Chapter 07.pdf

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.

Chapter 7 Rate-of-Return Analysis Concept of Rate of Return Note: Symbol convention---The symbol i* represents the break-even 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. In that case, we should apply the net net investment test. If the project fails the net investment test, we need to calculate the return on invested capital (RIC) at a given MARR. We will set IRR = RIC for a mixed investment.

7.1

$1, 24 245 = $1, 00 000(1 + i) i = 24.5%

7.2 $24, 500 = $514.55( P / A, i, 60 60) i = 0.7917% per month r = 0.7917% ×12 = 9.5% 12 ia = (1 + 0.007917) − 1 = 9.93% per year

7.3 $950 = $35( P / A, 4%, 8) + F( P / F, 4 %, 8) 0.7307 F = $714.35 F = $977.64 7.4 $10, 887, 60 600 = $1, 65 650( F / P, i, 40 40) 6, 598.55 = (1 + i) 40 i = 24.59% 7.5 $2.5 = $1.2( P / A,i ,10) i = 46.98%

7.6

,1)) − $40,000( P / F , i, 2) PW ( i ) = −$50,000 + $100,000( P / F , i,1

=

0

i = 44.72%

Page | 1


7.7 $104, 20 200, 00 000 = $30, 00 000( F / P, i, 54 54) 3, 473.33 = (1 + i) 54 i = 16.30%

Investment Classification and Calculation of i* 7.8 (a) Simple investment: Project A, D (b) Non-simple (b) Non-simple investment: investment: Project B (c) •

Project A: PW (i) i

•

$22, 000 + $30, 000( P / A, i, 3) − $10, 000( P / G, i, 3)

= −

*

=

0

=

94.95%

Project B: PW (i ) = −$28, 000 + $32, 000( P / A, i, 2) − $22, 000( P / F, i, 3) = 0 i

•

*

=

55.29 .29% an and − 46.46 .46%

Project C: PW (i) = $35, 000 − $18, 000( P / A, i, 3)

0 i = 25.26% → Borrowing rate of return =

*

•

Project D: PW (i)

$56, 500 + $2, 500( P / F, i,1) + $6, 459( P / F, i, 2) +$78, 345( P/ F, i, 3)

= − =

i*

=

0

16.47%

(d) Project B has no unique rate of return. Given that project B is nonsimple, the true rate of return cannot be calculated until an MARR is specified. Page | 2


7.9

The equivalent annual cash flow for the first cash flow cycle ($500, $900, $800, $700) will be AE (i ) = [$500( P / F , i,1) + $900( P / F , i, 2) + $800( P, F, i,3)

+

$700( P, F, i, 4)]( A / P, i, 4)

Then, the present worth of the infinite cash flow series is expressed as AE ( i) PW ( i) = −$1,000 + i

$ − 1,000 +

=

[$500( P / F , i,1) + ⋅ ⋅ ⋅ + $700( P / F , i,4)]( A / P, i,4) i

0

=

i

*

=

67.9%

7.10 (a) Classification of investment projects: Simple projects: A,B, and E • Non-simple projects: C and D •

(b) −

$150 +

Let X =

$60 $150 + 1 + i (1 + i) 2

1 (1 + i)

=

0

, then,

$150 + $60 X + $150 X 2 = 0 X 1 = 0.8198, X 2 = −1.2198 −

i*

=

21.98%

(c) Find i* by plotting the NPW as a function of interest rate: Project

i*

A B C D E

21.98% 15.99% 19.59% 0% or 79.37% 23.52% Page | 3


7.11 (a) Classification of investment projects: • Simple projects: A, B, and D • Non-simple projects: C (b) Project A: • Project B: • Project C: • Project D: •

i*

5% i* = 17.23% i* = 18.99% i* = 34.76% =

(c) Use the PW plot command provided in Cash Flow Analyzer, or you may use Excel’s Chart Wizard. 7.12 (a) $90,000 + ($27,000 − $8,000)( P / A, i,6) + $10,000( P / F , i,6) = 0

−

Solving for i yields i* = 9.4208% . (b) With the geometric expense series −

$90, 000 + $27, 000( P / A, i, 6) − $8, 000( P / A1, 7%, i, 6)

+

$10, 000( P / F , i, 6) = 0

Solving for i* yields i * = 7.1745% . (c) To maintain i*

=

9.4208%

PW ( i ) = −$90,000 + $27,000( P / A1 , g ,9.4208%,6)

$8,000( P / A1 ,7%,9.4208%,6) + $10,000( P / F ,9.4208%,6)

− =

0

Solving for g yields g

=

2.20%

7.13 (a) Rate of return calculation: * • Project A: i = 23.75% * • Project B: i = 16.19%

Page | 4


(b) PW Plot $50,000 $40,000 $30,000 ) $20,000 $ ( $10,000 W P $0

A B

($10,000) 0

5

10

15

20

25

30

35

40

45

50

($20,000) ($30,000) Interest Rate (%)

The interest rate that makes two projects equivalent is 37.93%. 7.14 (a) Cash flow sign rules: Projects A B C D E F

Number of Sign Changes 1 2 1 1 2 3

Possible Number of i* 0, 1 0, 1, 2 0, 1 0, 1 0, 1, 2 0, 1, 2, 3

By the accumulated sign test, there is only one positive i* for B, E, and F.

(b) Use the PW plot command provided in Cash Flow Analyzer, or you may use Excel’s Chart Wizard. (c) Project A: i* = 100% * • Project B: i = 482.92% and − 59.69% * • Project C: i = 20.04% * • Project D: i = 40.57% * • Project E: i = 265.42% and -86.18% * • Project F: i = 209.46% •

7.15 (a) From the Cash Flow Analyzer IRR = 20% Page | 5


(b) Use the PW plot command provided in Cash Flow Analyzer, or you may use Excel’s Chart Wizard. (c) Since MARR(15%) < IRR = 20%, accept the project!

Mixed Investments 7.16 (a) •

Project 1: i*1 = 110%, i*2

60%

= −

Project 2: i* = 136.22% * * • Project 3: i 1 = 23.85%, i 2 •

83.85%

= −

(b) Investment classification: •

Project 1: nonsimple and mixed investment, IRR = 75% PB(75%,12%) 0

$1, 000 PB(75%,12%)1 = −$1, 000(1 + 0.75) + $2, 500 = $750 PB(75%,12%) 2 = $750(1 + 0.12) −840 = $0 = −

•

Project 2: simple and pure investment, IRR = 136.22%

•

Project 3: nonsimple and mixed investment, IRR = 22.14% PB(22.14%,12%) 0

$1, 000 PB(22.14%,12%)1 = −$1, 000(1 + 0.2214) + $1, 400 = $178.6 PB(22.14%,12%) 2 = $178.6(1 + 0.12) − 200 = $0 = −

(c) If MARR = 12%, all the projects are acceptable. 7.17 (a) −

$150 +

Let X =

$150 $36 + 1 + i (1 + i )2

1 (1 + i)

=

0

, then,

Page | 6


$150 + $150 X + $36 X 2 = 0 X 1 = 0.8333, X 2 = −5 * i = 20% −

(b) Simple projects: A, B, and D • Non-simple projects: C and E •

(c) Apply the cash flow sign rule: Projects Number of Sign Changes A 1 B 1 C 2 D 1 E 2

Possible Number of i* 0, 1 0, 1 0, 1, 2 0, 1 0, 1, 2

Actual i* 20% 15.32% No return 17.70% 12.63%,41.42%

(d) Project B: IRR B = 15.32% • Project C: IRRC = None •

Project D: IRR D = 17.70% • Project E: IRR E = 12.29% •

Note that, since project E is a mixed investment, we need to find the IRRs for E by using external interest rate of 12%. (e) Apply the net investment test for project E.

0 1 2 3 4 5

Project Balances Project E (i*=12.63%) $200 $325 -$134 -$651 -$533 $0

So, project E fails the net investment test. Projects A, B, and D pass the net investment test, indicating that they are pure investment. Project C has no meaningful rate of return. 7.18 IRR = 9.69% .

Page | 7


7.19 (a) Project 1: i* = 82.21% * * • Project 2: i 1 = 14.64%, i 2 •

•

=

210.28%

Project 3: i* = 100%

(b) Apply the net investment test: •

Project 1: PB(82.21%) 0

$8, 500 PB(82.21%)1 = −$8, 500(1 + 0.8221) + $10, 000 = −$5, 487.85 PB(82.21%) 2 = −$5, 487.85(1 + 0.8221) + $10, 000 = 0 (-,-,0), pure investment •

= −

Project 2: PB(14.64%) 0

$5, 000 PB(14.64%)1 = −$5, 000(1 + 0.1464) + $10, 000 = $4, 268 PB(14.64%) 2 = $4, 268(1 + 0.1464) + $30, 000 = $34, 892.80 PB(14.64%) 3 = $34, 892.8(1 + 0.1464) −$40, 000 = 0 (-,+,0), mixed investment •

= −

Project 3: PB(100%) 0

= −

$1,500 PB(100%)1 = −$1, 500(1 +1) + $6, 000 = $3, 000 PB(100%) 2

=

$3,000(1 + 1) − $6,000 = 0 (-,+,0), mixed investment

(c) Project 1: IRR1 = 82.21% • Project 2: IRR2 = −2.04% • Project 3: IRR3 = −57.14% •

(d) Only project 1 is acceptable. 7.20 (a) Pure investment: B (b) Mixed investments: A and C Page | 8


(c) Project A: IRR A = 140.56% , Project B: IRR B IRRC = 12.24% at an external rate of 12%

=

21.86% , Project C:

(d) All three projects are acceptable. 7.21 (a) There are two sign changes in cash flow, indicating the possibility of multiple * i s. * * i 1 = 10%, i 2 = 20 Apply the net investment test: PB(10%) 0

$1, 000, 000 PB(10%)1 = −$1,000,000(1.1) + $2,300,000 = $1,200,000 PB(10%) 2 = $1, 200, 000(1.1) − $1, 320, 000 = 0 = −

(-,+,0), mixed investment

∴

(b) At an external interest rate of 12%, IRR = 12.14%. (c) The project is acceptable. 7.22 (a) i*1 = 6%, i*2

=

0%

(b) Apply the net investment test: •

Project A PB(6%) 0

$1, 000 PB(6%)1 = −$1, 000(1.06) + $2, 060 = $1, 000 PB(6%) 2 = $1, 000(1.06) − $1, 060 = 0 ∴ (-,+,0), a mixed investment •

= −

Project B A simple and pure investment

•

Project C

Page | 9


PB(17.66%) 0

= −

$1, 500

PB(17.66%)1

= −

PB(17.66%) 2

= −

PB(17.66%) 3

= −

$1, 500(1.1766) + $1, 000 764.90(1.1766) − $800

$764.90

= −

$1, 700

= −

$1, 700(1.1766) + $2, 000 = 0 ∴ (-,-,-,0), a pure investment Project A is the only mixed investment. (c) Projects B & C are acceptable. 7.23 (a) Project A: i*1 = 10%, i*2

=

100% , Project B: i*1 = 350.34%, i*2

80.83%

= −

(b) Pure investment: C, mixed investments: A, B, D and E (c) Project A: IRR A = 13.57% , Project B: IRR B IRRC = 18% , Project D: IRR D

=

=

342.16% , Project C:

31.07% , Project E: IRR E = 19.67%

(d) All projects are acceptable. 7.24 (a) Use the PW plot command provided in Cash Flow Analyzer. From the plot, we get i*1 = 43.47%, i*2 = 77.67% (b) Apply the net investment test: $8,000

PB(43.47%)0

= −

PB(43.47%)1

= −

PB(43.47%) 2

= −

PB(43.47%)3

=

$8,000(1 + 0.4347) + $10,000 = −$1,477.98 $1,477.98(1 + 0.4347) + $30,000 = $27,879.5

$27,879.5(1 + 0.4347) − $40,000 = 0

(c) Since IRR = -16.30 % < 18%, the project is not acceptable. 7.25 (a) Apply the net investment test using i* = 10% : PB(10%) 0 = −$150, 000 PB(10%)1 = −$150, 000(1.1) + $465, 000 = $300, 000 PB(10%) 2 = $300, 000(1.1) − $330, 000 = 0 ∴ (-,+,0), mixed investment Page | 10


(b) By cash flow analyzer IRR = 6.30%. So the project is not acceptable. 7.26 (d)

IRR Analysis 7.27 PW (10%) = −$2,000 + $1,000( P / F ,10%,1)

$1,200( P / F ,10%,2) + $ X ( P / F ,10%,3)

+

0 X = $132 =

7.28 The present worth of the project cash flow is PW (i ) = −$10 M

+

$1.8 M ( P / A, i,8) + $1 M ( P / F, i,8)

=

0

Since IRR = 10.18% > MARR, accept the project.

7.29 The present worth of the project cash flow is PW (i)

= −

$15, 000 + $14, 520( P / F, i, 2) + $3,993( P / F, i,3)

=

0

Since IRR = 10% = MARR, the project breaks even.

7.30 (a) Since IRR = 10% and PW(10%) = 0, we have, PW (10%) = −$2,500 + $700( P / F ,10%,1) + $900( P / F ,10%, 2) +

X ( P / F ,10%, 3) = 0

X = $1,490.5

(b) Since IRR > 8%, the project is acceptable. 7.31 PW (13%) = −$2, 000 + $500( P / A,13%,3) + $ X ( P / F,13%,5)

=

0

X = $1,509.74

Page | 11


7.32 

Let X be the annual rent per apartment unit. Then the net cash flow table is: N

Capital Investment

0

-14,500,000

Revenue

Maintenance

Manager

Net Cash Flow -14,500,000

1

50 X

-350,000

-85,000

50X - 435,000

2

50 X

-400,000

-85,000

50X - 485,000

3

50 X

-450,000

-85,000

50X - 535,000

4

50 X

-500,000

-85,000

50X - 585,000

50 X

-550,000

-85,000

50X +15,365,000

5

16,000,000

Through the Excel Solver function by setting PW(15%) = 0, X = $49,473.35.

7.33 

Let X be the annual savings in labor.

PW (12%) = −$35, 000 + ( X

−

$4, 000)( P / A,12%, 6) + $5, 000( P / F,12%, 6) = 0

X = $11,896.82

7.34 

Net cash flow table: n Land Building Equipment Revenue Expenses Net Cash Flow 0 -$1.50 -$3 -$4.50 1 -$4 -$4.00 2 $3.50 -$1.40 $2.10 3 $3.68 -$1.47 $2.21 4 $3.86 -$1.54 $2.32 5 $4.05 -$1.62 $2.43 6 $4.25 -$1.70 $2.55 7 $4.47 -$1.79 $2.68 8 $4.69 -$1.88 $2.81 9 $4.92 -$1.97 $2.95 10 $5.17 -$2.07 $3.10 11 $5.43 -$2.17 $3.26 12 $5.43 -$2.17 $3.26 13 $5.43 -$2.17 $3.26 14 $2 $1.40 $0.50 $5.43 -$2.17 $7.16

Page | 12




Rate of return calculation: PW (i ) = −$4.5 − $4( P / F , i,1) + $2.1( P / F , i, 2) +  +

$7.16( P / F , i ,14) = 0 i



*

=

24.85%

Since this is a simple investment, IRR = 24.85%. At MARR = 15%, the project is economically attractive.

7.35 PW (18%) = −$150, 000 + ( X

50, 000)( P / A,18%,10) +15, 000( P / F ,18%,10) = 0 X = $82,739.26 −

7.36 (a) PW (i ) = −$30 − $9( P / F , i,1) + $18( P / F , i , 2) + $20( P / F , i , 3) +

$18( P / F , i, 4) + $10( P / F , i, 5) + $5( P / F , i, 6)

0 This is a simple investment. Therefore, IRR = i* =

=

40.20% .

Since IRR is higher than MARR, the project is acceptable.

(b) IRR = 45.47% (c) IRR = 27.30%

Comparing Alternatives Notes to Instructors: To apply the IRR decision rule in comparing mutually exclusive investments, we need to conduct an incremental analysis. If the incremental cash flow series represents a mixed investment, we should apply the net investment test and make the selection by calculating the return on invested capital (or true internal rate of return). Computational procedure for finding the RIC is not shown in this presentation. Excel or Cash Flow Analyzer could be used.

Page | 13


7.37 (a) Project A: IRR = 16.01% Project B: IRR = 18.18% (b) Project A and B are acceptable. (c) Since the incremental cash flow displays a nonsimple investment, we may abandon the IRR analysis and make a selection based on the NPW criterion. PW (15%) A

$2, 673.21 PW (15%) B = $7, 991.29 PW(15%) A-B = −$5,318.07 < 0 =

Select project B.

7.38 Option 1: Buy a certificate. Option 2: Purchase a bond, and assume that MARR = 5% . n

0 1 2 3 4 5

Net Cash Flow Option 2 Option 1 – Option 2 -$10,000 $0 -650 650 650 -650 650 -650 650 -650 10,650 4736.24

Option 1 -$10,000 0 0 0 0 15,386.24

The rate of return on incremental investment is *

i1

−

2 =

25.49% > 5%

Option 1 is a better choice.

7.39 Determine the cash flow on incremental investment: Net Cash Flow n

Project A

Project B

B-A

0

-$2,500

-$3,600

-$1,100

1

1,600

2,600

1,000

2

1,840

2,200

360

*

i B

−

A =

18.52% > 15% , select project B.

Page | 14


7.40 (a) IRR on the incremental investment: Net Cash Flow Project A1 Project A2 -11,000 -13,000 5,000 6,200 5,000 6,200 5,000 6,200

n

0 1 2 3 *

i A 2

−

A1

=

A2 – A1 -$2,000 1,200 1,200 1,200

36.31%

(b) Since it is an incremental simple investment, IRR A2-A1 = 36.31% > 10% . Therefore, select project A2. 7.41 (a) IRR on the incremental investment: n

0 1 2 3

A1 -$15,000 7,500 7,500 7,500

Net Cash Flow A2 -$20,000 8,000 15,000 5,000

A2 – A1 -$5,000 500 7,500 -2,500

Since the incremental cash flow displays a mixed investment with i*1 = 64.50% and i*2 = 6.67%, we need to find the return on invested capital (RIC). The true rate of return on incremental investment (A2-A1) at a MARR of 10% is 7.36%, which is smaller than MARR, so we select A1. (b) PW (10%) A1

= −

15, 000 + $7,500( P / A,10%, 3) = $3, 651

PW (10%) A2

= −

20, 000 + $8, 000( P / F,10%,1) + $15, 000( P/ F,10%2)

+

$5, 000( P / F ,10%, 3) = $3, 426

Select project A1.

7.42 (c)

Page | 15


7.43 (a) IRR for incremental investment: Project A -$300 0 690

n

0 1 2 *

i B

=

A

−

Net Cash Flow Project B -$800 1,150 40

B-A -$500 1,150 -650

0% or30%

Since this is a mixed incremental investment, we need to find the IRR using an external interest rate of 15%. IRR B

A

−

=

16.96% > 15% Project B is preferred.

(b) Use the PW plot command provided in Cash Flow Analyzer. 7.44 Incremental cash flows (Model A – Model B): n

0 1 2 3 4

A–B -$2,376 0 0 0 2,500 IRR A B = 1.28% −

If MARR < 1.28%, Model A is preferred.

7.45 PW ( i) A PW ( i) B

$8,000 + ($900 − $150)( P / A, i,20) + $500( P / F , i,20) = 0

= −

$12,000 + ($1,100 − $100)( P / A, i,20) + $600( P / F , i,20) = 0

= −

With MARR at 12%, IRR for Model A: 7.12%, IRR for Model B: 5.65%. Neither model is attractive.

Page | 16


7.46 (a) The least common multiple project lives = 6 years n

Project A -$100 60 50 50-100 60 50 50

0 1 2 3 4 5 6

→

Net Cash Flow Project B -$200 120 150-200 120 150-200 120 150

Analysis period 6 years

B–A -$100 60 -100 170 -110 70 100

By the net investment test, the incremental investment is a pure investment, so IRR = 15.98%. Since IRR B

−

A =

15.98% > 15% , select project B.

(b) Incremental analysis between C and D: n

Project C -$4,000 2,410 2,930

0 1 2

IRR C D −

=

Net Cash Flow Project D -$2,000 1,400 1,720

C–D -$2,000 1,010 1,210

7.03% < 15% , Project D is preferred.

(c) Incremental analysis between E and F: n

0 1 2

Project E -$2,000 3,700 1,640

Net Cash Flow Project F -$3,000 2,500 1,500

F –E -$1,000 -1,200 -140

No IRR, indicating E dominates F, so select E.

Page | 17


7.47 Let A0 = current practice, A1 = just-in-time system, A2 = stock-less supply system. Comparison between A0 and A1:



A0

n

0 1-8

A1 -$3,000,000 -3,800,000

0 -9,000,000 *

i A1

−

A0

=

IRR A1

−

=

A0

A1 – A0 -$3,000,000 5,200,000

173.28% > 10%

A1 is a better choice.

Comparison between A1 and A2:



A2 -$6,000,000 -1,900,000

n

0 1-8

*

i A 2

−

A1

=

A1 -$3,000,000 -3,800,000

IRR A2

−

A1

=

A2 – A1 -$3,000,000 1,900,000

62% > 10%

A2 is a better choice. That means that the stockless supply system is the final choice.

7.48 (a) •

Project A vs. Project B n

Project A -$1,000 900 500 100 50

0 1 2 3 4 *

i B •

−

A

=

IRR B

−

A

=

Net Cash Flow Project B -$1,000 600 500 500 100

B–A $0 -300 0 400 50

21.27% > 12% , select B.

Project B vs. Project C n

0 1 2

Project B -$1,000 600 500

Net Cash Flow Project C -$2,000 900 900

C–B -$1,000 300 400 Page | 18


3 4 i

500 100

* C − B

=

900 900

IRR C B −

=

400 800

26.36% > 12% , select C.

(b) $1, 000 = $300( P / A, i , 4) i = 7.71% BRR = 7.71% .

(c) Since BRR is less than MARR, project D is acceptable. (d) n

Net Cash Flow Project E -$1,200 400 400 400 400

Project C -$2,000 900 900 900 900

0 1 2 3 4 i

* C−E

=

IRR C

−

E

=

C–E -$800 500 500 500 500

50.23% > 12% , select C.

7.49 (a) *

i1

=

54.52%,

i2

*

=

57.61%, and i3*

=

38.41%

(b) •

Project 1 versus Project 2: n

0 1 2

Project 1 -$1,500 700 2,500

Project 2 -$5,000 7,500 600

2–1 -$3,500 6,800 -1,900

The incremental cash flow represents a mixed investment, we calculate the RIC at MARR of 15%: IRR 2-1 = 47.08% > 15%, so select project 2.

Page | 19

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. •

Project 2 versus Project 3: Project 2 -$5,000 7,500 600

n

0 1 2

Project 3 -$2,200 1,600 2,000

2–3 -$2,800 5,900 -1,400

This incremental cash flow (2 - 3) is another mixed investment; we need to calculate the RIC at 15% which is 67.24% > 15%, so we choose project 2. Select Project 2.

7.50 (a) IRR B

=

25.99%

(b) PW (15%) A

$10, 000 + $5,500( P / A,15%, 3) = $2,558

= −

(c) Incremental analysis: Net Cash Flow Project A Project B -$10,000 -$20,000 5,500 0 5,500 0 5,500 40,000

n

0 1 2 3

Since IRR B

−

A =

B–A -$10,000 -5,500 -5,500 34,500

24.24% > 15%, select project B.

7.51 Project B 7.52 Model C 7.53 All projects would be acceptable because individual ROR exceed the MARR. Based on the incremental analysis, we observe the following relationships: IRR A 2 IRR A3 IRR A3

−

−

−

10% < 15% (Select A1) A1 = 18% > 15% (Select A3) A 2 = 23% > 15% (Select A3) A1 =

Therefore, A3 is the best alternative.

Page | 20


7.54 From the incremental rate of return table, we can deduce the following relationships: IRR A 2

−

IRR A3

−

A1

A 2

IRR A 4 IRR A5 IRR A6

−

=

8.9% < 15% (Select A1)

=

42.7% > 15% (Select A3)

0% < 15% (Select A3) = 20.2% > 15% (Select A5) = 36.3% > 15% (Select A6)

A3 =

−

A4

−

A5

It is necessary to determine the preference relationship among A1, A3, and A6. IRR A3 IRR A6

−

−

16.66% > 15% (Select A3) = 20.18% > 15% (Select A6)

A1 = A3

A6 is the best alternative.

7.55 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. Category Investment cost Salvage value Annual labor savings Annual power cost Net annual savings

Power Saw Model A Model B $6,000 $4,000 400 600 1,296 1,725 400 420 896 1,305 Net Cash Flow Model A Model B -$4,000 -$6,000 896 1,305 896 1,305

n

0 1 2

Model C $7,000 700 1,944 480 1,464

Model C -$7,000 1,464 1,464



20 IRR PW(10%) 

400+896 22.02% $3,688

600+1,305 21.34% $5,199

700+1,464 20.46% $5,568

Model A versus Model B: PW (i ) B A = −$2, 000 + $409( P / A, i, 20) + $200( P / F, i, 20) −

0 = 19.97% > 10% =

IRR B

−

A

Select Model B.

Page | 21




Model B versus Model C: PW ( i)C B = −$1,000 + $159( P / A, i,20) + $100( P / F , i,20) −

0 = 15.02% > 10%

=

IRR C B −

Select Model C. The PW rule also selects Model C, as indicated in the table above.

Unequal Service Lives 7.56 With the least common multiple of 6 project years,

n

0 1 2 3 4 5 6

Net Cash Flow Project B -$10,000 8,000 8,000 – 10,000 8,000 8,000 – 10,000 8,000 8,000

Project A -$5,000 3,000 4,000 4,000-5,000 3,000 3,000 3,000

B–A -$5,000 5,000 -6,000 9,000 -5,000 4,000 4,000

Since the incremental cash flow series is a mixed investment, we calculate the RIC at 15%. IRR B-A = 25.67% > 15%, so choose project B. We can easily verify the result by using the NPW: PW(15%) B

−

$5, 000 + $5, 000( P / F ,15%,1)

A = −

+ +

$4, 000( P / F ,15%, 6) = $1,587.86

7.57 (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 3 years and project A2 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. Page | 22


A2 – A1 -$5,000 0 0 15,000

n

0 1 2 3 IRR A2

A1

44.22%

=

−

The MARR must be less than 44.22% for Project A1 to be preferred.

Short Case Studies ST 7.1 We assume monthly service fee = $40, the number of monthly transactions per scanner = 1,000, and a service period of 5 years with no salvage value. Here we assume an effective annual return on investment is 20%. This means the required monthly return would be 1.53%. Let R be the required monthly revenue per scanner. PW(1.53%) = −$50 + [ R − ($40 + $100)]( P / A,1.53%, 60) = 0 R = $141.28

It is interesting that mere $1.28 additional revenue over its O&M brings 1.53% monthly return on investment. ST 7.2 (a) Analysis period of 40 years(unit: thousand $): • Without “mothballing” cost: PW ( i ) = −$1,500,000 + $138,000( P / A1 ,0.05%, i,40) i* •

=

0

=

8.95%

With “mothballing” cost of $0.75 billion: PW(i) = −$1,500, 000 + $138, 000( P / A1, 0.05%, i, 40) $750, 000( P / F , i, 40) = 0 −

i

*

=

8.77% and -18.60%

Page | 23


Now this is a mixed investment, so we need to find out the true rate of return (RIC), which requires an MARR. If we assume a MARR of 7%, the RIC will be about 8.70%. For a 40-year analysis period, the drop in IRR with the mothballing cost is about 2.79%, which is relatively insignificant. (b) Analysis period of 25 years (unit: thousand $): • Without “mothballing” cost: PW ( i ) = −$1,500,000 + $138,000( P / A1 ,0.05%, i,25) i •

*

=

0

=

7.84%

With “mothballing” cost of $0.75 billion: PW(i ) = −$1, 500, 000 + ($207, 000 −$69, 000)( P / A1, 0.05%, i, 25) $750, 000( P / F , i, 25) =0 −

i

*

=

6.80% and -18.17%

Once again this is a mixed investment, so we need to assume a MARR. At MARR = 7%, the RIC is about 6.78%. Clearly, the project is no longer profitable. For a 25-year analysis period, the drop of IRR with the mothballing cost is about 13.52%, which is relatively significant. ST 7.3 (a)

Assumptions required We need to assume there are no cash flows for the first three years if B&E Cooling decides to defer the decision. • Assume the firm will be in business for an indefinite period. •

•

We 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 now, Option 2 will be adopted for an indefinite period at the end of 8 years.

(b) Investment decision: •

Rate of return analysis: We need incremental analysis for these two options and then we find that the incremental cash flow is a mixed investment case over a period of 40 years.

Page | 24


Period

Option 1

Option 2

Option 1 ‐ Option 2

0

‐6000

‐6000

1

9000

9000

2

9000

9000

3

9000

‐5000

14000

4

9000

4000

5000

5

9000

4000

5000

6

9000

4000

5000

7

9000

4000

5000

8

5000

4000

1000

9

4000

4000

0

10

4000

4000

0

11

4000

1000

3000

12

4000

4000

0

13

4000

4000

0

14

4000

4000

0

15

4000

4000

0

16

1000

4000

‐3000

17

4000

4000

0

18

4000

4000

0

19

4000

1000

3000

20

4000

4000

0

21

4000

4000

0

22

4000

4000

0

23

4000

4000

0

24

1000

4000

‐3000

25

4000

4000

0

26

4000

4000

0

27

4000

1000

3000

28

4000

4000

0

29

4000

4000

0

30

4000

4000

0

31

4000

4000

0

32

1000

4000

‐3000

33

4000

4000

0

34

4000

4000

0

35

4000

1000

3000

36

4000

4000

0

Page | 25


37

4000

4000

0

38

4000

4000

0

39

4000

4000

0

40

1000

4000

‐3000

unit:$1,000

We use Cash Flow Analyzer with MARR=15% and calculate RIC1 2 = 154% > 15% , so option 1 is a better choice. . −

ST 7.4 Current Pump(A) Larger Pump(B) $0 -$1,600,000 $10,000,000 $20,000,000 $10,000,000 $0

n

0 1 2

IRR =

B-A -$1,600,000 $10,000,000 -$10,000,000 25% 400%

The incremental cash flows result in multiple rates of return (25% and 400%), so we need to find the RIC. At MARR = 20%, the RIC is 4.17%, which is less than 20%, so we stick with the smaller pump. With the NPW analysis, we reach the same conclusion. PW (20%) B-A

$1.6 M + $10 M ( P / F ,20%,1) − $10 M ( P / F ,20%,2)

= −

$0.21 < 0 Reject the larger pump.

= −

ST 7.5 (a)

Clearly the fellow engineer’s advice seems genuine, but the scale of investment as well as of timing of cash flows will affect the true rate of return. To determine whether or not the higher cost investment (project 2) can be justified, we need to perform an incremental analysis.

(b) n

0 1 2

Project 2 – Project 1 -$10,000 +$23,000 -$13,200 Page | 26


This incremental cash flow series represents a nonsimple investment with two rates of return at * i 2 1 = 10% or 20% −

It is also a mixed investment. We may develop the RIC as a function of MARR.. Let i = RIC (or true IRR) and assume i < 1.3 PB(i, MARR)0

= −

$10, 000

PB(i, MARR)1

= −

$10, 000(1 + i) + $23, 000

$13,000 − 10,000 i PB(i, MARR) 2 = ($13, 000 − 10, 000 i)(1 + MARR) =

$13,200 = 0 −

(Note that if i > 1.3, there will be no feasible solution.) Rearranging the terms in PB(i,MARR)2 gives an expression of IRR as a function of MARR. 1.32 IRR = 1.3 − 1 + MARR For example, at MARR = 15% IRR B

A

−

=

15.2% > 15%

Select project 2.

Page | 27

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