 Fixed $0.20 $0.20

Unit cost $16.50 $18.28

No. The outsourcing option would cost $1.78 more for each unit. Note that the fixed cost of $20,000 (or $0.20 per unit based on 100,000 production volume) remains unchanged under either option.




b) Break-even price = $4.80 - $1.78 = $3.02 per unit 9.4)

(a) $255,000Breakeven volume 150,000 units

$3(0.1) $2(0.5) $1(0.4)= =

+ +

(b) Total marginal contribution: $3(0.1) +$2(0.5) +$1(0.4) = $1.70 Operating income:

$3(20,000) $2(100,000) $1(80,000) $255,000 $85,000+ + − = (c)

$255,000Breakeven volume 159,375$3(0.1) $2(0.4) $1(0.5)

= =+ +

Operating income:

$3(20,000) $2(80,000) $1(100,000) $255,000 $65,000+ + − = 9.5)

a) Marginal tax rates:

Without project With project

Taxable income $350,000 $530,000Income taxes $119,000 $180,200

marginal tax rate without the project = 34%

marginal tax rate with the project = 34% b) Average tax rates:

without the project = $119,000/$350,000= 34%

with the project = $180,200/$530,000= 34%

9.6)

a) Marginal tax rates with the project:

2



n Revenue Depreciation Taxable income Combined income Marginal rate1 $120,000 $21,000 $99,000 $399,000 34%2 $120,000 $33,600 $86,400 $386,400 34%3 $120,000 $20,160 $99,840 $399,840 34%4 $120,000 $12,096 $107,904 $407,904 34%5 $120,000 $12,096 $107,904 $407,904 34%6 $120,000 $6,048 $113,952 $413,952 34%

b) Average tax rates

n Combined income Combined income taxes Average tax rate1 $399,000 $135,660.00 34%2 $386,400 $131,376.00 34%3 $399,840 $135,945.60 34%4 $407,904 $138,687.36 34%5 $407,904 $138,687.36 34%6 $413,952 $140,743.68 34%

9.7) Incremental tax rate calculation:

Year 1 Year 2Revenue $200,000 $200,000Operating Costs $100,000 $100,000Depreciation $10,000 $16,000Taxable income $90,000 $84,000

Year 1 Year 2

Taxable income without project $500,000 $500,000Income taxes (34%) $170,000 $170,000

Taxable income with project $590,000 $584,000Income taxes (34%) $200,600 $198,560

Incremental taxable income $90,000 $84,000Incremental income taxes $30,600 $28,560Incremental tax rate 34% 34%

Comments: Note that the marginal tax rates over the project life remain unchanged because the additional income from the new project is not large enough to push the company into a higher tax bracket.

3



4

9.8) Taxable income from the project during year 1:

a) & b) Increment in income tax due to the project during year 1:

Year 1

Taxable income without project $195,000

Income taxes $59,300

Taxable income with project $255,000

Income taxes $82,700

Incremental taxable income $60,000

Incremental income taxes $23,400

Incremental tax rate 39%

9.9) Incremental tax calculations:

b) Additional income tax calculation:

a) Additional taxable income due to project:

Year 1 Year 2 Year 3

Annual revenue $90,000 $90,000 $90,000

Operating cost $25,000 $25,000 $25,000

Depreciation $16,667 $22,222 $3,704

Taxable income $48,333 $42,778 $61,296

Year 1 Year 2 Year 3Taxable income without project $400,000 $400,000 $400,000Income taxes $136,000 $136,000 $136,000

Taxable income with project $448,333 $442,778 $461,296Income taxes $152,433 $150,544 $156,841

Incremental taxable income $48,333 $42,778 $61,296Incremental income taxes $16,433 $14,544 $20,841Incremental tax rate 34% 34% 34%

1 0.20($100,000) $20,000Taxable income $80,000 $20,000 $60,000

D = == − =




9.10) (a), (b), and (c)

Input OutputTax Rate(%)= 40 PW(i) = $299,551 >0, acceptable

MARR(%)= 12 IRR(%) = 97%

0 1 2 3 4 5 6Income StatementRevenues (savings) $300,000 $300,000 $300,000 $300,000 $300,000 $300,000Expenses: Labor $100,000 $100,000 $100,000 $100,000 $100,000 $100,000 Materials 50,000 50,000 50,000 50,000 50,000 50,000 Depreciation 20,000 32,000 19,200 11,520 11,520 5,760

Taxable Income $130,000 $118,000 $130,800 $138,480 $138,480 $144,240Income Taxes (40%) 52,000 47,200 52,320 55,392 55,392 57,696

Net Income $78,000 $70,800 $78,480 $83,088 $83,088 $86,544

Cash Flow StatementOperating Activities: Net Income 78,000$ 70,800$ 78,480$ 83,088$ 83,088$ 86,544$ Depreciation 20,000$ 32,000$ 19,200$ 11,520$ 11,520$ 5,760$ Investment Activities: Investment (100,000)$ Salvage 0 Gains Tax 0

Net Cash Flow ($100,000) $98,000 $102,800 $97,680 $94,608 $94,608 $92,304

5



9.11) Investment in industrial robot:

0 1 2 3 4 5Income StatementRevenues (savings) $125,000 $125,000 $125,000 $125,000 $125,000Expenses: Depreciation 35,725 61,225 43,725 31,225 11,163

Taxable Income $89,275 $63,775 $81,275 $93,775 $113,838Income Taxes (35%) 31,246 22,321 28,446 32,821 39,843

Net Income $58,029 $41,454 $52,829 $60,954 $73,994

Cash Flow StatementOperating Activities: Net Income $58,029 $41,454 $52,829 $60,954 $73,994 Depreciation $35,725 $61,225 $43,725 $31,225 $11,163Investment Activities: Investment ($250,000) Salvage $50,000 Gains Tax $5,928

Net Cash Flow ($250,000) $93,754 $102,679 $96,554 $92,179 $141,085

PW(15%)= $95,498 Accept the investmentIRR= 29% >15%(MARR)

6



9.12) $60,000( / ,15%,10)$301,128

X P A==

9.13) Investment in a new trench excavator:

0 1 2 3 4 5Income StatementRevenues (savings) $50,000 $50,000 $50,000 $50,000 $50,000Expenses: Required annual digging (ft) 8,000 8,000 8,000 8,000 8,000 Number of hours to operate 500 500 500 500 500 Operating cost (@$10/hr) $5,000 $5,000 $5,000 $5,000 $5,000 Depreciation $60,000 $96,000 $57,600 $34,560 $17,280

Taxable Income ($15,000) ($51,000) ($12,600) $10,440 $27,720Income Taxes (35%) ($5,250) ($17,850) ($4,410) $3,654 $9,702

Net Income ($9,750) ($33,150) ($8,190) $6,786 $18,018

Cash Flow StatementOperating Activities: Net Income ($9,750) ($33,150) ($8,190) $6,786 $18,018 Depreciation $60,000 $96,000 $57,600 $34,560 $17,280Investment Activities: Investment ($300,000) Salvage $100,000 Gains Tax ($22,904)

Net Cash Flow ($300,000) $50,250 $62,850 $49,410 $41,346 $112,394

IRR= 1.6% BV= $34,560PV(15%)= (96,773)$ < 0 Not Acceptable

7



9.14) Tucson Solar Company: (a)

Income StatementRevenues (savings) $66,000 $70,000 $74,000 $80,000 $64,000 $50,000Expenses: Operating Expenses 29,000 28,400 32,000 38,800 31,000 25,000 Depreciation 10,800 17,280 10,368 6,221 6,221 3,110


Net Income $17,030 $15,808 $20,561 $22,736 $17,406 $14,228

Cash Flow StatementOperating Activities: Net Income 17,030$ 15,808$ 20,561$ 22,736$ 17,406$ 14,228$ Depreciation 10,800$ 17,280$ 10,368$ 6,221$ 6,221$ 3,110$ Investment Activities: Investment (54,000)$ Salvage 8,000$ Gains Tax (2,800)$

Net Cash Flow ($54,000) $27,830 $33,088 $30,929 $28,957 $23,627 $22,539NPV= $62,469 AE(12%)= $15,194

(b) ( )AE(12%) $62,469 ( / /12%,6)$15,194

A P= ×

=

8



9.15) Investment in energy management system: N = 9 years

Input OutputTax Rate(%) = 35 PW(i) = $1,998

MARR(%) = 10 IRR(%) = 11.19%

0 1 2 3 4 5 - 8 9Income Statement

Energy Savings $10,000 $10,000 $10,000 $10,000 $10,000 $10,000Expenses: Depreciation 16,665 22,225 7,405 3,705 0 0

Taxable Income ($6,665) ($12,225) $2,595 $6,295 $10,000 $10,000Income Taxes (2,333) (4,279) 908 2,203 3,500 3,500

Net Income ($4,332) ($7,946) $1,687 $4,092 $6,500 $6,500

Cash Flow StatementOperating Activities: Net Income (4,332)$ (7,946)$ 1,687$ 4,092$ 6,500$ 6,500$ Depreciation 16,665$ 22,225$ 7,405$ 3,705$ -$ -$ Investment Activities: Investment (50,000)$ Salvage 0 Gains Tax (0)

Net Cash Flow ($50,000) $12,333 $14,279 $9,092 $7,797 $6,500 $6,500

9



9.16) Investment decision based on after-tax IRR:

Input OutputTax Rate(%) = 40 PW(i) = ($0)

MARR(%) = 12 IRR(% )= 12.00%

0 1 2 3 4 5Income Statement

Revenues (savings) $130,000 $130,000 $130,000 $130,000 $130,000Expenses: O&M costs $20,000 $20,000 $20,000 $20,000 $20,000 Depreciation 116,920 155,928 51,953 25,994 0

Taxable Income ($6,920) ($45,928) $58,047 $84,006 $110,000Income Taxes (40%) (2,768) (18,371) 23,219 33,602 44,000

Net Income ($4,152) ($27,557) $34,828 $50,404 $66,000

Cash Flow StatementOperating Activities: Net Income (4,152)$ (27,557)$ 34,828$ 50,404$ 66,000$ Depreciation 116,920$ 155,928$ 51,953$ 25,994$ -$ Investment Activities: Investment (350,794)$ Salvage 0 Gains Tax 0

Net Cash Flow ($350,794) $112,768 $128,371 $86,781 $76,398 $66,000

10



9.17) Investment in Mazda automatic screw machine:


MARR(%) = 15 IRR(%) = 33.74%

0 1 2 3 4 5 6Income Statement

Revenues (savings) $38,780 $38,780 $38,780 $38,780 $38,780 $38,780Expenses: Depreciation 9,817 16,825 12,016 8,581 6,135 3,064


Net Income $17,378 $13,173 $16,059 $18,120 $19,587 $21,430

Cash Flow StatementOperating Activities: Net Income 17,378$ 13,173$ 16,059$ 18,120$ 19,587$ 21,430$ Depreciation 9,817$ 16,825$ 12,016$ 8,581$ 6,135$ 3,064$ Investment Activities: Investment (68,701)$ Salvage 3,500$ Gains Tax 3,505$

Net Cash Flow ($68,701) $27,195 $29,998 $28,074 $26,700 $25,722 $31,499 Since PW(15%) > 0, accept the investment.

11



9.18)

Income Statement0 1 2 3

Income StatementRevenue $16,651 $16,651 $16,651Expenses: Depreciation 10,600 16,960 5,088

Taxable Income $6,051 ($310) $11,563Income Taxes $2,118 ($108) $4,047

Net Income $3,933 ($201) $7,516Cash Flow StatementCash from operation Net Income 3,933$ (201)$ 7,516$ Depreciation 10,600$ 16,960$ 5,088$ Investment / Salvage (53,000)$ 22,000$ Security Deposit 1,500$ (1,500)$ Gains Tax (577)$

Net Cash Flow (actual) (51,500)$ 14,533$ $16,759 32,527$

PW (10%) = ($0)

The required lease payment should be $16,651 per year, payable at the end of each year. If the ACLC schedules each lease payment to be made at the beginning of each year, the required lease payment should be much lower, or precisely $15,137 per year.

12



9.19) Investment decision based on after-tax IRR:

Input Output

Tax Rate(%) = 40 PW(i) = $137,306MARR(%) = 15 IRR(% )= 48.13%


Revenues (savings) $200,000 $200,000 $200,000 $200,000 $200,000Expenses: O&M costs 80,000 80,000 80,000 80,000 80,000 Depreciation 25,000 25,000 25,000 25,000 25,000


Net Income $57,000 $57,000 $57,000 $57,000 $57,000

Cash Flow StatementOperating Activities: Net Income 57,000 57,000 57,000 57,000 57,000 Depreciation 25,000 25,000 25,000 25,000 25,000 Investment Activities: Investment (150,000)$ Salvage 25,000 Gains Tax 0

Net Cash Flow ($150,000) $82,000 $82,000 $82,000 $82,000 $107,000

13



9.20) (a), (b), and (c) (Unit:$000)

0 1 2 3 4 5-7 8 9 10-11 12Income StatementRevenue 84,000$ 84,000$ 84,000$ 84,000$ 140,000$ 224,000$ 224,000$ 224,000$ 224,000$ Expenses: Production costs 45,000$ 45,000$ 45,000$ 45,000$ 75,000$ 120,000$ 120,000$ 120,000$ 120,000$ Depreciation : Building 529$ 1,154$ 1,154$ 1,154$ 1,154$ 1,154$ 1,154$ 1,154$ 1,106$ Equipment 14,290$ 24,490$ 17,490$ 12,490$ 8,930$ 4,460$

Taxable Income 24,181$ 13,356$ 20,356$ 25,356$ 54,916$ 98,386$ 102,846$ 102,846$ 102,894$ Income Taxes (40%) 9,672$ 5,342$ 8,142$ 10,142$ 21,966$ 39,354$ 41,138$ 41,138$ 41,158$

Net Income 14,509$ 8,014$ 12,214$ 15,214$ 32,950$ 59,032$ 61,708$ 61,708$ 61,736$

Cash Flow StatementOperating Activities: Net Income 14,509$ 8,014$ 12,214$ 15,214$ 32,950$ 59,032$ 61,708$ 61,708$ 61,736$ Depreciation 14,819$ 25,644$ 18,644$ 13,644$ 10,084$ 5,614$ 1,154$ 1,154$ 1,106$ Investment Activities: Land (5,000)$ 8,000$ Building (45,000)$ 30,000$ Equipment (100,000)$ 10,000$ Gains Tax Land (35%) (1,050)$ Building (40%) 731$ Equipment (40%) (4,004)$

Net Cash Flow ($150,000) $29,328 $33,658 $30,858 $28,858 $43,034 $64,646 $62,862 $62,862 $106,519

PW(15%) = $81,880 IRR = 24.30% Note 1: In a strict sense, capital gains are only realized for the sale of land. Note 2: It is assumed that the building will be disposed of at the end of December of the 12th year.

14



15

9.21) Investment in 3-D computerized car-styling system


MARR(%) = 12 IRR(%) = 95.53%


Revenues (savings) $350,000 $350,000 $350,000 $350,000 $350,000Expenses: O&M costs $80,000 $80,000 $80,000 $80,000 $80,000 Depreciation 36,000 57,600 34,560 20,736 10,368


Net Income $140,400 $127,440 $141,264 $149,558 $155,779

Cash Flow StatementOperating Activities: Net Income 140,400$ 127,440$ 141,264$ 149,558$ 155,779$ Depreciation 36,000$ 57,600$ 34,560$ 20,736$ 10,368$ Investment Activities: Investment (180,000)$ Salvage 5,000$ Gains Tax 6,294$

Net Cash Flow ($180,000) $176,400 $185,040 $175,824 $170,294 $177,442



9.22) a) Equal repayment of the principal:

Loan Repayment Loann Interest Principal Balance0 $300,0001 $36,000 $50,000 $250,0002 $30,000 $50,000 $200,0003 $24,000 $50,000 $150,0004 $18,000 $50,000 $100,0005 $12,000 $50,000 $50,0006 $6,000 $50,000 0

b) Equal repayment of the interest:

Loan Repayment Loan

n Interest Principal Balance0 $300,0001 $36,000 $300,0002 $36,000 $300,0003 $36,000 $300,0004 $36,000 $300,0005 $36,000 $300,0006 $36,000 $300,000 0

c) Equal annual installment:

$300,000( / ,12%,6) $72,968A A P= =

Loan Repayment Loan

n Interest Principal Balance0 $300,0001 $36,000 $36,968 $263,0322 $31,564 $41,404 $221,6283 $26,000 $46,373 $175,2554 $21,031 $51,937 $123,3185 $14,798 $58,170 $65,1486 $7,818 $65,148 0

$26,595

16



9.23)

InputTax Rate(%) = 40

MARR(%) = 15

0 1 2Income StatementRevenues (savings) $35,000 $35,000Expenses: Operation cost $5,000 $5,000 Depreciation 6,666 4,445 Debt interest 2,000 1,048

Taxable Income $21,334 $24,507Income Taxes (40%) 8,534 9,803

Net Income $12,800 $14,704

Cash Flow StatementOperating Activities: Net Income 12,800$ 14,704$ Depreciation 6,666$ 4,445$ Investment Activities: Investment (20,000)$ Salvage 6,000$ Gains Tax 1,156$ Financing Activities: Borrowed funds 20,000$ Principal repayment (9,524)$ (10,476)$

Net Cash Flow -$ 9,942$ 15,829$

OutputPW(i) = $20,614

17



18

9.24) (a), (b), and (c)

Income Statement0 1 2 3 4 5

Income StatementRevenue $95,000 $95,000 $95,000 $95,000 $95,000Expenses: Depreciation 30,000 48,000 28,800 17,280 8,640 Interest (12%) 10,800 9,100 7,196 5,063 2,675

Taxable Income $54,200 $37,900 $59,004 $72,657 $83,685Income Taxes (35%) $18,970 $13,265 $20,651 $25,430 $29,290

Net Income $35,230 $24,635 $38,353 $47,227 $54,395Cash Flow StatementCash from operation: Net Income 35,230$ 24,635$ 38,353$ 47,227$ 54,395$ Depreciation 30,000$ 48,000$ 28,800$ 17,280$ 8,640$ Investment / Salvage (150,000)$ 10,000$ Gains Tax 2,548$ Loan repayment 90,000$ (14,167)$ (15,867)$ (17,771)$ (19,903)$ (22,292)$

Net Cash Flow (actual) ($60,000) $51,063 $56,768 $49,382 $44,603 $53,291

PW (20%) = $93,479IRR = 82.19%



19

9.25) Income statement approach: (a), (b), and (c)

Input OutputTax Rate(% )= 40 PW(i) = $240,033

MARR(%) = 18 IRR(%) = 54.45%(a)

0 1 2 3 4 5 - 7 8 9Income StatementRevenues: Additional revenue $120,000 $120,000 $120,000 $120,000 $120,000 $120,000 $120,000 Labor & materials savings 50,000 50,000 50,000 50,000 50,000 50,000 50,000 Expenses: Depreciation 50,015 85,715 61,215 43,715 31,255 15,610 Debt interest 22,500 15,000 7,500 Taxable Income 97,485 69,285 101,285 126,285 138,745 154,390 170,000 Income Taxes 38,994 27,714 40,514 50,514 55,498 61,756 68,000 Net Income $58,491 $41,571 $60,771 $75,771 $83,247 $92,634 $102,000

Cash Flow StatementOperating Activities: Net Income 58,491 41,571 60,771 75,771 83,247 92,634 102,000 Depreciation 50,015 85,715 61,215 43,715 31,255 15,610 - Investment Activities: Investment (350,000) Salvage Gains Tax Financing Activities: Borrowed funds 250,000 Principal repayment (83,333) (83,333) (83,333)

Net Cash Flow ($100,000) $25,173 $43,953 $38,653 $119,486 $114,502 $108,244 $102,000

10

$120,00050,000

170,00068,000

$102,000

102,000-

20,000 (8,000)

$114,000



20

9.26) (a) and (b)

Input OutputTax Rate(%) = 35 PW(i) = ($1,318,770)

MARR(%) = 18 AE(i)= ($421,713.40)

0 1 2 3 4 5Income StatementRevenues (savings)Expenses: Depreciation $285,800 $489,800 $349,800 $249,800 $89,300 Debt interest (10%) 80,000 66,896 52,482 36,626 19,185

Taxable Income (365,800) (556,696) (402,282) (286,426) (108,485) Income Taxes (35%) (128,030) (194,844) (140,799) (100,249) (37,970)

Net Income

Cash FlowOperating A Net Incom DeprecInvestme Investme Salvage Gains TaFinancing Act Borrowed f Principal r

Net Cash

AEC(18%)

($237,770) ($361,853) ($261,483) ($186,177) ($70,515)

Statementctivities:e (237,770) (361,853) (261,483) (186,177) (70,515)

iation 285,800 489,800 349,800 249,800 89,300 nt Activities:

nt (2,000,000) 200,000

x 117,425 ivities:

unds 800,000 epayment (131,038) (144,142) (158,556) (174,412) (191,853)

Flow (1,200,000)$ (83,008)$ (16,194)$ (70,239)$ (110,789)$ 144,357$ (c) This is a service project. The equivalent annual cost is

= $1,318,770( A / P,18%,5)= $421,713.40



21

9.27) (a) and (b)


MARR(%) = 14 IRR(%) = 43.24%

0 1 2 3 4 5 6 7 8Income Statement

Revenues (savings) $60,000 $60,000 $60,000 $60,000 $60,000 $60,000 $60,000 $60,000Expenses: O&M cost 8,000 8,000 8,000 8,000 8,000 8,000 8,000 8,000 Depreciation 15,719 26,939 19,239 13,739 9,823 9,812 9,823 4,906 Debt interest (10%) 4,000 3,650 3,265 2,842 2,377 1,865 1,301 682

Taxable Income 32,281 21,411 29,496 35,419 39,800 40,323 40,876 46,412 Income Taxes (40%) 12,912 8,564 11,798 14,168 15,920 16,129 16,350 18,565

Net Income $19,369 $12,846 $17,697 $21,251 $23,880 $24,194 $24,525 $27,847

Cash Flow StatementOperating Activities: Net Income 19,369 12,846 17,697 21,251 23,880 24,194 24,525 27,847 Depreciation 15,719 26,939 19,239 13,739 9,823 9,812 9,823 4,906 Investment Activities: Investment (110,000) Salvage 10,000 Gains Tax (4,000) Financing Activities: Borrowed funds 40,000 Principal repayment (3,498) (3,848) (4,232) (4,656) (5,121) (5,633) (6,196) (6,816)

Net Cash Flow ($70,000) $31,590 $35,938 $32,704 $30,335 $28,582 $28,373 $28,152 $31,937



9.28) (a) with no borrowed funds:


MARR(%) = 9

0 1 2 3 4 5Income StatementRevenues (savings) $7,500 $7,500 $7,500 $7,500 $7,500Expenses: Depreciation 3,000 4,800 2,880 1,728 864 Debt interest - - - - -

Taxable Income 4,500 2,700 4,620 5,772 6,636 Income Taxes (35%) 1,575 945 1,617 2,020 2,323

Net Income $2,925 $1,755 $3,003 $3,752 $4,313

Cash Flow StatementOperating Activities: Net Income 2,925 1,755 3,003 3,752 4,313 Depreciation 3,000 4,800 2,880 1,728 864 Investment Activities: Investment (15,000) Salvage 3,000 Gains Tax (445) Financing Activities: Borrowed funds - Principal repayment - - - - -

Net Cash Flow ($15,000) $5,925 $6,555 $5,883 $5,480 $7,732

(b) With borrowed funds:

22



Input Output

Tax Rate(%) = 35 PW(9%) = $10,629MARR(%) = 9

0 1 2 3 4Income StatementRevenues (savings) $7,500 $7,500 $7,500 $7,500 $7,500Expenses: Depreciation 3,000 4,800 2,880 1,728 864 Debt interest (9%) 1,350 1,124 879 611 318

Taxable Income 3,150 1,576 3,741 5,161 6,318 Income Taxes (35%) 1,103 552 1,309 1,806 2,211

Net Income $2,048 $1,024 $2,432 $3,355 $4,107

Cash Flow StatementOperating Activities: Net Income 2,048 1,024 2,432 3,355 4,107 Depreciation 3,000 4,800 2,880 1,728 864 Investment Activities: Investment (15,000) Salvage 3,000 Gains Tax (445) Financing Activities: Borrowed funds 15,000 Principal repayment (2,506) (2,732) (2,978) (3,246) (3,538)

Net Cash Flow $0 $2,542 $3,092 $2,334 $1,837 $3,988

5

) Which alternative to choose? The debt financing option is more attractive.

(c

23



9.29) Net cash flow

Input OutputTax Rate(%) = 40 PW(12%) = $77,275

MARR(%) = 12

0 1 2 3 4Income StatementRevenues (savings) $60,000 $60,000 $60,000 $60,000 $60,000Expenses: Depreciation 25,000 40,000 24,000 14,400 7,200 Debt interest (10%) 12,500 10,453 8,200 5,723 2,998

Taxable Income 22,500 9,547 27,800 39,877 49,802 Income Taxes (40%) 9,000 3,819 11,120 15,951 19,921

Net Income $13,500 $5,728 $16,680 $23,926 $29,881

Cash Flow StatementOperating Activities: Net Income 13,500 5,728 16,680 23,926 29,881 Depreciation 25,000 40,000 24,000 14,400 7,200 Investment Activities: Investment (125,000) Salvage 50,000 Gains Tax (14,240) Financing Activities: Borrowed funds 125,000 Principal repayment (20,475) (22,522) (24,774) (27,252) (29,977)

Net Cash Flow $0 $18,025 $23,206 $15,906 $11,074 $42,864

5

9.30)

PW(18%) $3,500 $6,343( / ,18%,1) $9, 454( / ,18%,15)$22,134 0

P F P F= − + + += >

Accept the investment.

24



(unit: $000) 0 1 2 3 4 5 6 7Income StatementRevenues (savings) $40,000 $40,000 $40,000 $40,000 $40,000 $40,000 $40,000Expenses: O&M $30,000 $30,000 $30,000 $30,000 $30,000 $30,000 $30,000 Depreciation 5,002 8,572 6,122 4,375 3,122 3126 3,122 Debt interest (9%) 2,835 2,835 2,835 2,835 2,835 2,835 2,835

Taxable Income 2,163 (1,407) 1,043 2,790 4,043 4,039 4,043 Income Taxes (38%) 822 (535) 396 1,060 1,536 1,535 1,536

Net Income $1,341 ($872) $647 $1,730 $2,507 $2,504 $2,507Cash Flow StatementOperating Activities: Net Income 1,341 (872) 647 1,730 2,507 2,504 2,507 Depreciation 5,002 8,572 6,122 4,375 3,122 3,126 3,122 Investment Activities: Investment (35,000) Financing Activities: Borrowed funds 31,500 Principal repayment

Net Cash Flow ($3,500) $6,343 $7,700 $6,769 $6,105 $5,629 $5,630 $5,629

8 9 10 11 12 13 14 15Income StatementRevenues (savings) $40,000 $40,000 $40,000 $40,000 $40,000 $40,000 $40,000 $40,000Expenses: O&M 30,000 30,000 30,000 30,000 30,000 30,000 30,000 30,000 Depreciation 1,561 Debt interest (9%) 2,835 2,835 2,835

Taxable Income 5,604 7,165 7,165 10,000 10,000 10,000 10,000 10,000 Income Taxes (38%) 2,130 2,723 2,723 3,800 3,800 3,800 3,800 3,800

Net Income $3,474 $4,442 $4,442 $6,200 $6,200 $6,200 $6,200 $6,200Cash Flow StatementOperating Activities: Net Income 3,474 4,442 4,442 6,200 6,200 6,200 6,200 6,200 Depreciation 1,561 Investment Activities: Investment Salvage 5250 Gains Tax (1,996) Financing Activities: Borrowed funds Principal repayment (31,500)

Net Cash Flow $5,035 $4,442 ($27,058) $6,200 $6,200 $6,200 $6,200 $9,454

25



9.31)

• Option 1: Lease (a lease paid at the start of each period)

( )leasePW(12%) $144,000(1 0.40) 1 ( / ,12%,29)$779,484

P A= − − +

= −

• Option 2: Purchase

- Note 1: It is assumed that the property is placed in service during January.

1 30

2 29

& (11.5 /12)(1/ 39)($650,000) $15,972to (12 /12)(1/ 39)($650,000) $16,667

D DD D

= == =

- Note 2: Property tax calculation:

($800,000)(0.05) = $40,000

Input OutputTax Rate(% )= 40 PW(i) = ($931,551)

MARR(%) = 12

0 1 2 3 4-- 28 29 30Income StatementRevenues:Expenses: Depreciation 15,972$ 16,667$ 16,667$ 16,667$ 16,667$ 15,972$ Property tax 40,000$ 40,000$ 40,000$ 40,000$ 40,000$ 40,000$ Taxable Income (55,972)$ (56,667)$ (56,667)$ (56,667)$ (56,667)$ (55,972)$ Income Taxes (22,389)$ (22,667)$ (22,667)$ (22,667)$ (22,667)$ (22,389)$ Net Income (33,583)$ (34,000)$ (34,000)$ (34,000)$ (34,000)$ (33,583)$

Cash Flow StatementOperating Activities: Net Income (33,583)$ (34,000)$ (34,000)$ (34,000)$ (34,000)$ (33,583)$ Depreciation 15,972$ 16,667$ 16,667$ 16,667$ 16,667$ 15,972$ Investment Activities: Investment (land) (150,000)$ Investment (structure) (650,000)$ Salvage 215,000$ Gains Tax 34,556$ Net Cash Flow (800,000)$ (17,611)$ (17,333)$ (17,333)$ (17,333)$ (17,333)$ 231,944$

26



purchasePW(12%) $931,551= −

• Option 3: Remodel

- Note 1: Depreciation base: Remodeling cost = $300,000 1 30

2 29

& (11.5 /12)(1/ 39)($300,000) $7,372to (12 /12)(1/ 39)($300,000) $7,692

D DD D

= == =

- Note 2: Cost basis for property tax calculation:

Land + building + remodeling cost = $660,000

Input Output

Tax Rate(% )= 40 PW(i) = ($494,425)MARR(%) = 12

0 1 2 3 4 29 30Income StatementRevenues:Expenses: Depreciation 7,372$ 7,692$ 7,692$ 7,692$ 7,692$ 7,372$ Property tax 33,000$ 33,000$ 33,000$ 33,000$ 33,000$ 33,000$ Lease fee (Parking lot) $9,000 $9,500 $10,000 $10,500 $23,000 $23,500Taxable Income (49,372)$ (50,192)$ (50,692)$ (51,192)$ (63,692)$ (63,872)$ Income tax (19,749)$ (20,077)$ (20,277)$ (20,477)$ (25,477)$ (25,549)$ Net Income (29,623)$ (30,115)$ (30,415)$ (30,715)$ (38,215)$ (38,323)$ Cash Flow StatementOperating Activities: Net Income (29,623)$ (30,115)$ (30,415)$ (30,715)$ (38,215)$ (38,323)$ Depreciation 7,372$ 7,692$ 7,692$ 7,692$ 7,692$ 7,372$ Investment Activities: Investment (Remodeling) (300,000)$

Salvage 30,000$ 15,949$

Net Cash Flow (300,000)$ (22,251)$ (22,423)$ (22,723)$ (23,023)$ (30,523)$ 14,997$

remodelPW(12%) $494,425=−

Option 3 is the least costly alternative.

27



9.32) Comparison by annual equivalent cost (all units in thousand dollars):

Plant A Plant B Plant C Book Value ( 20) $380.61 $423.80 $470.56

Salvage Value $853.00 $949.80 $1,054.60Taxable gains $472.39 $526 $584.04

Gains tax (39%) $184.23

n =

$205.14 $227.78Net Proceeds from sale $668.77 $744.66 $826.82

Plant A • Capital recovery cost with return:

A1 = ($8,530 − $668.77)( A / P,12%,20) + $668.77(0.12) = $1,132.70 • After-tax O&M cost:

2 (1 0.39)($1,964) $1,198.04A = − = • Depreciation tax shield:

[ ]3 0.39($8,530) 0.0375( / ,12%,1) ( / ,12%,20)$172.22

A P F A= +

=

P

• Total equivalent annual cost:

A = $1,132.70 + $1,198.04 − $172.22 = $2,158.52 • Unit cost:

$2,158,52050,000,000kWh

= $0.04317 / kWh

Plant B • Capital recovery cost with return:

1 ($9,498 $744.66)( / ,12%,20) $744.66(0.12) $1,261.25A A P= − + = • After-tax O&M cost:

A2 = (1− 0.39)($1,744) = $1,063.84

28



• Depreciation tax shield:

[ ]3 0.39($9,498) 0.0375( / ,12%,1) ( / ,12%,20)$191.76

A P F A= +

=

P


$1, 261.25 $1,063.84 $191.76 $2,133.33A = + − = • Unit cost:

$2,133,330 $0.04267 / kWh50,000,000kWh

=

Plant C • Capital recovery cost with return:

1 ($10,546 $826.82)( / ,12%,20) $826.82(0.12) $1,400.41A A P= − + = • After-tax O&M cost:

2 (1 0.39)($1,632) $995.52A = − = • Depreciation tax shield:

[ ]3 0.39($10,546) 0.0375( / ,12%,1) ( / ,12%,20)$212.92

A P F A= +

=

P


$1, 400.41 $1,995.52 $212.92 $3,183.01A = + − = • Unit cost:

$3,183,010 $0.06366 / kWh50,000,000kWh

=

Plant B is the most economical.

29



9.33)

(a) H&H’s cost of leasing in present worth:

lease

after-tax lease expense = (1 - 0.40)($11,000)= $6,600

PW(15%) $6,600 $6,600( / ,15%,3)

$21,670

P A= − −

= −

(b) H&H’s cost of owning in present worth:

• PW of after-tax maintenance expenses:

1 $1,200(1 0.40)( / ,15%, 4)$2,055

P P= − −= −

A

• PW of after-tax loan repayment

2 $13,169( / ,15%, 4)$37,597

P P A= −= −

• PW of tax credit (shield) on depreciation and interest:

Combined Tax Savings1 $8,000 $4,800 $12,800(0.40) $5,1202 $12,800 $3,796 $16,596(0.40) $6,6383 $7,680 $2,671 $10,351(0.40) $4,1404 $2,304 $1,411 $3,715(0.40) $1,486

n nn D I====

3 $5,120( / ,15%,1) $6,638( / ,15%, 2)

$4,140( / ,15%,3) $1, 486( / ,15%, 4)$13,043

P P F P FP F P F

= +

+ +=

• PW of net proceeds from sale:

30



4

total depreciation amount = $30,784book value = $9,216

taxable gain = $10,000 - $9,216= $784

gains tax = (0.40)($784) = $314net proceeds from sale = $10,000 - $314

= $9,686= $9,686( / ,15%, 4)= $5,538

P P F

buy 1 2 3 4(15%) $21,071PW P P P P= + + + = −

(c) Should the truck be leased or purchased? The borrow–buy option is a better choice.

Input OutputTax Rate(%) = 40 PW(15%) = ($21,072)

MARR(%) = 15

0 1 2 3 4Income StatementRevenues (savings)Expenses: O&M $1,200 $1,200 $1,200 $1,200 Depreciation 8,000 12,800 7,680 2,304 Debt interest (12%) 4,800 3,796 2,671 1,411

Taxable Income (14,000) (17,796) (11,551) (4,915) Income Taxes (40%) (5,600) (7,118) (4,620) (1,966)

Net Income ($8,400) ($10,678) ($6,931) ($2,949)

Cash Flow StatementOperating Activities: Net Income (8,400) (10,678) (6,931) (2,949) Depreciation 8,000 12,800 7,680 2,304 Investment Activities: Investment (40,000) Salvage 10,000 Gains Tax (314) Financing Activities: Borrowed funds 40,000 Principal repayment (8,369) (9,374) (10,499) (11,758)

Net Cash Flow $0 ($8,769) ($7,252) ($9,750) ($2,717)

31



9.34) Note: Since the operating revenues will be the same for both options, we will only consider the cost of ownership. (a) PW (incremental) cost of owning the equipment:

• PW of after-tax O&M

1 $50,000(1 0.40)( / ,15%, 4)$85,649

P P= − −= −

A

• PW of after-tax loan repayment:

2 $37,857( / ,15%, 4)

$108,080P P A= −= −


Combined Tax Savings1 $24,000 $12,000 $36,000(0.40) = $14,4002 $38,400 $9,414 $47,817(0.40) = $19,1263 $23,040 $6,570 $29,610(0.40) = $11,8444 $6,912 $3,441 $10,353(0.40) = $4,141

n nn D I

3 $14, 400( / ,15%,1) $19,126( / ,15%, 2)

$11,844( / ,15%,3) $4,141( / ,15%, 4)$37,139

P P F P FP F P F

= +

+ +=


4

total depreciation amount = $92,352book value = $27,648

taxable gain = $20,000 - $27,648 = ($7,648)loss credit = (0.40)($7,648) = $3,059

net proceeds from sale = $20,000 + $3,059= $23,059 = $23,059( /P P ,15%,4)= $13,184

F

buy 1 2 3 4(15%) $143,406PW P P P P= + + + = −

32



Input Output Tax Rate(%) = 40 PW(15%) = ($143,405) MARR(%) = 15 0 1 2 3 4 Income Statement Revenues (savings) $0 $0 $0 $0 Expenses: O&M $50,000 $50,000 $50,000 $50,000 Depreciation 24,000 38,400 23,040 6,912 Debt interest (10%) 12,000 9,414 6,570 3,441 Taxable Income (86,000) (97,814) (79,610) (60,353) Income Taxes (40%) (34,400) (39,126) (31,844) (24,141) Net Income ($51,600) ($58,688) ($47,766) ($36,212) Cash Flow Statement Operating Activities: Net Income (51,600) (58,688) (47,766) (36,212) Depreciation 24,000 38,400 23,040 6,912 Investment Activities: Investment (120,000) Salvage 20,000 Gains Tax 3,059 Financing Activities: Borrowed funds 120,000 Principal repayment (25,857) (28,442) (31,286) (34,415) Net Cash Flow $0 ($53,457) ($48,730) ($56,012) ($40,656)

(b) PW (incremental) cost of leasing the equipment: • PW of after-tax operating cost:

1 $40,000(1 0.40)( / ,15%, 4)

$68,519P P= −=

A

• PW of after-tax leasing

33



2

1 2

$44,000(1 0.40) $44,000(1 0.40)( / ,15%,3)$86,677

$155,196

P P

P P P

= − + −== +

=

A

(c) Should ICI buy or lease the equipment? The buying option is a better choice.

9.35)

(a) OMC’ PW cost of leasing (payments at start of year):

leasingPW(15%) $22,000(0.60)( / ,15%,3)*(1.15)

=$34,658.98

P A=

(b) OMC’ PW cost of owning:

• PW of after-tax maintenance expenses:

1 $6,000(1 0.40)( / ,15%,3)$8, 219.52

P P A= −=

• PW cost of after-tax loan repayment:

2 $40,386( / ,15%,3)

$92, 209.32P P A==


n Dn In Combined Tax Savings1 $13,861 $11,640 $25,501(0.40) = $10,2002 $23,755 $8,191 $31,946(0.40) = $12,7783 $8,483 $4,327 $12,810(0.40) = $5,124

P3 = $10,200(P / F ,15%,1) + $12,778(P / F ,15%,2)+$5,124(P / F ,15%,3)= $21,901


34



total depreciation amount = $46,099book value = $50,901taxable loss = $45,000 - $50,901 = -$5,901

tax credit = (0.40)($5,901) = $2,360net proceeds from sale = $45,000 + $2,360

= $47,360P4 = $47,360(P / F ,15%,3)

= $31,140

buy 1 2 3 4PW(15%) $47,387.84P P P P= + − − =

Input OutputTax Rate(%) = 40 PW(15%) = ($47,388)

MARR(%) = 15

0 1 2 3Income StatementRevenues (savings) $0 $0 $0Expenses: O&M $6,000 $6,000 $6,000 Depreciation 13,861 23,755 8,483 Debt interest (12%) 11,640 8,191 4,327

Taxable Income (31,501) (37,946) (18,810) Income Taxes (40%) (12,600) (15,178) (7,524)

Net Income ($18,901) ($22,768) ($11,286)

Cash Flow StatementOperating Activities: Net Income (18,901) (22,768) (11,286) Depreciation 13,861 23,755 8,483 Investment Activities: Investment (97,000) Salvage 45,000 Gains Tax 2,360 Financing Activities: Borrowed funds 97,000 Principal repayment (28,746) (32,195) (36,059)

Net Cash Flow $0 ($33,786) ($31,208) $8,498

9.36) (a) and (b)

35



36

0 1 2Income StatementRevenue $114,000 $114,000Expenses: O&M $56,490 $59,315 Depreciation $11,000 $8,800 Interest $5,000 $2,619

Taxable Income $41,510 $43,266Income Taxes(40%) $16,604 $17,306

Net Income $24,906 $25,960

$24,906 $25,96011,000 $8,800

$29,768$2,173

($23,810) ($26,190)

12,096 $40,51011,520 $36,743

34,34509.75%

Cash Flow StatementCash from operation Net Income Depreciation $Investment / Salvage ($55,000)Gains TaxLoan repayment $50,000

Net Cash Flow (actual) ($5,000) $Net Cash Flow (constant) ($5,000) $

PW (18%) = $ IRR' (%) = 3

Sample calculation:

• Note that both depreciation and interest expenses are not responsive to inflation.

• O & M Expense in year 1: $53,800(1 + 0.05) = $56,490 • Salvage value in year 2: $27,000(1 + 0.05)2

= $29,768



9.37) (a) and (b)

(a) Project Cash Flows with Inflation 0 1 2 3 4 5 6

Income StatementRevenue $162,750 $170,888 $179,432 $188,403 $197,824 $207,715Expenses: O&M 92,400 97,020 101,871 106,965 112,313 117,928 Depreciation 26,000 41,600 24,960 14,976 14,976 7,488 Interest 11,700 11,700

Taxable Income 32,650 20,568 52,601 66,463 70,535 82,298 Income Taxes (40%) 13,060 8,227 21,040 26,585 28,214 32,919

Net Income $19,590 $12,341 $31,561 $39,878 $42,321 $49,379Cash Flow StatementCash from operation Net Income 19,590 12,341 31,561 39,878 42,321 49,379 Depreciation 26,000 41,600 24,960 14,976 14,976 7,488 Cash from investing activities: Investment / Salvage (130,000) 26,802 Gains Tax (10,721) Working CapitalCash from financing activities: Loan repayment 130,000 (130,000)

Net Cash Flow (actual) $0 $45,590 ($76,060) $56,521 $54,854 $57,297 $72,948

PW (18%) = $98,771

37



38

(b) Income Statement (without inflation)

0 1 2 3 4 5 6Income StatementRevenue $155,000 $155,000 $155,000 $155,000 $155,000 $155,000Expenses: O&M $88,000 $88,000 $88,000 $88,000 $88,000 $88,000 Depreciation $26,000 $41,600 $24,960 $14,976 $14,976 $7,488 Interest $11,700 $11,700

Taxable Income $29,300 $13,700 $42,040 $52,024 $52,024 $59,512Income Taxes $11,720 $5,480 $16,816 $20,810 $20,810 $23,805

Net Income $17,580 $8,220 $25,224 $31,214 $31,214 $35,707Cash Flow StatementCash from operation Net Income $17,580 $8,220 $25,224 $31,214 $31,214 $35,707 Depreciation $26,000 $41,600 $24,960 $14,976 $14,976 $7,488Investment / Salvage ($130,000) $20,000Gains Tax ($8,000)Working CapitalLoan repayment $130,000 ($130,000)

Net Cash Flow (actual) $0 $43,580 ($80,180) $50,184 $46,190 $46,190 $55,195

PW (12.38%) = $92,781



(c) Present value gain (or loss) due to inflation:

no inflation

with inflation

0.18 0.05 12.38%1 0.05

PW(12.38%) $92,781PW(18%) $98,771present value gain = $98,771 - $92,781

=$5,990

i −′ = =+

==

(d) Present value gain due to borrowing:

Net Financing cost NET

n Principal Interest(A/T) Loan flow 0 $130,000 $130,000 1 -$7,020 -$7,020 2 -$130,000 -$7,020 -$137,020

Note: Interest payment (before tax) = $130,000(0.09) = $11,700 Interest payment (after-tax) = $11,700(1 – 0.40) = $7,020

PW(18%) $130,000 $7,020( / ,18%,1)

$137,020( / ,18%,2)

$25,642.79

Loan P FP F

= −−

=

39



9.38) (a), (b)

0 1 2 3Income Statement inflationRevenue (Savings) 5% $84,000 $88,200 $92,610Expenses: O&M Depreciation $21,435 $36,735 $13,118 InterestTaxable Income $62,565 $51,465 $79,493Income Taxes (40%) $25,026 $20,586 $31,797Net Income $37,539 $30,879 $47,696

Cash Flow StatementCash from operation Net Income $37,539 $30,879 $47,696 Depreciation $21,435 $36,735 $13,118Cash from investing activities: Investment / Salvage ($150,000) $80,000 Gains Tax ($515) Working capital 8% ($10,000) ($800) ($864) $11,664Cash from financing activities: Loan repaymentNet Cash Flow (actual) ($160,000) $58,174 $66,750 $151,962Net Cash Flow (constant) ($160,000) $54,881 $59,407 $127,590

PW (20%) = 22,773$ PW (13.21%) = 22,765$ ←rounding error

Accept the project.

(c). The project is acceptable

40



9.39)

0 1 2 3 4Income StatementRevenue $33,000 $36,300 $39,930 $43,923Expenses: O&M 11,000 12,100 13,310 14,641 Depreciation 10,000 10,000 10,000 10,000

Taxable Income 12,000 14,200 16,620 19,282 Income Taxes 6,000 7,100 8,310 9,641

Net Income $6,000 $7,100 $8,310 $9,641Cash Flow StatementCash from operation Net Income 6,000 7,100 8,310 9,641 Depreciation 10,000 10,000 10,000 10,000 Investment / Salvage (40,000) Gains Tax

Net Cash Flow (actual) ($40,000) $16,000 $17,100 $18,310 $19,641Net Cash Flow (constant) ($40,000) $14,545 $14,132 $13,757 $13,415

IRR'(%) = 15.06%

41



9.40)

0 1 2 3 4 5 6 7 8Income StatementRevenue $20,000 $20,000 $20,000 $20,000 $20,000 $20,000 $20,000 $20,000Expenses: O&M 8,000$ 8,000$ 8,000$ 8,000$ 8,000$ 8,000$ 8,000$ 8,000$ Depreciation 7,145$ 12,245$ 8,745$ 6,245$ 4,465$ 4,460$ 4,465$ 2,230$

Taxable Income $4,855 ($245) $3,255 $5,755 $7,535 $7,540 $7,535 $9,770Income Taxes $1,699 ($86) $1,139 $2,014 $2,637 $2,639 $2,637 $3,420

Net Income $3,156 ($159) $2,116 $3,741 $4,898 $4,901 $4,898 $6,351

Cash Flow StatementCash from operation Net Income 3,156$ (159)$ 2,116$ 3,741$ 4,898$ 4,901$ 4,898$ 6,351$ Depreciation 7,145$ 12,245$ 8,745$ 6,245$ 4,465$ 4,460$ 4,465$ 2,230$ Investment / Salvage (50,000)$ 5,000$ Gains Tax (1,750)$ Working capital (10,000)$ 10,000$

Net Cash Flow ($60,000) $10,301 $12,086 $10,861 $9,986 $9,363 $9,361 $9,363 $21,831

PW (18%)= ($14,523) IRR(%) = 10.18%

42



(b) Project's IRR with inflation

Income Statement (no inflation)0 1 2 3 4 5 6 7 8

Income StatementRevenue $21,600 $23,328 $25,194 $27,210 $29,387 $31,737 $34,276 $37,019Expenses: O&M 8,480 8,989 9,528 10,100 10,706 11,348 12,029 12,751 Depreciation 7,145 12,245 8,745 6,245 4,465 4,460 4,465 2,230

Taxable Income 5,975 2,094 6,921 10,865 14,216 15,929 17,782 22,038 Income Taxes 2,091 733 2,422 3,803 4,976 5,575 6,224 7,713

Net Income $3,884 $1,361 $4,499 $7,062 $9,240 $10,354 $11,559 $14,325Cash Flow StatementCash from operation Net Income 3,884 1,361 4,499 7,062 9,240 10,354 11,559 14,325 Depreciation 7,145 12,245 8,745 6,245 4,465 4,460 4,465 2,230 Investment / Salvage (50,000) 7,387 Gains Tax (2,585) Working capital (10,000) (800) (864) (933) (1,008) (1,088) (1,175) (1,269) 17,138 Loan repayment

Net Cash Flow (actual dollars) ($60,000) $10,229 $12,742 $12,311 $12,299 $12,617 $13,639 $14,754 $38,495Net Cash Flow (constant dollars) ($60,000) $9,741 $11,558 $10,635 $10,119 $9,885 $10,178 $10,485 $26,054

PW (12.38%)= ($2,902.01) IRR' (%)= 11.11% < 12.38%, so it is not acceptable.

9.41) (a) Real after-tax yield on bond investment:

• Nontaxable municipal bond:

0.09 0.03 5.825%1 0.03municipali −′ = =+

43



44

• Taxable corporate bond:

0.12(1 0.3) 0.03 5.245%1 0.03municipali − −′ = =+

(b) Given i = 6%, and f = 3%

2.91%savingsi′ =

2.91%municipali′ > 2.91%corporateiSince and

′ >

After- tax Cash F lown Municipal Corporate Incremental0 -$10,000 -$10,000 $01 $900 $840 -$602 $900 $840 -$603 $900 $840 -$604 $900 $840 -$605 $900 $840 -$60

, both bond investments are better than the savings account. Now to compare two mutually exclusive bond investment alternatives, we need to perform an incremental analysis.

We cannot find the rate of return on incremental investment, as returns from municipal bond dominate those from corporate bond in every year. Municipal bond is a clear choice for any value of MARR.



9.42) (a), (b), and (c) Engine A

0 1 2 3 4 5Income StatementRevenueExpenses: O&M $135,000 $145,800 $157,464 $170,061 $183,666 Depreciation 12,000 12,000 12,000 12,000 12,000 Taxable Income (147,000) (157,800) (169,464) (182,061) (195,666) Income Taxes (58,800) (63,120) (67,786) (72,824) (78,266) Net Income ($88,200) ($94,680) ($101,678) ($109,237) ($117,400)

Cash Flow StatementCash from operation Net Income (88,200) (94,680) (101,678) (109,237) (117,400) Depreciation 12,000 12,000 12,000 12,000 12,000 Investment / Salvage (100,000) 40,000 Gains Tax $0Net Cash Flow ($100,000) ($76,200) ($82,680) ($89,678) ($97,237) ($65,400)

PW (20%) = ($345,989) AE (20%) = ($115,692) FW (20%) = ($860,932)

Engine B0 1 2 3 4 5

Income StatementRevenueExpenses: O&M $86,400 $93,312 $100,777 $108,839 $117,546 Depreciation 24,000 24,000 24,000 24,000 24,000

Taxable Income (110,400) (117,312) (124,777) (132,839) (141,546) Income Taxes (44,160) (46,925) (49,911) (53,136) (56,618)

Net Income ($66,240) ($70,387) ($74,866) ($79,703) ($84,928)Cash Flow StatementCash from operation Net Income (66,240) (70,387) (74,866) (79,703) (84,928) Depreciation 24,000 24,000 24,000 24,000 24,000 Investment / Salvage (200,000) 80,000 Gains Tax 0

Net Cash Flow ($200,000) ($42,240) ($46,387) ($50,866) ($55,703) $19,072

PW (20%)= ($316,048) AE (20%)= ($105,680) FW (20%)= ($786,429) Select B.

45



9.43) (a) & (b) Actual and constant dollar analysis:

0 1 2Income StatementRevenue $126,000 $132,300Expenses: O&M 62,400 64,896 Depreciation 12,000 9,600

Taxable Income 51,600 57,804 Income Taxes 15,480 17,341

Net Income 36,120 40,463 Cash Flow StatementCash from operation Net Income 36,120 40,463 Depreciation 12,000 9,600 Investment / Salvage (60,000) 40,000 Working capital (5,000) (200) 5,200 Gains Tax (480)

Net Cash Flow (actual) ($65,000) $47,920 $94,783Net Cash Flow (constant) ($65,000) $44,370 $81,261

IRR'(%) = 51.04%

(c) Given 8%, 15%f i= =

0.15 0.08 6.48%1 0.08

i −′ = =+

(Inflation-free MARR)

Since IRR’> 6.48%, the project is a profitable one.

46



9.44)

(a) & (b) Project cash flows in actual and constant dollars:

0 1 2 3 4 5 6Income StatementRevenue $84,800 $89,888 $95,281 $100,998 $107,058 $113,482Expenses: O&M Depreciation $20,000 $32,000 $19,200 $11,520 $11,520 $5,760 Interest


Net Income $38,880 $34,733 $45,649 $53,687 $57,323 $64,633Cash Flow StatementCash from operation Net Income $38,880 $34,733 $45,649 $53,687 $57,323 $64,633 Depreciation $20,000 $32,000 $19,200 $11,520 $11,520 $5,760Investment / Salvage ($100,000) $42,556Gains Tax ($17,022)Working capitalLoan repayment

Net Cash Flow (actual $) ($100,000) $58,880 $66,733 $64,849 $65,207 $68,843 $95,927Net Cash Flow (constant $) ($100,000) $55,547 $59,392 $54,448 $51,650 $51,443 $67,625

PW (18%) = $136,553IRR'(%) = 51.53%

47



(c) The effects of project financing under inflation:

(c) Income Statement 0 1 2 3 4 5 6

Income StatementRevenue $84,800 $89,888 $95,281 $100,998 $107,058 $113,482Expenses: O&M Depreciation $20,000 $32,000 $19,200 $11,520 $11,520 $5,760 Interest $12,000 $10,521 $8,865 $7,010 $4,933 $2,606


Net Income $31,680 $28,420 $40,330 $49,481 $54,363 $63,070Cash Flow StatementCash from operation Net Income $31,680 $28,420 $40,330 $49,481 $54,363 $63,070 Depreciation $20,000 $32,000 $19,200 $11,520 $11,520 $5,760Investment / Salvage ($100,000) $42,556Gains Tax ($17,022)Working capitalLoan repayment $100,000 ($12,323) ($13,801) ($15,457) ($17,312) ($19,390) ($21,717)

Net Cash Flow (actual $) $0 $39,357 $46,619 $44,072 $43,688 $46,493 $72,646Net Cash Flow (constant $) $0 $37,130 $41,491 $37,004 $34,605 $34,743 $51,213

PW (18%) = $163,425

48



(d) The present value loss due to inflation:

(d) Income Statement (no inflation) 0 1 2 3 4 5 6

Income StatementRevenue $80,000 $80,000 $80,000 $80,000 $80,000 $80,000Expenses: O&M Depreciation $20,000 $32,000 $19,200 $11,520 $11,520 $5,760 Interest


Net Income $36,000 $28,800 $36,480 $41,088 $41,088 $44,544Cash Flow StatementCash from operation Net Income $36,000 $28,800 $36,480 $41,088 $41,088 $44,544 Depreciation $20,000 $32,000 $19,200 $11,520 $11,520 $5,760Investment / Salvage ($100,000) $30,000Gains Tax ($12,000)Working capitalLoan repayment

Net Cash Flow (actual $) ($100,000) $56,000 $60,800 $55,680 $52,608 $52,608 $68,304

PW (11.32%) = $140,656Present value loss=$136,553-$140,656=($4,103)

Present value loss = $136,553 - $140,656 = ($4,103)

(e) Required additional before-tax annual revenue in actual dollars (equal amount) to make-up the inflation loss.

$4,103( / ,18%,6) $1,9551 0.40

A P=

−

49



9.45)

(a) & (b) The project cash flows and IRR with no inflation:

Income Statement 0 1 2 3 4 5 6 7 8 9 10Revenue $80,000 $80,000 $80,000 $80,000 $80,000 $80,000 $80,000 $80,000 $80,000 $80,000Expenses: O&M $3,000 $3,000 $3,000 $3,000 $3,000 $3,000 $3,000 $3,000 $3,000 $3,000 Labor $15,000 $15,000 $15,000 $15,000 $15,000 $15,000 $15,000 $15,000 $15,000 $15,000 Material $9,000 $9,000 $9,000 $9,000 $9,000 $9,000 $9,000 $9,000 $9,000 $9,000 Energy $4,500 $4,500 $4,500 $4,500 $4,500 $4,500 $4,500 $4,500 $4,500 $4,500 Depreciation : Building Milling machine $15,719 $26,939 $19,239 $13,739 $9,823 $9,812 $9,823 $4,906 $0 $0 Jigs & dies $3,333 $4,445 $1,481 $741 $0 $3,333 $4,445 $1,481 $741 $0Taxable Income $29,448 $17,116 $27,780 $34,020 $38,677 $35,355 $34,232 $42,113 $47,759 $48,500Income Taxes $10,307 $5,991 $9,723 $11,907 $13,537 $12,374 $11,981 $14,740 $16,716 $16,975Net Income $19,141 $11,125 $18,057 $22,113 $25,140 $22,981 $22,251 $27,373 $31,043 $31,525Cash Flow StatementCash from operation Net Income $19,141 $11,125 $18,057 $22,113 $25,140 $22,981 $22,251 $27,373 $31,043 $31,525 Depreciation Building Milling machine $15,719 $26,939 $19,239 $13,739 $9,823 $9,812 $9,823 $4,906 $0 $0 Jigs & dies $3,333 $4,445 $1,481 $741 $0 $3,333 $4,445 $1,481 $741 $0Investment / Salvage Building Milling machine ($110,000) $10,000 Jigs & dies ($10,000) $300 (Replacement) ($10,000) $300Gains Taxes: Building Milling machine ($3,500) Jigs & dies ($105) ($105)Net Cash Flow ($120,000) $38,193 $42,509 $38,777 $36,593 $25,158 $36,126 $36,519 $33,760 $31,784 $38,220

PW (11.32%) = $90,992 IRR (%) = 28.40%

50



51

(c) & (d) (c) and (d) Income Statement (with inflation)

0 1 2 3 4 5 6 7 8 9Revenue $85,600 $91,592 $98,003 $104,864 $112,204 $120,058 $128,463 $137,455 $147,077 $15Expenses: O&M $3,090 $3,183 $3,278 $3,377 $3,478 $3,582 $3,690 $3,800 $3,914 Labor $15,750 $16,538 $17,364 $18,233 $19,144 $20,101 $21,107 $22,162 $23,270 $2 Material $9,360 $9,734 $10,124 $10,529 $10,950 $11,388 $11,843 $12,317 $12,810 $1 Energy $4,635 $4,774 $4,917 $5,065 $5,217 $5,373 $5,534 $5,700 $5,871 Depreciation : Building Milling machine $15,719 $26,939 $19,239 $13,739 $9,823 $9,812 $9,823 $4,906 $0 Jigs & dies $3,333 $4,445 $1,481 $741 $0 $3,333 $4,445 $1,481 $741

Taxable Income $33,713 $25,979 $41,599 $53,181 $63,592 $66,468 $72,021 $87,088 $100,470 $10Income Taxes $11,800 $9,093 $14,560 $18,613 $22,257 $23,264 $25,207 $30,481 $35,165 $3

Net Income $21,913 $16,887 $27,040 $34,568 $41,335 $43,204 $46,814 $56,607 $65,306 $7Cash Flow StatementCash from operation Net Income $21,913 $16,887 $27,040 $34,568 $41,335 $43,204 $46,814 $56,607 $65,306 $7 Depreciation Building Milling machine $15,719 $26,939 $19,239 $13,739 $9,823 $9,812 $9,823 $4,906 $0 Jigs & dies $3,333 $4,445 $1,481 $741 $0 $3,333 $4,445 $1,481 $741Investment / Salvage Building Milling machine ($110,000) $1 Jigs & dies ($10,000) $300 (Replacement) ($10,000)Gains Taxes: Building Milling machine ( Jigs & dies ($105)

Net Cash Flow (actual) ($120,000) $40,965 $48,271 $47,760 $49,048 $41,353 $56,349 $61,082 $62,994 $66,047 $7Net Cash Flow (constant) ($120,000) $38,647 $42,960 $40,100 $38,850 $30,899 $39,724 $40,620 $39,523 $39,093 $4

PW (11.32%) = $108,411 IRR' (%) = 30.53%

107,372

$4,0324,4333,322

$6,048

$0$0

9,5378,338

1,199

1,199

$0$0

0,000

$300

$3,500)($105)

7,8943,496



52

0.25(0.12)(1 0.40) 0.072(0.072)(0.40) (0.25)(0.60)0.1788

e

d

i

k

== − =

= +=

_ &

_

_

( )

0.0435 1.2(0.115 1293

0.0435 1.0(0.115

0.0435 0.75(0.115 .09713

e f M f

e AT T

e MS

e Walmart

i r r r

i

i

i

(e). The economic gain in present worth due to inflation = $108,411- $90,992 = $17,419.

9.46) After-tax cost of debt:

a) (0.12)(1 – 0.25) = 0.09 or (9%) b) (0.14)(1 – 0.34) = 0.924 or (9.24%)

c) (0.15)(1 – 0.40) = 0.09 or (9%)

9.47) 0.07 + 1.7(0.14 – 0.07) = 18.9% 9.48)

9.49)

9.50)

i

0.0435) 0.

0.0435) 0.115

0.0435) 0

β= + −

= +

= +

= +

0.2 0.05 (0.15 0.05)1.5

− =

− =

− =

ββ= + −=



9.51) (a) The net after-tax cash flows for each financing option:

• Option 1: Equity Financing (Retained earnings)


MARR(%) = 18 IRR(%) = 42.46%


Revenues (savings) $174,000 $174,000 $174,000 $174,000 $174,000 $174,000Expenses: O&M costs $22,000 $22,000 $22,000 $22,000 $22,000 $22,000 Depreciation $28,580 $48,980 $34,980 $24,980 $17,860 $8,930 Debt interest


Net Income $75,286 $62,842 $71,382 $77,482 $81,825 $87,273

Cash Flow StatementOperating Activities:Net Income $75,286 $62,842 $71,382 $77,482 $81,825 $87,273Depreciation $28,580 $48,980 $34,980 $24,980 $17,860 $8,930Investment Activities: Investment ($200,000) Salvage $30,000 Gains Tax $2,219 Working capital ($25,000) $25,000Financing Activities: Borrowed funds Principal repayment

Net Cash Flow ($225,000) $103,866 $111,822 $106,362 $102,462 $99,685 $153,422

53



• Option 2: Debt Financing at 12% MARR(%) = 18 IRR(%) = 263.36%


Revenues (savings) $174,000 $174,000 $174,000 $174,000 $174,000 $174,000Expenses: O&M costs $22,000 $22,000 $22,000 $22,000 $22,000 $22,000 Depreciation 28,580 48,980 34,980 24,980 17,860 8,930 Debt interest 24,000 21,043 17,730 14,020 9,866 5,212


Net Income $60,646 $50,006 $60,567 $68,930 $75,807 $84,093

Cash Flow StatementOperating Activities:Net Income 60,646$ 50,006$ 60,567$ 68,930$ 75,807$ 84,093$ Depreciation 28,580$ 48,980$ 34,980$ 24,980$ 17,860$ 8,930$ Investment Activities: Investment (200,000)$ Salvage 30,000$ Gains Tax 2,219$ Working capital (25,000)$ 25,000$ Financing Activities: Borrowed funds 200,000 Principal repayment (24,645) (27,603) (30,915) (34,625) (38,780) (43,433)

Net Cash Flow (25,000)$ 64,581$ 71,383$ 64,632$ 59,285$ 54,887$ 106,809$

PW(18%)= 214,469$

54



55

• Option 3: Lease Financing


MARR(%) = 18 IRR(%) = 101.06%


Revenues (savings) $174,000 $174,000 $174,000 $174,000 $174,000 $174,000Expenses: Lease Payment $55,000 $55,000 $55,000 $55,000 $55,000 $55,000 O&M costs 22,000 22,000 22,000 22,000 22,000 22,000 Debt interest

Taxable Income ($55,000) $97,000 $97,000 $97,000 $97,000 $97,000 $152,000Income Taxes (39%) ($21,450) 37,830 37,830 37,830 37,830 37,830 59,280

Net Income ($33,550) $59,170 $59,170 $59,170 $59,170 $59,170 $92,720

Cash Flow StatementOperating Activities:Net Income (33,550)$ 59,170$ 59,170$ 59,170$ 59,170$ 59,170$ 92,720$ DepreciationInvestment Activities: Investment Salvage Gains Tax Working capital (25,000)$ 25,000$ Financing Activities: Borrowed funds Principal repayment

Net Cash Flow (58,550)$ 59,170$ 59,170$ 59,170$ 59,170$ 59,170$ 117,720$



56

1 $174,000(1 0.39)( / ,18%,6)$371, 236

P P A

(b) Vermont’s PW cost of owning the equipment by borrowing:

• PW of total after-tax revenue:

= −=

2 $25,000 $25,000( / ,18%,6)$15,739

P P F

• PW cost of working capital drain:

= −=

3 $22,000(1 0.39)( / ,18%,6)$46,938

P P A

• PW cost of operating expense: = −=

1 2 3Net cost $214, 469$94,090

P P P

• PW cost of owning by borrowing:

= − + − −=

1 0 ,000(1 0.39)($138, 467

P P A

(c) Vermont’s PW cost of leasing the equipment:

• PW cost of after-tax leasing

$55,000( .39) $55 / ,18%,5)= − + −

=

(d) Buy the tipping machine.



9.52) (a),(b),(c) & (d): Assumption: The building will be placed in service in January.

-2 -1 0 1 2 3 4 5 6 7 8Revenues:Sales unit 200,000 200,000 200,000 200,000 200,000 200,000 200,000 200,000Unit price $400 $420 $441 $463 $486 $511 $536 $563Sales volume $80,000,000 $84,000,000 $88,200,000 $92,610,000 $97,240,500 $102,102,525 $107,207,651 $112,568,034Expenses: Fixed costs $8,500,000 $8,925,000 $9,371,250 $9,839,813 $10,331,803 $10,848,393 $11,390,813 $11,960,354 Variable costs $52,000,000 $54,600,000 $57,330,000 $60,196,500 $63,206,325 $66,366,641 $69,684,973 $73,169,222 Depreciation : Building $258,017 $269,231 $269,231 $269,231 $269,231 $269,231 $269,231 $269,231 Equipment $2,715,100 $4,653,100 $3,323,100 $2,373,100 $1,696,700 $1,694,800 $1,696,700 $847,400 Amortization $1,000,000 $1,000,000 $1,000,000 $1,000,000 $1,000,000 $1,000,000 $1,000,000 $1,000,000Taxable Income $15,526,884 $14,552,670 $16,906,420 $18,931,357 $20,736,441 $21,923,460 $23,165,934 $25,321,828Income Taxes (38%) $5,900,216 $5,530,014 $6,424,439 $7,193,916 $7,879,848 $8,330,915 $8,803,055 $9,622,295Net Income $9,626,668 $9,022,655 $10,481,980 $11,737,441 $12,856,594 $13,592,545 $14,362,879 $15,699,533Cash Flow StatementOperating Avtivities: Net Income $9,626,668 $9,022,655 $10,481,980 $11,737,441 $12,856,594 $13,592,545 $14,362,879 $15,699,533 Depreciation $2,973,117 $4,922,331 $3,592,331 $2,642,331 $1,965,931 $1,964,031 $1,965,931 $1,116,631 Amortization $1,000,000 $1,000,000 $1,000,000 $1,000,000 $1,000,000 $1,000,000 $1,000,000 $1,000,000Investment activities Opportunity cost* ($3,040,000) Land ($2,500,000) $4,500,000 Building ($3,500,000) ($7,000,000) $3,000,000 Equipment ($19,000,000) $3,500,000Gains Taxes Land ($760,000) Building $2,035,801 Equipment ($1,330,000)Working capital ($9,600,000) ($480,000) ($504,000) ($529,200) ($555,660) ($583,443) ($612,615) ($643,246) $13,508,164Net Cash Flow (actual) ($5,540,000) ($3,500,000) ($35,600,000) $13,119,784 $14,440,986 $14,545,111 $14,824,112 $15,239,081 $15,943,961 $16,685,564 $42,270,128Net Cash Flow (constant) ($5,540,000) ($3,333,333) ($32,290,249) $11,333,363 $11,880,635 $11,396,475 $11,061,980 $10,830,130 $10,791,500 $10,755,663 $25,950,192

PW (15%,n =-2) $21,153,974 PW (15%,n = 0)PW(F /P ,15%,2) $27,976,130.45 IRR' = 21.47%AE(15%)= PW(A /P ,15%,8)= $6,234,483.18Unit profit per production= $31.17

Note: If the firm decides not to invest in the project, the firm could write off the R&D expenditure. The amount of write-off will be (0.38)($8,000,000) = $3,040,000. If the firm decides to undertake this project, then an opportunity cost of $3,040,000 will be

incurred.

57



(e)

-2 -1 0 1 2 3 4 5 6 7 8Revenues:Sales unit 132,016 132,016 132,016 132,016 132,016 132,016 132,016 132,016Unit price $400 $420 $441 $463 $486 $511 $536 $563Sales volume $52,806,477 $55,446,801 $58,219,141 $61,130,098 $64,186,603 $67,395,933 $70,765,729 $74,304,016Expenses: Fixed costs $8,500,000 $8,925,000 $9,371,250 $9,839,813 $10,331,803 $10,848,393 $11,390,813 $11,960,354 Variable costs $34,324,210 $36,040,420 $37,842,441 $39,734,564 $41,721,292 $43,807,356 $45,997,724 $48,297,610 Depreciation : Building $269,231 $269,231 $269,231 $269,231 $269,231 $269,231 $269,231 $269,231 Equipment $2,715,100 $4,653,100 $3,323,100 $2,373,100 $1,696,700 $1,694,800 $1,696,700 $847,400 Amortization $1,000,000 $1,000,000 $1,000,000 $1,000,000 $1,000,000 $1,000,000 $1,000,000 $1,000,000Taxable Income $5,997,936 $4,559,050 $6,413,119 $7,913,391 $9,167,577 $9,776,153 $10,411,262 $11,929,421Income Taxes $2,279,216 $1,732,439 $2,436,985 $3,007,089 $3,483,679 $3,714,938 $3,956,280 $4,533,180Net Income $3,718,721 $2,826,611 $3,976,134 $4,906,303 $5,683,898 $6,061,215 $6,454,982 $7,396,241Cash Flow StatementOperating Avtivities: Net Income $3,718,721 $2,826,611 $3,976,134 $4,906,303 $5,683,898 $6,061,215 $6,454,982 $7,396,241 Depreciation $2,984,331 $4,922,331 $3,592,331 $2,642,331 $1,965,931 $1,964,031 $1,965,931 $1,116,631 Amortization $1,000,000 $1,000,000 $1,000,000 $1,000,000 $1,000,000 $1,000,000 $1,000,000 $1,000,000Investment activities Opportunity cost* ($3,040,000) Land ($2,500,000) $4,500,000 Building ($3,500,000) ($7,000,000) $3,000,000 Equipment ($19,000,000) $3,500,000Gains Taxes Land ($760,000) Building $2,031,539 Equipment ($1,330,000)Working capital ($6,336,777) ($316,839) ($332,681) ($349,315) ($366,781) ($385,120) ($404,376) ($424,594) $8,916,482Net Cash Flow (actual) ($5,540,000) ($3,500,000) ($32,336,777) $7,386,212 $8,416,261 $8,219,149 $8,181,852 $8,264,709 $8,620,870 $8,996,318 $29,370,893Net Cash Flow (constant) ($5,540,000) ($3,333,333) ($29,330,410) $6,380,488 $6,924,078 $6,439,919 $6,105,424 $5,873,574 $5,834,944 $5,799,107 $18,031,180

PW (15%,n =-2) = $0 PW (15%,n =0) =PW(F /P ,15%,2) $0.00

58



(f) Income Statement 2006 2007 2008 2009 2010 2011 2012 2013 2014 2015 2016

-2 -1 0 1 2 3 4 5 6 7 8Revenues:Sales unit 200,000 200,000 200,000 200,000 200,000 200,000 200,000 200,000Unit price $400 $388 $376 $365 $354 $343 $333 $323Sales volume $80,000,000 $77,600,000 $75,272,000 $73,013,840 $70,823,425 $68,698,722 $66,637,760 $64,638,628Expenses: Fixed costs $8,500,000 $8,925,000 $9,371,250 $9,839,813 $10,331,803 $10,848,393 $11,390,813 $11,960,354 Variable costs $52,000,000 $50,440,000 $48,926,800 $47,458,996 $46,035,226 $44,654,169 $43,314,544 $42,015,108 Depreciation : Building $269,231 $269,231 $269,231 $269,231 $269,231 $269,231 $269,231 $269,231 Equipment $2,715,100 $4,653,100 $3,323,100 $2,373,100 $1,696,700 $1,694,800 $1,696,700 $847,400 Amortization $1,000,000 $1,000,000 $1,000,000 $1,000,000 $1,000,000 $1,000,000 $1,000,000 $1,000,000Taxable Income $15,515,670 $12,312,670 $12,381,620 $12,072,701 $11,490,465 $10,232,129 $8,966,473 $8,546,536Income Taxes $5,895,954 $4,678,814 $4,705,015 $4,587,626 $4,366,377 $3,888,209 $3,407,260 $3,247,684Net Income $9,619,715 $7,633,855 $7,676,604 $7,485,075 $7,124,088 $6,343,920 $5,559,213 $5,298,852Cash Flow StatementOperating Avtivities: Net Income $9,619,715 $7,633,855 $7,676,604 $7,485,075 $7,124,088 $6,343,920 $5,559,213 $5,298,852 Depreciation $2,984,331 $4,922,331 $3,592,331 $2,642,331 $1,965,931 $1,964,031 $1,965,931 $1,116,631 Amortization $1,000,000 $1,000,000 $1,000,000 $1,000,000 $1,000,000 $1,000,000 $1,000,000 $1,000,000Investment activities Opportunity cost* ($3,040,000) Land ($2,500,000) $4,500,000 Building ($3,500,000) ($7,000,000) $3,000,000 Equipment ($19,000,000) $3,500,000Gains Taxes Land ($760,000) Building $2,031,539 Equipment ($1,330,000)Working capital ($9,600,000) $288,000 $279,360 $270,979 $262,850 $254,964 $247,315 $239,896 $7,756,635Net Cash Flow (actual) ($5,540,000) ($3,500,000) ($35,600,000) $13,892,046 $13,835,546 $12,539,914 $11,390,255 $10,344,983 $9,555,266 $8,765,040 $26,113,657Net Cash Flow (constant) ($5,540,000) ($3,333,333) ($32,290,249) $12,000,471 $11,382,538 $9,825,351 $8,499,584 $7,351,986 $6,467,380 $5,650,023 $16,031,520

PW (15%,n =-2) = $8,660,609 PW (15%) = PW(F /P ,15%,2) $11,453,656 IRR' = 15.48%AE(15%) = PW(A /P ,15%,8) $2,552,448

59



9.53) (a) The net cash flow from the cogeneration project with bond financing:

0 1 2 3 4 5 6 7 8 9 10 11 12Income StatementRevenue Electricity bill $6,120,000 $6,120,000 $6,120,000 $6,120,000 $6,120,000 $6,120,000 $6,120,000 $6,120,000 $6,120,000 $6,120,000 $6,120,000 $6,120,000 Excess power $480,000 $480,000 $480,000 $480,000 $480,000 $480,000 $480,000 $480,000 $480,000 $480,000 $480,000 $480,000Expenses: O&M $500,000 $500,000 $500,000 $500,000 $500,000 $500,000 $500,000 $500,000 $500,000 $500,000 $500,000 $500,000 Misc. $1,000,000 $1,000,000 $1,000,000 $1,000,000 $1,000,000 $1,000,000 $1,000,000 $1,000,000 $1,000,000 $1,000,000 $1,000,000 $1,000,000 Standby power $6,400 $6,400 $6,400 $6,400 $6,400 $6,400 $6,400 $6,400 $6,400 $6,400 $6,400 $6,400 Fuel $1,280,000 $1,280,000 $1,280,000 $1,280,000 $1,280,000 $1,280,000 $1,280,000 $1,280,000 $1,280,000 $1,280,000 $1,280,000 $1,280,000Other Overhaul $1,500,000 $1,500,000 $1,500,000 $1,500,000

$100,000 $100,000 $100,000 $100,000 Depreciation Unit $500,000 $950,000 $855,000 $770,000 $693,000 $623,000 $590,000 $590,000 $591,000 $590,000 $591,000 $295,000 Inter Equipment $100,000 $160,000 $96,000 $57,600 $57,600 $28,800 Interest (9%) $945,000 $945,000 $945,000 $945,000 $945,000 $945,000 $945,000 $945,000 $945,000 $945,000 $945,000 $945,000

Taxable Income $2,268,600 $1,758,600 $317,600 $2,041,000 $2,118,000 $616,800 $2,278,600 $2,278,600 $677,600 $2,278,600 $2,277,600 $973,600Income Taxes $816,696 $633,096 $114,336 $734,760 $762,480 $222,048 $820,296 $820,296 $243,936 $820,296 $819,936 $350,496

Net Income $1,451,904 $1,125,504 $203,264 $1,306,240 $1,355,520 $394,752 $1,458,304 $1,458,304 $433,664 $1,458,304 $1,457,664 $623,104Cash Flow StatementCash from operation Net Income $1,451,904 $1,125,504 $203,264 $1,306,240 $1,355,520 $394,752 $1,458,304 $1,458,304 $433,664 $1,458,304 $1,457,664 $623,104 Depreciation Unit $500,000 $950,000 $855,000 $770,000 $693,000 $623,000 $590,000 $590,000 $591,000 $590,000 $591,000 $295,000 Inter Equipment $100,000 $160,000 $96,000 $57,600 $57,600 $28,800 $0 $0 $0 $0 $0 $0Investment / Salvage Unit ($10,000,000) $1,000,000 Inter Equipment ($500,000)Gains Tax $490,320Loan repayment $10,500,000 ($10,500,000)

Net Cash Flow (actual) $0 $2,051,904 $2,235,504 $1,154,264 $2,133,840 $2,106,120 $1,046,552 $2,048,304 $2,048,304 $1,024,664 $2,048,304 $2,048,664 ($8,091,576)

PW (27%) = $5,954,443

(a) The net cash flow from the cogeneration project with bond financing:

Standby power(overhaul)

60



(b). The maximum annual lease amount that ACC is willing to pay is $907,673.

(b) The maximum annual lease amount that ACC is willing to pay is $907,664:0 1 2 3 4 5 6 7 - 11 12

Income StatementRevenue Electricity bill $6,120,000 $6,120,000 $6,120,000 $6,120,000 $6,120,000 $6,120,000 $6,120,000 $6,120,000 Excess power $480,000 $480,000 $480,000 $480,000 $480,000 $480,000 $480,000 $480,000Expenses: O&M $500,000 $500,000 $500,000 $500,000 $500,000 $500,000 $500,000 $500,000 Misc. $1,000,000 $1,000,000 $1,000,000 $1,000,000 $1,000,000 $1,000,000 $1,000,000 $1,000,000 Standby power $6,400 $6,400 $6,400 $6,400 $6,400 $6,400 $6,400 $6,400 Overhead $1,280,000 $1,280,000 $1,280,000 $1,280,000 $1,280,000 $1,280,000 $1,280,000 $1,280,000 Lease $907,673 $907,673 $907,673 $907,673 $907,673 $907,673 $907,673 $907,673Taxable Income $2,905,927 $2,905,927 $2,905,927 $2,905,927 $2,905,927 $2,905,927 $2,905,927 $2,905,927Income Taxes $1,046,134 $1,046,134 $1,046,134 $1,046,134 $1,046,134 $1,046,134 $1,046,134 $1,046,134Net Income $1,859,793 $1,859,793 $1,859,793 $1,859,793 $1,859,793 $1,859,793 $1,859,793 $1,859,793Cash Flow StatementCash from operation Net Income $1,859,793 $1,859,793 $1,859,793 $1,859,793 $1,859,793 $1,859,793 $1,859,793 $1,859,793Net Cash Flow (actual) $0 $1,859,793 $1,859,793 $1,859,793 $1,859,793 $1,859,793 $1,859,793 $1,859,793 $1,859,793

PW (27%) = $6,496,872

61



Chapter 10 Handling Project Uncertainty 10.1)

(a) AEC(10%) (25,000 5,000)( / ,10%,6) 0.1(5,000) 3,000

$8,092A P= − + +

=

(b) AEC(10%) (25,000 5,000)( / ,10%,5) 0.1(5,000) 3,000

$8,776A P= − + +

=

(c) AEC(10%) (25,000 5,000)( / ,10%,6) 0.1(5,000) 3,300

$8,392A P= − + +

=

10.2) (a) Project cash flows based on most-likely estimates:

Tax Rate(%)= 40 PW(i) = $11,342 >0MARR(%)= 15% IRR(%) = 20%


Taxable Income $17,000 $6,200 $17,720 $24,632 $29,816Income Taxes 6,800 2,480 7,088 9,853 11,926

Net Income $10,200 $3,720 $10,632 $14,779 $17,890

Cash Flow StatementOperating Activities: Net Income 10,200$ 3,720$ 10,632$ 14,779$ 17,890$ Depreciation 18,000$ 28,800$ 17,280$ 10,368$ 5,184$ Investment Activities: Investment (90,000)$ Salvage 20,000 Gains Tax -3,853

Net Cash Flow ($90,000) $28,200 $32,520 $27,912 $25,147 39,221$



(b) Required annual savings (X): $34,930

Input OutputTax Rate(%)= 40 PW(i) = $0

MARR(%)= 20% IRR(%) = 20%


Taxable Income $16,930 $6,130 $17,650 $24,562 $29,746Income Taxes 6,772 2,452 7,060 9,825 11,898

Net Income $10,158 $3,678 $10,590 $14,737 $17,848

Cash Flow StatementOperating Activities: Net Income 10,158$ 3,678$ 10,590$ 14,737$ 17,848$ Depreciation 18,000$ 28,800$ 17,280$ 10,368$ 5,184$ Investment Activities: Investment (90,000)$ Salvage 20,000 Gains Tax -3,853

Net Cash Flow ($90,000) $28,158 $32,478 $27,870 $25,105 39,179$

(c)

Input Output Not acceptableTax Rate(%)= 40 PW(i) = ($8,771) <0

MARR(%)= 15% IRR(%) = 11%


Taxable Income $7,000 ($3,800) $7,720 $14,632 $19,816Income Taxes 2,800 (1,520) 3,088 5,853 7,926

Net Income $4,200 ($2,280) $4,632 $8,779 $11,890

Cash Flow StatementOperating Activities: Net Income 4,200$ (2,280)$ 4,632$ 8,779$ 11,890$ Depreciation 18,000$ 28,800$ 17,280$ 10,368$ 5,184$ Investment Activities: Investment (90,000)$ Salvage 20,000 Gains Tax (3,853)

Net Cash Flow ($90,000) $22,200 $26,520 $21,912 $19,147 33,221$

2



10.3)

(a) Project’s IRR if the investment is made now:

PW( ) $600,000 $250,000( / , ,5) 030.77%

i Pi

A i= − +=

=

(b) Let X denote the revised annual cash flow:

PW(30.77%) $600,000 ( / ,30.77%,4)( / ,30.77%,1)0

X P A P F= − +=

X = $366,885.54

10.4) (a) Economic building height • The optimal building height is 5 Floors. 5% < i < 20% :

Net Cash Flows

n 2 Floors 3 Floors 4 Floors 5 Floors

0 ($500,000) ($750,000) ($1,250,000) ($2,000,000)

1 $199,100 $169,200 $149,200 $378,150

2 $199,100 $169,200 $149,200 $378,150

3 $199,100 $169,200 $149,200 $378,150

4 $199,100 $169,200 $149,200 $378,150

5 $799,100 $1,069,200 $2,149,200 $3,378,150

Sensitivity Analysis

PW(i) as a Function of Interest Rate

i (%)

2 Floors

3 Floors

4 Floors

5 Floors

Best

Floor Plan

5 $832,115 $687,721 $963,010 $1,987,770 5

6 $787,037 $635,264 $873,001 $1,834,680 5

7 $744,141 $585,441 $787,722 $1,689,448 5

8 $703,298 $538,091 $706,879 $1,551,593 5

9 $664,388 $493,067 $630,199 $1,420,666 5

10 $627,298 $450,230 $557,428 $1,296,250 5

11 $591,924 $409,452 $488,330 $1,177,957 5

3



12 $558,167 $370,612 $422,686 $1,065,427 5

13 $525,937 $333,599 $360,291 $958,321 5

14 $495,148 $298,309 $300,953 $856,326 5

15 $465,720 $264,644 $244,495 $759,148 5

16 $437,580 $232,512 $190,751 $666,513 5

17 $410,657 $201,829 $139,565 $578,166 5

18 $384,885 $172,516 $90,792 $493,867 5

19 $360,205 $144,496 $44,298 $413,393 5

20 $336,557 $117,701 ($46) $336,533 2

(b) Effects of overestimation on resale value:

n 2 Floors 3 Floors 4 Floors 5 Floors0 ($500,000) ($750,000) ($1,250,000) ($2,000,000)1 $199,100 $169,200 $149,200 $378,1502 $199,100 $169,200 $149,200 $378,1503 $199,100 $169,200 $149,200 $378,1504 $199,100 $169,200 $149,200 $378,1505 $739,100 $979,200 $1,949,200 $3,078,150

Net Cash Flows

Best i (%) Floor Plan15 $435,890 $219,898 $145,060 $609,995 5

PW(i) as a Function of Interest Rate

Present Worth as a Function of Number of Floors

Resale value 2 3 4 5 Base $465,720 $264,644 $244, 495 $759,148

10% error $435,890 $219,898 $145,060 $609,995

Difference $29,831 $44,746 $99, 435 $149,153

4



10.5) (a) With infinite planning horizon: We assume that both machines will be available with the same cost in the future.

5

Model A

Financial Datan 0 1 2 3 4 5-7 8

Depreciation $9,289 $15,919 $11,369 $8,119 $5,805 $2,899Book value $65,000 $55,712 $39,793 $28,425 $20,306 $14,502 $0Market value $5,000Gain/Loss $5,000Operation Cost $7,500 $7,500 $7,500 $7,500 $7,500 $7,500

Cash Flow StatementInvestment ($65,000)+(.30)*(Depreciation) $2,787 $4,776 $3,411 $2,436 $1,741 $870-(1-0.30)*(OpeNet proceeds

Net Cash Flow

Model B

Financial Dat9 10

DepreciationBook value $0 $0Market value $8,000Gain/Loss $8,000Operation Cost $5,800 $5,800

Cash Flow StaInvestment+(.30)*(Deprec $0 $0-(1-0.30)*(Ope ($4,060) ($4,060)Net proceeds $5,600

Net Cash Flow ($4,060) $1,540

ration cost) ($5,250) ($5,250) ($5,250) ($5,250) ($5,250) ($5,250)from sale $3,500

($65,000) ($2,463) ($474) ($1,839) ($2,814) ($3,509) ($880)

PW (10%) = ($77,306) AE (10%) = ($14,491)

an 0 1 2 3 4 5-7 8

$11,932 $20,449 $14,604 $10,429 $7,457 $3,724$83,500 $71,568 $51,119 $36,515 $26,085 $18,629 $0

$5,800 $5,800 $5,800 $5,800 $5,800 $5,800

tement($83,500)

iation) $3,580 $6,135 $4,381 $3,129 $2,237 $1,117ration cost) ($4,060) ($4,060) ($4,060) ($4,060) ($4,060) ($4,060)from sale

($83,500) ($480) $2,075 $321 ($931) ($1,823) ($2,943)

PW (10%) = ($88,214) AE (10%) = ($14,356)

Model B is preferred over Model A.



(b) Break-even annual O&M costs for machine A: Let X denote a before-tax annual operating cost for model.

Model A

Financial Datan 0 1 2 3 4 5 6 7 8

Depreciation $9,289 $15,919 $11,369 $8,119 $5,805 $5,805 $5,805 $2,899Book value $65,000 $55,712 $39,793 $28,425 $20,306 $14,502 $8,697 $2,893 $0Market value $5,000Gain/Loss $5,000Operation Cost $7,309 $7,309 $7,309 $7,309 $7,309 $7,309 $7,309 $7,309

Cash Flow StatementInvestment ($65,000)+(.30)*(Depreciation) $2,787 $4,776 $3,411 $2,436 $1,741 $1,741 $1,741 $870-(1-0.30)*(Operation cost) ($5,116) ($5,116) ($5,116) ($5,116) ($5,116) ($5,116) ($5,116) ($5,116)Net proceeds from sale $3,500

Net Cash Flow ($65,000) ($2,330) ($341) ($1,706) ($2,681) ($3,375) ($3,375) ($3,375) ($747)

PW (10%) = ($76,593) AEC(10%) = $14,357

Since we are comparing two mutually exclusive alternatives with unequal service lives, we could use AEC instead of PW. With the annual O&M cost of $7,309, the AEC of the both model would be the same. If we use the PW as a base to compare, we need to assume a least common multiple of 40 years.

6



(c) With a shorter service life, Model A is preferred over Model B.

Financial Datan 0 1 2 3 4 5

Depreciation $9,289 $15,919 $11,369 $8,119 $2,902Book value $65,000 $55,712 $39,793 $28,425 $20,306 $17,404Market value $13,000Gain/Loss ($4,404)Operation Cost $7,500 $7,500 $7,500 $7,500 $7,500

Cash Flow StatementInvestment ($65,000)+(.30)*(Depreciation) $2,787 $4,776 $3,411 $2,436 $871-(1-0.30)*(Operation cost) ($5,250) ($5,250) ($5,250) ($5,250) ($5,250)Net proceeds from sale $14,321

Net Cash Flow ($65,000) ($2,463) ($474) ($1,839) ($2,814) $9,942

PW (10%) = ($64,763) AEC(10%) = $17,084

Model B

Financial Datan 0 1 2 3 4 5

Depreciation $11,932 $20,449 $14,604 $10,429 $3,728Book value $83,500 $71,568 $51,119 $36,515 $26,085 $22,357Market value $18,500Gain/Loss ($3,857)Operation Cost $5,800 $5,800 $5,800 $5,800 $5,800

Cash Flow StatementInvestment ($83,500)+(.30)*(Depreciation) $3,580 $6,135 $4,381 $3,129 $1,118-(1-0.30)*(Operation cost) ($4,060) ($4,060) ($4,060) ($4,060) ($4,060)Net proceeds from sale $19,657

Net Cash Flow ($83,500) ($480) $2,075 $321 ($931) $16,716

PW (10%) = ($72,238) AEC(10%) = $19,056

7



10.6) Sensitivity graph as a function of number of units produced and sales price per unit.

v

x 1500 2000 2500 3000 4000 4500 5000

10 $202,880 $281,700 $360,520 $439,340 $596,980 $675,800 $754,620

15 $384,448 $523,790 $663,133 $802,475 $1,081,160 $1,220,503 $1,359,845

20 $566,015 $765,880 $965,745 $1,165,610 $1,565,340 $1,765,205 $1,965,070

25 $747,583 $1,007,970 $1,268,358 $1,528,745 $2,049,520 $2,309,908 $2,570,295

30 $929,150 $1,250,060 $1,570,970 $1,891,880 $2,533,700 $2,854,610 $3,175,520

35 $1,110,718 $1,492,150 $1,873,583 $2,255,015 $3,017,880 $3,399,313 $3,780,745

$0

$500,000

$1,000,000

$1,500,000

$2,000,000

$2,500,000

$3,000,000

$3,500,000

$4,000,000

10 15 20 25 30 35Sales price(X)

NPW

($)

V=1500V=2000V=2500V=3000V=3500V=4000V=4500V=5000

8



9

0 1 2 3 4 5 6 7 8Income StatementRevenue:

Bill savings $3,000,000 $3,000,000 $3,000,000 $3,000,000 $3,000,000 $3,000,000 $3,000,000 $3,000,000 Mile Savings 1,250,000 $1,250,000 $1,250,000 $1,250,000 $1,250,000 $1,250,000 $1,250,000 $1,250,000Expenses: Depreciation 2,000,000 3,200,000 1,920,000 1,152,000 1,152,000 576,000

Taxable Income 2,250,000 1,050,000 2,330,000 3,098,000 3,098,000 3,674,000 4,250,000 4,250,000

Income Tax (38%) 855,000 399,000 885,400 1,177,240 1,177,240 1,396,120 1,615,000 1,615,000

Net Income $1,395,000 $651,000 $1,444,600 $1,920,760 $1,920,760 $2,277,880 $2,635,000 $2,635,000Cash Flow StatementCash From Operation: Net Income 1,395,000 651,000 1,444,600 1,920,760 1,920,760 2,277,880 2,635,000 2,635,000 Depreciation 2,000,000 3,200,000 1,920,000 1,152,000 1,152,000 576,000 0 0Investment&Salvage -10,000,000

Net Cash Flow -10,000,000 3,395,000 3,851,000 3,364,600 3,072,760 3,072,760 2,853,880 2,635,000 2,635,000

PW (18%) = $3,204,044

(a) Project cash flows based on most-likely estimates:

10.7)



(b) Sensitivity analysis:

Percentage deviation

Savings In T.B

PW(18%)

Savings In D.M

PW(18%)

-30% $2,100,000 $928,762 $875,000 $2,256,010 -20% 2,400,000 1,687,189 1,000,000 2,572,021 -10% 2,700,000 2,445,616 1,125,000 2,888,032

0 (base) 3,000,000 3,204,044 1,250,000 3,204,044 +10% 3,300,000 3,962,471 1,375,000 3,520,055 +20% 3,600,000 4,720,898 1,500,000 3,836,066 +30% 3,900,000 5,479,325 1,625,000 4,152,078

(c) Sensitivity diagrams

$0

$1,000,000

$2,000,000

$3,000,000

$4,000,000

$5,000,000

$6,000,000

-30% -20% -10% 0 (base) 10% 20% 30%Percentage deviation

PW(1

8%)

Savings in T.BSavings in D.M.

10



10.8) • PW of net investment:

0 $3,500,000 $500,000 $700,000 $4,700,000P = − − − = −

• PW of after-tax revenue:

1 $6,000(365) (1 0.30)( / ,10%,25)$13,915,041

P X P AX

= −=

• PW of after-tax operating costs:

2 ($430,000 $220,000 )(1 0.30)( / ,10%,25)$2,732,177 1,397,858

P X PX

= − + −= − −

A

• PW of tax credit (shield) on depreciation:

Depreciation Combined n Building Furniture Tax savings 1 $86,006 $100,030 $55,811

2 $89,744 $171,430 $78,352

3 $89,744 $122,430 $63,652

4 $89,744 $87,430 $53,152

5 $89,744 $62,510 $45,676

6 $89,744 $62,440 $45,655

7 $89,744 $62,510 $45,676

8 $89,744 $31,220 $36,289

9--24 $89,744 0 $26,923

25 $86,006 0 $25,802

3 $55,811( / ,10%,1) $78,352( / ,10%, 2)$25,802( / ,10%, 25)

$394,763

P P F P FP F

= +

+ +=


Cost Salvage Book Gains Gains Property

basis value value (losses) Taxes

Furniture $700,000 $0 $0 $0 $0

Building $3,500,000 $0 $1,263,889 ($1,263,889) ($379,167)

Land $500,000 $1,693,177 $500,000 $1,193,177 $357,953

11



Net proceeds from sale $1,693,177 $379,167 $357,953 = + −

4

$1,714,391$1,714,391( / ,10%, 25)$158, 231

P P F===

0 1 2 3 4PW(10%)$6,879,183 12,517,183

0

P P P P PX

= + + + +

= − +=

X = 54.96% 10.9)

(a) With an assumption of the unit price of $100, the PW(12%) for options A and B would be the same when unit cost is set at $9.35. Here we also assumed that the old machine is sold off and the current net after-tax salvage value in the amount of $15,000 is credited for Option B. If the firm retains the old machines for other use, then the unit cost would be $8.93.

Option A 1000 1 2 3

Income StatementRevenue:

$2,500,000 $2,500,000 $2,500,000

Expenses: Operating cost 225,000 225,000 225,000 Depreciation 0 0 0

Taxable Income 2,275,000 2,275,000 2,275,000Income Tax (40%) 910,000 910,000 910,000

Net Income $1,365,000 $1,365,000 $1,365,000Cash Flow StatementCash From Operation: Net Income 1,365,000 1,365,000 1,365,000 Depreciation 0 0 0Investment&Salvage 6,000Gain taxes -2,400

Net Cash Flow 0 1,365,000 1,365,000 1,368,600

PW (12%) = $3,281,062

12



Option B 100 unit cost= 9.3452277240 1 2 3


$2,500,000 $2,500,000 $2,500,000



Net Income $1,359,822 $1,359,822 $1,359,822Cash Flow StatementCash From Operation: Net Income 1,359,822 1,359,822 1,359,822 Depreciation 0 0 0Investment&Salvage 25,000Gain taxes (10,000) 0

Net Cash Flow 15,000 1,359,822 1,359,822 1,359,822

PW (12%) = $3,281,062 . (b) If we assume the unit price of $100, the PW(12%) for options A and C would be $3,342,656, when unit cost is set at $7.22.

Option C

0 1 2 3Income StatementRevenue:

$2,500,000 $2,500,000 $2,500,000Expenses: Operating cost 168,750 168,750 168,750 Depreciation 7,860 13,470 4,810


Net Income $1,394,034 $1,390,668 $1,395,864Cash Flow StatementCash From Operation: Net Income 1,394,034 1,390,668 1,395,864 Depreciation 7,860 13,470 4,810Investment&Salvage (55,000) 15,000Gain taxes 5,545Net Sale of old machine 15,000

Net Cash Flow (40,000) 1,401,894 1,404,138 1,421,218

PW (12%) = $3,342,656

13



Option B 100 unit cost= 7.2192388030 1 2 3


$2,500,000 $2,500,000 $2,500,000



Net Income $1,391,711 $1,391,711 $1,391,711Cash Flow StatementCash From Operation: Net Income 1,391,711 1,391,711 1,391,711 Depreciation 0 0 0Investment&SalvageGain taxes 0

Net Cash Flow 0 1,391,711 1,391,711 1,391,711

PW (12%) = $3,342,656 Need to add the net proceeds from sale.

(c). Option C is the most economical. . 10.10) Useful life of the old bulb:

14,600 /(19 365) 2.1 years× = For computational simplicity, let’s assume the useful life of 2 years for the old bulb. Then, the new bulb will last 4 years. Let X denote the price for the new light bulb. With an analysis period of 4 years, we can compute the equivalent present worth cost for each option as follows:

old

new

PW(15%) (1 0.40)[$61.90 $61.90( / ,15%,2)]$65.23

PW(15%) (1 0.40)( $16)

P F

X

= − +== − +

The break-even price for the new bulb will be

0.6 9.6 $65.23X + =

X = $92.72

14



Since the new light bulb costs only $60, it is a good bargain. 10.11)

• PW of net investment:

0 $350,000P = −

• PW of after-tax rental revenue:

1 (1 0.30)( / ,10%,20)$5.95952

P X P AX

= −=

• PW of after-tax operating costs:

2 (1 0.30)$18,000( / ,10%,20)$107,271

P P= − −= −

A

• PW of tax credit (shield) on depreciation: (In this problem, we assume that the purchasing cost of $350,000 does not include any land value. Therefore, the entire purchasing cost will be the cost basis for depreciation purpose.)

Depreciation Combined

n Building Tax savings 1 $8,601 $8,601 (0.30) =$2580

2-19 $8,974 $8,974 (0.30) =$2692 20 $8,601 $8,601 (0.30) = $2580

3 $2,580( / ,10%,1) $2,692( / ,10%,18)( / ,10%,1)$2,580( / ,10%, 20)$22,800

P P F P A P FP F

= ++=


Total depreciation $178,734= Book value $350,000 $178,734= −

$171, 266=

Salvage value 20$350,000(1.05) $928,654= =

Taxable gain $928,654 $171, 266= − $757,388=

Gains tax $757,388(0.30) $227, 216= =

15



Net proceeds from sale $928,654 $227, 216= − $701, 438=

4 $701,438( / ,10%,20)$104,264

P P F==

The break-even rental

0 1 2 3 4(10%)$330, 207 $5.95952

0

PW P P P P PX

= + + + +

= − +=

X = $55, 408

16



10.12) Let X denote the additional annual revenue (above $16,000) for model A that is

required to break even. • After-tax cash flows for Model A:

End of Year Cash flows

elements 0 1 2 3 4-5 6 Investment -$80,000

(0.40) nD 6,400 10,240 6,144 3,686 1,843

-(0.60)O& nM -13,200 -13,200 -13,200 -13,200 -13,200

+(0.60) nR 9,600 9,600 9,600 9,600 9,600

+0.6X +0.6X +0.6X +0.6X +0.6XNet proceeds 12,000Net cash flow -$80,000 $2,800

+0.6X$6,640+0.6X

$2,544+0.6X

$86 +0.6X

$10,234+0.6X

PW(20%) $80,000 0.6 ( / , 20%,6)

$2,800( / , 20%,1) $10, 234( / , 20%,6)$68,077 $1.9953

A X P AP F P F

X

= − ++ + += − +

• After-tax cash flows for Model B:

End of Year Cash flows elements 0 1 2 3 4-5 6 Investment -$52,000

(0.40) nD 4,160 6,656 3,994 2,397 1,198

-(0.60)O& nM -10,200 -10,200 -10,200 -10,200 -10,200

+(0.60) nR 0 0 0 0 0

Net proceeds 9,000Net cash flow -$52,000 -$6,040 -$3,544 -$6,206 -$7,803 -$2

17



PW(20%) $52,000 $6,040( / , 20%,1)$2( / , 20%,6)

$69,985

B P FP F

= − −+ −= −

Now let PW(20%) PW(20%)A B= and solve for X.

$68,077 1.9953 $69,985X− + = − X = $957−

The required break-even annual revenue for model A is then $16,000 $15,043X+ =

10.13) Let X denote the annual number of copies to break-even. If the selling price is set at $0.50 per copy,

A/T annual revenue (0.6)[$0.1 ($0.50 $0.15)]X= + − 0.27X=

A/T O&M cost (0.60)[$300,000(12) $0.20 ]X= − + $2,160,000 0.12X= − −

Depreciation tax credit (0.40)[$85,714( / ,12%,1)$26,775( / ,12%,8)]( / ,12%,10)(0.40)(408,576)( / ,12%,10)$28,923

P FP F A P

A P

= ++==

CR(12%) $600,000( / ,12%,10)$100,000( / ,12%,10)

$100,500Gains tax $40,000 (or AE(12%) = $40,000( / ,12%,10) = $2,279)AE(12%) 0.27 $2,160,000 0.12 $28,923 $100,500 $2, 279

0.15 $2, 233,8560

A PA F

A FX XX

= −+= −== − − + − −= −=

X = 14,892,373 copies per year or 40,801 copies per day

Note: This is assuming 365 days per year and making no use of “25 printing days per month” information in the problem.

10.14)

18



[ ] (0.15 5%) (0.25 15%) (0.35 22%)(0.15 30%) (0.1 40%)

20.7%

E return = × + × + ×+ × + ×

=

2 2 2

2 2

(0.15 (5 20.7) ) (0.25 (15 20.7) ) (0.35 (22 20.7) )(0.15 (30 20.7) ) (0.1 (40 20.7) )

95.919.79%

σ

σ

= × − + × − + × −

+ × − + × −==

2

10.15) We can calculate the mean and variance for each periods with the three- point estimate.

Period (n)

Pessimistic Most Likely

Optimistic E[An] Var[An]

0 ($10,000) ($8,000) ($7,000) ($8,167) 250,0001 $5,000 $12,000 $15,000 $11,333 2,777,7782 $4,000 $10,000 $13,000 $9,500 2,250,000

2

2 4

$11,333 $9,500[PW(10%)] $8,1671.1 1.1

$9,986.972,777,778 2, 250,000[PW(10%)] 250,000

1.1 1.14,082, 464.57

E

V

= − + +

=

= + +

=

0 9,986.97 4.94284,082,464

z −= = −

NORMDIST(-4.9428,0,1,1)= 0.0000385%

10.16)

19



ligh t

m oderate

h igh

P W (12% ) $8, 000, 000 $1, 300, 000( / ,12% , 3)

$4, 877 , 619P W (12% ) $8, 000, 000 $2, 500, 000( / ,12% , 4)

$406, 627P W (12% ) $8, 000, 000 $4, 000, 000( / ,12% , 4)

$4,149, 397[P W (12% )] $4, 877 , 619(0 .20

P A

P A

P A

E

= − +

= −= − +

= −= − +

== − ) $406, 627 (0 .40)

$4,149, 397 (0 .40)

$521, 584

−+

=

10.17) (a)

2 3

2 2 2

2 4 6

$500 $1,500 $800[PW(10%)] $1,0001.1 1.1 1.1

$1, 295.27200 300 100[PW(10%)]1.1 1.1 1.1

100,173.8

E

V

= − + + +

=

= + +

=

(b)

$1,295.27 $1,295.27 0100,173.8

z −= =

NORMDIST(0,0,1,1) = 50%

(c)

certainty equivalent 2 3

$500 $1,500 $800PW(25%) $1,0001.25 1.25 1.25

$769.6 0

= − + + +

= >

Yes, it would be justified. 10.18)

20



(a)

2 3 4

2 2 2 2 22

2 4 6 8 10

$120 $150 $150 $110 $100[PW(10%)] $3001.1 1.1 1.1 1.1 1.1

$182.9810 15 20 25 30[PW(10%)] 201.1 1.1 1.1 1.1 1.1

1,501

E

V

= − + + + + +

=

= + + + + +

=

5

(b) Certainty equivalent value:

certainty equivalent 2 3 4

$120 $150 $150 $110 $100PW(18%) $3001.18 1.18 1.18 1.18 1.18

$101 0

= − + + + + +

= >

5

Yes, it would be justified. 10.19) (a)

2 3 4 5

2 2 2 2 22

2 4 6 8 1

$5 $8 $12 $10 $5[PW(12%)] $181.12 1.12 1.12 1.12 1.12

$10.588 9 10 5 3[PW(12%)] 0

1.12 1.12 1.12 1.12 1.12166.155

E

M

V

= − + + + + +

=

= + + + + +

=

0

(b)

0 10.58 0.8208166.155

z −= = −

NORMDIST(-0.8208,0,1,1) = 20.59%

(c) Using Excel’s Goal Seek function to get i.

21



2 3 4 5

$5 $8 $12 $10 $5[PW( )] $18(1 ) (1 ) (1 ) (1 ) (1 )

0

32.38%

E ii i i i

i

= − + + + + ++ + + + +

=

=

i

10.20) (a)

2

$2, 200 $1,800[PW(10%)] $2,5001.1 1.1

$987.6 0, acceptable.

E = − + +

= >

(b)

2 22

2 4

200 300[PW(10%)] 1001.1 1.1

104,529

104,529 $323.31

V

σ

= + +

=

= =

(c)

2 22

2 4

200 300[PW(10%)] 1001.1 1.1

104,529

104,529 $323.312 $987.6 2($323.31)

$340.98

V

σµ σ

= + +

=

= =− = −

=

Using Excel NORMDIST(-2,0,1,1) = 2.275%

10.21) (a) The mean return for projects

22



[ ] (0.1 20%) (0.2 0%) (0.25 10%)(0.3 15%) (0.1 20%) (0.05 40%)

9%[ ] (0.1 35%) (0.2 10%) (0.25 15%)

(0.3 25%) (0.1 40%) (0.05 50%)12.25%

A

B

E return

E return

= ×− + × + ×+ × + × + ×

== ×− + ×− + ×+ × + × + ×

=

(b) The variance of return for projects

2

(c) It is not a clear case, because but also

e principle of maxi

10.22)

2 2 2

2 2

2 2

2 2

2 2

2

(0.1 ( 20 9) ) (0.2 (0 9) )

(0.25 (10 9) ) (0.3 (15 9) )(0.1 (20 9) ) (0.05 (40 9) )171.50

(0.1 ( 35 12.25) ) (0.2 ( 10 12.25) )

(0.25 (15 12.25) ) (0.3 (25 12.25) )(0.1 (40 12.25) ) (0.05 (

A

B

σ

σ

= × − − + × −

+ × − + × −

+ × − + × −=

= × − − + × − −

+ × − + × −

+ × − + × 250 12.25) )521.19

−=

[ ] [ ]B AE return E return> B AVar Var> .

If you make decision solely based on th mization of expected return, you may prefer project B. )

(a

[ ][ ][ ]

[ ]

[PE W(12%)] (0.3) $150,000 $35,000( / ,12%,5)

(0.5) $150,000 $40,000( / ,12%,5)

(0.2) $150,000 $50,000( / ,12%,5)$4,006.6

[PW(12%)] (0.3) $180,000 $45,000( / ,12%,5)

(0.5) $180,000 $55,000( / ,12%,

A

B

P A

P A

P A

E P A

P A

= − +

+ − +

+ − +

= −

= − +

+ − +[ ][ ]

5)

(0.2) $180,000 $67,000( / ,12%,5)$16,100

P A+ − +

=

23



2 2

2

2 2

2

(0.3 ( 23,832 4,006.6) ) (0.5 ( 5,808 4,006.6) )

(0.2 (30,240 4,006.6) )354,097,713(0.3 ( 17,785 16,100) ) (0.5 (18,263 16,100) )

(0.2 (61,520 16,100) )759,392,532

A

B

σ

σ

= × − + + × − +

+ × +=

= × − − + × −

+ × −=

2

2

(b).Project A has a higher probability to lose money. 0 4,006.6 0.2129354,097,713Az +

= =

NORMDIST(0.2129,0,1,1)=58.43%

0 16,100 0.58428

759,392,532Bz −= = −

10.23) (a) The PW distribution for project 1:

NORMDIST(-0.58428,0,1,1)= 27.95%

Event Joint (10%) PW(x,y) probability

($20,$10) 0.18 $400($20,$20) 0.12 0($40,$10) 0.42 2,400($40,$20) 0.28 1,600

) The mean and variance of the PW for Project 1:

1,528) (0.41,528) (0.2

830,016

Var

+ +

= −

+ −

+ −=

(b

1

2 21

2

2

[PW(10%)] $400(0.18) $0(0.12) $2, 400(0.42)$1,600(0.28)$1,528

[PW(10%)] (400 1,528) (0.18) (0 1,528) (0.12)

(

E =+=

+ −

2,400 2)(1,600 8)

(c) The mean and variance of the PW for Project 2:

24



2

2 22

2

2

[PW(10%)] $0(0.24) $400(0.20) $1,600(0.36)$2,400(0.20)$1,136

[PW(10%)] (0 1,136) (0.24) (400 1,136) (0.20)

(1,600 1,136) (0.36)(2,400 1,136) (0.20)815,104

E

Var

= + ++=

= − + −

+ −

+ −=

(d)

It is not a clear case, because but also .

If Juan makes the decision solely based on the principle of maximization of expected value, she may prefer contract A.

10.24)

Expected value criterion: Assume that the opportunity cost rate is 7.5%.

Option 2:

1 2E E> 1 2Var Var>

Option 1:

E R = + +1[ ] $2, 450(0.25) $2,000(0.45) $1,675(0.30)$150( / ,7.5%,1)$1,854

F P−=

2[ ] $25,000(0.075) $1,875E R = =

maximization. 10.25)

(a) E

E

= ++ −== + +

Option 2 is the better choice based on the principle of expected value

1

2

[PW] ($2,000)(0.20) ($3,000)(0.60)($3,500)(0.20) $1,000$1,900

[PW] ($1,000)(0.30) ($2,500)(0.40) ($4,500)(0.30)$800$1,850−=

25



Project 1 is preferred over Project 2. )

ar Var

(b2 2

12

2 22

2

[PW] (2,000 1,900) (0.20) (3,000 1,900) (0.60)

(3,500 1,900) (0.20)1, 240,000

[PW] (1,000 1,850) (0.30) (2,500 1,850) (0.40)

(4,500 1,850) (0.30)2, 492,500

Var

Var

= − + −

+ −=

= − + −

+ −=

Project 1 is still preferred, because V 1 2[PW] [PW]<

but .

10.26)

(a) Mean and variance calculations:

, 000)(0.40)( $10, 000)(0.30)$13, 000

[PW ] (100, 000 40, 000) (0.20)

(50, 000 40, 000) (0.40)(0 40, 000) (0.40)

A

A

E

Var

= ++=

+ −=

= −

+ −

+ −

2

2

2

[PW ] (40, 000 13, 000) (0.30)

(10, 000 13, 000) (0.40)( 10, 000 13, 000) (0.30)381, 000, 000

BVar = −

+ −

+ − −=

Contract A has no chance of losing money, so the executive should certainly choose A rather than no contract at all. Between A and B, the choice is less clear-

cut, because but also . If he makes the decision solely

based on the principle of maximization of expected value, he may prefer contract A.

1 2[PW] [PW]E E>

2

2

2

[PW ] ($100, 000)(0.20) ($50, 000)(0.40)(0)(0.40)$40, 000

[PW ] ($40, 000)(0.30) ($10BE = +

1, 400,= 000, 000

A BE E> A BVar Var>

26



(b) Assuming that the contracts are statistically independent from each other,

Joint event

(PW PW )A B>

Joint Probability

($100,000,$40,000) (0.20)(0.30) = 0.06($100,000,$10,000) (0.20)(0.40) = 0.08

($100,000,-$10,000) (0.20)(0.30) = 0.06($50,000,$40,000) (0.40)(0.30) = 0.12($50,000,$10,000) (0.40)(0.40) = 0.16

($50,000,-$10,000) (0.40)(0.30) = 0.12($0,-$10,000) (0.40)(0.30) = 0.12

0.72Σ =

10.27)

(a

[AE(10%)] ($5,000)(0.20) ($8,000)(0.30)00)(0.30) ($12 0.20)

$10,924$19,725

[AE(10 725)

72

A

A

A P

E

Var

= −+

= ++

+=

5)725725

Machine B:

) Machine A:

CR(10%) ($60,000 $22,000)( / ,10%,6)(0.10)($22,000)$10,924=

2 (0.20)

($10,0+ ,000)(

%)] (15,924 19,A = −

(18,924 19,+ − 2 (0.30)2) (0.30)2) (0.20)

(20,924 19,+ −

(22,924 19,+ −5,560,000=

27



2

2

CR(10%) $35,000( / ,10%, 4)$11,042

[AE(10%)] ($8,000)(0.10) ($10,000)(0.30)($12,000)(0.40) ($14,000)(0.20)$11,042$22, 442

[AE(10%)] (19,042 22,442) (0.10)

(21,042 22, 442) (0.30)(23,042 22,442)

B

B

B

A P

E

Var

=== ++ ++=

= −

+ −

+ − 2

2

(0.40)(25,042 22, 442) (0.20)3, 240,000+ −=

(b) Prob[AE(10%) AE(10%) ]:A B>

Joint event

(O&M ,O&M )A B (AE ,AE )A B

Joint Probability

($10,000, $8,000) ($20,924, $19,042) (0.30)(0.10) = 0.03 ($12,000, $8,000) ($22,924, $19,042) (0.20)(0.10) = 0.02 ($12,000, $10,000) ($22,924, $21,042) (0.20)(0.30) = 0.06 0.11Σ =

10.28)

(a) Mean and variance calculation (Note: For a random variable Y, which can be expressed as a linear function of another random variable X (say, Y = , where is a constant) the variance of Y can be calculated as a function of

varian of X,

aXa

ce 2[ ] [ ]Var Y a Var X= .

28



2 2

2 2

2

5,0 %,500, 5%

$8,000( / ,15%,2)$1,266.54

[PW] 1,000 ( / ,15%,1) 1,000

( / ,15%,2) 1,500A

P F

V P F

P F

+=

= +

+

7,988,336=

(b) Comparing risky projects

2 2 2

3,042,588[PW] 2,000 ( / ,15%,1) 1,500

( /BV P F

P

=

= +

+ 2 2,15%,2) 2,000F

[PW] $E = − 00 $4,000( / ,15P A+ 2)$1,

A

= 2.84[PW] $1BE = − 000 $6,000( / ,1P F+ ,1)

Project A Project B [ ]E PW $1,503 $1,267[ ]V PW 3,042,588 7,988,336

Project A is preferred over project B, because [ ]AV PW < [ ]BV PW

and

10.29) Select (b). 10.30)

Cash flow for this investment Assum

[ ] [ ]A BE PW E PW> .

ption: MARR is 15%.

29



0 1 2 3 4 5InRev

come Statementenue:

Unit price $50 $50 $50 $50 $50Demand(units) 2,000 2,000 2,000 2,000 2,000Sales Revenue $100,000 $100,000 $100,000 $100,000 $100,000

s: Unit $15 $15 $15 $15 $15 Variable c $30,000 $30,000 $30,000 $30,000 $30,000 Fixed cost $10,000 $10,000 $10,000 $10,000 $10,000 Depreciation $17,863 $30,613 $21,863 $15,613 $5,581

Taxable Income $42,138 $29,388 $38,138 $44,388 $54,419Income Tax (40%) $16,855 $11,755 $15,255 $17,755 $21,768

Net Income $25,283 $17,633 $22,883 $26,633 $32,651Cash Flow StatementCash From Operation: Net Income $25,283 $17,633 $22,883 $26,633 $32,651

Depreciation $17,863 $30,613 $21,863 $15,613 $5,581Investment ($125,000)Salvage $40,000Gains tax ($2,613)

Net Cash Flow ($125,000) $43,145 $48,245 $44,745 $42,245 $75,620NPV(15%)= $40,168

Sensitivity analysis for five key input variables

Expense

variable costost

Deviation -20% -15% -10% -5% 0% 5% 10% 15% 20%

Unit price ($57) $9,999 $20,055 $30,111 $40,168 $50,225 $60,281 $70,337 $80,393

Demand(units) $12,010 $19,049 $26,088 $33,130 $40,168 $47,208 $54,247 $61,286 $68,325

Variable cost $52,236 $49,219 $46,202 $43,186 $40,168 $37,152 $34,135 $31,118 $28,101

Fixed cost $44,191 $43,185 $42,179 $41,175 $40,168 $39,163 $38,157 $37,151 $36,145

Salvag l $40,765 $41,361 $41,957 $42,553 e va ue $37,782 $38,378 $38,974 $39,573 $40,168

Sensit aph for the BMC’s is u r

ivity gr transm sion-ho sings p oject

30



$0

$10,000

$20,000

$30,000

$40,000

$50,000

$60,000

$70,000

$80,000

$90,000

-20% -15% -10% -5% 0% 5% 10% 15% 20%

Deviation from base curve(%)

PW

($)

unit price

demand

variable cost

fixed cost

salvage value

10.31)

(a)

2

0.25(3% ) 0.5(5% ) 0.25(7% ) 5%

0.1 0.05 0.1(0.05) 0.155

Salvage_year 2=$6,000(1.05) =$6,615G ain tax_year 2=$1,443 (tax credit)W orking capital_year 2=$3,308

PW (15.5% ) $26, 000 (0.6 0.4($7, 666))(

f

i i f i f

X

= + + =

′ ′= + + = + + =

= − + + / ,15.5% ,1))(0.6 0.4($5,112) $6, 615 $1, 443 $3, 308)( / ,15.5% , 2))0.96924 $13, 294.1

P FX P

X+ + + + += −

F

(b) & (c) X = $15,000, PW(15.5%) $1,583=

31



Income Statement inflation 0 1 2Revenue (Savings) 5% $15,000 $15,750Expenses: O&M Depreciation 7,666$ 5,112$ Interest

Taxable Income $7,334 $10,6385Income Taxes (40%) $2,934 $4,25

Net Income $4,400 $6,383

Cash Flow StatementCash from operation Net Income $4,400 $6,383 Depreciation $7,666 $5,112Cash from investing activities: Investment / Salvage 5% (23,000)$ 6,615$ G 1,443$ W ital 5% (3,000)$ 3,308$

s g activities:

Net Cash Flow (actual) ($26,000) $12,066 $22,860Net Cash Flow (constant) -26,000 11,492 20,735

PW (15.5%) = 1,583$

X = $25,000,

ains Taxorking cap

Ca h from financin Loan repayment

PW(15.5%) $11,501=

Income Statement inflation 0 1 2Revenue (Savings) 5% $25,000 $26,250Expenses: O&M Depreciation 7,666$ 5,112$ Interest


Net Income $10,400 $12,683

Cash Flow StatementCash from operation Net Income $10,400 $12,683 Depreciation $7,666 $5,112Cash from investing activities: Investment / Salvage 5% (23,000)$ 6,615$ Gains Tax 1,443$ Working capital 5% (3,000)$ 3,308$ Cash from financing activities: Loan repayment

Ne Cash Flow t (actual) ($26,000) $18,066 $29,160tNe Cash Flow (constant) (26,000) 17,206 26,449

PW (15.5%) = 11,501$

X = $35,000, PW(15.5%) $21, 418=

32



Income Statement inflation 0 1Revenue

2(Savings) 5% $35,000 $36,750

Expenses: O&M Depreciation 7,666$ 5,112$ Interest


Net Income $16,400 $18,983

Cash Flow StatementCash from operation Net Income $16,400 $18,983 Depreciation $7,666 $5,112Cash from investing activities: Investment / Salvage 5% (23,000)$ 6,615$ Gains Tax 1,443$ Working capital 5% (3,000)$ 3,308$ Cash from financing activities: Loan repayment

Net Cash Flow (actual) ($26,000) $24,066 $35,460Net Cash Flow (constant) (26,000) 22,920 32,163

PW (15.5%) = 21,418$

X = $15,000, PW(15.5%) $1,583= X = $25,000, PW(15.5%) $11,501= X = $35,000, PW(15.5%) $21, 418=

10.32) Since the amount of annual labor savings is the same for both alternatives, this labor savings factor is not considered in the following analysis.

(a) After-tax cash

2 2

2

[PW(15.5%)] $1,583(0.2) $11,501(0.5) $21,418(0.3)$12,492.5

[PW(15.5%)] (1,583 12,492.5) (0.2) (11,501 12,492.5) (0.5)(21,418 12,492.5) (0.3)48,194,339.25

E

Var

= + +=

= − + −

+ −=

flows:

33



After-Tax Cash Flows n Lectra System Tex System 0 -$136,150 -$195,5001 117,927 149,0752 124,462 158,4593 117,491 148,4494 113,308 142,4435 113,308 142,4436 122,171 146,939

PW(12%) $350,189 $415,383AE(12%) $85,175 $101,032

etter choice.

(b) Let X and Y denote the annual material savings for the Lectra system and Tex

Based on the most-likely estimates, the Tex system is the b

system, respectively.

Af ter-Tax Cash Flows n Lect Tra System ex System 0 -$136,150 -$195,5001 0.6X-20 0.6Y,073 -15,3252 0.6X-13 0.6,538 Y-5,9413 0.6X-20 0.6Y,508 -15,9514 0.6X-24 0.6Y,691 -21,9565 0.6X-24 0.6Y,691 -21,9566 0.6X-15 0.6Y,828 -17,461

Lectra System

[ ] $224,000

$52,824 0.6 [ ]$81,576

[AE(12%)] (0

E X

:

LectraAE(12%) $52,824 0.6X= − +

Lectra

[ ] 2,124,000,000[AE(12%)]

Var X

20.6) [ ]Lectra

764,640,00

E E X

Var

=

= − +=

=

=

Var X=

34



35

Tex System:

Tex

Tex

2Tex

AE(12 )[ ] $[ ] 1

[AE(12% [$

[AE(12%)] (0.6) [ ]618,638,400

Y

V Y

% $63,368 0.6259,400,718,440,000

)] $63,368 0.6 ]92,272

E Yar

E E Y

Var Var Y

= −=== −=

==

+

+



36

(a) Incremental project cash flows (FMS - CMT):

No. of part types 3,000 3,000 3,000 3,000 3,000 3,000 3,000 3,000 3,000 3,000 No. of pieces per year 544,000 544,000 544,000 544,000 544,000 544,000 544,000 544,000 544,000 544,000

Year 0 1 2 3 4 5 5 7 8 9 10

Income StatementRevenue (Savings): Labor $462,400 $462,400 $462,400 $462,400 $462,400 $462,400 $462,400 $462,400 $462,400 $462,400

Material 233,920 233,920 233,920 233,920 233,920 233,920 233,920 233,920 233,920 233,920 Overhead 1,200,000 1,200,000 1,200,000 1,200,000 1,200,000 1,200,000 1,200,000 1,200,000 1,200,000 1,200,000 Tooling 170,000 170,000 170,000 170,000 170,000 170,000 170,000 170,000 170,000 170,000 Inventory 109,500 109,500 109,500 109,500 109,500 109,500 109,500 109,500 109,500 109,500Expenses: Depreciation 928,850 1,591,850 1,136,850 811,850 580,450 579,800 580,450 289,900

Taxable Income 1,246,970 583,970 1,038,970 1,363,970 1,595,370 1,596,020 1,595,370 1,885,920 2,175,820 2,175,820Income Tax (40%) 498,788 233,588 415,588 545,588 638,148 638,408 638,148 754,368 870,328 870,328

Net Income $748,182 $350,382 $623,382 $818,382 $957,222 $957,612 $957,222 $1,131,552 $1,305,492 $1,305,492

Cash Flow StatementCash From Operation: Net Income 748,182 350,382 623,382 818,382 957,222 957,612 957,222 1,131,552 1,305,492 1,305,492 Depreciation 928,850 1,591,850 1,136,850 811,850 580,450 579,800 580,450 289,900Investment&Salvage (6,500,000) 500,000

Gains Tax (40%) (200,000)

Net Cash Flow (6,500,000)$ 1,677,032$ 1,942,232$ 1,760,232$ 1,630,232$ 1,537,672$ 1,537,412$ 1,537,672$ 1,421,452$ 1,305,492$ 1,605,492$

PW (15%) = $1,756,225

10.33)



(b) &(c) Sensitivity analysis:

AOC = annual overhead cost VLC = variable labor cost / part AIC = annual inventory cost ATC = annual tooling cost VMC = variable material cost / part

Deviation -30% -20% -10% 0% 10% 20% 30%AOC $3,517,813 $2,930,617 $2,343,421 $1,756,225 $1,169,030 $581,834 -$5,362VLC $2,395,095 $2,182,138 $1,969,182 $1,756,225 $1,543,268 $1,330,313 $1,117,356AIC $1,784,682 $1,775,196 $1,765,711 $1,756,225 $1,746,740 $1,737,225 $1,727,769ATC $2,027,239 $1,936,901 $1,846,563 $1,756,225 $1,665,888 $1,575,550 $1,485,212VMC $2,296,807 $2,116,613 $1,936,419 $1,756,225 $1,576,032 $1,395,838 $1,215,644

37



(d) Best and worst scenarios:

• Best case: Material cost = $1.00 per part, annual inventory cost = $25,000

(15%) $1,939,611FMS CMTPW − =

• Worst case: Material cost = $1.40 per part, annual inventory cost = $100,000

(15%) $1,058,516FMS CMTPW − =

(e) Mean and variance:

• : $1,595,123 [ (15%)FMS CMTE PW − ]

]• : 46,073,274,329 [ (15%)FMS CMTVar PW −

(f) In no situation, the FMS would be a more expensive investment option than the CMT.

38



Chapter 11 Replacement Decisions 11.1)

Tax Rate(%) = 0.00% PW(i) = ($20,065)MARR(%) = 10.00% AE(%) = ($6,329.8)

0 1 2 3 4Income Statement

Revenues (savings)Expenses: O&M $2,500 $3,000 $3,500 $4,000 Depreciation $0 $0 $0 $0

Taxable Income ($2,500) ($3,000) ($3,500) ($4,000)Income Taxes (%) 0 0 0 0

Net Income ($2,500) ($3,000) ($3,500) ($4,000)

Cash Flow StatementOperating Activities: Net Income (2,500)$ (3,000)$ (3,500)$ (4,000)$ Depreciation $0 $0 $0 $0Investment Activities: Investment (12,000)$

Salvage 3,000$ Gains Tax $0.00

Net Cash Flow ($12,000) ($2,500) ($3,000) ($3,500) ($1,000)

PW(10%) $12,000 $2,500( / ,10%,1) $1,000( / ,10%,4)$20,065

AEC(10%) $20,065( / ,10%,4)

$6,329.8

P F P A

A P

= − − ⋅⋅⋅⋅ ⋅ −= −=

=

11.2)

A. Original cost: The printing machine was purchased for $20,000 B. Market value: The old machine’s market value is estimated at $10,000. C. Book value: If the machine sold now its book value is $14,693. D. Trade in allowance: This amount is the same as the market value.



The market value is the most relevant information, but the defender’s current book value is also relevant as this will be the basis to determine the gains or losses related to disposal of the defender.

11.3) Option 1: Keep the defender

PW(12%) $8,000( / ,12%,3) $2,500( / ,12%,3)$17,434.9

AEC(12%) $17,434.9( / ,12%,3)

$7,259.1

D

D

P A P F

A P

= − += −=

=

Option 2: Replace the defender with the challenger

PW(12%) $5,000 $6,000( / ,12%,3) $5,500( / ,12%,3)

$15,495.9AEC(12%) $15,495.9( / ,12%,3)

$6,451.9

C

C

P A P F

A P

= − − +

= −=

=

The replacement should be made now.

11.4)

Option 1: Sold today

1PW(15%) $2,500= Option 2:

2PW(15%) $8,000 $3,000( / ,15%,5)$2,056.6

P A= − +=

Option 3:

3PW(15%) $20,000 $3,000( / ,15%,5) $6,000( / ,15%,5)$1,604.8

P F P A= − + +=

Select Option 1. (or Select 1 & 3 as these two options may not be viewed as mutually exclusive alternatives.)

2



11.5)

(a) Purchase cost = $22,000, market value = $6,000, sunk cost = $22,000 - $6,000 = $16,000

(b) opportunity cost = $6,000 (c)

PW(15%) $6,000 $2,500 $5,000( / ,15%,1)

($5,500 $3,500)( / ,15%,2)$14,360.2

AEC(15%) $14,360.2( / ,15%,2)

$8,832.96

P FP F

A P

= − − −− −= −=

=

(d)

PW(15%) $8,500 $5,000( / ,15%,1) $5,500( / ,15%,2)$6,000( / ,15%,3) $9,500( / ,15%,4)($7,500 $2,000)( / ,15%,5)

$29,117.84AEC(15%) $29,117.84( / ,15%,5)

$8,686.30

P F P FP F P F

P F

A P

= − − −− −− −= −=

=

11.6)

(a) Opportunity cost = $30,000

(b) Assume that the old machine’s operating cost is $35,000 per year. Then the new machine’s operating cost is zero per year. The cash flows associated with the retaining the defender for two more years are

0 1 2Cash Flows: -$30,000 -$35,000 -$25,000

n

PW(12%) $30,000 $35,000( / ,12%,2) $10,000( / ,12%,2)

$81,180AEC(12%) $81,180( / ,12%,2)

$48,034

D

D

P A P F

A P

= − − += −=

=

(b) Cash flows for the challenger: Year 0: -$175,000; Years 1-7: 0; Year 8: $8,000

3



PW(12%) $175,000 $8,000( / ,12%,8)$171,769

AEC(12%) $171,769( / ,12%,8)

$34,578

C

C

P F

A P

= − += −=

=

(d) Since AEC AECD C> , we should replace the defender now.

11.7)

(a) Initial cash outlay for the new machine = $144,000 (b) Cash flows for the defender: Year 0: -$13,000 Years 1-5: 0

(c)

[ ]AEC(15%) $13,000( / ,15%,5) $3,877.9AEC(15%) ($144,000 $40,000)( / ,15%,7) $40,000(0.15)

$60,000$29,003 (savings)

D

C

A PA P

= =

= − +

−= −

Replace the defender now.

11.8)

(a) Cash flows

Year: 0 1 2 3 4 5 Defender -$10K 0 0 0 0 $5KChallenger -$75K $33K $33K $33K $33K $33K

(b)

PW(10%) $10 $5 ( / ,10%,5) $6,895.5PW(10%) $75 $33 ( / ,10%,5) $50,096.4

D

C

K K P FK K P A

= − + = −= − + =

Should replace the defender.

11.9)

(a) and (b) Cash flows:

Year: 0 1 2 3 4 5 Defender -$6,000 $24,500 $24,500 $26,000 Challenger -$38,500 $31,500 $31,500 $31,500 $31,500 $38,500

4



• Revenue for defender = ($22 - $15) × 3,500 = $24,500 per year

(c)

• Revenue for challenger = ($22 - $13) × 3,500 = $31,500 per year

[ ]

[ ]

(15%) ($6,000 $1,500)( / ,15%,3) $1,500(0.15) $24,500$22,304

(15%) ($38,500 $7,000)( / ,15%,5) $7,000(0.15) $31,500$21,053.55

D

C

AE A P

AE A P

= − − + +

=

= − − + +

=

Keep the defender for now.

1.10) The economic service life is 4 years.

1

Annual changes in MV (1,000)$ Annual increases in O&M 1,500$ Interest rate 12%

n Market Value O&M Costs CR(12%) OC(12%) AEC(12%)

0 $20,0001 $10,000 $20,000 $12,400 $20,000 $32,4002 $9,000 $21,500 $7,589 $20,708 $28,2963 $8,000 $23,000 $5,956 $21,387 $27,3434 $7,000 $24,500 $5,120 $22,038 $27,1585 $6,000 $26,000 $4,604 $22,662 $27,266

1.11)

At , the economic service life is 1 year. 1

12%i =

5



Interest rate 12%

n Market ValueO&M Costs CR(12%) OC(12%) AEC(12%)

0 $30,0001 $25,800 $5,000 $7,800 $5,000 $12,8002 $16,000 $6,500 $10,204 $5,708 $15,9113 $10,000 $10,000 $9,527 $6,980 $16,5074 $5,000 $12,500 $8,831 $8,135 $16,9665 $0 $14,800 $8,322 $9,184 $17,506

11.12) Economic service life is 6 years.

Annual changes in MV -25%Annual increases in O&M 15%Interest rate 15%


0 $18,0001 $10,000 $1,000 $10,700 $1,000 $11,7002 $7,500 $1,150 $7,584 $1,070 $8,6533 $5,625 $1,323 $6,264 $1,143 $7,4064 $4,219 $1,521 $5,460 $1,218 $6,6785 $3,164 $4,749 $4,900 $1,742 $6,6426 $2,373 $2,011 $4,485 $1,773 $6,2587 $1,780 $6,813 $4,166 $2,228 $6,3948 $1,335 $2,660 $3,914 $2,260 $6,1749 $1,001 $3,059 $3,713 $2,307 $6,020

10 $751 $3,518 $3,550 $2,367 $5,91611 $563 $4,045 $3,416 $2,436 $5,85212 $422 $4,652 $3,306 $2,512 $5,81813 $317 $5,350 $3,215 $2,595 $5,810

6



11.13) (a) At , the economic service life for the defender is 2 years: 12%i =

Annual changes in MVAnnual increases in O&MInterest rate 12%

Market O&M n Value Costs CR(12%) OC(12%) AEC(12%)

0 $7,7001 $4,300 $3,200 $4,324 $3,200 $7,5242 $3,300 $3,700 $2,999 $3,436 $6,4353 $1,100 $4,800 $2,880 $3,840 $6,7204 $0 $5,850 $2,535 $4,261 $6,796

* 2 and AEC $6,435D DN = = .

(b) & (c) 10 yearsCN =

AEC $31,000( / ,12%,10) $1,000 $2,500( / ,12%,10)

$6,344C A P A F= + −

=

Since AEC AEC , the defender should be replaced now.D C>

11.14) Correction: Two different O&M figures were given for the new machine in the first printing. The correct figure is $4,200 for the first year, increasing at an annual rate of $500.

AEC $53,500( / ,12%,5) $12,000( / ,12%,5)$4, 200 $500( / ,12%,5)$18,039.80

AEC $8,500( / ,12%,5) $8,700$11,057.98

C

D

A P A FA G

A P

= −

+ +== +=

Since don’t purchase the challenger. AEC AEC ,C > D

7



11.15)

Defender: Economic service year is 2 years

Interest rate 15%


0 $5,000 $1,2001 $4,000 $2,000 $1,750 $3,380 $5,1302 $3,000 $3,500 $1,680 $3,436 $5,1163 $2,000 $5,000 $1,614 $3,886 $5,5004 $1,000 $6,500 $1,551 $4,410 $5,9615 $0 $8,000 $1,492 $4,942 $6,434

Challenger: Economic service year is 4 years

Interest rate 15%


0 $10,0001 $6,000 $2,000 $5,500 $2,000 $7,5002 $5,100 $2,800 $3,779 $2,372 $6,1513 $4,335 $3,600 $3,131 $2,726 $5,8574 $3,685 $4,400 $2,765 $3,061 $5,8265 $3,132 $5,200 $2,519 $3,378 $5,897

Since AEC AECD C< , we should not replace the defender now. If no technological advances are expected in the next few years, the defender should be used for at least 2 more years. However, it is not necessarily best to replace the defender at the end of its economic year either.

8



Marginal analysis:

1. Opportunity cost at the end of year two, which is equal to the market value then, or $3,000

2. Operating cost for the third year: $5,000 3. Salvage value of the defender at the end of year three: $2,000

The cost of using the defender for one more year from the end of its economic service life is

3 $3,000( / ,15%,1) $5,000 $2,000$6,450

F F P= +=

−

Compare this cost with AEC $5,826C∗ = of the challenger.

Since keeping the defender for the 3rd year is more expensive than replacing it with the challenger, do not keep the defender beyond its economic service life.

11.16)

It is assumed that the required service period is very long.

AEC $7,000( / ,12%,6) $3,000 $2,500( / ,12%,6)$4,394.4

AEC $24,000( / ,12%,12) $1,500 $2,000( / ,12%,12)$5,290.8

D

C

A P A F

A P A F

= + −== + −

=

We should continue to use the old machine. The economic advantage is $5,290.8-$4,394.4 = $896.4 per year.

11.17) (a) and (b)

Defender Challenger

0 -$5,000 -$10,0001 -$3,000 -$2,0002 -$4,500 -$3,0003 -$4,000 -$0

n

( )

2 3

AEC (15%) $10,000 $2,000( / ,15%,1) $3,000( / ,15%,2) ( / ,15%,3)$6,135.29

$3,000 $4,500 $4,000AEC (15%) ($5,000 )( / ,15%,3)1.15 1.15 1.15

$5,974.94

C

D

P F P F A P

A P

= + +

=

= + + +

=

Do not replace the defender now.

9



11.18) (a) opportunity cost = $0 (b) The cash flows are:

Year: 0 1 2 3 4 5Defender $0 -$3K -$3K -$3K -$3K -$3K

Challenger -$10K 0 0 0 0 0C - D -$10K $3K $3K $3K $3K $3K

(c)

PW( ) $10,000 $3,000( / , ,5)0

C Di P− A i= − +=

We find . Since the replacement should be made now.

* 15.24%i = * ,i MARR>

11.19) (a)

AE(12%) $2,000( / ,12%,3) $10,000 $7,000$2,167.4

AE(12%) $14,000( / ,12%,5) $12,500 $5,000 $4,000( / ,12%,5)$4,246

D

C

A P

A P A F

= − + −== − + − +

=

Yes, the new machine should be purchased now. (b) Let

( / ,12%,5) $7,500 $4,000( / ,12%,5) $2,167.4P A P A F− + + =

We find $21,493.15P = 11.20)

Assume that the old system has a current market value of P. AEC ( / ,14%,5) $20,000AEC ($200,000 $18,000)( / ,14%,10) (0.14)($18,000) $5,000

$42, 411.86

D

C

P A PA P

= += − + +

=

Let AEC AECD C= and solve for P. We find that P = $76,942. If the resale value of the defender is higher than $76,941.73, the installation of the new system is justified.

10



11.21)

AEC(12%) $60,000( / ,12%,10) $18,000$28,619

AEC(12%) ($200,000 $20,000)( / ,12%,10) $20,000(0.12) $4,000$38,260

D

C

A P

A P

= +== − + +

= Since AEC AECD C< , do not replace the defender. 11.22)

For the challenger, we have:

AEC $50,000( / ,10%,12) $3,000 $6,000 $3,000( / ,10%,12)

$4,198C A P A F= + − −

=

For the defender, we need to find its economic life. Since the annual operating cost is constant and the salvage value declines as it ages, the annual equivalent cost is a decreasing function of the holding period. This means that the economic life is equal to its physical life, as illustrated in the following table.

With i = 10% ( 6 years, $4,213).D DN AEC= =

10%

Market O&M n value Costs CR(10%)OC(10%) AEC(10%)

0 $2,0001 $1,500 $3,800 $700 $3,800 $4,5002 $1,125 $3,800 $617 $3,800 $4,4173 $844 $3,800 $549 $3,800 $4,3494 $633 $3,800 $495 $3,800 $4,2955 $475 $3,800 $450 $3,800 $4,2506 $356 $3,800 $413 $3,800 $4,213

Interest rate

Since AEC AEC ,D C> the new machine should be purchased.

11



11.23)

Option 1

Option 2

AEC $15,000 $48,000( / ,12%,10) $12,000 $5,000( / ,12%,10)

$35,210.32AEC ($84,000 $6,000)( / ,12%,10) $24,000 $9,000( / ,12%,10)

$37, 291.91

A P A F

A P A F

= + + −

== − + −

=

Since Option 1 should be selected. Option 1 Option 2AEC AEC< 11.24)


0 1 2 3 4Income Statement

Revenues (savings)Expenses: O&M $2,500 $3,000 $3,500 $4,000 Depreciation $2,400 $3,840 $2,304 $691

Taxable Income ($4,900) ($6,840) ($5,804) ($4,691)Income Taxes (%) -1,960 -2,736 -2,322 -1,876

Net Income ($2,940) ($4,104) ($3,482) ($2,815)

Cash Flow StatementOperating Activities: Net Income (2,940)$ (4,104)$ (3,482)$ (2,815)$ Depreciation $2,400 $3,840 $2,304 $691Investment Activities: Investment (12,000)$

Salvage 3,000$ Gains Tax ($94)

Net Cash Flow ($12,000) ($540) ($264) ($1,178) $782

AEC(8%) $13,087( / ,8%,4)

$3,951

A P=

=

12



11.25)

Defender Cash flows for the defender: Year 0: -$13,000 Years 1-5: 0

AEC(12%) $13,000(1 0.40)( / ,12%,5) $2,164D A P= − =

Challenger

Input OutputTax Rate(%) = 40.00% PW(i) = 70,499$

MARR(%) = 12.00% AE(%) = $15,448

0 1 2 3 4 5 6 7Income Statement

Revenues (savings) $60,000 $60,000 $60,000 $60,000 $60,000 $60,000 $60,000Expenses: O&M $0 $0 $0 $0 $0 $0 $0 Depreciation $20,578 $35,266 $25,186 $17,986 $12,859 $12,845 $6,430

Taxable Income $39,422 $24,734 $34,814 $42,014 $47,141 $47,155 $53,570Income Taxes (%) 15,769 9,894 13,926 16,806 18,856 18,862 21,428

Net Income $23,653 $14,841 $20,889 $25,209 $28,284 $28,293 $32,142

Cash Flow Statement

Operating Activities: Net Income 23,653$ 14,841$ 20,889$ 25,209$ 28,284$ 28,293$ 32,142$ Depreciation $20,578 $35,266 $25,186 $17,986 $12,859 $12,845 $6,430Investment Activities: Investment (144,000)$

Salvage 40,000$ Gains Tax ($10,859)

Net Cash Flow (144,000)$ 44,231$ 50,106$ 46,074$ 43,194$ 41,144$ 41,138$ 67,713$

Replace the defender now as the challenger would provide an annual savings equivalent to $15,448 whereas the defender would cost $2,164 annually.

13



11.26)

(a) Cash Flows

Defender

Tax Rate(%) = 35.00% PW(i) = ($1,059)MARR(%) = 12.00% AEC(%) = ($294)


Revenues (savings)Expenses: Depreciation $5,625 $5,625 $5,625 $5,625 $5,625

Taxable Income ($5,625) ($5,625) ($5,625) ($5,625) ($5,625)Income Taxes (%) (1,969) (1,969) (1,969) (1,969) (1,969)

Net Income ($3,656) ($3,656) ($3,656) ($3,656) ($3,656)

Cash Flow StatementOperating Activities: Net Income (3,656)$ (3,656)$ (3,656)$ (3,656)$ (3,656)$ Depreciation $5,625 $5,625 $5,625 $5,625 $5,625Investment Activities: Investment (10,000)$ Salvage $5,000 Gains Tax -$1,750

Net Cash Flow ($10,000) $1,969 $1,969 $1,969 $1,969 $5,219

14



Challenger

Tax Rate(%) = 35.00% PW(i) = $20,801

MARR(%) = 12.00% AE(%) = $5,770


Revenues (savings) $33,000 $33,000 $33,000 $33,000 $33,000Expenses: Depreciation $10,718 $18,368 $13,118 $9,368 $3,349

Taxable Income $22,283 $14,633 $19,883 $23,633 $29,651Income Taxes (%) $7,799 $5,121 $6,959 $8,271 $10,378

Net Income $14,484 $9,511 $12,924 $15,361 $19,273

Cash Flow StatementOperating Activities: Net Income $14,484 $9,511 $12,924 $15,361 $19,273 Depreciation $10,718 $18,368 $13,118 $9,368 $3,349Investment Activities: Investment (75,000)$ Salvage $0 Gains Tax $7,028

Net Cash Flow (75,000)$ $25,201 $27,879 $26,041 $24,729 $29,651

(b) Yes, should replace the defender

15



11.27) At and tax rate = 40%, the economic service life is 1 year. 10%i =

Tax Rate 40%MARR 10% Permitted Annual Depreciation Amounts over theHolding Holding Period Total BookPeriod 1 2 3 4 5 6 7 8 Depreciatio Value

0 $30,0001 $4,287 $4,287 $25,7132 $4,287 $3,674 $7,961 $22,0403 $4,287 $7,347 $2,624 $14,258 $15,7434 $4,287 $7,347 $5,247 $1,874 $18,755 $11,2465 $4,287 $7,347 $5,247 $3,747 $1,340 $21,968 $8,033

Annual O&M Costs over the Holding Period Total PWHolding Total PW o of A/TPeriod 1 2 3 4 5 6 7 8 O&M CostsO&M Costs

0 `

1 $5,000 $4,545 $2,7272 $5,000 $6,500 $9,917 $5,9503 $5,000 $6,500 $10,000 $17,431 $10,4584 $5,000 $6,500 $10,000 $12,500 $25,968 $15,5815 $5,000 $6,500 $10,000 $12,500 $14,800 $35,158 $21,095

Expected Net A/T A/T Operating Costs (in PW) Total

Holding Market Taxable Gains Market over the Holding Period OC(10%) CR(10%) AEC(10%)

Period Value Gains Tax Value O&M CostsTax Shield Total OC01 $25,800 $87 $35 $25,765 $2,727 $1,559 $1,168 $1,285 $7,235 $8,520

($6,040) ($2,416)($5,743) ($2,297)($6,246) ($2,498)($8,033) ($3,213)

2 $16,000 $18,416 $5,950 $2,773 $3,177 $1,831 $8,516 $10,347

3 $10,000 $12,297 $10,458 $4,776 $5,682 $2,285 $8,348 $10,633

4 $5,000 $7,498 $15,581 $6,076 $9,505 $2,998 $7,848 $10,847

5 $0 $3,213 $21,095 $6,921 $14,174 $3,739 $7,388 $11,127

11.28) (a) Interest Defender: 10%i = The depreciation schedule when the defender was placed in service: D1 = $3,573, D2 = $6,123, D3 = $4,373, D4 = $3,125, D5 = $2,231, D6 = $2,231, D7 = $2,231, D8 = $1,116.



Tax Rate 35%MARR 10% Permitted Annual Depreciation Amounts over theHolding Holding Period Total BookPeriod 1 2 3 4 5 6 7 8 Depreciation Value

0 $7,8091 $2,231 $2,231 $5,5782 $2,231 $2,231 $4,462 $3,3473 $2,231 $2,231 $2,231 $6,693 $1,1164 $2,231 $2,231 $2,231 $1,116 $7,809 $0

Annual O&M Costs over the Holding Period Total PWHolding Total PW of of A/TPeriod 1 2 3 4 5 6 7 8 O&M Costs O&M Costs

0 `1 $3,200 $2,909 $1,8912 $3,200 $3,700 $5,967 $3,8793 $3,200 $3,700 $4,800 $9,573 $6,2234 $3,200 $3,700 $4,800 $5,850 $13,569 $8,820

Expected Net A/T A/T Operating Costs (in PW) TotalHolding Market Taxable Gains Market over the Holding Period OC(10%) CR(10%) AEC(10%)Period Value Gains Tax Value O&M Costs Tax Shield Total OC

0 $7,700 $8,2851 $4,300 ($1,278) ($447) $4,747 $1,891 $710 $1,181 $1,299 $3,843 $5,1422 $3,300 ($47) ($16) $3,316 $3,879 $1,355 $2,523 $1,454 $2,920 $4,3743 $1,100 ($16) ($6) $1,106 $6,223 $1,942 $4,281 $1,721 $2,806 $4,5274 $0 $0 $0 $0 $8,820 $2,209 $6,611 $2,086 $2,464 $4,549

Note that the cost of retaining the defender on after-tax basis is $8,285, instead of $7,700. The scheduled depreciation amount during the fourth year of ownership is $3,125. Since the asset will be disposed of during the recovery period, the allowed depreciation amount will be (0.5) ($3,125) = $1,561. Then, the book value becomes $9,370, instead of $7,809. With the market value of $7,700, there will be a loss of $1,670. The tax credit on this loss will be $1,670(0.35) = $584.50. Finally, the net proceeds from sale of old asset will be $8,285 (= $7,700 + $584.50).

The defender’s remaining useful (economic) life is 2 more years with an AEC value of $4,374, i.e., . 2,AEC $4,374D DN = =

17



(b) 10 yearsCN =AEC $4,319C =

Input Output


0 1 2 3 4 5 6 7 8 9 10Income Statement

Revenues (savings)Expenses: O&M $1,000 $1,000 $1,000 $1,000 $1,000 $1,000 $1,000 $1,000 $1,000 $1,000 Depreciation $4,430 $7,592 $5,422 $3,872 $2,768 $2,765 $2,768 $1,383 $0 $0

Taxable Income ($5,430) ($8,592) ($6,422) ($4,872) ($3,768) ($3,765) ($3,768) ($2,383) ($1,000) ($1,000)Income Taxes (%) -1,900 -3,007 -2,248 -1,705 -1,319 -1,318 -1,319 -834 -350 -350

Net Income ($3,529) ($5,585) ($4,174) ($3,167) ($2,449) ($2,447) ($2,449) ($1,549) ($650) ($650)

Cash Flow StatementOperating Activities: Net Income (3,529)$ (5,585)$ (4,174)$ (3,167)$ (2,449)$ (2,447)$ (2,449)$ (1,549)$ (650)$ (650)$ Depreciation $4,430 $7,592 $5,422 $3,872 $2,768 $2,765 $2,768 $1,383 $0 $0Investment Activities: Investment (31,000)$

Salvage 2,500$ Gains Tax ($875)

Net Cash Flow (31,000)$ $900 $2,007 $1,248 $705 $319 $318 $319 ($166) ($650) 975$

(c) Marginal analysis:

• From n = 2 to n = 3:

$3,316(1.10) $1,106 $4,280 $6,824.6 $4,319− + = >

Keep the defender for two years, which happens to be the same as the economic service life as calculated before. (In general, you should not expect this to happen all the time.)

18



11.29) Corrections: In the first printing, the book values for years 5 and 6 were stated incorrectly. The correct figures should be: B5 = $576 and B6 = 0.

(a) Economic service life = 6 years with MARR is 15% Tax Rate 40% Investment $10,000MARR 15% Book value $10,000 Permitted Annual Depreciation Amounts over theHolding Holding Period Total BookPeriod 1 2 3 4 5 6 7 8 Depreciation Value

0 $10,0001 $2,000 $2,000 $8,0002 $2,000 $3,200 $5,200 $4,8003 $2,000 $3,200 $1,920 $7,120 $2,8804 $2,000 $3,200 $1,920 $1,152 $8,272 $1,7285 $2,000 $3,200 $1,920 $1,152 $1,152 $9,424 $5766 $2,000 $3,200 $1,920 $1,152 $1,152 $576 $10,000 $0


0 `1 $1,500 $1,304 $7832 $1,500 $2,100 $2,892 $1,7353 $1,500 $2,100 $2,700 $4,668 $2,8014 $1,500 $2,100 $2,700 $3,400 $6,612 $3,9675 $1,500 $2,100 $2,700 $3,400 $4,200 $8,700 $5,2206 $1,500 $2,100 $2,700 $3,400 $4,200 $4,900 $10,818 $6,491


01 $5,300 ($2,700) ($1,080) $6,380 $783 $696 $87 $100 $5,120 $5,2202 $3,900 ($900) ($360) $4,260 $1,735 $1,664 $72 $44 $4,170 $4,2143 $2,800 ($80) ($32) $2,832 $2,801 $2,168 $632 $277 $3,564 $3,8414 $1,800 $72 $29 $1,771 $3,967 $2,432 $1,535 $538 $3,148 $3,6865 $1,400 $824 $330 $1,070 $5,220 $2,661 $2,559 $763 $2,824 $3,5886 $600 $600 $240 $360 $6,491 $2,761 $3,730 $986 $2,601 $3,587

19



(b) Economic service life = 5 years with MARR is 10%

Tax R ate 40% Investm ent $10,000

M A R R 10% B ook value $10,000

P erm itted A nnual D epreciation A m ounts over the

H olding H olding P eriod Total B ook

P eriod 1 2 3 4 5 6 7 8 D ep. V alue

0 $10,000

1 $2,000 $2,000 $8,000

2 $2,000 $3,200 $5,200 $4,800

3 $2,000 $3,200 $1,920 $7,120 $2,880

4 $2,000 $3,200 $1,920 $1,152 $8,272 $1,728

5 $2,000 $3,200 $1,920 $1,152 $1,152 $9,424 $576

6 $2,000 $3,200 $1,920 $1,152 $1,152 $576 $10,000 $0

A nnual O & M C osts over the H olding P eriod Total P W

H olding Total P W of of A /T

P eriod 1 2 3 4 5 6 7 8 O & M C ostsO & M C

0 `

1

osts

$1,500 $1,364 $818

2 $1,500 $2,100 $3,099 $1,860

3 $1,500 $2,100 $2,700 $5,128 $3,077

4 $1,500 $2,100 $2,700 $3,400 $7,450 $4,470

5 $1,500 $2,100 $2,700 $3,400 $4,200 $10,058 $6,035

6 $1,500 $2,100 $2,700 $3,400 $4,200 $4,900 $12,824 $7,694

E xpected N et A /T A /T O perating C osts (in P W ) Total

H olding M arket Taxable G ains M arket over the H olding P eriod O C (10% ) C R (10% ) A E C (10% )

P eriod V alue G ains Tax V alue O & M C osts Tax S hield Total O C

0

1 $5,300 ($2,700) ($1,080) $6,380 $818 $727 $91 $100 $4,620 $4,720

2 $3,900 ($900) ($360) $4,260 $1,860 $1,785 $74 $43 $3,733 $3,776

3 $2,800 ($80) ($32) $2,832 $3,077 $2,362 $715 $287 $3,166 $3,453

4 $1,800 $72 $29 $1,771 $4,470 $2,677 $1,793 $566 $2,773 $3,339

5 $1,400 $824 $330 $1,070 $6,035 $2,963 $3,072 $810 $2,463 $3,273

6 $600 $600 $240 $360 $7,694 $3,093 $4,601 $1,056 $2,249 $3,306

11.30) (a) At i = 10%, the economic service life = 7 years:

20







P eriod 1 2 3 4 5 6 7 8 9 10 D epreciation V alue

0

1 $3,000 $3,000 $27,000

2 $3,000 $3,000 $6,000 $24,000

3 $3,000 $3,000 $3,000 $9,000 $21,000

4 $3,000 $3,000 $3,000 $3,000 $12,000 $18,000

5 $3,000 $3,000 $3,000 $3,000 $3,000 $15,000 $15,000

6 $3,000 $3,000 $3,000 $3,000 $3,000 $3,000 $18,000 $12,000

7 $3,000 $3,000 $3,000 $3,000 $3,000 $3,000 $3,000 $21,000 $9,000

8 $3,000 $3,000 $3,000 $3,000 $3,000 $3,000 $3,000 $3,000 $24,000 $6,000

9 $3,000 $3,000 $3,000 $3,000 $3,000 $3,000 $3,000 $3,000 $3,000 $27,000 $3,000

10 $3,000 $3,000 $3,000 $3,000 $3,000 $3,000 $3,000 $3,000 $3,000 $3,000 $30,000 $0



P eriod 1 2 3 4 5 6 7 8 O & M C osts O & M C osts

0 `

1 $3,000 $2,727 $1,636

2 $3,000 $3,450 $5,579 $3,347

3 $3,000 $3,450 $3,968 $8,560 $5,136

4 $3,000 $3,450 $3,968 $4,563 $11,676 $7,006

5 $3,000 $3,450 $3,968 $4,563 $5,247 $14,934 $8,961

6 $3,000 $3,450 $3,968 $4,563 $5,247 $6,034 $18,340 $11,004

7 $3,000 $3,450 $3,968 $4,563 $5,247 $6,034 $6,939 $21,901 $13,141

8 $3,000 $3,450 $3,968 $4,563 $5,247 $6,034 $6,939 $7,960 $25,615 $15,369

9 $3,000 $3,450 $3,968 $4,563 $5,247 $6,034 $6,939 $7,960 $9,177 $29,506 $17,704

10 $3,000 $3,450 $3,968 $4,563 $5,247 $6,034 $6,939 $7,960 $9,177 $10,554 $33,576 $20,145




0

1 $20,000 ($7,000) ($2,800) $22,800 $1,636 $1,091 $545 $600 $10,200 $10,800

2 $18,000 ($6,000) ($2,400) $20,400 $3,347 $2,083 $1,264 $729 $7,571 $8,300

3 $16,000 ($5,000) ($2,000) $18,000 $5,136 $2,984 $2,152 $865 $6,625 $7,491

4 $14,000 ($4,000) ($1,600) $15,600 $7,006 $3,804 $3,202 $1,010 $6,103 $7,113

5 $12,000 ($3,000) ($1,200) $13,200 $8,961 $4,549 $4,412 $1,164 $5,752 $6,916

6 $10,000 ($2,000) ($800) $10,800 $11,004 $5,226 $5,778 $1,327 $5,488 $6,815

7 $8,000 ($1,000) ($400) $8,400 $13,141 $5,842 $7,299 $1,499 $5,277 $6,776

8 $6,000 $0 $0 $6,000 $15,369 $6,402 $8,967 $1,681 $5,099 $6,779

9 $4,000 $1,000 $400 $3,600 $17,704 $6,911 $10,793 $1,874 $4,944 $6,818

10 $2,000 $2,000 $800 $1,200 $20,145 $7,373 $12,772 $2,079 $4,807 $6,886

(b) At i = 25%, the economic service life = 10 years

21







P eriod 1 2 3 4 5 6 7 8 9 10 D epreciation V alue

0

1 $3,000 $3,000 $27,000

2 $3,000 $3,000 $6,000 $24,000

3 $3,000 $3,000 $3,000 $9,000 $21,000

4 $3,000 $3,000 $3,000 $3,000 $12,000 $18,000

5 $3,000 $3,000 $3,000 $3,000 $3,000 $15,000 $15,000

6 $3,000 $3,000 $3,000 $3,000 $3,000 $3,000 $18,000 $12,000

7 $3,000 $3,000 $3,000 $3,000 $3,000 $3,000 $3,000 $21,000 $9,000

8 $3,000 $3,000 $3,000 $3,000 $3,000 $3,000 $3,000 $3,000 $24,000 $6,000

9 $3,000 $3,000 $3,000 $3,000 $3,000 $3,000 $3,000 $3,000 $3,000 $27,000 $3,000

10 $3,000 $3,000 $3,000 $3,000 $3,000 $3,000 $3,000 $3,000 $3,000 $3,000 $30,000 $0



P eriod 1 2 3 4 5 6 7 8 O & M C osts O & M C osts

0 `

1 $3,000 $2,400 $1,440

2 $3,000 $3,450 $4,608 $2,765

3 $3,000 $3,450 $3,968 $6,640 $3,984

4 $3,000 $3,450 $3,968 $4,563 $8,509 $5,105

5 $3,000 $3,450 $3,968 $4,563 $5,247 $10,228 $6,137

6 $3,000 $3,450 $3,968 $4,563 $5,247 $6,034 $11,810 $7,086

7 $3,000 $3,450 $3,968 $4,563 $5,247 $6,034 $6,939 $13,265 $7,959

8 $3,000 $3,450 $3,968 $4,563 $5,247 $6,034 $6,939 $7,960 $14,600 $8,760

9 $3,000 $3,450 $3,968 $4,563 $5,247 $6,034 $6,939 $7,960 $9,177 $15,832 $9,499

10 $3,000 $3,450 $3,968 $4,563 $5,247 $6,034 $6,939 $7,960 $9,177 $10,554 $16,965 $10,179




0

1 $20,000 ($7,000) ($2,800) $22,800 $1,440 $960 $480 $600 $14,700 $15,300

2 $18,000 ($6,000) ($2,400) $20,400 $2,765 $1,728 $1,037 $720 $11,767 $12,487

3 $16,000 ($5,000) ($2,000) $18,000 $3,984 $2,342 $1,641 $841 $10,648 $11,488

4 $14,000 ($4,000) ($1,600) $15,600 $5,105 $2,834 $2,271 $962 $9,998 $10,959

5 $12,000 ($3,000) ($1,200) $13,200 $6,137 $3,227 $2,910 $1,082 $9,547 $10,629

6 $10,000 ($2,000) ($800) $10,800 $7,086 $3,542 $3,544 $1,201 $9,205 $10,406

7 $8,000 ($1,000) ($400) $8,400 $7,959 $3,793 $4,166 $1,318 $8,933 $10,251

8 $6,000 $0 $0 $6,000 $8,760 $3,995 $4,766 $1,432 $8,710 $10,141

9 $4,000 $1,000 $400 $3,600 $9,499 $4,156 $5,344 $1,543 $8,523 $10,066

10 $2,000 $2,000 $800 $1,200 $10,179 $4,285 $5,895 $1,651 $8,366 $10,017

22



(c) At i = 0%, the economic service life = 4 years:

• Capital recovery cost:

gaingain tax ( )

net proceeds (1 )(1 0.40)(22,000 2000 ) 0.40(30,000 3,000 )

(1 )

4,800 2, 400

4,800 2, 400

n n

m n n

m n m n

m n m n

S Bt S B

t S t Bn n

I t S t BCRn

nn

n

= −= −= − +

= − − + −− − −

=

+=

= +

• Equivalent annual O&M cost:

( )

( )( )

( )

1

1

1

1&

1

1

A/T O&M (1 ) 3,000 1.15

1,800 1.15

1,800 1.15

1.151,800

1,800 (1.15 1) (1.15 1)12,0001.15 1

nm

n

n n

nO M

n n

n

n n

t

AEn

n

n n

−

−

−

=

−

=

⎡ ⎤= − ⎣ ⎦

=

=

=

− −= =

−

∑

∑

• Depreciation tax credit:

1( )

, where $3,000

1, 200

nm nn

D n

t DAE D

n=

×= =

=

∑

23



• Minimum total annual equivalent cost:

4,800 (1.15 1)AEC 2,400 12,000 1,200

4,800 (1.15 1)12,000 1,200

1.15 7,20012,000 1,200

n

n

n

n n

n n

n n

−= + + −

−= + +

= − +

Using Excel, we find the economic service life at 5n = years.

n AEC

1 $ 7,800 2 $ 5,535 3 $ 4,884 4 $ 4,647 5 $ 4,587 6 $ 4,626 7 $ 4,731 8 $ 4,889 9 $ 5,091

10 $ 5,335

24



11.31) Economic service life With i = 12% and tax rate = 40%: Economic service life = 1 year

Tax Rate 40% Investment $30,000MARR 12% Book value $30,000 Permitted Annual Depreciation Amounts over theHolding Holding Period Total BookPeriod 1 2 3 4 5 6 7 8 Depreciation Value

01 $6,000 $6,000 $24,0002 $6,000 $4,800 $10,800 $19,2003 $6,000 $9,600 $2,880 $18,480 $11,5204 $6,000 $9,600 $5,760 $1,728 $23,088 $6,9125 $6,000 $9,600 $5,760 $3,456 $1,728 $26,544 $3,456


0 `1 $5,000 $4,464 $2,6792 $5,000 $6,500 $9,646 $5,7883 $5,000 $6,500 $10,000 $16,764 $10,0584 $5,000 $6,500 $10,000 $12,500 $24,708 $14,8255 $5,000 $6,500 $10,000 $12,500 $14,800 $33,106 $19,863


01 $25,800 $1,800 $720 $25,080 $2,679 $2,143 $536 $600 $8,520 $9,1202 $16,000 ($3,200) ($1,280) $17,280 $5,788 $3,673 $2,114 $1,251 $9,600 $10,8513 $10,000 ($1,520) ($608) $10,608 $10,058 $6,024 $4,034 $1,680 $9,347 $11,0264 $5,000 ($1,912) ($765) $5,765 $14,825 $7,283 $7,541 $2,483 $8,671 $11,1545 $0 ($3,456) ($1,382) $1,382 $19,863 $8,115 $11,749 $3,259 $8,105 $11,364

25



11.32) Replacement Analysis Correction: In the first printing, the current book value is listed as $6,248. It should be stated as $5,623. Replacement analysis:

(a) Keep the defender

Financial Data n -4 -3 -2 -1 0 1 2 3 4 5 6

Depreciation $2,572 $4,408 $3,148 $2,248 $1,607 $1,606 $1,607 $803Book value $18,000 $15,428 $11,020 $7,871 $5,623 $4,016 $2,410 $803 ($0) ($0) ($0)Salvage value $2,500O&M cost $3,000 $3,000 $3,000 $3,000 $3,000 $3,000

Cash Flow Statement(-0.7)*(O&M cost) ($2,100) ($2,100) ($2,100) ($2,100) ($2,100) ($2,100)+(.3)*(Depreciation) $482 $482 $482 $241 $0 $0Investment ($7,000)Net proceeds from sale 1,750

Net Cash Flow $0 $0 $0 $0 ($7,000) ($1,618) ($1,618) ($1,618) ($1,859) ($2,100) ($350)

PW (8%) = ($14,186) AEC(8%) = $3,069

(b) Replace the defender

Financial Data n 1 2 3 4 5 6 7 8 9 10-11 12

Depreciation $3,144 $5,388 $3,848 $2,748 $1,965 $1,962 $1,965 $981 $0 $0 $0Book value $22,000 $18,856 $13,468 $9,621 $6,873 $4,908 $2,946 $981 $0 $0 $0 $0Salvage value $2,000O&M cost $1,500 $1,500 $1,500 $1,500 $1,500 $1,500 $1,500 $1,500 $1,500 $1,500 $1,500

Cash Flow Statement+(.3)*(Depreciation) $943 $1,616 $1,154 $824 $589 $589 $589 $294 $0 $0 $0(-0.7)*(O&M cost) ($1,050) ($1,050) ($1,050) ($1,050) ($1,050) ($1,050) ($1,050) ($1,050) ($1,050) ($1,050) ($1,050)Investment ($22,000)Net proceeds from sale $1,400

Net Cash Flow ($22,000) ($107) $566 $104 ($226) ($461) ($461) ($461) ($756) ($1,050) ($1,050) $350

PW (8%) = ($24,301) AEC(8%) = $3,225

Keep the defender. Note: The salvage value of the defender is reduced by the removal cost at the end of its service life ($2,500 = $4,000 - $1,500).

26



11.33) (a), (b), and (c):

(a) , (b) and ( c )

Option 1 : Keep the defender

Financial Data n 0 1 2 3 4 5

Depreciation $800 $800 $800 $800 $800Book value $4,000 $3,200 $2,400 $1,600 $800 $0Expected Market value $0 $0 $0 $0 $0 $0O&M cost $0 $0 $0 $0 $0

Cash Flow Statement(-0.6)*(O&M cost) 0 0 0 0+(.4)*(Depreciation) 320 320 320 320 320Investment (1,600)Net proceeds from sale

Net Cash Flow ($1,600) $320 $320 $320 $320 $320

PW (10%) = ($387) AEC(10%) = $102

Option 2 : Replace the defender

Financial Data n 0 1 2 3 4 5Depreciation $1,429 $2,449 $1,749 $1,249 $446Book value $10,000 $8,571 $6,122 $4,373 $3,124 $2,678Expected Market value $0 $0 $0 $0 $0 $0Savings in O&M cost $3,000 $3,000 $3,000 $3,000 $3,000

Cash Flow Statement+(.4)*(Depreciation) $572 $980 $700 $500 $178(0.6)*(Savings in O&M cost) $1,800 $1,800 $1,800 $1,800 $1,800Investment ($10,000)Net proceeds from sale $1,071

Net Cash Flow ($10,000) $2,372 $2,780 $2,500 $2,300 $3,049

PW (10%) = ($205) AEC(10%) = $54

0

Should replace the defender.

27



11.34) Replacement analysis: Let X denote the current market value of the old call-switching system:

AEC(14%) $20,000(0.60) 0.60 ( / ,14%,5)

$12,000 (0.6 )(0.29128)$12,000 0.1748

AEC(14%) $156,291( / ,14%,10)

$29,963

defender

challenger

X A P

XX

A P

= +

= += +=

=

To justify the new call-switching system now, we must have

AEC(14%) >AEC(14%)

$12,000 0.1748 $29,962

$102,757

defender challenger

X

X

+ >

>

Challenger: Challenger


Depreciation $40,000 $64,000 $38,400 $23,040 $23,040Book value $200,000 $160,000 $96,000 $57,600 $34,560 $11,520Salvage valueO&M cost $5,000 $5,000 $5,000 $5,000 $5,000

Cash Flow Statement-(0.6)*(O&M cost) ($3,000) ($3,000) ($3,000) ($3,000) ($3,000)+(.4)*(Depreciation) $16,000 $25,600 $15,360 $9,216 $9,216Investment ($200,000)Net proceeds from sale

Net Cash Flow ($200,000) $13,000 $22,600 $12,360 $6,216 $6,216

Financial Data n 6 7 8 9 10

Depreciation $11,520 $0 $0 $0 $Book value $0 $0 $0 $0 $0Salvage value $18,000O&M cost $5,000 $5,000 $5,000 $5,000 $5,000

Cash Flow Statement-(0.6)*(Savings in O&M cost) ($3,000) ($3,000) ($3,000) ($3,000) ($3,000)+(.4)*(Depreciation) $4,608 $0 $0 $0 $0InvestmentNet proceeds from sale $10,800

Net Cash Flow $1,608 ($3,000) ($3,000) ($3,000) $7,800

0

28



11.35)

Defender:

Tax Rate 35% Investment $12,725MARR 18% Book value $15,000 Permitted Annual Depreciation Amounts over theHolding Holding Period Total BookPeriod 1 2 3 4 5 6 7 8 Depreciation Valu

0 $15,0001 $4,000 $4,000 $11,0002 $4,000 $4,000 $8,000 $7,0003 $4,000 $4,000 $4,000 $12,000 $3,000


0 `1 $4,500 $3,814 $2,4792 $4,500 $5,300 $7,620 $4,9533 $4,500 $5,300 $6,100 $11,333 $7,366


01 $5,200 ($5,800) ($2,030) $7,230 $2,479 $1,186 $1,292 $1,525 $7,786 $9,3112 $3,500 ($3,500) ($1,225) $4,725 $4,953 $2,192 $2,761 $1,764 $5,960 $7,7243 $1,200 ($1,800) ($630) $1,830 $7,366 $3,044 $4,322 $1,988 $5,340 $7,328

e

Note: The opportunity cost of retaining the defender is as follows:

• Current market value = $11,500 • Current book value = $15,000 • Losses = ($11,500 - $15,000) = ($3,500) • Loss tax credit = $3,500(0.35) = $1,225 • Cost of retaining the defender = $11,500+ $1,225 = $12,725

29



Challenger:


Depreciation $6,216 $10,653 $7,608 $5,433 $3,885Book value $43,500 $37,284 $26,631 $19,023 $13,589 $9,705Salvage value

O&M cost $1,500 $1,500 $1,500 $1,500 $1,500

Cash Flow Statement

-(0.65)*(O&M cost) ($975) ($975) ($975) ($975) ($975)+(.35)*(Depreciation) $2,176 $3,729 $2,663 $1,902 $1,360Investment ($43,500)Net proceeds from sale

Net Cash Flow ($43,500) $1,201 $2,754 $1,688 $927 $385

Financial Data n 6 7 8 9 10Depreciation $3,880 $3,885 $1,940 $0 $0Book value $5,825 $1,940 ($0) ($0) ($0)Salvage value $3,500O&M cost $1,500 $1,500 $1,500 $1,500 $1,500

Cash Flow Statement

-(0.65)*(O&M cost) ($975) ($975) ($975) ($975) ($975)+(.35)*(Depreciation) $1,358 $1,360 $679 $0 $0Investment

Net proceeds from sale $2,275

Net Cash Flow $0 $383 $385 ($296) ($975) $1,300

PW (18%) = ($38,619)

AEC(18%) = $8,593

Optimal time to replace: Since the remaining useful life for the defender is 3 years, which is the same as the physical life, keep the defender for 3 years.

30



11.36) Decision: Do not replace the defender now.

Keep the defender


Depreciation $0 $0 $0 $0 $0Book value $0 $0 $0 $0 $0 $0Market value $8,500Operation Cost $8,700 $8,700 $8,700 $8,700 $8,700

Cash Flow Statement+(.35)*(Depreciation) $0 $0 $0 $0 $0Opportunity cost ($5,525)-(1-0.35)*(Operation cost) ($5,655) ($5,655) ($5,655) ($5,655) ($5,655)

Net Cash Flow ($5,525) ($5,655) ($5,655) ($5,655) ($5,655) ($5,655)

PW (12%) = ($25,910) AEC(12%) = $7,188

Replace the defender

Financial Data n 0 1 2 3 4 5Depreciation $7,645 $13,102 $9,357 $6,682 $2,386Book value $53,500 $45,855 $32,753 $23,396 $16,713 $14,327Market value $53,500 $12,000Operation Cost $4,200 $4,700 $5,200 $5,700 $6,200

Cash Flow StatementInvestment ($53,500)Net proceeds from sale $12,815+(.35)*(Depreciation) $2,676 $4,586 $3,275 $2,339 $835-(1-0.35)*(Operation cost) ($2,730) ($3,055) ($3,380) ($3,705) ($4,030)

Net Cash Flow ($53,500) ($54) $1,531 ($105) ($1,366) $9,620

PW (12%) = ($47,813) AEC(12%) = $13,264

31



11.37)

Option 1: Option 1

Financial Data n 0 1 2 3 4 5Depreciation $6,859 $11,755 $8,395 $5,995 $4,286Book value $48,000 $41,141 $29,386 $20,990 $14,995 $10,709Current Market value $6,000O&M cost $27,000 $27,000 $27,000 $27,000 $27,000

Cash Flow Statement-(0.60)*(O&M cost) ($16,200) ($16,200) ($16,200) ($16,200) ($16,200)+(.40)*(Depreciation) $2,744 $4,702 $3,358 $2,398 $1,715Opportunity cost ($3,600)Investment ($48,000)Net proceeds from sale

Net Cash Flow ($51,600) ($13,456) ($11,498) ($12,842) ($13,802) ($14,485)

Financial Data n 6 7 8 9 10Depreciation $4,282 $4,286 $2,141 $0 $0Book value $6,427 $2,141 $0 $0 $0Salvage value $5,000O&M cost $27,000 $27,000 $27,000 $27,000 $27,000

Cash Flow Statement-(0.60)*(O&M cost) ($16,200) ($16,200) ($16,200) ($16,200) ($16,200)+(.40)*(Depreciation) $1,713 $1,715 $856 $0 $0InvestmentNet proceeds from sale 3,000

Net Cash Flow ($14,487) ($14,485) ($15,344) ($16,200) ($13,200)

PW (12%) = ($129,093) AEC(12%) = $22,847

32



Option 2:

Option 2

0 1 2 3 4 5Depreciation $12,004 $20,572 $14,692 $10,492 $7,501Book value $84,000 $71,996 $51,425 $36,733 $26,242 $18,740O&M cost $24,000 $24,000 $24,000 $24,000 $24,000

Cash Flow Statement-(0.60)*(O&M cost) ($14,400) ($14,400) ($14,400) ($14,400) ($14,400)+(.40)*(Depreciation) $4,801 $8,229 $5,877 $4,197 $3,000Investment ($84,000)Net proceeds from sale

Net Cash Flow ($84,000) ($9,599) ($6,171) ($8,523) ($10,203) ($11,400)

6 7 8 9 10Depreciation $7,493 $7,501 $3,746 $0 $0Book value $11,248 $3,746 ($0) ($0) ($0)Salvage value $9,000O&M cost $24,000 $24,000 $24,000 $24,000 $24,000

Cash Flow Statement-(0.60)*(O&M cost) ($14,400) ($14,400) ($14,400) ($14,400) ($14,400)+(.40)*(Depreciation) $2,997 $3,000 $1,499 $0 $0InvestmentNet proceeds from sale $5,400

Net Cash Flow ($11,403) ($11,400) ($12,901) ($14,400) ($9,000)

PW (12%) = ($140,744)AEC(12%) = $24,910

Select Option 1.

33



34

11.38) The remaining useful life of the defender is 1 year. Its annual equivalent cost is $1,666. When the defender is replaced now by the challenger, its equivalent annual cost is $2,191, indicating that the defender should be kept for now. (a) Economic service life = one year Tax Rate 30% Investment $1,050MARR 12% Book value $0 Permitted Annual Depreciation Amounts over theHolding Holding Period Total BookPeriod 1 2 3 4 5 6 7 8 Depreciation Value

01 $0 $02 $0 $03 $0 $04 $0 $05 $0 $0

Annual O&M Costs over the Holding Period Total PWHolding Total PW of of A/TPeriod 1 8 O&M Costs O&M Costs

01 $ $1,696 $1,1882 $ $3,530 $2,4713 $ $5,452 $3,8164 $ $7,422 $5,1955 $ $9,351 $6,546

Expec TotalHolding Mark C(12%) CR(12%) AEC(12%)Period Val

01 $1 $1,330 $336 $1,6662 $1 $1,462 $291 $1,7533 $1,589 $333 $1,9224 $1,710 $346 $2,0565 $1,816 $291 $2,107

2 3 4 5 6 7`

1,9001,900 $2,3001,900 $2,300 $2,7001,900 $2,300 $2,700 $3,1001,900 $2,300 $2,700 $3,100 $3,400

ted Net A/T A/T Operating Costs (in PW)et Taxable Gains Market over the Holding Period Oue Gains Tax Value O&M Costs Tax Shield Total OC

,200 $1,200 $360 $840 $1,188 $0 $1,188,000 $1,000 $300 $700 $2,471 $0 $2,471$500 $500 $150 $350 $3,816 $0 $3,816$0 $0 $0 $0 $5,195 $0 $5,195$0 $0 $0 $0 $6,546 $0 $6,546

(b) Replace the defender


Depreciation $1,200 $1,920 $1,152 $691 $346Book value $6,000 $4,800 $2,880 $1,728 $1,037 $691Salvage value $1,000O&M cost $1,100 $1,300 $1,500 $1,700 $1,800

Cash Flow StatementInvestment ($6,000)Net proceeds from sale $907+(.30)*(Depreciation) $360 $576 $346 $207 $104-(1-0.30)*(O&M cost) ($770) ($910) ($1,050) ($1,190) ($1,260)

Net Cash Flow ($6,000) ($410) ($334) ($704) ($983) ($249)

PW (12%) = ($7,899)AEC(12%) = $2,191

Note: The purchase of a new machine will result in the combined savings in delays, operation and repairs in the amount of $200 a year, so that the O&M cost for the new machine will be reduced by $200 each year. For example, $1300 - $200 = $1,100 for n = 1.



11.39) (a) and (b): Quintana should purchase the new equipment. Financial Data n 0 1 2 3 4 5 6 7 8 9 10

Depreciation $21,435 $36,735 $26,235 $18,735 $13,395 $13,380 $13,395 $6,690 $0 $0Book value $150,000 $128,565 $91,830 $65,595 $46,860 $33,465 $20,085 $6,690 $0 $0 $0Market value $150,000Savings $30,000 $30,000 $30,000 $30,000 $30,000 $30,000 $30,000 $30,000 $30,000 $30,000

Cash Flow Statement+(1-0.40)*(Savings) $18,000 $18,000 $18,000 $18,000 $18,000 $18,000 $18,000 $18,000 18,000 18,000+(.4)*(Depreciation) $8,574 $14,694 $10,494 $7,494 $5,358 $5,352 $5,358 $2,676 0 0Investment ($150,000)

Net Cash Flow ($150,000) $26,574 $32,694 $28,494 $25,494 $23,358 $23,352 $23,358 $20,676 $18,000 $18,000

PW (10%) = $3,889 AE (10%) = $633

(b) DefenderFinancial Data n 0 1 2 3 4 5 6 7 8 9 10Depreciation $12,000 $12,000 $12,000 $12,000 $12,000 $12,000 $0 $0 $0 $0Book value $72,000 $60,000 $48,000 $36,000 $24,000 $12,000 $0 $0 $0 $0 $0Current market value

Cash Flow Statement+(.4)*(Depreciation) $4,800 $4,800 $4,800 $4,800 $4,800 $4,800 0 0 0 0Investment ($28,800)

Net Cash Flow ($28,800) $4,800 $4,800 $4,800 $4,800 $4,800 $4,800 $0 $0 $0 $0

PW (10%) = ($7,895) AE (10%) = ($1,285)

35



(c): Purchase the new equipment; (d): Retain the old machine (c) Defender with a current market value of $ 45,000Financial Data n 0 1 2 3 4 5 6 7 8 9 - 10

Depreciation $12,000 $12,000 $12,000 $12,000 $12,000 $12,000 $0 $0 $0Book value $72,000 $60,000 $48,000 $36,000 $24,000 $12,000 $0Current market value $45,000

Cash Flow Statement+(.4)*(Depreciation) $4,800 $4,800 $4,800 $4,800 $4,800 $4,800Investment ($55,800)

Net Cash Flow ($55,800) $4,800 $4,800 $4,800 $4,800 $4,800 $4,800

PW (10%) = ($34,895) AEC(10%) = $5,679

(d) Challenger with an extended service life of 12 yearsFinancial Data n 0 1 2 3 4 5 6 7 8 9-12Depreciation $21,435 $36,735 $26,235 $18,735 $13,395 $13,380 $13,395 $6,690 $0Book value $150,000 $128,565 $91,830 $65,595 $46,860 $33,465 $20,085 $6,690 $0 $0Savings $15,000 $15,000 $15,000 $15,000 $15,000 $15,000 $15,000 $15,000 $15,000

Cash Flow Statement+(.4)*(Depreciation) $8,574 $14,694 $10,494 $7,494 $5,358 $5,352 $5,358 $2,676 $0+(0.60)*(Savings) $9,000 $9,000 $9,000 $9,000 $9,000 $9,000 $9,000 $9,000 $9,000Investment ($150,000)

Net Cash Flow ($150,000) $17,574 $23,694 $19,494 $16,494 $14,358 $14,352 $14,358 $11,676 $9,000

PW (10%) = ($45,390) AEC(10%) = $6,662

36



(e) and (f): (e) Defender (Model A) Decision: Replace Model A withOriginal Investment = $150,000 Model B

n 0 1 2 3 4 5 6 7 8 (f) It is rather difficult to predictDepreciation $26,235 $18,735 $13,395 $13,380 $13,395 $6,690 $0 $0 what technological advances wouldBook value $91,830 $65,595 $46,860 $33,465 $20,085 $6,690 $0 $0 $0 be made on a typical equipment in theCurrent market value $0 future. If the industrial engineer hadCash Flow Statement all the information available in one or two years, he+(.4)*(Depreciation) $10,494 $7,494 $5,358 $5,352 $5,358 $2,676 $0 $0 could defer the replacement Investment ($36,732) decision. Since Model A was

already placed in service, theNet Cash Flow ($36,732) $10,494 $7,494 $5,358 $5,352 $5,358 $2,676 $0 $0 amount of $ 150,000 expended is a

sunk cost, and it should not be PW (10%) = ($8,480) AEC(10%) = $1,590 considered in future replacement decisions.

Challenger (Model B)n 0 1 2 3 4 5 6 7 8 9 10

Depreciation $42,870 $73,470 $52,470 $37,470 $26,790 $26,760 $26,790 $13,380

Book value $300,000 $257,130 $183,660 $131,190 $93,720 $66,930 $40,170 $13,380 $0

Savings $75,000 $75,000 $75,000 $75,000 $75,000 $75,000 $75,000 $75,000 $75,000 $75,000Cash Flow Statement+(.4)*(Depreciation) $17,148 $29,388 $20,988 $14,988 $10,716 $10,704 $10,716 $5,352 $0 0

+(0.60)*(Savings) $45,000 $45,000 $45,000 $45,000 $45,000 $45,000 $45,000 $45,000 $45,000 45,000

Investment ($300,000)

Net Cash Flow ($300,000) $62,148 $74,388 $65,988 $59,988 $55,716 $55,704 $55,716 $50,352 $45,000 $45,000

PW (10%) = $63,079 AE (10%) = $10,266

37



11.40) Option 1: Keep the defender

Option 1 : Keep the defender

Financial Data n 0 1 2 3 4 5 6 7 8Depreciation $8,930 $8,920 $4,460

Book value $13,387 $4,460Market value $40,000Setup cost $16,000 $16,000 $16,000 $16,000 $16,000 $16,000 $16,000 $16,000Operating cost $15,986 $16,785 $17,663 $18,630 $19,692 $20,861 $22,147 $23,562

Cash Flow Statement

+(.4)*(Depreciation) $3,568 $1,784Opportunity cost ($31,141)-(1-0.40)*(Setup) ($9,600) ($9,600) ($9,600) ($9,600) ($9,600) ($9,600) ($9,600) ($9,600)-(1-0.40)*(Operating cost) ($9,592) ($10,071) ($10,598) ($11,178) ($11,815) ($12,517) ($13,288) ($14,137)

Net Cash Flow ($31,141) ($15,624) ($17,887) ($20,198) ($20,778) ($21,415) ($22,117) ($22,888) ($23,737)

PW (12%) = ($130,228)AEC(12%) = $26,215

Note: Opportunity cost (Investment required to keep the defender) $40,000 – 0.4($40,000 – ($13,387+0.5($8,930)) = $31,141

38



Option 2: Purchase a used machine

Option 2 : Purchase a used machine

Financial Data n 0 1 2 3 4 5 6 7 8Depreciation $21,178 $36,294 $25,920 $18,510 $13,234 $13,219 $13,234 $6,610Book value $148,200 $127,022 $90,728 $64,808 $46,298 $33,063 $19,844 $6,610 $0Salvage value $0 $0Setup cost $15,000 $15,000 $15,000 $15,000 $15,000 $15,000 $15,000 $15,000Operating cost $11,500 $11,950 $12,445 $12,990 $13,590 $14,245 $14,950 $15,745Savings 36,000 36,000 36,000 36,000 36,000 36,000 36,000 36,000

Cash Flow Statement

+(1-0.40)*(Savings) $21,600.00 $21,600.00 $21,600.00 $21,600.00 $21,600.00 $21,600.00 $21,600.00 $21,600.00+(.4)*(Depreciation) $8,471.11 $14,517.67 $10,368.07 $7,404.07 $5,293.70 $5,287.78 $5,293.70 $2,643.89Investment ($148,200.00)-(1-0.40)*(Setup) ($9,000.00) ($9,000.00) ($9,000.00) ($9,000.00) ($9,000.00) ($9,000.00) ($9,000.00) ($9,000.00)-(1-0.40)*(Operating cost) ($6,900.00) ($7,170.00) ($7,467.00) ($7,794.00) ($8,154.00) ($8,547.00) ($8,970.00) ($9,447.00)

Net Cash Flow ($148,200) $14,171 $19,948 $15,501 $12,210 $9,740 $9,341 $8,924 $5,797

PW (12%) = ($84,215)AEC(12%) = $16,953

39



Option 3: Keep the defender one year and switch to a brand new machine

Option 3 : Keep the old machine for one more year (Assume that the market value will be $ 30,000.)Financial Data n 0 1 2 3 4 5 6 7Depreciation $8,930 $4,460Book value $13,138 $8,927Market value $40,000 $30,000Setup cost $16,000Operating cost $16,785Cash Flow Statement

+(.4)*(Depreciation) $1,784Opportunity cost ($31,041)-(1-0.40)*(Setup) ($9,600)-(1-0.40)*(Operating cost) ($10,071)Net proceeds from sale $21,570

Net Cash Flow ($31,041) $3,683

Option 3 : Purchase a new machine after 1 year

Financial Data n 1 2 3 4 5 6 7 8 9Depreciation $28,644 $49,090 $35,059 $25,036 $17,900 $17,880 $17,900 $8,940Book value $200,450 $171,806 $122,715 $87,657 $62,621 $44,720 $26,840 $8,940 $0Market value $200,450 $0 $0Setup cost $15,000 $15,000 $15,000 $15,000 $15,000 $15,000 $15,000 $15,000Operating cost $10,350 $10,755 $11,200 $11,691 $12,231 $12,821 $13,455 $14,171Savings $36,000 $36,000 $36,000 $36,000 $36,000 $36,000 $36,000 $36,000Cash Flow Statement

+(1-0.40)*(Savings) $21,600.00 $21,600.00 $21,600.00 $21,600.00 $21,600.00 $21,600.00 $21,600.00 $21,600.00+(.4)*(Depreciation) $11,457.72 $19,636.08 $14,023.48 $10,014.48 $7,160.07 $7,152.06 $7,160.07 $3,576.03Investment ($200,450.00)-(1-0.40)*(Setup) ($9,000.00) ($9,000.00) ($9,000.00) ($9,000.00) ($9,000.00) ($9,000.00) ($9,000.00) ($9,000.00)-(1-0.40)*(Operating cost) ($6,210.00) ($6,453.00) ($6,720.00) ($7,014.60) ($7,338.60) ($7,692.60) ($8,073.00) ($8,502.60)

Net Cash Flow ($200,450) $17,848 $25,783 $19,903 $15,600 $12,421 $12,059 $11,687 $7,673Combined cash flow ($31,041) ($196,767) $17,848 $25,783 $19,903 $15,600 $12,421 $12,059 $11,687 $7,673

PW (12%) = ($133,410) AEC(12%) = $25,038

40



Conclusion: Option 2 is the least cost option. (b) Answer is not provided. (Note: Analysis similar to Part(a) except that all cash flows will be truncated over 5 years.)

41



Chapter 12 Benefit-Cost Analysis 12.1)

$117,400( / ,6%,5)$494,535.76$5,000 $48,830( / ,6%,5)$210,691.49

$494,535.76BC(6%)$210,691.492.35 1

B P A

C P

=== +=

=

= >

A

This project is justifiable based on the benefit-cost analysis. 12.2)

$250,000( / ,6%, 25) $50,000( / ,6%, 25)$3,207,500$1,200,000 $100,000( / ,6%,25)$2,478,340

$3,207,500BC(6%)$2,478,3401.29 1

B P A P F

C P A

= +== +=

=

= >

12.3) (a) analysis: BC( )i

• Design A:

$400,000' $50,000( / ,8%,15) $427,974

$85,000( / ,8%,15) $727,557.5

BC(8%)'$727,557.5

$400,000 $427,9740.88 1

IC P AB P A

BI C

== == =

=+

=+

= <



• Design B:

$300,000' $80,000( / ,8%,15) $684,758

$85,000( / ,8%,15) $727,557.5

BC(8%)'$727,557.5

$300,000 $684,7580.74 1

IC P AB P A

BI C

== == =

=+

=+

= <

• Incremental analysis: Fee collections in the amount of $85,000 will be the same for both alternatives. Therefore, we will not be able to compute the ratio. If this happens, we may select the best alternative based on either the least cost

BC( )i

( ')I C+ criterion or the incremental criterion. B'C( )i Using the incremental B' criterion, C( )i

'B'C(8%)

0 ($427,974 $684,758)$100,000

2.56 1

A BB C

I−

∆ − ∆∆ =

∆− −

=

= >

Select Design A.

(b) Incremental analysis (A-C):

'B'C(8%)

0 ($427,974 $556,366)$50,000

2.57 1

A CB C

I−

∆ − ∆∆ =

∆− −

=

= >

Select Design A. 12.4)

• Building X:

2



$1,960,000( / ,10%, 20)$16,686,585$8,000,000 $240,000( / ,10%, 20)$4,800,000( / ,10%, 20)$9,329,766$16,686,585BC(10%)$9,329,766

1.79 1

X

X

X

B P A

C PP F

=== +−=

=

= >

A

• Building Y: $1,320,000( / ,10%,20)$11, 237,904$12,000,000 $180,000( / ,10%, 20)$7, 200,000( / ,10%,20)$12, 462, 207$11, 237,904BC(10%)$12, 462, 2070.90 1

Y

Y

Y

B P A

C PP F

=== +−=

=

= <

A

Since Building Y is not desirable at the outset, we don’t need an incremental analysis. Building X becomes the better choice. 12.5) Incremental ( )BC i analysis:

Proposals Incremental Present worth A1 A2 A3 A3-A1 A2-A1

I $100 $300 $200 $100 $200 B $400 $700 $500 $100 $300 C’ $100 $200 $150 $50 $100

B'C( )i 3 1.7 1.75 0.50 1

Select either A1 or A2.

3



12.6) Incremental analysis: BC

Equivalent Present Worth Design Incremental

Design A Design B Design C C-B A-B B $7,824 $7,070 $5,656 -$1,414 $754I $2,440 $880 $1,600 $720 $1,560

C’ $3,865 $3,394 $2,922 -$472 $471BC(10%) 1.24 1.65 1.25 -5.7 0.37

Select Design B.

12.7)

(a) The benefit-cost ratio for each alternative:

• Alternative A: ($1,000,000 $250,000 $350,000$100,000)( / ,10%,50)$16,855,185$8,000,000 $200,000( / ,10%,50)$9,982,963

BC(10%) 1.69 1A

BP A

C P

= + ++== +== >

A

A

• Alternative B:

($1, 200,000 $350,000 $450,000$200,000)( / ,10%,50)$21,812,592$10,000,000 $250,000( / ,10%,50)$12,478,704

BC(10%) 1.75 1B

BP A

C P

= + ++== +== >

4



• Alternative C: ($1,800,000 $500,000 $600,000$350,000)( / ,10%,50)$32, 223,147$15,000,000 $350,000( / ,10%,50)$18, 470,185

BC(10%) 1.74 1C

BP A

C P

= + ++== +== >

A

(b) Select the best alternative based on : BC( )i

$21,812,592 $16,855,185(10%)$12,478,704 $9,982,963

1.99 (Select B.)

B ABC −−

=−

=

$32,223,147 $21,812,592(10%)$18,470,185 $12,478,7041.74 (Select C.)

C BBC −−

=−

=

Select C.

Comments: You could select the best alternative based on : B'C( )i

A B C I $8,000,000 $10,000,000 $15,000,000

C’ $1,982,963 $2,478,704 $3,470,185 B’C(10%) 1.86 1.93 1.92

($21,812,592 $16,855,185) (2,478,704 1,982,963)' (10%)

$10,000,000 $8,000,0002.23 (Select B.)

B AB C −

− − −=

−=

($32,223,147 $21,812,592) ($3,470,185 $2,478,704)' (10%)

$15,000,000 $10,000,0001.88 (Select C.)

C BB C −

− − −=

−=

5



12.8) • Option 1-The “long” route:

Users’ annual cost 22= miles $0.25× per mile 400,000× cars $2, 200,000=

Sponsor’s annual cost $21,000,000( / ,10%, 40)A P= $140,000+ $2, 287, 448=

• Option 2-Shortcut: Users’ annual cost 10= miles $0.25× per mile 400,000× cars

$1,000,000= Sponsor’s annual cost $45,000,000( / ,10%, 40)A P= $165,000+

$4,766,674=

• Incremental analysis (option 2-option 1): Incremental users’ benefit $2, 200,000 $1,000,000= −

2 1

$1,200,000$1,200,000(10%)

$4,766,674 $2,287,4480.48 1

BC −

=

=−

= <

Assuming no do-nothing alternative, select option 1.

12.9) Multiple alternatives:

Projects

PW of Benefits

PW of Costs

Net PW

B/C ratio

A1 $40 $85 -$45 0.47 A2 $150 $110 $40 1.36 A3 $70 $25 $45 2.80 A4 $120 $73 $47 1.64

Since the BC ratio for project A1 is less than 1, we eliminate it from our comparison. Incremental Analysis: ordering (A3, A4, A2)

• A3 versus A4:

4 3$120 $70BC( )$73 $25

1.04 1

A Ai −−

=−

= >

Select A4.

6



• A2 versus A4:

2 4$150 $120BC( )$110 $73

0.81 1

A Ai −−

=−

= <

Select A4.

12.10) Given garbage amount/day8%,i = 10%,g = 300= tons

(a) The operating cost of the current system in terms of $/ton of solid waste:

• Annual garbage collection required (assuming 365 days): Total amount of garbage 300= tons 365× days

109,500= tons/year • Equivalent annual operating and maintenance cost:

1PW(8%) $905, 400( / ,10%,8%, 20)$20,071,500

AEC(8%) $20,071,500( / ,8%, 20)$2,044,300

P A

A P

= −= −==

• Operating cost per ton:

cost per ton $2,044,300109,500

=

ton $18.67=

(b) The economics of each solid-waste disposal alternative in terms of $/ton of solid waste:

• Site 1:

1

1

AEC(8%) $4,053,000( / ,8%, 20)$342,000( / ,10%,8%, 20)( / ,8%, 20)($13, 200 $87,600)$1,084,773.719

A PP A A P

=+− +=

Cost per ton $1,084,773.719 /109,500=

per ton $9.91=

7



• Site 2:

2

1

AEC(8%) $4,384,000( / ,8%, 20)$480,000( / ,10%,8%, 20)( / ,8%, 20)($14,700 $99,300)$1, 417,042.61

A PP A A P

=+− +=

Cost per ton $1, 417,042.61/109,500=

per ton $12.94=• Site 3:

3

1

AEC(8%) $4,764,000( / ,8%, 20)$414,000( / ,10%,8%, 20)( / ,8%, 20)($15,300 $103,500)$1,301,871.57

A PP A A P

=+− +=

Cost per ton $1,301,871.57 /109,500=

per ton $11.89=• Site 4:

4

1

AEC(8%) $5, 454,000( / ,8%, 20)$408,000( / ,10%,8%, 20)( / ,8%, 20)($17,100 $119, 400)$1,340,928.66

A PP A A P

=+− +=

Cost per ton $1,340,928.66 /109,500=

per ton $12.25=Site 1 is the most economical choice.

(c) Incremental analysis: BC

Present System

Site1 Site2 Site3 Site4

B 0 $989.67 $1,119.26 $1,166.39 $1,340.17

I 0 $4,053.0 $4,384.0 $4,764.0 $5,454.0

C' $20,071.48 $7,581.68 $10,640.95 $9,177.82 $9,044.81

Reduction in C’ over the present

system

$12,489.80 $9,430.53 $10,893.66 $11,026.67

8



• Site 1 vs. Site 2:

2 1($1,119.26 $9,430.53) ($989.67 $12,489.80)BC(8%)

$4,384 $4,053$2,929.68

3318.85 1

Select Site 1.

−

+ − +∆ =

−−

=

= − <

• Site 1 vs. Site 3:

3 1($1,166.39 $10,893.66) ($989.67 $12,489.80)BC(8%)

$4,764 $4,053$1,419.42

7111.996 1

Select Site 1.

−

+ − +∆ =

−−

=

= − <

• Site 1 vs. Site 4

4 1($1,340.17 $11,026.67) ($989.67 $12, 489.80)BC(8%)

$5, 454 $4,053$1,112.63$1, 401

0.7942 1Select Site 1.

−

+ − +∆ =

−−

=

= − <

The ultimate choice: Site 1. 12.11) Suggestions: Ask students to visit the Atlanta Airport website

(http://www.atlanta-airport.com) to obtain the current and projected airport operational statistics such as number of aircraft landings/takeoffs and passengers. If we just focus on some of the primary benefits and costs, we may identify the following elements:

9



http://www.atlanta-airport.com/

• Sponsor’s costs: (1) Required capital investments in airport expansion, (2) Additional O&M costs associated with the expanded airport operation.

• Sponsor’s Revenue: (1) Incremental landing/takeoff fees due to additional traffic volumes, and (2) Increased parking and concession revenues due to additional passenger traffics.

• Users’ Benefits: (1) Savings due to reduced waiting costs (value of travel time) (2) Savings on fuel costs for airliners due to reduced taxiing, landing and departure times, (3) Reduced air and noise pollution

• Users’ Disbenefits: (1) Relocation of residents and commercial buildings due to airport expansion

Once these values are quantified, we compute the following for each option:

• Step 1: Users’ net benefits=Users’ benefits-Users’ disbenefits. • Step 2: Sponsor’s net costs=Sponsor’s costs-Sponsor’s revenue.

Then, identify the option(s) with Users’ net benefits exceeding the sponsor’s net cost. Select the option with the largest differential net benefits. If the initial analysis based on the primary benefits and costs does not lead to any clear-cut choice, the analysis may be broadened to include the secondary benefits such as the regional economic impact studies.

10



Chapter 13 Understanding Financial Statements 13.1) (2) Income statement; (1) balance sheet; (3) cash flow statement; (4) operating

activities; (5) investing activities, and (6) financing activities; (7) capital account (paid-in capital)

13.2) (7), (8), (1), (11), (3), (9) 13.3) (a)

• Current assets = $150,000 + $200,000 + $150,000 + $50,000 + $30,000 = $580,000 • Current liabilities = $50,000 + $100,000 + $80,000 = $230,000 • Working capital = $580,000 - $230,000 = $350,000 • Shareholder’s equity = $100,000 + $150,000 + $150,000 + $70,000 = $470,000

(b) EPS = $500,000/10,000 = $50 per share (c) Par value = $15; capital surplus = $150,000/10,000 = $15;

Market price = $15 + $15 = $30 per share 13.4) (a) Working capital = Current assets – Current liabilities; Working capital

requirements = Changes in current assets – Changes in current liabilities; WC req. = (+$100,000 - $20,000) – (+$30,000 - $40,000) = $90,000, indicating that additional financing is needed to fund the increase in current assets.

(b) Taxable income = $1,500,000 - $650,000 - $150,000 - $20,000 = $680,000 (c) Net income = $680,000 - $272,000 = $408,000

(d) Net cash flow:

• Operating activities = net income + depreciation – WC = $408,000 + $200,000 - $90,000 = $518,000

• Investing activities = equipment purchase = ($400,000) • Financing activities = borrowed fund = $200,000 • Net cash flow = $518,000 - $400,000 + $200,000 = $318,000

13.5) (b) 13.6) (b) 13.7) (d)



13.8) (b)

ROE= Profit margin × Asset turnover × Financial leverage

(a) 0.1668 (b) 0.1900 (c) 0.1447 (d) 0.1152

13.9) Income Statement:

A B C D E F

$900,000 $585,000 $315,000 $270,000 $108,000 $162,000

Balance Sheet:

0 1 2 3 4 5

$160,000 $120,000 $320,000 $600,000 $900,000 $1,500,000

6 7 8 9

$450,000 $700,000 $100,000 $700,000 $800,000

• From Current ratio Total current assets = 2.4 × $250,000 = $600,000 ----------------------------------- ③ Plant and equipment, net = $1,500,000-$600,000=$900,000----------------------- 4 • From Quick ratio Inventory = $600,000 - (1.12 × $250,000) = $320,000 -----------------------------② • From Inventory Turnover Net Revenue = (($320,000 +$280,000)/2) × 6.0 =$1,800,000 Cost of goods sold = $1,800,000- $900,000= $900,000 ------- A • From DSO Accounts receivable = 24.3333 × ($1,800,000 ÷365) = $120,000 ------------------ ① Cash = ③-(②+①) = $160,000 -----------------------------------------------------------ⓞ • From interest expense of income statement

2



Bond = $450,000 ----------------------------------- ⑥ 250,000 + ⑥ = $700,000 --------------------------⑦

• From Debt to Equity ratio Total Equity ⑩ = $700,000 ÷ 0.875 =$ 800,000 ------------- ⑩ Total assets or Total liabilities and equity = ⑦ + ⑩ = $1,500,000 ------⑤

• From Return on total assets Net income F = 14%× ($1,350,000) - ($45,000)(0.6)=$162,000 • From F, D =F ÷ 0.6 = $270,000,

E = D× (0.4) =$108,000 C = D+45,000 = $315,000 B=$900,000-C = $585,000

• From EPS Stock Outstanding = F ÷ 4.05 = 40,000 shares

Common stock = $2.50 × 40,000 = $100,000 -------------------------------⑧ Retained Earnings = ⑩ - ⑧ = $700,000 ------------------------------------9

13.10)

Given Olson’s EPS = $8 per share; Cash dividend = $4 per share; Book value per share = $80; Changes in the retained earnings = $24 million; Total debt = $240 million; Find debt ratio = total debt/total assets

• Net IncomeEPS $8X

= =

Where X = the number of outstanding shares

• Total shareholders' equityBook value $80X

= =

• Retained earnings = Net income – Cash dividend; Net income = 8X from EPS relationship and the total cash dividend = 4X, so we rewrite 8X – 4X = $24 million, or X = 6 million shares

• From the book value per share, we know that total shareholders’ equity

= 80X, or $480 million; Total assets = Total liabilities + Total shareholders’ equity = $240 million + $480 million = $720 million

• Debt ratio = $240 million/$720 million = 0.33

3



13.11)

(a) Debt ratio = Total debt $2,047,599 123.46%Total assets $1,658,528

= =

(b) Times-interest-earned ratio= EBIT -$1,038,770 1,622.27%Interest expense $64,032

= = −

(c) Current ratio = Current assets $617,266 70.54%Current liabilities $875,065

= =

(d) Quick ratio = Current assets - Inventory $617,266-34,502 66.60%Current liabilities $875,065

= =

(e) Inventory turnover = Net sales $637,235 26.14 timesavg. Inventory ($34,502+$14,256)/2

= =

(f) DSO = A/R $71,014 40.68avg. sales/day $637,235/365

= = days

(g) Total assets turnover ratio = Net sales $637,235 0.3842Total assets $1,658,528

= = times

(h) Profit margin on sales = Net Income -$1,104,867 173.38Net sales $637,235

= = − %

(i) Return on total assets =

Net Income+Interst expense(1-tax rate) -$1,104,867+64,032(0.6) 56.97%avg. total assets ($1,658,528+2,085,362)/2

= = −

with an assumption of 40% tax rate.

(j) Return on common equity = Net Income -$1,104,867 34.47%

avg. shareholder's equity (-$389,701+$324,968)/2= =

(k) Price-earning ratio = Price per share $2.83 3.59Earnings per share $0.7877

= = −−

Net income $1,104,867Earnings per share=Avg. number of shares outstanding 1, 402,619

$0.7877

−=

= −

(l) Book value per share = total stockholders' equity-preferred stock -$389,071 0.277

Shares outstanding 1,402,619= =

4



13.12)

(a) Debt ratio = Total debt $1,118,455 79.39%Total assets $1,408,785

= =

(b) Times-interest-earned ratio= EBIT $290,235 946.69%Interest expense $30,658

= =

(c) Current ratio = Current assets $515,173 179.88%Current liabilities $286,394

= =

(d) Quick ratio = Current assets - Inventory $515,173-$311,464 71.13%Current liabilities $286,394

= =

(e) Inventory turnover = Net revenue $3,198,084 16.37timesavg. Inventory ($311,464+$79,181)/2

= =

(f) DSO = A/R $126,634 14.45avg. sales/day $3,198,084/365

= = days

(g) Total assets turnover ratio = Net revenue $3,198,084 2.27Total assets $1,408,785

= = times

(h) Profit margin on sales = Net Income $157,368 4.92Net sale $3,198,084

= = %

(i) Return on total assets = Net Income+Interst expense(1-tax rate) $157,368+30,658(0.6) 15.94%

avg. total assets ($1,408,785+758,780)/2= =

We assume a tax rate of 40%.

(j) Return on common equity = Net Income $157,368 55.23%

avg. share holder's equity ($290,330+$279,493)/2= =

(k) Price-earning ratio = Price per share $65 19.30Earnings per share $3.37

= =

$157,368Earnings per share=46,738

$3.37=

(l) Book value per share = Total stockholders' equity - Preferred stock $290,330 6.21

Shares outstanding 46,738= =

13.13) Not provided

5



13.14) Not provided.

6



fundamentals of engineering economics instructor’s manual

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