978-0073523439 Chapter 5

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