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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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