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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Example 8-7 Incremental Analysis using Pair-wise Comparisons B 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 = -$1000 + $117 (P/A, IRRD, 20) IRRD = 9.94% < 10% “Do Nothing” is preferred & is still “Defender.” Copyright Oxford University Press 2017

Example 8-7 Incremental Analysis D 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 = -$2000 + $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)+(639-410)(P/A, IRRA-B,20) IRRA-B = 9.63% < 10% “Alternative B” is preferred & is still the “Defender.” Copyright Oxford University Press 2017

Example 8-7 Incremental Analysis D 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)+(761-410)(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)+(785-410)(P/A, IRRE-B,20) IRRE-B = 0.67% < 10% “Alternative B” is still preferred & is final selection. Copyright Oxford University Press 2017

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