1. Engineering Economic Analysis
Chapter 8
Choosing the Best Alternative
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 Incremental Analysis Graphical Solutions
Elements in Comparing Mutually Exclusive Alternatives
Doing a Set of Incremental Challenger-Defender Comparisons
Choosing an Analysis Method
Copyright Oxford University Press 2017

3. Learning Objectives
Use graphical technique to choose between mutually exclusive alternatives
Define incremental analysis
Use spreadsheets in incremental analysis
Copyright Oxford University Press 2017

4. Vignette: Selecting the Best Pavement
Rigid concrete vs. flexible asphalt
What initial cost and what rehabilitation interval & cost?
Asphalt often smoothest & quietest ride with greatest satisfaction for motorists
Asphalt subject to rutting & shoving
Concrete with design lives of 20 to 40 years; asphalt with lives of 10 years
More pavement & overlay choices now available
Copyright Oxford University Press 2017

5. Vignette: Selecting the Best Pavement
Should transportation agencies be indifferent about rigid vs. flexible pavement? Why?
What engineering economic principles would you apply to choose pavement type? What is most effective measure for public forum?
How can pavement life cycle be used in choosing pavement type? How does initial vs. life-cycle costs affect choice?
What uncertainties are likely in estimating initial cost and rehabilitation interval & cost?
If federal money is significant share of initial cost, how might this distort decision making by state agency responsible for rehabilitation costs?
Copyright Oxford University Press 2017

6. What does “Mutually Exclusive“ mean when there are multiple alternatives?
You must do all
You can do as many as you have money for
You can do at most one
They are expensive
I don’t know
Copyright Oxford University Press 2017

7. What does “Mutually Exclusive“ mean when there are multiple alternatives?
You must do all
You can do as many as you have money for
You can do at most one
They are expensive
I don’t know
Copyright Oxford University Press 2017

8. Choosing Best of Mutually Exclusive Alternatives
Identify all alternatives
Build NPW, EUAW, or EAC graph
Show all alternatives
Plot on the same axes
Line of best values  which alternative is best over each range
Determine intersection points
Create choice table
Copyright Oxford University Press 2017

9. Example 8-1 Graphical Solution
High
Low
Cost
$13,400
$10,310
Capacity
150
100
Benefit
$4000/year
$3300/year
Life
5 years
Highest PW is best
Choice table
0 – 4.3% dashed high capacity line
4.3 – 18% solid low capacity line
Above 18% PW = 0 for do nothing best
Copyright Oxford University Press 2017

10. Example 8-1 5-Button for Line Intersections & Incremental Analysis
IRR’s for low & high have PW = 0
Intersection of low & high
Increment = spend $3090 more now for $700 more annual income
Where low & high curves intersect PWhigh − low = 0
For incremental analysis at given interest rate
Pairwise challenger-defender comparisons are made
Copyright Oxford University Press 2017

11. Example 8-2 Graphical Solution
Machine X
Machine Y
Initial cost
$200
$700
Annual benefit
$95
$120
Salvage value
$50
$150
Life (years) 6 12
Machine X has better PW
Except i < 1.32%
1.32% found using IRR & 12-yr period
Choice table
0 – 1.32% choose Y
> 1.32% choose X
Copyright Oxford University Press 2017

12. Example 8-3 Graphical Solution like 8-2 but EUAW instead of PW
Machine X
Machine Y
Initial cost
$200
$700
Annual benefit
$95
$120
Salvage value
$50
$150
Life (years) 6 12
X’s EUAW better
Except i < 1.32%
Then machine Y
EUAW curves closer to linear than for PW
Copyright Oxford University Press 2017

13. Example 8-3 5-Button for Intersection & Incremental Analysis
Use GOAL SEEK to find interest rate with 0 difference in EUAWs
Cell B13 contains =B10 so interest rate same for both EUAWs
Copyright Oxford University Press 2017

14. Example 8-4 Graphical Solution
Brass
Stainless
Titanium
Cost
$100,000
$175,000
$300,000
Life (years) 4 10 25
Lowest EUAC is best
Choice table
0 – 6.3% titanium
6.3 – 15.3% stainless
> 15.3% brass
Copyright Oxford University Press 2017

15. Example 8-4 5-Button for Intersection & Incremental Analysis
B3 to B6 =B$2 and B9 to B12 =B$8
Use GOAL SEEK to change B2 so that H5 = 0
Use GOAL SEEK to change B8 so that H12 = 0
Copyright Oxford University Press 2017

16. Example 8-5 Graphical Solution
From graph want rates at intersections for Y & Z, Z & X, X & do nothing
Machine X
Machine Y
Machine Z
Initial cost
$2000
$7000
$4250
Uniform annual benefit
650
1100
1000
Useful life, in years 6 12 8
Copyright Oxford University Press 2017

17. Example 8-5 5-Button for Intersections & Incremental Analysis
B3 to B6 =B$2 and B10 to B13 =B$9
Use GOAL SEEK to change B2 so that J5 = 0
Use GOAL SEEK to change B8 so that J13 = 0
Copyright Oxford University Press 2017

18. Incremental Analysis Can be used with IRR, PW, EUAC, & EUAW
[Higher cost alternative] = [Lower cost alternative] + [Increment between]
“Defender” is best alternative identified so far; “challenger” is next higher cost alternative to evaluate
N mutually exclusive alternatives
Need (N-1) “challenger/defender” comparisons
From [N(N−1)/2] possible comparisons
Copyright Oxford University Press 2017

19. Example 8-7 Incremental Analysis using Pair-wise ComparisonsB C D E
Initial Cost
$4000
$2000
$6000
$1000
$9000
Uniform Annual Benefit
639
410
761
117
785
Rearrange the alternatives in order of increasing cost D B A C E
Initial Cost
$1000
$2000
$4000
$6000
$9000
Uniform Annual Benefit
117
410
639
761
785
Calculate IRR of least expensive alternative to see if better than “Do Nothing” at MARR of 10%
PWD = 0 = -$ $117 (P/A, IRRD, 20)
IRRD = 9.94% < 10%
“Do Nothing” is preferred & is still “Defender.”
Copyright Oxford University Press 2017

20. Example 8-7 Incremental AnalysisD B A C E
Initial Cost
$1000
$2000
$4000
$6000
$9000
Uniform Annual Benefit
117
410
639
761
785
Calculate ∆IRR of next alternative, B, over “Do Nothing” alternative
PWB = 0 = -$ $410 (P/A, IRRB, 20)
IRRB = 19.96% > 10%
“Alternative B” is preferred & is the new “Defender.”
Calculate ∆IRR of next alternative, A, over defender B.
PWA-B = 0 = -($4000-$2000)+( )(P/A, IRRA-B,20)
IRRA-B = 9.63% < 10%
“Alternative B” is preferred & is still the “Defender.”
Copyright Oxford University Press 2017

21. Example 8-7 Incremental AnalysisD B A C E
Initial Cost
$1000
$2000
$4000
$6000
$9000
Uniform Annual Benefit
117
410
639
761
785
Calculate ∆IRR of next alternative, C, over defender B.
PWC-B = 0 = -($6000-$2000)+( )(P/A, IRRC-B,20)
IRRC-B = 6.08% < 10%
“Alternative B” is preferred & is still the “Defender.”
Calculate ∆IRR of next alternative, E, over defender B.
PWE-B = 0 = -($9000-$2000)+( )(P/A, IRRE-B,20)
IRRE-B = 0.67% < 10%
“Alternative B” is still preferred & is final selection.
Copyright Oxford University Press 2017

22. Choosing an Analysis Method
If MARR known
PW or equivalent annual easier than incremental IRR
If MARR not known or known approximately
Graphical approach to choice table often enough
Spreadsheets for intersections that define limits
Graphical approach shows when alternatives have close equivalent values
 pick either
Rate of return analysis often easier to explain
Many firms have preferred measure use it
Copyright Oxford University Press 2017