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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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of McGraw-Hill Education.
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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.
