5 Solutions

7/31/2019 5 Solutions

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CHAPTER 5 Solutions Manual

For

Basics of Engineering Economy, 1e

Leland Blank, PhD, PETexas A&M University

andAmerican University of Sharjah, UAE

Anthony Tarquin, PhD, PEUniversity of Texas at El Paso

PROPRIETARY MATERIAL.

The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may bedisplayed, reproduced or distributed in any form or by any means, without the prior writtenpermission of the publisher, or used beyond the limited distribution to teachers andeducators permitted by McGraw-Hill for their individual course preparation. If you are astudent using this Manual, you are using it without permission.

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Chapter 5

5.1 (a) Jim forgot the impact of interest, that is, the time value of money, which willincrease the annual recovery amount.

(b) Use Equation [5.3] to find the capital recovery amount with S = 0.

CR = AW = -20,000(A/P,15%,5)= -20,000(0.29832)= $-5966 per year

5.2 (a) Use Equation [5.3] for CR per year.

CR = -3,800,000(A/P,12%,12) + 250,000(A/F,12%,12)= -3,800,000(0.16144) + 250,000(0.04144)= $-603,112

(b) Use Equation [5.2].AW = CR + A of AOC

= -603,112 350,000 25,000(A/G,12%,12)= -603,112 350,000 - 25,000(4.1897)= $-1,057,855

One way to use a spreadsheet follows:CR: single cell entry = -PMT(12%,12,-3800000,250000)AW: Enter increasing gradient into B3:B14 and combine the PMT functions.

= -PMT(12%,12,-3800000,250000) - PMT(12%,12,NPV(12%,B3:B14))

5.3 The CR is lower than the $500,000 estimated 5 years ago.

CR = -3,150,000(A/P,10%,13) + 300,000(A/F,10%,13)= -3,150,000(0.14078) + 300,000(0.04078)= $-431,223

5.4 (a) AW = CR + (A of AOC and overhaul)= -115,000(A/P,7%,7) + 45,000(A/F,7%,7)

- 9500 - 3200(P/F,7%,4)(A/P,7%,7)= -115,000(0.18555) + 45,000(0.11555) 9500 3200(0.7629)(0.18555)

= $-26,091 per year

(b) No since $26,091 > $20,000, and $20,000 was an optimistic estimate.

5.5 Alternative A: At least, you need estimates of cost to provide same service for years4 and 5, for example, repurchase of A for 2 years, plus any

salvage (or other amounts) during these 2 years.Alternative B: You need market value estimate after 5 years.

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5.6 (a) Monetary terms in $ million. From AW, project clearly makes 10% per year.

AW = -170(A/P,10%,20) + 85(1-0.22.5)= -170(0.11746) + 65.875

= $+45.90 ($+45.9 million)

(b) AW goes negative between 35% and 40% per year.

@ 20%: AW = -170(A/P,20%,20) + 65.875 = $+30.96 million@ 30%: AW = -170(A/P,30%,20) + 65.875 = $+14.60 million@ 40%: AW = -170(A/P,40%,20) + 65.875 = $-2.21 million

A spreadsheet solution with x-y scatter graph follows.

5.7 (a) AW = -175,000(A/P,i%,5) + 35,000 + 10,000(A/G,i%,5) indicates the turn tounjustified just above 15% per year.

@12%: AW = -175,000(0.27741) + 35,000 + 10,000(1.7746) = $+4199@14%: AW = -175,000(0.29128) + 35,000 + 10,000(1.7399) = $+1425@15%: AW = -175,000(0.29832) + 35,000 + 10,000(1.7228) = $+22@16%: AW = -175,000(0.30541) + 35,000 + 10,000(1.7060) = $-1387

(b) Spreadsheet indicates just above 155 at the point where AW = 0.

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5.8 Use LCM of 8 years to calculate AW by Equation [5.1].

AWA = -517,510(A/P,8%,8)= $-517,510(0.17401)

= $-90,052

AWB = -812,100(A/P,8%,8)= $-812,100(0.17401)= $-141,314

By spreadsheet, use = -PMT(8%,8,PW) functions.

5.9 Calculate AW values to select machine R.

AWR= -250,000(A/P,9%,3) + 20,000(A/F,9%,3) - 40,000

= -250,000(0.39505) + 20,000(0.30505) 40,000= $-132,662

AWS = -370,500(A/P,9%,5) + 20,000(A/F,9%,5) - 50,000= -370,500(0.25709) + 20,000(0.16709) 50,000= $-141,910

By spreadsheet, enter single cell functions.R: = -PMT(9%,3,-250000,20000) 40000 Display: $-132,663S: = -PMT(9%,5,-370500,20000) - 50000 Display: $-141,911

5.10 Calculate AW values to now select machine S by a small margin.

AWR= [-250,000-60,000(P/F,9%,2)](A/P,9%,3) + 20,000(A/F,9%,3) - 40,000= [-250,000-60,000(0.8417)](0.39505) + 20,000(0.30505) 40,000= $-152,612

AWS = [-370,500-25,000(P/F,9%,2)-25,000(P/F,9%,4)](A/P,9%,5)+ 20,000(A/F,9%,5) - 50,000

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= [-370,500-25,000(0.8417)-25,000(0.7084)](0.25709)+ 20,000(0.16709) 50,000

= $-151,873

A spreadsheet solution follows. Cash flows are entered; select machine S.

5.11 For n = 4 years, S = 20 of first cost. Aw is lowest for 200 atm alternative.

AW200 = -310,000(A/P,6%,4) + 0.2(310,000)(A/F,6%,4) 95,000= -310,000(0.28859) + 62,000(0.22859) - 95,000= $-170,290

AW500 = -600,000(A/P,6%,4) + 0.2(600,000)(A/F,6%,4) 60,000= -600,000(0.28859) + 120,000(0.22859) - 60,000= $-205,723

AWcnt = $-190,000

A spreadsheet solution follows, also selecting 200 atm.

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5.12 (a) AW evaluation indicates chamber 490G to be more economic.

AWD103 = -400,000(A/P,10%,3) + 40,000(A/F,10%,3) 4000= -400,000(0.40211) + 40,000(0.30211) 4000= $-152,760

AW490G = -250,000(A/P,10%,2) + 25,000(A/F,10%,2) 3000= -250,000(0.57619) + 25,000(0.47619) 3000= $-135,143

(b) At P = $-300,000 and S = $30,000:AWD103 = -300,000(A/P,10%,3) + 30,000(A/F,10%,3) 4000

= -300,000(0.40211) + 30,000(0.30211) 4000= $-115,570

At P = $-500,000 and S = $50,000:

AWD103 = -500,000(A/P,10%,3) + 50,000(A/F,10%,3) 4000= -500,000(0.40211) + 50,000(0.30211) 4000= $-189,950

The cheaper model, P = $-300,000 will change the decision to D103.

By spreadsheet for part (b) use the PMT functions at different P values.490G: = -PMT(10%,2,-250000,25000)-3000 Display: $-135,143D103 @ P=-300,000: = -PMT(10%,3,-300000,30000)-4000 Display: $-115,571D103 @ P=-500,000: = -PMT(10%,3,-500000,50000)-4000 Display: $-189,952

5.13 AWforklift = CR AOC salary AW of pallets= -30,000(A/P,8%,12) + 8000(A/F,8%,12) 1000 - 32,000

-500(10)[1+(P/F,8%,2)+(P/F,8%,4)+(P/F,8%,6)+(P/F,8%,8)+(P/F,8%,10)](A/P,8%,12)

= -30,000(0.13270) + 8000(0.05270) 33,000-5000[1+(0.8573)+(0.7350)+(0.6302)+(0.5403)+(0.4632)](0.13270)

= $-39,363

AWwalkies = -(2)2,000(A/P,8%,4) 2(150) 55,000- 800(10)[1+(P/F,8%,2)](A/P,8%,4)

= -4000(0.30192) 55,300 8000[1+(0.0.8573)](0.30192)= $-60,994

Select forklift alternative (#1). A spreadsheet evaluation is easier to perform.

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5.14 From Example 4.2, bond P = $-4750; I = $150 each 6 months; n = 20; i = 3.35% per6 months.

AW = -4750(A/P,3.35%,20) + 150 + 5000(A/F,3.35%,20)= -4750(0.06941) + 150 + 5000(0.03591)

= $-.15

The return is approximately that desired based on AW value.

5.15 (a) AWjboy = -85,000(A/P,10%,3) + 40,000(A/F,10%,3) 30,000= -85,000(0.40211) + 40,000(0.30211) 30,000= $-52,095

AWweye = -97,000(A/P,10%,3) + 42,000(A/F,10%,3) 27,000= -97,000(0.40211) + 42,000(0.30211) 27,000= $-53,316

Select robot Joeboy.

(b) Determine P for Watcheye that makes the two AW relations equal.

AWjboy = AWweye-52,095 = P(A/P,10%,3) + 42,000(A/F,10%,3) 27,000-52,095 = P(0.40211) -14,311

P = $-93,964

If the first cost of Watcheye is reduced to lower than this amount, it is selected.

A spreadsheet solution, using GOAL SEEK follows. (If GOAL SEEK is not to be (inChapter 8) or has not been covered by the instructor, the manual solution is better toassign. Or, use trial and error on the spreadsheet.

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5.16 (a) Determine amount needed at end of year 20, followed by A to accumulatethis future amount.

CC = P = A/i = 24,000/0.08 = $300,000

F = A(F/A,8%,20)300,000 = A(45.7620)A = $6556 per year

By spreadsheet, enter = -PMT(8%,20,,300000) to display $6556.

(b) P = 24,000(P/A,8%,30)= 24,000(11.2578)= $270,187

270,187 = A(F/A,8%,20)

= A(45.7620)A = $5904 per year

$652 per year less is required for 30-year payout.

By spreadsheet, enter two functions.P after 20 years: = -PV(8%,30,24000)Display: $270,187A over 20 years: = -PMT(8%,20,,270187) Display: $5904

5.17 Find F in year 12; treat it as a CC value; find A forever.

F12 = 4(F/P,12%,11) -1(F/P,12%,9) -3(F/P,12%,8) -3(F/P,12%,7)+1(F/P,12%,6) +4(F/P,12%,5) +6(F/P,12%,4) +8(F/P,12%,3)+10(F/P,12%,2) +12(F/A,12%,2) + 38

= 4(3.4785) -1(2.7731) -3(2.4760) -3(2.2107)+1(1.9738) +4(1.7623) +6(1.5735) +8(1.4049)+10(1.2544) +12(2.1200) + 38

= $102.768

A = CC(i) = F12(i) = 102.768(0.08)= $ 8.22 per year ($8,221,440 per year)

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A spreadsheet solution follows.

5.18 Use procedure in section 4.4, step 3, to find equivalent A over one life cycle ofthe 2 recurring series ($-800 and $-2000). These will continue forever.

AW800 = {-800[(P/F,5%,2)+(P/F,5%,4)+(P/F,5%,6)+(P/F,5%,8)+(P/F,5%,10)]}(A/P,5%,10)

= {-800[(0.9070)+(0.8227)+(0.7462)+(0.6768)+(0.6139)]}(0.12950)= $-390

AW2000 = {-2000[(P/F,5%,3)+(P/F,5%,6)+(P/F,5%,9)]}(A/P,5%,10)= {-2000[(0.8638)+(0.7462)+(0.6446)]}(0.12950)= $-584

Total AW = -390 584= $-974 per year

A spreadsheet solution follows.

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5.19 Perpetual AW is equal to AW over one life cycle.

AW = {[-150,000(P/A,10%,4) - 25,000(P/G,10%,4)](P/F,10%,2)- [225,000(P/A,10%,4)(P/F,10%,6)]}(A/P,10%,10)

= {[-150,000(3.1699) 25,000(4.3781)](0.8264)

-[225,000(3.1699)(0.5645)]}(0.16275)= $-144,198

5.20 Monetary terms are in $ million. Determine AW values to select license.

AWcontract = -2 + 2.5= $0.5 ($500,000)

AWlicense = -2(A/P,12%,10) - 0.2 + 1.5= $0.946 ($946,000)

AWin-house = -30(0.12) 5 + 9= $0.4 ($400,000)

5.21 AWbrush = -400,000(A/P,8%,10) + 50,000(A/F,8%,10)- 50,000 + 5000(A/G,8%,10)

= -400,000(0.14903) + 50,000(0.06903) 50,000 + 5000(3.8713)= $-86,804

AWblast= -300,000(0.08) 50,000= $-74,000

Select the bead blasting alternative.

5.22 Monetary terms are $ million. Effective i = (1.025)4 1 = 10.38%. Select A.

AWA = -10(A/P,10.38%,5) + 0.7(A/F,10.38%,5) 0.8= -10(0.26636) + 0.7(0.16256) 0.8= $-3.35 ($-3.35 million)

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AWB = -50(0.1038) 0.6= $-5.79 ($-5.79 million)

Problems for Test Review and FE Exam Practice

5.23 Answer is (d).

5.24 Answer is (b).

5.25 AW2 = -550,000(A/P,6%,15) + 100,000= $-43,372

Answer is (b).

5.26 Answer is (c).

5.27 Answer is (d).

5.28 Answer is (a).

5.29 AW = -40,000(A/P,15%,4) + 40,000(1-.05(4))(A/F,15%,4) 5000= -40,000(0.35027) + 32,000(0.20027) 5000= $-12,602

Answer is (d).

5.30 PW = 50,000 + (20,000/0.12)(P/F,12%,15)= 50,000 + 166666.67(0.1827)= $80,450

AW = 80,450(0.12) = $9654Answer is (c).

5.31 Save for 240 months at 6/12 = 0.5% per month

F240 = 1000(F/A,0.5%,240)= 1000(462.04090)= $462,041

F360 = 462,041(F/P,0.5%,120)= 462,041(1.8194)= $840,637

A = 840,637(0.005)= $4203 per month

Answer is (c).

5.32 Answer is (b).

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