978-0073523439 Chapter 9 Part 1

Engineering Economy, 8th edition

Leland Blank and Anthony Tarquin

Chapter 9

Benefit/Cost Analysis and Public Sector Economics

Understanding B/C Concepts

9.1 The primary purpose of public sector projects is to provide services for the

9.2 (a) It is best to take a specific viewpoint in determining costs, benefits and disbenefits

9.3 (a) Bridge across Ohio River – public

9.4 (a) Municipal bonds – public

9.5 A disbenefit commonly has an indirect impact on a person, business, etc., while a cost is a

9.6 (a) $600,000 annual income to area businesses from tourism created by new freshwater

reservoir/recreation area – citizen or business owner; benefit

9.7 (a) Public

(b) Public

9.8 Two advantages of a PPP are: (1) greater efficiency in the private sector, and (2) could be an

Project B/C Value

9.9 The salvage value is placed in the denominator because it is a recovery of cost, which is a

consequence to the government. The salvage value is subtracted from costs.

9.12 C = 45,000(0.025)

9.13 C = 190,000(A/P,6%,20) + 21,000

9.14 C = P(A/P,7%,50) + 200,000

9.15 C = 4000(300)(A/P,6%,10) + 3,200,000

9.16 (a) AW of B–D = 820,000 – 135,000

= $685,000

9.17 (a) Determine AW of benefits and costs

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(b) Not justified, since B/C < 1.0; but it is a required project based on noneconomic

criteria — health of the citizenry

9.18 AW of C = 6,500,000(0.08) + 130,000

= $650,000

9.19 (a) In $1000 units

AW of C = 13,000(A/P,10%,20) + 400

(b) Let P = minimum first cost allowed

9.20 AW of C = 1,000,000(A/P,6%,30)

9.21 C = P(A/P,8%,10) + 150,000

9.22 Put all cash flows in same units of $/year

AW of benefits = 3,800,000(A/P,10%,20)

9.23 Determine annual worth values

B = 100,000(0.06) + 100,000

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(b) Modified B/C ratio = (B – D + S)/C

= (106,000 – 60,000 + 90,000)/170,822

= 0.80

9.24 Convert annual benefits in years 6 through infinity to an A value in years 1 through 5.

B = (A/0.08)(A/F,8%,5)

9.25 B = 40(4,000,000) = $160 million per year

9.26 B = 41,000(33 –18) = $615,000

9.27 In $10,000 units

PW of ∆NCF = 5(P/A,10%,6) + 2(P/G,10%,6) +5(P/F,10%,6)

Two Alternative Comparison

9.28 Alternative X should be selected because the B/C ratio on the incremental cash flows

9.29 PW of cost of Retention = 880,000 + 92,000(P/A,8%,20)

Channel has higher equivalent total cost

PW of ∆C = 3,194,543 – 1,783,265

9.30 DT will have the larger equivalent total costs

9.31 PW1 of costs = 600,000 + 50,000(P/A,8%,20)

Alternative 2 has the larger total cost

∆C = PW2 – PW1 = 1,787,267 – 1,090,905

9.32 DN is an option. Rank alternatives by increasing PW of total costs: DN, 1, 2

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∆C = 1,780,000 – 840,900

= $939,100

∆B/C = 830,000/939,100

= 0.88 eliminate 2

9.33 Location 1 vs DN: B = $520,000

9.34 CN = 1,100,000(0.06) + 480,000

= $546,000

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= 1.67

Select location S

9.35 Use PW values for B and C

CConv = $200,000

9.36 Compare on a per household basis

(a) Program 1 vs. DN

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Select program 1

(b) Program 1: B/C = 1.08 (from above) Acceptable

Program 2:

= 0.83 Not acceptable

9.37 DN is not an option; compare DA vs. CS; use PW values

By spreadsheet: Select domed arena by a very slim margin with ∆B/C = 1.0003. (Note:

rows are truncated in image of spreadsheet.)

Multiple ( > 2) Alternatives

9.38 (a) Incremental analysis is needed between alternatives Z (challenger) and Y

(b) Incremental analysis is not needed for independent project; select Y and Z

9.39 Calculate total costs and rank sites; benefits are direct

Determine B/C for all sites initially

Site A: B/C = (B–D)/C

9.40 Eliminate K and M because B/C < 1.0; compare L vs. J incrementally

9.41 (a) First calculate PW of benefits for each alternative from the PW of cost and B/C ratio

values, then calculate ∆B/C ratios.