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Engineering Economic Analysis

Engineering Economic Analysis. Chapter 8  Incremental Analysis. Y. X. $15. $10. Benefit-Cost Graph. Year X Y Y - X 0 -$10 -$20 $-10 1 15 28 13 MARR = 6% Y is preferred at RoR 40% over X at 50% Increment earns at 30%. Rejection. $20.

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Engineering Economic Analysis

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  1. Engineering Economic Analysis Chapter 8  Incremental Analysis rd

  2. Y X $15 $10 Benefit-Cost Graph Year X Y Y - X 0 -$10 -$20 $-10 1 15 28 13 MARR = 6% Y is preferred at RoR 40% over X at 50% Increment earns at 30% Rejection $20 rd

  3. Benefit-Cost Graph • MARR = 6% A B C 20-yearsFirst cost $2000 $4000 $5000UAB 410 639 700PWBenefits 4703 7329 8029 8029 C 7329 B i=6% NPW=0 4703 A 2000 4000 5000 rd

  4. Incremental Analysis • A B C MARR = 10% • First cost $18K $25K $15K Life 25 yearsUAB 1055 2125 1020 Salvage ~ 0IRR (%) 7 9 8 • IRRA-C = (UIRR 3000 35 25 0) -7.86% C > A • IRRB-C = (UIRR 10E3 1105 25 0) 10.04% B > C rd

  5. Incremental Analysis MARR = 8% A B C First Cost 1000 2000 3000UA Benefits 150 150 0 Salvage value 1000 2700 5600Life 5 6 7IRR 15% 11.81% 9.33% Take in order of increasing first cost: A > 15% RoRB-A(IRR ‘(-1000 0 0 0 0 –1000 2850)) 9.8% => B > A RoRC-B(IRR –1000 –150 -150 -150 –150 -150 -2850 5600)) returns 6.75% => B is best. rd

  6. Incremental RoR Analysis • A B C D 5-year life • First cost 100 130 200 330Annual income 100 90.78 160 164.55Annual cost 73.62 52.00 112.52 73.00 • IRR (%) 10 15 6 12 • B – A ~ (UIRR 30 12.4 5 0) 30.35% • C – B ~ (UIRR 70 8.7 5 0) - 14% • D – B ~ (UIRR 200 52.77 5 0)  10% • If MARR > 15% Do Nothing 15% > MARR > 0% Select B rd

  7. Problem 8-27 • A  (-1300 100 130 160 190 220 250 280 310 340 370) • B  (-1300 10 60 110 160 210 260 310 360 410 460) • B – A  (mapcar #'- b a))  (0 -90 -70 -50 -30 -10 10 30 50 70 90) • (sum *)  0 => 0% => a is better than b for any positive rate • Check I: (cum+ '(0 -90 -70 -50 -30 -10 10 30 50 70 90)) • (0 -90 -160 -210 -240 -250 -240 -210 -160 -90 0)=> no positive rate of return for the B – A increment exists • Check II: (list-pgf a 8) --> 150.31; (list-pgf b 8) --> 65.94 • (mapcar #'- a c)  A – C  (0 -160 -130 -100 -70 -40 -10 20 50 80 110) • (cum-add *)  (0 -160 -290 -390 -460 -500 -510 -490 -440 -360 -250) There is no positive rate of return for which A is better than C => Reject A • Check: (list-pgf c 8)  $444.62 > $ 150.31 • (mapcar #' - c d)  (0 -190 -140 -90 -40 10 60 110 160 210 260) • (cum-add *)  (0 -190 -330 -420 -460 -450 -390 -280 -120 90 350) => UIRR(IRR '(-190 -140 -90 -40 10 60 110 160 210 260) 0.95)  8.97% > 8% • => C is better than D and is best. Check: (list-pgf d 8) --> 420.69 rd

  8. Problem 8-2 XYX - YFirst Cost -100 -50 -50UAB 31.5 16.5 15Life (years) 4 4 4 RoR 9.93% 12.11% 7.71% Which is better if a) MARR = 6%? Xb) MARR = 9% Yc) MARR = 10% Yd) MARR = 14% Do Nothing rd

  9. Problem 8-3 A B B-AFirst Cost -100 -150 -50 UAB 30 43 13Life (years) 5 5 IRR (%) 15.24 13.34 9.43 (UIRR 100 30 5 0)  15.24% for A(UIRR 150 43 5 0)  13.34% for B(UIRR 50 13 5 0)  9.43% for B – AWhich is better if a) MARR = 6%? Bb) MARR = 8%? B c) MARR = 10% A d) MARR = 16% Do Nothing rd

  10. Example 8-6 • MARR = 6% A B C D E • First Cost 4K 2K 6K 1K 9KUAB 639 410 761 117 785Life 20 20 20 20 20 • (UIRR 1 0.117 20) 9.84 > 6% D is better than MARR • (UIRR 1 0.293 20)  29.12% => B > D • (UIRR 2 0.229 20)  9.62% => A > B • (UIRR 2 0.122 20)  1.97% => A > C • (UIRR 5 0.146 20)  -4.65% => A > E Choose A rd

  11. Problem 8-8 (UIRR 200 70 5 50) 26.05% => Select Neutralization rd

  12. Problem 8-? • MARR = 6% A B C 20-year analysis • First cost $10K $15K $20KUAB 1625 1625 1890Life 10 20 20 • (UIRR 10e3 1625 10) 9.96% A • (UIRR 15e3 1625 20) 8.84% B • (UIRR 20e3 1890 20) 7.01% C • (UIRR 5e3 0 10 10e3)  7.18% B – A 10-year(UIRR 5e3 265 20 0) 0.56%C – B 20-year rd

  13. Problem 8-14 • MARR = 8% A B C no replacementFirst cost 1000 2000 3000UAB 150 150 0Salvage 1000 2700 5600Life (yrs) 5 6 7 RoR 15% 11.83% 13.3%(IRR '(-1000 0 0 0 0 -1000 2850))  9.8% B > A (IRR '(-1000 -150 -150 -150 -150 -150 -2850 5600)) • 6.75% for C – B < 8% => Reject C; • Conclude B is best. rd

  14. Problem 8-15 Year X Y Y- X0 -10 -20 -101 15 28 13IRR (%) 50 40 30 Over what range of MARR is Y preferred over X? Y is better for MARR < 30%X is better for 30% < MARR < 50%Do Nothing for MARR > 50%. rd

  15. Problem 8-19 • Replace B and C when needed. Use MARR = 8% A B C • First cost $100 $150 $200UAB 10 17.62 55.48Life (years) ∞ 20 5 • Capitalized Costs Analysis NPWA = 10/0.08 – 100 = $25 • NAWB = 17.62 -150(A/P, 8%, 20) = $2.34 perpetuity NPWB = 2.34/0.08 = $29.28 NAWC = 55.48 -200(A/P, 8%, 5) = $5.39 or perpetuity NPWC = 5.39/0.08 = $67.36 *** C rd

  16. Problem 8-21 MARR = 12% • n 0 1 2 3 4 A $-20K 10K 5K 10K 6K B -20K 10K 10K 10K 0C -20K 5K 5K 5K 15K • (IRR '(-20 10 5 10 6)) 21.35% A • (IRR '(-20 10 10 10 0)) 23.38% B • (IRR '(-20 5 5 5 15)) 14.98% C • (IRR '(-5 0 6))  9.54% A – B Reject A • (IRR '(-5 -5 -5 15))  0.0% C – B Reject CChoose B rd

  17. Problem 8-25 • A B C D First Cost 100K 130K 200K 330KUAB 26.38K 38.78K 47.48K 91.55KLife 5 5 5 5 • RoR 10% 15% 6% 12% • At a MARR of 8%, which to choose? Reject C & Do Nothing • B dominates A as its return is greater for a larger investment. • D – B => (UIRR 200 52.55 5)  9.84% => D is best. • RoRD-B= 9.84% > 8% MARR => Select D. rd

  18. Problem 8-32 • Option A $30,976 tax free retirement annuity • Option B $359.60/month for test of life or 20 years • Option C $513.80/month for next 10 years What to d • 12 * 359.6 = $4315.20, 12 * 513.80 = 6165.60 • B – A (UIRR 30976 4315.20 20)  12.64% • C – A (UIRR 30976 6165.60 10)  14.97% • B – C (IRR '(-1850.4 -1850.4 -1850.4 -1850.4 -1850.4 -1850.4 -1850.4 -1850.4 -1850.4 -1850.4 4315.2 4315.2 4315.2 4315.2 4315.2 4315.2 4315.2 4315.2 4315.2 4315.2) 0.8)  8.836% • MARR < 8.836% Choose B8.836 < MARR < 14.9% Choose C at 9%14.9% < MARR < i% Choose A 30976(A/P, i%, 20) rd

  19. Problem 8-27 • MARR = 6% A B C D First Cost, $ 2K 5K 4K 3KAnn-Benefits 800 500 400 1300 Salvage 2K 1.5K 1.4K 3KLife 5 6 7 4 RoR 40% -2.4% 1% 43.3% Reject B & C • (UIRR 2 0.8 5 2)  40%; (UIRR 5 0.5 1.5 6 1.5)  -2.38% • (UIRR 4 0.4 7 1.4)  0.98%; (UIRR 3 1.3 4 3)  43.33% • (IRR '(-1000 500 500 500 3500 -2800)) 51.9% D - A • (cum-add '(-1000 500 500 500 3500 -2800))  • (-1000 -500 0 500 4000 1200) => unique positive • RoR • 2800(P/F, 6%, 1) = $2641.51 (IRR '(-1000 500 500 500 858.49 0))  41.1% rd

  20. Problem 8-29 • MARR = 8% Atlas ZippyFirst cost $6700 $16,900AO&M cost 1500 1200UAB 4000 4500Salvage 1000 3500Life (years) 3 6 • (UIRR 6700 2500 3 1000) 12.134% for Atlas(UIRR 16900 3300 6 3500)  8.983% for Zippy • Atlas cf: -6700 2500 2500 -3200 2500 2500 3500Zippy cf: -16900 3300 3300 3300 3300 3300 6800(IRR '(-10200 800 800 6500 800 800 3300))  6.802% Select Atlas rd

  21. Problem 8-33 • A B B-A C First Cost $100K $300K 200K $500K Annual Benefit 30K 66K 33K 80KProfit Rate (%) 30% 22% 18% 16% • MARR = 20% thus eliminating C. • B – A cash flow is -200K 36K returns a profit rate of 18%. • Thus best to choose A as the $200K difference can be making MARR money at 20%. rd

  22. Problem 8-34 • A B C D $30K Budget • First cost $10K $18K $25K $30K MARR = 15%AB 4K 6K 7.5K 9K • AOC 2K 3K 3K 4K • A earns 2K + 0.15 * 20K = $5K / year • B earns 3K + 0.15 * 12K = $4800 / year • C earns 4.5K + 0.15 * 5K = $5200 / year *** • D earns 5K + 0.15 * 0 = $5K • Choose C rd

  23. Problem 8-35 • 24-month lease costing $267/month for $9400 car which can be bought for 24 equal monthly payments at 12% APR. Assume car salvage value is $4700. Lease or buy? • 9400(A/P, 1%, 24) = $442.49 • 4700 = (442.49 – 267)(F/A, i%, 24) • (F/A, i%, 24) = 26.78215 < 1% 0.94% => 11.28% APR • => Lease at rate above 11.28%. rd

  24. ME; MARR = 9%; Life 10 years • A B C D E • First cost $4K $5K $2K $3K $6KUAB $797 $885 $259 $447 $1063 • (UIRR 2000 259 10 0)  5% Reject • (UIRR 3000 447 10 0)  8% Reject D • (UIRR 4000 797 10 0)  15% Accept A • (UIRR 1000 88 10 0) -% => Reject B • 5. (UIRR 2K 266 10 0)  5.52% => Reject C; A is best. rd

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