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Solutions to end-of-chapter problems
Engineering Economy, 8th edition
Leland Blank and Anthony Tarquin
Chapter 5
Present Worth Analysis
Types of Projects
5.2 (a) Service alternatives all have the same revenues; revenue alternatives each have different
Alternative Comparison - Equal Lives
5.5 (a) For independent projects, select all that have PW ≥ 0.
5.7 PWX = -45,000 – 8000(P/A,12%,5) + 2000(P/F,12%,5)
5.8 PWA = -80,000 – 30,000(P/A,12%,3) + 15,000(P/F,12%,3)
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PWB = -120,000 – 8,000(P/A,12%,3) + 40,000(P/F,12%,3)
= -120,000 – 8,000(2.4018) + 40,000(0.7118)
= $-110,742
Select Method B
Spreadsheet functions: For PWA: = -PV(12%,3,-30000,15000) - 80000
For PWB: = -PV(12%,3,-8000,40000) - 120000
5.9 (a) PWFull = $-122,000,000
PWSmall = -80,000,000 -100,000,000(P/F,6%,20) - 25,000(P/A,6%,20)
5.10 PWold = -1200(3.50)(P/A,15%,5)
= -4200(3.3522)
5.11 PW1 = -900,000 – 560,000(P/F,20%,2) – 79,000(P/A,20%,10)
Select option 2 - subcontracting
5.12 PWsingle = -4000 - 4000(P/A,12%,4)
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= $-16,149
PWsite = $-15,000
Buy the site license
5.13 Units are $ million
PWponds = -13 – 2.1(P/A,10%,5)
5.14 Compare each alternative against DN and select all with PW ≥ 0. Monetary units are $1000.
PWA = -1200 + 200(P/A,15%,10) + 5(P/F,15%,10)
5.15 PWA = -80,000 – [30,000(P/A,12%,3) + 4000(P/G,12%,3)] + 15,000(P/F,12%,3)
5.16 Municipal water: Cost/mth = -5(30)(2.90)/1000 = $0.435
Drinking bottled water varies from 34 to 83 times more expensive than drinking tap water.
Alternative Comparison - Different Lives
5.17 PWDDM = -164,000 – 55,000(P/A,20%,4) – 164,000(P/F,20%,2)
5.18 PWM = -205,000 – 29,000(P/A,10%,4) – 203,000(P/F,10%,2)
+ 2000(P/F,10%,4)
Select material FF
5.19 (a) PW1 = -400,000 – 140,000(P/A,15%,6)-360,000(P/F,15%,3)+ 40,000(P/F,15%,6)
Select method 2
(b) Incorrect PW1 over 3 years, PW2 is correct in (a)
5.20 Study period = LCM = 8 years
PWA = -15,000 – 6000(P/A,10%,8) – 12,000(P/F10%,4) + 3000(P/F,10%,8)
Select Alternative A
5.21(a) PWBFP = -203,000 – 90,500(P/A,6%,10) – 182,700(P/F,6%,5) + 20,300(P/F,6%,10)
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= $-977,416
Centrifuges have a slightly lower PW
(b) PWBFP = -203,000 – 90,500(P/A,6%,8) – 182,700(P/F,6%,5) + 20,300(P/F,6%,8)
Centrifuges still have a slightly lower PW, but by a smaller amount
5.22 PWA = -400 + (360 – 100)(P/A,20%,3)
PWB = -510 + (235 – 140)(P/A,20%,10) + 22(P/F,20%,10)
PWC = -660 + (400 – 280)(P/A,20%,5)
PWD = -820 + (605 – 315)(P/A,20%,8) + 80(P/F,20%,8)
PWE = -900 + (790 – 450)(P/A,20%,4) + 95(P/F,20%,4)
5.23 PWC = -40,000 – [7000(P/A,10%,10) + 1000(P/G,10%,10)] + 9000(P/F,10%,10)
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Select alternative D
5.24 PWK = -160,000 – 7000(P/A,2%,16) –120,000(P/F,2%,8) + 40,000(P/F,2%,16)
5.25 (a) Factor solution. Must use a 6-year evaluation period for equal-service analysis.
Plan A: effective i/year = (1 + 0.03)2 – 1 = 6.09%
PWC = -1,500,000 – 500,000(P/F,3%,4) – 1,500,000(P/F,3%,6)
- 500,000(P/F,3%,10)
(b) Spreadsheet: Cash flow details not included. PW functions and displays are:
5.26 Set the PWS relation equal to $-33.16, and solve for the first cost
XS (a positive number) with repurchase in year 5. In $1 million units,
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XS = $14.576 ($14,576,000)
Select seawater option for any first cost ≤ $14.576 million
Future Worth Comparison
5.27 FWsolar = -12,600(F/P,10%,4) – 1400(F/A,10%,4)
FWline = -11,000(F/P,10%,4) – 800(F/A,10%,4)
5.28 Calculate FW at LCM of 6 years
FWD = -62,000[(F/P,15%,6) + (F/P,15%,3)] – 15,000(F/A,15%,6) + 8,000[(F/P,15%,3) + 1]
5.29 FW20% = -100(F/P,10%,15) – 80(F/A,10%,15)
5.30 (a) FWA = -40,000[(F/P,10%,8) + (F/P,10%,6) + (F/P,10%,4) + (F/P,10%,2)]
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- 9000(F/A,10%,8)
= -40,000[(2.1436) + (1.7716) + (1.4641) + (1.2100)] – 9000(11.4359)
= $-366,495
FWB = -80,000[(F/P,10%,8) + (F/P,10%,4)] - 6000(F/A,10%,8)
= -80,000[(2.1436) + (1.4641)] - 6000(11.4359)
= $-357,231
FWC = -130,000(F/P,10%,8) - 4000(F/A,10%,8) + 13,000
= -130,000(2.1436) - 4000(11.4359) + 13,000
= $-311,412
Select method C
(b) Find the PW values from the FW values with n = 8 years. Select method C
5.31 FWpurchase = -150,000(F/P,15%,6) + 12,000(F/A,15%,6) + 65,000
Capitalized Cost
5.32 CC = -78,000 - 3500(A/F,8%,5)/0.08
5.35 CC = -1,700,000 – 350,000(A/F,6%,3)/0.06
5.36 P-1 = 10,000/0.10 = 100,000
(b) Find future value in year 25 and multiply by i
5.38 (a) Find future value in year 19, then multiply by i
F19 = 10,000(F/P,12%,19) + 30,000(F/P,12%,16) + 8000(F/A,12%,5)(F/P,12%,11)
5.39 CC = -250,000,000 – 800,000/0.08 – [950,000(A/F,8%,10)]/0.08
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-[75,000(0.17046)]/0.08
= -250,000,000 – 10,000,000 – 819,731 – 159,806
= $-260,979,538
5.40 For alternatives E and F, find AW, then divide by i using CC = AW/i
AWE = -50,000(A/P,10%,2) – 30,000 + 5000(A/F,10%,2)
= -50,000(0.57619) – 30,000 + 5000(0.47619)
5.41 Use C to identify the contractor option.
(a) CCC = -5 million/0.12 = $-41.67 million
(b) Find Pg and A of the geometric gradient (g = 2%), then CC.
Exercises for Spreadsheets
Solution shows functions. Select option 3.
5.43 (a) Use PV and logical IF functions, as in top half of the spreadsheet.
5.44 Select alternative A
5.45 Selection changes from Ferguson (5 years) to Halgrove (8 or 10 years)
5.46 Selection is D in both cases, but by a larger margin for the maximum life plan.
5.47 (a) Develop the spreadsheet with 2% increases in M&O for purchase and a constant
Additional Problems and FE Exam Review Questions
5.48 Answer is (c)
5.52 CC = -70,000 – 70,000(A/F,10%,10)/0.10
Machine X Machine Y
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Annual revenue, $/year 80,000 75,000
Salvage value, $ 10,000 25,000
Life, years 3 6
5.53 PWX = -146,000 + (80,000-15,000)(P/A,10%,6) - 136,000(P/F,10%,3) +
5.54 FWX = -146,000(F/P,10%,6) + (80,000-15,000)(F/A,10%,6) - 136,000(F/P,10%,3) + 10,000
Problems 5.57 through 5.61 are based on the following cash flows for alternatives A and B
at an interest rate of 10% per year:
Alternative A B
5.57 LCM is ∞; first find AW and then divide by i
5.58 CCB = -750,000 - 10,000/0.10
Solution to Case Study, Chapter 5
There is not always a definitive answer to case study exercises. Here are example responses
COMPARING SOCIAL SECURITY BENEFITS
1. Total payments are shown in row 30 of the spreadsheet.
2. Future worth values at 6% per year are shown in row 29.
3. Plots of FW values by year are shown in the (x-y scatter) graph below.
4. Develop all feasible plans for the couple and use the summed FW values to determine which
is largest.
Spouse #1 Spouse #2 FW, $
A A 1,707,404
