Chapter 05

Page | 1

Chapter 5 Present-Worth Analysis Identifying Cash Inflows and Outflows

5.1 Revenue = 45× 2,000 × 5 = 450,000 (No. of engineer:5)

n Inflow Outflow Net flow 0 $300,000 -$300,000 1 $450,000 175,000 275,000 2 $450,000 175,000 275,000 … … … … 8 $450,000 175,000 275,000

5.2 (a) Cash inflows: (1) savings in labor, $45,000 per year, (2) salvage value,

$3,000 at year 5. (b) Cash outflows: (1) capital expenditure = $30,000 at year 0, (2) operating

costs = $5,000 per year. (c) Estimating project cash flows:

n Inflow Outflow Net flow 0 $30,000 -$30,000 1 $45,000 5,000 40,000 2 45,000 5,000 40,000 3 45,000 5,000 40,000 4 45,000 5,000 40,000 5 45,000+3,000 5,000 43,000

Payback Period

5.3 (a) Conventional payback period: $100,000/$25,000 = 4 years

(b) Discounted payback period at i = 15%:

n Net Cash Flow Cost of Funds (15%) Cumulative Cash

Flow 0 -$100,000 -$100,0001 $25,000 -$100,000(0.15) =-$15,000 -$90,0002 $25,000 -$90,000(0.15) =-$13,500 -$78,5003 $25,000 -$11,775 -$65,2754 $25,000 -$9,791 -$50,0665 $25,000 -$7,510 -$32,5766 $25,000 -$4,886 -$12,4637 $25,000 -$1,869 $10,668

Contemporary Engineering Economics, Fifth Edition, by Chan S. Park. ISBN: 0-13-611848-8 © 2011 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. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Page | 2

Payback period = 6.54 years

5.4 (a) Conventional payback period: 0.75 years

(b) Discounted payback period = 0.85 years, assuming continuous payments:

n Net Cash Flow Cost of Funds (15%) Cumulative Cash Flow

0 -$30,000 -$30,0001 +$40,000 -$30,000(0.15) = -$4,500 +$5,500

5.5

(a) Conventional payback period

Project A B C D Payback period 4.38 years 3 years 3.38 years 0.8 years

(b) It maybe viewed as a combination of two separate projects, where the first

investment is recovered at the end of year 1 and the investment that made in year 3 and 4 will be recovered at the end of year 7.

(c) Discount payback period

Project A B C D

Payback period 6.55 years 5.43 years 4.05 years 0.88 years

5.6 (a) Conventional payback period: 4.38 years (b) Discounted payback period at i = 15%: No payback

NPW Criterion

5.7 (12%) $100,000 $30,000( / ,12%,5) $8,143.3PW P A= − + =

5.8 (12%) $250,000 $20,000 $90,000( / ,12%,5) $75,000( / ,12%,5)PW P A P F= − − + +

= $96,986.87 Since PW(12%) > 0, this purchase should be justified.

5.9

(a) (10%) $3,500 $4, 200( / ,10%,4) $631.34APW P F= − + = −


Page | 3

PW (10%)B = −$6,500 + $1,600(P / F ,10%,1) + $1,800(P / F ,10%,2)+$1,500(P / F ,10%,3)+ $2,200(P / F ,10%,4) = −$928.25

PW (10%)C = −$2,800 − $1,800(P / F ,10%,1) − $900(P / F ,10%,2)+$2,500(P / F ,10%,3)+$3,500(P / F ,10%,4) = −$911.33

PW (10%)D = −$4,300 − $1,000(P / F ,10%,1) + $1,900(P / F ,10%,2)+$2,300(P / F ,10%,3)+$1,500(P / F ,10%,4)= −$886.30

(b) Draw the graphs using Excel or other software.

5.10 (a)

PW(15%) $250,000 $40,000( / ,15%,35) $50,000( / ,15%,35)$15,040 0

P A P F= − + += <

(b) PW(20%) $250,000 $40,000( / , 20%,35) $50,000( / , 20%,35)

$50,254P A P F= − + +

= −

(c) PW( ) $250,000 $40,000( / , ,35) $50,000( / , ,35) 0

15.93%i P A i P F ii= − + + ==

∴ Correct answer is (d) as all of the statements are correct.

5.11 Given: Estimated remaining service life = 25 years, current rental income = $250,000 per year, O&M costs = $85,000 for the first year increasing by $5,000 thereafter, salvage value= $50,000,and MARR 12%= . Let A0 be the maximum investment required to break even.

0(12%) [$250,000( / ,12%,25) $25,000( / ,12%,20)

$27,500( / ,12%,15) $30,250( / ,12%,10)$33, 275( / ,12%,5) $50,000]( / ,12%,25)$85,000( / ,12%, 25) $5,000( / ,12%, 25)0

PW A F A F AF A F AF A P FP A P G

= − + ++ ++ +− −=

∴ Solving for 0A = yields

0 $1,241, 461A =


Page | 4

5.12

n = 4 : P4 = $32,500 + $33,000(P / F ,12%,1) = $61,964.29n = 3: P3 = $32,500 + $61,964.29(P / F ,15%,1) = $86,381.99n = 2 : P2 = $33,400 + $86,381.99(P / F ,13%,1) = $109,844.24n = 1: P1 = $32,400 + $109,844.2(P / F ,11%,1) = $131,358.77n = 0 : P0 = −$42,000 + $131,358.8(P / F ,10%,1) = $77,417.07

The investment has a net present worth of $77,417.07 and would be acceptable at any price less than equal to that amount.

5.13 PW (15%) = −$500,000 − 2,200,00(P / F ,15%,1) + $2,700,000(P / A,15%,8)

+ $1,500,000(P / F ,15%,8) = $10,193,077

5.14 (18%) $3,500,000 [$1,550,000 $350,000 $150,000]( / ,18%,10)

$200,000( / ,18%,10)$1, 257,004

PW P AP F

= − + − −+

= Since PW is positive, the project is justified.

5.15 0(15%) $60,000( / ,15%,10)PW A P A= − + =0

0 $301,126A = Future Worth and Project Balance

5.16 US: $1,000(F/P, 4%, 5) = $1,216.65 Euro: [$1,000(0.735)(F/P, 5%, 5)]/0.88 = $1,065.99 Japan: [$1,000(97.5)(F/P, 6%, 5)]/110 = $1,186.15

∴ Therefore, A. Investing in the United States

5.17 (a)

(15%) $10,500 $4,300( / ,15%,1) $4,800( / ,15%,2)$5,500( / ,15%,3) $485

APW P F P FP F

= − + ++ =

(15%) $16,000 $3,000( / ,15%,1) $21,000( / ,15%, 2)$13,000( / ,15%,3) $5,818

BPW P F P FP F

= − − ++ =

(15%) $15,000 $7,000( / ,15%,1) $4,000( / ,15%, 2)$6,000( / ,15%,3) $9,834

CPW P F P FP F

= − −

+ =

(15%) $20,000 $7,500( / ,15%,1) $7,500( / ,15%,2)$8,500( / ,15%,3) $2,218

dPW P F P FP F

= − + ++ = −


Page | 5

(b) (15%) $485( / ,15%,3) $738AFW F P= = (15%) $5,818( / ,15%,3) $8,849BFW F P= = (15%) $9,834( / ,15%,3) $14,956CFW F P= = (15%) $2,218( / ,15%,3) $3,374DFW F P= − = −

∴ Except for Project D, all the projects are acceptable.

5.18 (a) ia = 12.68% : $5,000(F / P,12.68%,20) = $54,438.58 (b) i = 1% : $80(P / A,1%,120)(F / P,1%,240) = $60,737.33 (c) i = 1% : $50(F / A,1%,240) = $49,462.77 (d) ia = 12.68% : $15,000(F / P,12.68%,10) = $49,494.81

∴ Option B is the best choice.

5.19 (a) The original cash flows of the project are as follows.

n nA Project Balance 0 -$1,000 -$1,000 1 ($100) -$1,100 2 ($520) -$800 3 $460 -$500 4 ($600) 0

(b) 500$460$)1(800$)( 3 −=++−= iiPB

∴ %20=i

(c) Yes, the project is acceptable. The project would be indifferent at i = 20%, and therefore is attractive at i = 15%.

5.20 (15%) $1, 200,000( / ,15%,1) $900,000( / ,15%, 2)

$675,000( / ,15%,3) $506,250( / ,15%,4)$2,457, 281

APW P F P FP F P F

= ++ +

=

5.21 (a) First find the interest rate that is used in calculating the project balances.

We can set up the following project balance equations:


Page | 6

1 1

2 2

3

4 4

5 5

( ) $10,000(1 ) $11,000( ) $11,000(1 ) $8, 200( ) $8,200(1 ) $8,000 $1,840( ) $1,840(1 ) $3,792( ) $3,792(1 ) $7,550

PB i i APB i i APB i iPB i i APB i i A

= − + + = −= − + + = −= − + + = −= − + + == + + =

From 3( )PB i , we can solve for :i 20%i =

Substituting i into other ( )nPB i yields

n nA ( )nPB i 0 -$10,000 -$10,000 1 1,000 -11,000 2 5,000 -8,200 3 8,000 -1,840 4 6,000 3,792 5 3,000 7,550

(b) Conventional payback period = 3years ( or 2.5 years with continuous cash

flows)

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

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

1(24%) $1,000 $400( / , 24%,1) $800( / , 24%, 2)( / , 24%,3) 0

PW P F P FX P F

= − + ++ =

2(23%) $1,000 $300( / , 23%,1) ( / , 23%,2)$800( / , 23%,3) 0

PW P F Y P FP F

= − + ++ =

∴ X = $299.58, Y = $493.49

(b) Since 0%)24( 1 =PW , 0%)24( 1 =FW . (c) ⓐ = $593.49, ⓑ = $499.58, ⓒ = 17.91%

5.23

(a) The original cash flows of the project are as follows

n nA Project Balance 0 -$3,000 -$3,000 1 ($600) -$2,700 2 $1,470 -$1,500 3 ($1,650) 0 4 (-$300) -$300 5 $600 ($270)


Page | 7

(b)

• 500,1$470,1$)1(700,2$)( 2 −=++−= iiPB

∴ %10=i

• 65.167$)5%,10,/(270$%)10( == FPPW

5.24 (a)

-8000-6000-4000-2000

02000400060008000

10000120001400016000

1 2 3 4 5 6 7 8 9

PBABCD

(b) Project A B C D

2PB -$2,555 -$2,350 -$4,875 $3,660

Project D because it has only a positive balance at the end of year 2. Graphically you can easily verify the same results.

5.25 (a)

(12%) $2, 200( / ,12%,5) $500( / ,12%, 4)$700

$1,366.12

AFW F P F P= − −+ −

=

(12%) $4,500( / ,12%,5) $1,500( / ,12%, 4)$3,000

$4,802.03

BFW F P F P= − ++ +

=

(12%) $3, 200( / ,12%,5) $1, 200( / ,12%, 4)$2,000

$11,106.33

CFW F P F P= − +

+ +=


Page | 8

(12%) $5, 400( / ,12%,5) $1,500( / ,12%,4)$2, 400

$3,771.76

DFW F P F P= − ++ +

=

(12%) $7, 200( / ,12%,5) $2,500( / ,12%, 4)$2,800( / ,12%,3) $3,500( / ,12%, 2)$430.86

EFW F P F PP F F P

= − ++ +

= −

(b) Accept all projects except E.

5.26

(a) You may plot the future worth for each project as a function of interest rate, using Excel software.

(b)

n A B C D E 0 -$2,200.00 -$4,500.00 -$3,200.00 -$5,400.00 -$7,200.001 -$2,964.00 -$3,540.00 -$2,384.00 -$4,548.00 -$5,564.002 -$2,419.68 -$7,964.80 -$2,670.08 -$4,093.76 -$3,431.683 -$1,210.04 -$3,920.58 $1,009.51 -$1,585.01 -$343.484 $1,844.75 $1,608.95 $8,130.65 $1,224.79 -$384.705 $1,366.12 $4,802.03 $11,106.33 $3,771.76 -$430.86

(c) Note that PB(12%)5 = FW (12%).

5.27 From the annual cash flow and balance table, shown below, it will take 6.54 years approximately.

n Cash flow Balance 0 -$20,000.00 -$20,000.00 1 $5,000.00 -$18,000.00 2 $5,000.00 -$15,700.00 3 $5,000.00 -$13,055.00 4 $5,000.00 -$10,013.25 5 $5,000.00 -$6,515.24 6 $5,000.00 -$2,492.52 7 $5,000.00 $2,133.60 8 $5,000.00 $7,453.64 9 $5,000.00 $13,571.68 10 $5,000.00 $20,607.44


Page | 9

5.28 The present worth of the expenses already spent on the project:

PW (20%)expense = −$10M(F / A,20%,5) = −$74.416M

The present worth of the expected revenue:

revenue(20%) $100M( / , 20%,10) $419.25MPW P A= = ∴ Price of the project = -$74.416M + $419.25M - $30M = $314.83M

5.29 (a)

(12%) $61.89APW = (12%) $19.43BPW = − (12%) $16.80CPW =

(b)

FW (12%)A = $61.89(F / P,12%,6) = $122.17

FW (12%)B = −$19.43(F / P,12%,5) = −$34.24

FW (12%)C = $16.80(F / P,12%,3) = $23.61

(c)

FW (i)A = $150.36

FW (i)B = −$9.78

FW (i)C = $17.95

5.30 (a) Since a project’s terminal project balance is equivalent to its future worth,

we can easily find the equivalent present worth for each project by

PW (10%)A = $105(P / F ,10%,5)= $65.20

PW (20%)C = −$1,000(P / F ,20%,5)= −$401.88


Page | 10

(b) 0 0

1 0 1

2 1 2

3 2 3

4 3 4

5 4 5

(10%) $1,000(10%) (10%) (1 0.10) $1,000(10%) (10%) (1 0.10) $900(10%) (10%) (1 0.10) $690(10%) (10%) (1 0.10) $359(10%) (10%) (1 0.10) $105

PB APB PB APB PB APB PB APB PB APB PB A

= = −= + + = −= + + = −= + + = −= + + = −= + + =

From the project balance equations above, we derive

0A = 1 2 3 4$1,000, $100, $200, $300, $400,A A A A= = = = and 5 $500A = . (c)

5(20%) (20%) $1,000FW PB= = − (Note that the ending project balance is the future worth of the project.)

(d)

5

( ) ( ) ( / , , )

$198 /(1 )$79.57

B BPW i FW i P F i N

i

=

= +=

∴Solving for i yields 20%i =

5.31 (a)

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

A

B

C

PWPW P FPW

== ==

(b) Assume that 2 $500.A =

PB(12%)0 = −$1,000PB(12%)1 = −$1,000(1.12) + A1 = −$530PB(12%)2 = −$530(1.12) + $500 = X

Solving for X yields X = -$93.60.

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

Net Cash Flow n A B C 0 -$1,000 -$1,000 -$1,0001 $300 $500 $5902 $280 $500 $5003 $260 $299 -$1064 $240 $300 $1475 $220 $300 $100


Page | 11

(d)

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

A

B

C

FWFWFW

===

Capitalized Equivalent Worth

5.32 (a)

PW (13%) = $100,000(P / A,13%,5) + $130,000(P / A,13%,5)(P / F,13%,5)

+ ($150,000 / 0.13)(P / F,13%,10) = $939,804.35

(b) $939,804.35(0.13) $122,174.57A = =

5.33

Capitalized equivalent amount:

CE(10%) = $100,000 +$10,000

0.1+

$20,000(A / F,10%,4)

0.1= $243,094.16

5.34

Find the equivalent annual series for the first cycle:

A = [$100(P / A,15%,2) + $40(P / A,15%,2)(P / F,15%,2)

+ $20(P / F,15%,5)](A / P,15%,5) = $66.13

Capitalized equivalent amount:

CE =$66.13

0.15= $440.88

5.35 Given: r =5% compounded monthly, maintenance cost = $80,000 per year

120.05(1 ) 1 5.116%12ai = + − =

∴ $80,000(5.116%) $1,563,663.650.05116

CE = =


Page | 12

5.36 Given: Construction cost = $15,000,000, renovation cost = $3,000,000 every 15 years, annual &O M costs = $1,000,000 and 5%i = per year

(a) P1 = $15,000,000

P2 =$3,000,000( A / F ,5%,15)

0.05= $2,780,537.26

P3 = $1,000,000 / 0.05= $20,000,000

CE(5%) = P1 + P2 + P3

= $37,780,537.26

(b)

P1 = $15,000,000

P2 =$3,000,000( A / F ,5%,20)

0.05= $1,814,555.23

P3 = $1,000,000 / 0.05= $20,000,000

CE(5%) = P1 + P2 + P3

= $36,814,555.23

(c)

• 15-year cycle with 10% of interest: 1

2

3

1 2 3

$15,000,000$3,000,000( / ,10%,15)

0.1$944, 213.31$1,000,000 / 0.1$10,000,000

(10%)$25,944,213.31

PA FP

P

CE P P P

=

=

==== + +=

• 20-year cycle with 10% of interest:


Page | 13

1

2

3

1 2 3

$15,000,000$3,000,000( / ,10%,20)

0.1$523,788.74$1,000,000 / 0.1$10,000,000

(10%)$25,523,788.74

PA FP

P

CE P P P

=

=

==== + +=

As interest rate increases, CE value decreases.

5.37 Given: Cost to design and build = $830,000 , rework cost = $120,000 every 10 years, new type of gear = $80,000 at the end of 5th year, annual operating costs = $70,000 for the first 15 years and $100,000 thereafter

CE(7%) = $830,000 +$120,000( A / F ,7%,10)

0.07+$80,000(P / F ,7%,5)+$70,000(P / A,7%,15)

+$100,000

0.07(P / F ,7%,15)

= $2,166,448.62

Mutually Exclusive Alternatives

5.38

(a) Option A

n Net Cash Flow Cumulative CF 0 -$5,000 -$5,000 1 $2,000 -$3,000 2 $2,000 -$1,000 3 $1,000 $0 4 $2,000 $2,000

Option B

n Net Cash Flow Cumulative CF 0 -$3,500 -$3,500 1 $500 -$3,000 2 $2,000 -$1,000 3 $1,000 $0 4 $2,000 $2,000

∴ Cannot select.


Page | 14

(b)

(10%) $2,000( / ,10%, 4) $1,000( / ,10%,3) $5,000$588.42

(10%) $2,000( / ,10%, 4) $1,000( / ,10%,3)$1,500( / ,10%,1) $3,500$724.78

A

B

PW P A P F

PW P A P FP F

= − −== −− −=

∴ Select project B.

5.39

PW (12%)A = −$1,000 − $1,300(P / F ,12%,1)+ + $660(P / F ,12%,10)= $1,333.31

PW (12%)B = −$2,800 − $660(P / F ,12%,1)+ + $840(P / F ,12%,10)= $1,595.94

∴ Select project B.

5.40 (a)

(15%) $1,800 $1,150( / ,15%,1)$200( / ,15%, 4)

$24.85(15%) $1,500 $1, 200( / ,15%,1)

$100( / ,15%, 4)$590.22

A

B

PW P FP F

PW P FP F

= − + ++= −= − + ++=

∴Select Project B.

(b) (15%) $1,800( / ,15%, 4) $1,150( / ,15%,3)

$200$43.45

(15%) $1,500( / ,15%, 4) $1,200( / ,15%,3)$100$1,032.29

A

B

FW F P F P

FW F P F P

= − + ++= −= − + ++=

∴Select Project B.


Page | 15

(c)

PW (15%)C = −$4,000 + $1,500(P / F ,15%,1)+X (P / F ,15%,2)+$1,800(P / F ,15%,3) + X (P / F ,15%,4)= 1.3279X − $1,512.12

To be acceptable, it must satisfy the following condition:

PW (15%)C > 01.3279X − $1,512.12 > 0

X > $1,138.74

(d)

PW (18%)D = $1,400 − $450(P / A,18%,4)= $189.47 > 0

Yes, project D is acceptable.

(e) If interest rate is higher than 11.76%, project D is better. Otherwise, project E is better.

5.41 (a)

(12%) $17,500 $13,610( / ,12%,1) $14,930( / ,12%, 2)$14,300( / ,15%,3) $16,732.35

APW P F P FP F

= − + ++ =

(12%) $15,900 $13,210( / ,12%,1) $13,720( / ,12%,2)$13,500( / ,12%,3) $16, 441.18

BPW P F P FP F

= − + ++ =

∴ Select project A.

(b) (12%) $16,732.35( / ,12%,3) $23,507.74AFW F P= = (12%) $16,441.18( / ,12%,3) $23,098.67BFW F P= =


5.42 (a)

66.281,5$)2%,15,/(000,14$)1%,15,/(800$000,6$%)15(

=++−= FPFPPW A

46.302,2$)2%,15,/(400$)1%,15,/(500,11$000,8$%)15(

=++−= FPFPPW B



Page | 16

(b) Project A dominates project B at any interest rate between 0% and 50%.

5.43 • Model A:

CE(12%)A = $60,000 +$25,000(A / F,12%,5)

0.12= $92, 791.67

• Model B:

CE(12%)B = $150,000 +$180,000(A / F,12%,50)

0.12= $150,625.5

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

5.44

• Standard Lease Option:

PW (0.5%)SL = −$5,500 − $1,150(P / A,0.5%,24)+$1,000(P / F ,0.5%,24)= −$30,560.10

• Single Up-Front Option:

PW (0.5%)SU = −$31,500 + $1,000(P / F ,0.5%,24)= −$30,612.82

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

5.45 • Machine A:

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

$21,000( / ,13%,6)$101,891

PW P AP F

= − − ++= −

• Machine B:

PW (13%) = −$44,000 − $11,500(P / A,13%,6)= −$89,972

∴ Machine B is a better choice.


Page | 17

5.46 (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)(2,000)(0.09) = $1,579.17 -Motor B: 11.111(0.7457)(2,000)(0.09) = $1,491.39

• Equivalent cost:

PW (8%)A = −$1,200 − $1,579.17(P / A,8%,15)+$50(P / F ,8%,15)= −$14,710.11

PW (8%)B = −$1,600 − $1,491.39(P / A,8%,15)+$100(P / F ,8%,15)= −$14,334

∴ Motor B is preferred.

(b) With 1,000 operating hours: Annual energy cost:

-Motor A: 11.765(0.7457)(1,000)(0.09) $789.58= -Motor B: 11.111(0.7457)(1,000)(0.09) $745.69=

Equivalent cost:

(8%) $1,200 $789.58( / ,8%,15)$50( / ,8%,15)

$7,942.62(8%) $1,600 $745.69( / ,8%,15)

$100( / ,8%,15)$7,951.19

A

B

PW P AP F

PW P AP F

= − −+= −= − −+= −

∴ Motor A is now preferred.

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


Page | 18

• Model A:

00.854,19$)3%,12,/(000,15$

)2%,12,/(000,17$)1%,12,/(500,17$000,20$%)12(

=+

++−=

FP

FPFPPW cycle

69.985,33$)]3%,12,/(1[854,19$%)12(

=+= FPPW total

• Model B:

35.117,12$

)2%,12,/(000,18$)1%,12,/(500,25$000,25$%)12(

=

++−= FPFPPW cycle

02.478,29$)]4%,12,/()2%,12,/(1[35.117,12$%)12(

=++= FPFPPW total

∴ Model A is preferred.

5.48 (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 is 24 years.

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

• Project A1:

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

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

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

cycle

total

PW P A

P F

PW P A

= − −

+= −= − +

= −

Note that the effective interest rate for a 3-year cycle is

3(1.10) 1 33.10%− = • Project A2:

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

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

(10%) $3,167.22[1 ( / ,10%,8)( / ,10%,16)]

$5,334.03

cycle

total

PW P A

P F

PW P FP F

= − −

+= −= − ++= −

∴ Project A2 is preferred.


Page | 19

(c)

48.744,1$%)10( 1 −=APW

SFPSAPPW A

7513.006.546,2$)3%,10,/()3%,10,/(300$800,1$%)10( 2

+−=+−−=

Let 21 %)10(%)10( AA PWPW = , then $1,067S =

5.49 (a) Assuming a common service period of 15 years

• Project B1:

PW (12%)cycle = −$20,000 − $2,000(P / A,12%,5) + $2,000(P / F,12%,5)

= −$26,074.7

PW (12%)total = −$26,074.7[1+ (P / A, 76.23%,2)]

= −$49,266.29

Note: %23.761)12.1( 5 =−

• Project B2:

PW (12%)cycle = −$17,000 − $2,500(P / A,12%, 3) + $3,000(P / F,12%, 3)

= −$20,869.24

PW (12%)total = −$20,869.24[1+ (P / A, 40.49%, 4)]

= −$59,180.43

Note: %49.401)12.1( 3 =−

∴ Select project B1.

(b) • Project B1 with 2 replacement cycles:

PW (12%) = −$26,074.7 − $26,074.7(P / F,12%,5)

= −$40,870.19

• Project B2 with 4 replacement cycles where the 4th replacement ends at

the end of first operating year::


Page | 20

PW (12%) = −$20,869.24[1+ (P / F,12%,3) + (P / F,12%,6)]

− [$17,000 − ($2,500 − $9,000)(P / F,12%,1)](P / F,12%,9)

= −$54,519.76

∴ Project B1 is still a better choice.

5.50

PW (12%)K = −$25,000 −11,000(P / A,12%,10) + 3,000(P / F ,12%,10)= −$86,186.53

PW (12%)T = −$32,000 − 9,700(P / A,12%,10) + S(P / F ,12%,10)= −$86,807.16 + S(0.3220)

PW (12%)K = PW (12%)T

−$86,186.53 = −$86,807.16 + S(0.3220)S = $1,927.42

5.51 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.

(12%) $8,000 $3,500( / ,12%,3)$406.41

(12%) $406.41( / ,12%,3)$169.21

(12%) $15,000 $10,000( / ,12%, 2)$1,900.51

(12%) $1,900.51( / ,12%, 2)$1,124.53

A

A

B

B

PW P A

AE A P

PW P A

AE A P

= − +==== − +===

(a) • Option 1:

PW (12%)AAA =$169.21

0.12= $1,410.08

• Option 2:

PW (12%)BAA = $1,900.51+$169.21

0.12(P / F ,12%,2)

= $3,024.62

∴ Option 2 is a better choice.


Page | 21

(b) Let S be the salvage value of Model A at the end of year 2.

$8,000 $3,500( / ,12%,2) ( / ,12%,2) $1,900.51$2,084.82 (0.7972) $1,900.51

P A S P FS

− + + =− + =

Solving for S yields

$4,999.16S =

5.52 • Since either tower will have zero salvage value after 20 years, we may

select the analysis period of 35 years:

710,154$

)35%,11,/(000,2$000,137$%)11(

−=

−−= APPW ABid

222,157$

)35%,11,/(300,3$000,128$%)11(

−=

−−= APPW BBid

∴ Bid A is a better choice.

• If you assume an infinite analysis period, the present worth of each bid will be:

PW (11%)Bid A =[−$137,000 − $2,000(P / A,11%, 40)](A / P,11%,40)

0.11=− $157, 322

PW (11%)Bid B =[−$128,000 − $3,300(P / A,11%,35)](A / P,11%,35)

0.11=− $161,407

∴ Bid A is still preferred.

5.53 (a)

PW (15%)A1 = −$15,000 + $9,500(P / F ,15%,1)+$12,500(P / F ,15%,2) + $7,500(P / F ,15%,3)= $7,644.04

(b)

PW (15%)A2 = −$25,000 + X (P / A,15%,2)(P / F ,15%,1) = $9,300X = $24,263


Page | 22

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

project balance.

3(15%) $7,644.04( / ,15%,3)$11,625.63

PB F P==

(d) Select A2. It has the greater present worth.

5.54 (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 ( ) ( ) ,NFW i PB i=

(10%) $1,139(10%) $7, 245

A

D

FWFW

==

(c) Assuming a required service period of 8 years

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

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

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

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

B

C

PW P AP F P FP F P F

PW P AP F

= − −− −− −= −= − −−= −

∴ Select Project C.


Page | 23

5.55 • Option 1: Non-deferred plan

PW (12%) = −$300,000 − $49,000(P / A,12%,8)

=− $543,414.35

• Option 2: Deferred plan PW (12%) = −$140,000(P / F,12%,2) − $14,000(P / A,12%,6)(P / F,12%,2)

− $160,000(P / F,12%,5) − $21,000(P / A,12%,3)(P / F,12%,5)

− $180,000(P / F,12%,8)

=− $349,600.72


5.56 • Alternative A: Once-for-all expansion

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

$0.85 ( / ,15%, 25)$32,559,839

APW M M P AM P F

= − −+= −

• Alternative B: Incremental expansion

(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

BPW M M P FM P F M P F

M P AM P A P FM P A P F

= − −− +−−−= −

∴ Select alternative B.

5.57 • 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

(12%) ($120,000 $12,000)($12,000 $1,000)( / ,12%, 20)$1,000( / ,12%,40) $1,000( / ,12%, 40)$141,373

PWP F

P F P A

= − +− −+ −=



Page | 24

Short Case Studies

ST 5.1 • 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 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

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

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

$383, 292

PW P AP FP A P F

= − −+−= −

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

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

$30,000( / ,12%,6)$340,371

PW 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

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

PW P A= −= −

With the restricted assumptions above, option 2 appears to be best alternative. Notes to Instructors: This problem is deceptively simple. However, it can make the problem interesting with the following embellishments. • If the required service period is changed to 5 years, what would be the best

course of action? • If there are price differentials in the subcontracting option (say, $55,000 a year

for a 6-year contract, $60,000 for a 5-year contract, $70,000 a year for a 4-year contract and $75,000 a year for any contract lasting less than 4 years), what would be the best option?

• If both processes A and B would be available in the subsequent years, but the required investment and salvage value would be increasing at the annual rate of 10%, what would be the best course of action?

• If both processes A and B will be available in the subsequent years, but the required investment and salvage value (as well as the O & M costs) would be decreasing at the annual rate of 10%, what would be the best course of action?


Page | 25

ST 5.2 1

2

(8%) $680,000 $680,000( / ,8%,7)$1,076,773.47

(8%) $920,000

PW P F

PW

= − −= −= −


1

2

(8%) $680,000 $680,000( / ,8%,7)$1,076,773.47

(8%) $920,000 $920,000( / ,8%,16)$1,188,539.23

PW P F

PW P F

= − −= −= − −= −


ST 5.3 Note to Instructors: This case problem requires several pieces of

information. (1) No minimum attractive rate return figure is given for Tampa Electrics. (2) What would be a typical number of accidents in line construction work? (3) How does a typical electric utility handle the nesting problems? If there is some cleaning cost, how much and how often?

• First, we may calculate the equivalent present value (cost) for each

option without considering the accident costs and nesting problems.

Design Options Factors Option 1

Cross Arm Option 2

Triangular Option 3

Horizontal Line

Option 4 Stand Off

Investment: Construction cost $495,243 $396,813 $402,016 $398,000Accident cost Annual cost: Flashover repair $6,000 $3,000 $3,000 $3,000Cleaning nest Annual savings: Inventory 0 $4,521 $4,521 $4,521

Assuming that Tampa Electrics’ required rate of return would be 12%. The equivalent present value cost for each option is as follows:


Page | 26

PW (12%)1 = −$495,243− $6,000(P / A,12%,20)= −$540,060

PW (12%)2 = −$396,813− $3,000(P / A,12%,20)+$4,521(P / A,12%,20)= −$385,452

PW (12%)3 = −$402,016 − $3,000(P / A,12%,20)+$4,521(P / A,12%,20)= −$390,655

PW (12%)4 = −$398,000 − $3,000(P / A,12%,20)+$4,521(P / A,12%,20)= −$386,639

It appears that the triangular design configuration appears to be the most economical.

• If we consider the potential accident costs ($65,000 per accident) during

line construction week, it can change the outcome. If we expect only a couple of accidents, option 2 still appears to be the best. However, if you expect more than three accidents, the conventional cross-arm design appears to be economical. If the nest cleaning cost were factored into the analysis, the accident cost would be reduced to the extent of the annual cleaning cost, indicating the preference of the triangular design.

ST 5.4

Not provided


Chapter 05

Documents

years c

years d

payback period5

pearson education

upper saddle river

b cash outflows

b pw20

end of year