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Engineering Economic Analysis

This chapter explores replacement analysis and economic life in engineering and economic analysis. It covers topics such as physical impairment, altered requirements, technology obsolescence, financing considerations, and the process for deciding whether to replace or retain an asset. The chapter also discusses the different types of lives associated with replacement analysis and the factors and assumptions involved. It concludes with examples and calculations to demonstrate the application of replacement analysis in practice.

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Engineering Economic Analysis

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  1. Engineering Economic Analysis Chapter 13  Replacement Analysis rd

  2. … from the Deacon/s Masterpiece by Oliver Wendell Holmes … Now in building of chaises, I tell you what,- There is always somewhere a weakest spot,—- In hub, tire, felloe, in spring or thill,- In panel, or crossbar, or floor, or sill,- In screw, bolt, thoroughbrace,—lurking still,- Find it somewhere, you must and will,—- Above or below, or within or without,—- And that's the reason, beyond a doubt,- A chaise breaks down, but doesn't wear out. You see, of course, if you 're not a dunce,How it went to pieces all at once,--All at once, and nothing first,--Just as bubbles do when they burst.- rd

  3. Reasons for Replacement Analysis • Physical Impairment (Deterioration) • Altered Requirements (Obsolescence) • Technology (Obsolescence • Financing (taxes, lease versus buy) • Replacement study is designed to decide to retain or to replace now. • If to replace, then done. If not, then revisit annually. rd

  4. Types of Lives • Economic life – resulting in minimum equivalent uniform annual cost (EUAC) • Ownership life – date of acquisition and date of abandonment or replacement • Physical life – original acquisition to disposal • Useful life – time an asset is kept in productive services rd

  5. Replacement Factors • Recognition and acceptance of past errors unable to see future, bad estimates • Sunk costs (forget the past, and the book value) • Use existing asset value and outsider viewpoint • Economic life of best challenge • Remaining economic life of defender • Income tax considerations • (sunk cost relevant here) rd

  6. Replacement Assumptions • Infinite Horizon (repeatability) • No technological change • Do Nothing option is not feasible • (or Keep defender is the do nothing option) • Economic life of asset is the age at which costs are minimized. rd

  7. Identify Participants of Comparison Defender Marginal Cost Data Defender Marginal Cost Increasing Find lowest EUAC for Defender Find EUAC over Given Life no yes AT# 2 Compare lowest D-EUAC with C-EUAC at min cost life AT# 1 Compare next year marginal cost with C-EUAC Lacking defender data, compare D-EUAC over life with C-EUAC at minimum cost life Defender Best Challenger Available Not available rd

  8. Annualize the Costs • Find the annual cost for the cash flow below. • n 0 1 2 3 cf -15M -500K -825K 7.68M & -1.36M $7.68M $500K 825K $15M 1.36M rd

  9. Economic Life • You bought a car for $3K. Determine economic life at 12%. • Year 1 2 3 4Annual Expenses 950 1050 1150 1550Market Value 2250 1800 1450 1160 • Loss in Market Value 750 450 350 290Loss in interest 360 270 216 174Total Marginal costs 2060 1770 1716 2014 • Find EUAC of Total Marginal Costs. • (EUAC '(2060 1770 1716 2014) 12) rd

  10. Problem 13-12 • First cost = $1050K; Salvage value = $225K at any time; • O&M = $235K with $75K gradient. MARR = 10%. Find economic service life. • n CR O&M CR+O&M EUAC • 930 235 1165 1165 • 497.86 310 807.86 994.93 • 354.24 385 739.24 917.68 • 282.76 460 742.76 880.00 • 5 240.13 535 775.13 862.82 • 211.93 610 821.93 857.52 *** • 7 191.96 685 876.96 859.57 rd

  11. Example Defender has AOC of $60K per year, a life of 5 years with zero salvage. Present market value is $30K. Challenger costs $120K with AOC of $30K and a salvage after 5 years of $50K. Use before tax MARR of 20%. AT#3 PWD(20%) = -30K – 60K(P/A, 20%, 5) = -$209,436.73 PWC(20%) = -120K - 30K(P/A, 20%, 5) + 50K(P/F, 20%, 5) = -$189,624.49 Choose the challenger. rd

  12. Example • Defender bought 3 years ago for $120K with expected life of 10 years and $25K salvage value with AOC of $30K. Current book value is $80K and current life is 3 years. • Challenger available for $100K with trade-in of $70K for defender. Useful life is 10 years, salvage value is $20K and ASOC is $20K. Market value for defender is $70K. • Defender ChallengerFirst cost $70K $100KAOC 30K 20KSalvage 10K 20KLife (years) 3 10 rd

  13. Example First cost (sunk 5 years) $12,500Depreciation Straight Line (salvage = $2,500)Market Value $8,000Useful Life 10 yearsAnnual Operating Cost $3,000Annual Benefit $4,500Market Value after life $1,500Combined tax rate 34% n BTCF Dep TI Tax Rate (34%) ATCF0 8K 7500 500 -170 $7830 (sell)0 -8K - -500 170 -$7830 (keep) 5 1500 1500 -510 990 Cost Basis = 12,500 - 5 * (12,500 –2500) / 10 = $7500 $8000 – 7500 = $500 (depreciation recapture) rd

  14. Example 13-1 $7500(A/P) $900(A/G) $500+$400(A/G, 8%, n) n CRWR Maint Operating Total EUAC rd

  15. Example 13-1 rd

  16. Marginal Costs Challenger First Cost = $25K; Annual O&M = $2K with -$500 gradient; Annual Risk $5K for 3 yrs; then -1500 GradientMarket value ($18 13 9 6 4 3 2.5)K from 1 to 7 years; MARR = 15% n Lost Market Lost Interest O&M Risk Total MC EUAC1 $7K 0.15*25K=$3750 $2.0K $5.0K $17,750 17,750 2 5K 2700 2.5K 5.0K 15,200 16,5643 4K 1950 3.0K 5.0K 13,950 15,8114 3K 1350 3.5K 6.5K 14,350 15,518 5 2K 900 4.0K 8.0K 14,900 15,427 *** 6 1K 600 4.5K 9.5K 15,600 15,4477 0.5K 450 5.0K 11.0K 16,950 15,582 (AGP (List-pgf '(0 17750 15200 13950 14350 14900 15600 16950) 15) 15 7) $15,582.46 (EUAC cash-flow 15) rd

  17. Example 13-3 Defender's Market value= $15K; Annual O&M = $10K with a $1500 gradient; Market value ($14 13 12 11 10)K from 1 to 5 years; MARR = 15% n Lost Market Lost Interest O&M Total Marginal Cost1 $1K 0.15*15K=$2250 $10.0K $13,2502 1K 2100 11.5K 14,6003 1K 1950 13.0K 15,950 > $15,427 ** 4 1K 1800 14.5K 17,3005 1K 1650 16K 18,650 Note that the marginal costs are increasing each year => AT #1 rd

  18. Problem 13-4 • Maintenance costs are expected to be higher. Fully depreciated and other factors irrelevant. Cost this year is expected to be $800. • AT#1 one year, rd

  19. Problem 13-12 • First cost = $1050K; Salvage value = $225K at any time • O&M = $235K with $75K gradient. MARR = 10%. Find economic service life. • n CR O&M EO&M EUAC • 1 $930K 235K 235K 1165K • 2 497.857K 310K 270.71K 768.567K • 3 354.244K 385K 305.24K 659K • 4 282.763K 460K 338.59K 621.354K • 5 240.133K 535K 370.76K 610.893K *** • 211.926K 610K 401.77K 613.696K • CR = (P – S)(A/P, i%, n) + Si rd 12/19/2019 rd 19

  20. Big-J Construction Problem 13-21 MARR = 20% rd

  21. Problem 13-22 • MARR = 10% • Year D-MC D-EUAC C-EUAC • 1 $2500 $2500.00 $4500 • 2 2400 2452.38 3600 • 3 2300 2406.34 *** 3000 • 4 2550 2437.302600 *** • 5 2900 2513.09 2700 • 6 3400 2628.04 3500 • 7 4000 2772.65 4000 • What is the lowest EUAC of the defender? $2406.34 • What is minimum cost life of challenger? $2600 • c) When should defender be replaced with challenger? Never rd

  22. Problem 13-23 • n MCD-D EUAC-C MARR = 10% • $3000 $45002 3150 40003 3400 33004 3800 41005 4250 44006 4950 6000 • (EUAC '(3000 3150 3400 3800 4250 4950) 10)  • 0 3000 3071.43 3170.69 3306.29 3460.87 3653.87 • Shouldn't replace Defender at all, but had year 1 for D • > 3300 etc., replace then/now. rd

  23. Problem 13-25 • Old forklift MARR = 8% AT#1n 1 2 3 4 5-10Maint Cost 400 600 800 1000 1400/yearYear BTCF Depc TI Tax 40% ATCF 1 -400 0 -400 $160 -$240 • Challenger • Year BTCF Depc TI Tax 40% ATCF0 -6500 -$6500 1-10 -50 650 -700 $280 230 • EUAC = 6500(A/P, 8%, 10) – 230 = $968.69 – $230 • = $738.69 => keep old forklift for another year. rd

  24. Example • The economic service life given current market value of asset is $15K and expected cash flows shown below and MARR at 10% is • Year Salvage Value O&M Cost • $10K $50K • 2 8K 53K • 3 5K 60K • 4 0 K 68K • a) i year b) 2 years c) 3 years d) 4 years • EUAC1 = (15 – 10)K(A/P, 10%, 1) + 10K*0.1 + 50K = $56.50K • EUAC2 = (15 – 8)K (A/P, 10%, 2) + 8K*0.1 + 51.53K = 56.36K *** • EUAC3 = (15 – 5)K (A/P, 10%, 3) + 5K*0.1 + 54.02K = 58.54K rd

  25. Example • When should the defender be replaced? • a) now b) 1 year from now c) 2 years from now d) 3 years from now rd

  26. Economic Life • Asset bought at $8K at 8% cost of money. • n 1 2 3 4 5 6Aexpense 3K 3K 3.5K 4K 4.5K 5.52KMV 4700 3200 2200 1450 950 600 • Loss in MV 3300 1500 1000 750 500 350Loss in Int 640 376 256 176 116 76Total Margin 6940 4876 4756 4928 5116 5946 • (EUAC '(6940 4876 4756 4928 5116 5946) 8) • $5381.27 5 years rd

  27. PW of ATCF • Present Worth of After Tax data, find economic life of eachand when to replace defender.Year Defender Challenger • 1 14020 18630 2 28100 34575 3 43075 48130 4 -- 65320 5 -- 77910 • (mapcar #' AGP '(18630 34575 48130 65320 77910) • (list-of 5 12) (upto 5))  • (20865.60 20457.96 20038.88 21505.59 21613) • (mapcar #' AGP '(14020 28100 43075) (list-of 3 12)(upto 3)) • 15702.40 16626.72 17934.23 rd

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