1. Engineering Economic Analysis
Chapter 9
Other Analysis Techniques
Donald G. Newnan
San Jose State University
Ted G. Eschenbach
University of Alaska Anchorage
Jerome P. Lavelle
North Carolina State University
Neal A. Lewis
University of New Haven
Copyright Oxford University Press 2017

2. Chapter Outline Future Worth Analysis Benefit-Cost Ratio Analysis
Payback Period
Sensitivity & Breakeven Analysis
Graphing with Spreadsheets for Sensitivity & Breakeven Analysis
Doing What-If Analysis with Spreadsheets
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3. Learning Objectives
Apply future worth, benefit-cost ratio, payback period, & sensitivity analysis methods
Link future worth analysis to present worth & annual worth methods
Develop the benefit-cost ratio
Understand the concept of “payback period”
Conduct sensitivity & breakeven analyses
Use spreadsheets for sensitivity & breakeven analyses
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4. Vignette: Clean & Green
Buildings in U.S. account for
75% of electricity consumption
45% of energy use
40% of CO2 emissions
40% of raw materials use
13% of potable water consumption
Green Building Council (2016)
Designers & engineers have been developing environment-friendly construction materials & building techniques
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5. Vignette: Clean & Green
Advantages of “Green” buildings
Improve workers’ productivity
Reduce health & safety costs
Improve indoor environmental quality
Reduce energy & maintenance costs
Myth
Cost more to build
Harder to recoup investment in a timely manner
Increasing by 10% annually
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6. Future Worth Analysis
Alternatives compared using future worth at given time
Emphasizes final outcome
Firm’s final wealth
Retirement savings
Easily converted to/from PW, EAW, & EAC
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7. Example 9-1 Future Worth Calculation
A 20-year-old quits $35/week smoking habit. What in savings account paying 5% compounded semiannually at age 65?
Work-week habit costs $7/day. What if $2/day alternative substituted 4 days/week? (Happy Money says more happiness from weekly treat than daily habit)
Stop: Semiannual saving = $35 𝗑 26 = $910
Habit to treat: $5 saved on 4 days = $20/wk = $520 every 6 mo.
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8. Example 9-2 Future Worth Analysis
New Plant
Year
Remodel
$85,000
$850,000
200,000 1
250,000
1,200,000 2 3
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9. Benefit-Cost Ratio Analysis
At given MARR, an acceptable alternative satisfies
NPW = PW(Benefits) – PW(Costs) ≥ 0
EUAW = EUAB – EUAC ≥ 0
As a ratio:
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10. Benefit/Cost Ratio Example for Road
Benefits = + outcomes for public
Faster, safer travel
PW = $3.8M
Disbenefits = − outcomes for public
Traffic delays during construction
PW = −$0.9M
Costs = paid by government
Building & maintaining roads
PW = −$2M
B/C ratio = (3.8M − 0.9M)/2M = 1.45
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11. Benefit-Cost Ratio Analysis
Situation
Criterion
Neither input nor output fixed
One alternative: B/C ≥ 1
Two or more alternatives:
Incremental analysis of B/C ratios
Fixed input; constant C
Maximize B/C; highest B
Fixed output; constant B
Maximize B/C; lowest C
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12. Example 9-3 Benefit-Cost Ratio Analysis
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13. Example 9-4 Benefit-Cost Ratio AnalysisD E F
Initial Cost
$4000
$2000
$6000
$1000
$9000
$10,000
PW of Benefit
7330
4700
8730
1340
9000
9500
B/C
1.83
2.35
1.46
1.34
1.00
0.95
Rearrange the alternatives in order of increasing investment D B A C E
Initial Cost
$1000
$2000
$4000
$6000
$9000
PW of Benefit
1340
4700
7330
8730
9000
B/C
1.34
2.35
1.83
1.46
1.00
Calculate ΔB/ΔC of the incremental investments, if ΔB/ΔC≥1, choose the higher-cost alternative  do A
B-D
A-B
C-A
E-A
Initial Cost
$2000 – 1000
$4000 – 2000
$6000 – 4000
$9000 – 4000
PW Benefit
4700 – 1340
7330 – 4700
8730 – 7330
9000 – 7330
ΔB/ ΔC
3.36
1.32
0.70
0.33
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14. New Public Swimming Pool
Pool costs $1.2M to build & $200K/year to operate. Benefit is $450K/year. Using equivalent annual values at i =6% & n =25 years, what is B/C ratio?
1.53
1.81
2.25
4.79
I don’t know
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15. New Public Swimming Pool
Pool costs $1.2M to build & $200K/year to operate. Benefit is $450K/year. Using equivalent annual values at i =6% & n =25 years, what is B/C ratio?
1.53
1.81
2.25
4.79
I don’t know
EAC costs = 1.2M(A/P,6%,25) + 200K
= 93.9K + 200K = $293.9K
EAW benefits = 450K
B/C ratio = 450K / 293.9K = 1.53
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16. Graphical Representation of Benefit-Cost Ratio Analysis
Incremental B/C = 1 is 45⁰ line
Do A since
B/CD > 1
B/CB – D > 1
B/CA – B > 1
B/CC – A & B/CE – A < 1
Not B largest B/C
Not E largest project with B/C > 1
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17. Variations on Benefit-Cost Ratio
Government B/C ratio in Public Sector:
Present Worth Index in Private Sector:
All B/C ratios use same criterion: B/C ≥ 1
Various B/C ratios may differ in value, but they provide consistent recommendations
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18. Example 9-5 Government B/C Ratio Analysis
Right turn lanes
Left turn lanes
Incremental
Initial cost
$8.9M
$3M
Annual maintenance
150K
75K
Uniform annual benefit
1.6M
$2.2M
Construction disbenefit
900K
2.1M
Useful life, in years 15
i=10%
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19. Example 9-6 Present Worth Index (Cost ≡ Initial
Minimal
Total
Incremental
Initial cost
$8.9M
$3M
Annual maintenance
150K
75K
Uniform annual benefit
1.6M
$2.2M
Disbenefit
900K
2.1M
Useful life, in years 15
i=10%
Ratios differ from Exp. 9-5 because maintenance cost in numerator.
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20. Payback Period
Time required for project’s profit & benefits to equal project’s cost
Approximate economic analysis method
Ignores time value of money
Ignores all cash flows after payback
May not be consistent with equivalent worth & rate of return methods
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21. When Can Payback be Used
If payback time measured in months & benefits continue for years
Most common in start-up enterprises which are VERY short of capital
Most projects with good payback periods look good on other measures
Not always, so payback is unreliable
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22. Example 9-7 Payback Period
Year 1 2 3 4 5 A
−$1000
+200
+1200 B
−$2783
Payback Period
=2.33 years
Payback Period
=2.5 years
Alternative A
Alternative B
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23. Example 9-8 Payback Period
Product: n = 5, A = −4000
Machine A: P = −$10, S = $0
Machine B: P = −$15, S = $9000
Payback A = $10,000/($4000/yr) = 2.5 years
Payback B = $15,000/($4000/yr) = 3.75 years
Minimizing payback Machine A
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24. Example 9-8 TVM solution
Incremental IRR = 12.5%  B better at rates below 12.5%
Above i = 12.5% A better
At 15%, incremental PW = −$525  A better
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25. Payback: A new computer systems costs $20,000
Savings each year:
Year Savings
1 $8,000
2 $6,000
3 $4,000
4 $4,000
5 $3,000
Payback period is:
2.6 years
3.0 years
3.5 years
3.6 years
4.0 years
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26. Payback: A new computer systems costs $20,000
Savings each year:
Year Savings
1 $8,000
2 $6,000
3 $4,000
4 $4,000
5 $3,000
Payback period is:
2.6 years
3.0 years
3.5 years
3.6 years
4.0 years
After 3 years, savings total $18,000.
$2000 is needed to payback, out of $4000 received in year 4.
Payback = /4000 = 3.5 years.
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27. Discounted Payback Period
Adds time value of money to payback
Computes interest on cumulative investment or project balance
Still ignores cash flows after payback
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28. Example 9-9 Discounted Payback Period
Alternatives
Cost
Uniform Annual Benefit A
$10,000
$3000 B
$12,000
$3500
i=10%
A preferred because shorter discounted payback period
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29. Sensitivity & Breakeven Analysis
Sensitivity Analysis:
Projected expenditures & returns  which decision
To what extent do uncertainties/variations in data affect decision?
Breakeven Analysis
Form of sensitivity analysis changing 1 variable
When are 2 alternatives economically equivalent?
Common example is project life since often one of most uncertain values
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30. Example 9-10 Breakeven Analysis
Construction Cost
Full-capacity
$140,000
Two-stage
$100,000 now +
$120,000 n years from now
Breakeven n = years
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31. Example 9-11 Breakeven Analysis
Right turn lanes
Left turn lanes
Incremental
Initial cost
$8.9M
$3M
Uniform annual benefit
1.6M
$2.2M
Annual maintenance
$150K
75K
Disbenefit
$900K
2.1M
Useful life, in years 15
i=10%
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32. Example 9-12 Automatic graph from spreadsheet Construction Cost
Full-capacity
$140,000
Two-stage
$100,000 now +
$120,000 n years from now
Automatic graph
from spreadsheet
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33. Example 9-13 What-if Analysis Can Change Multiple Variables
Initial Estimate
Adjustment
First cost
$70,000
+10%
Units / year
1200
−20%
Net unit revenue
$25
−15%
Life, in years 8 −3
Interest rate
12%
None
Initial B/C = 2.13
but only 0.96 with adjustments
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