Engineering Economic Analysis Chapter 8 Incremental Analysis 10202021

Engineering Economic Analysis Chapter 8 Incremental Analysis 10/20/2021 rd 1

Benefit-Cost Graph Y $15 Year 0 1 X X -$10 15 MARR = 6% Rejection Y Y-X -$20 $-10 28 13 Y is preferred at Ro. R 40% over X at 50% Increment earns at 30% $10 10/20/2021 $20 rd 2

Benefit-Cost Graph MARR = 6% A First cost $2000 UAB 410 PWBenefits 4703 8029 7329 B $4000 639 7329 C 20 -years $5000 700 8029 C B 4703 A % i=6 =0 W NP 2000 4000 10/20/2021 5000 rd 3

Incremental Analysis First cost UAB IRR (%) A B C $18 K $25 K $15 K 1055 2125 1020 7 9 8 IRRA-C = (UIRR 3000 35 25 0) -7. 86% MARR = 10% Life 25 years Salvage ~ 0 C>A IRRB-C = (UIRR 10 E 3 1105 25 0) 10. 04% B > C 10/20/2021 rd 4

Incremental Analysis MARR = 8% First Cost UA Benefits Salvage value Life IRR A 1000 150 1000 5 15% B 2000 150 2700 6 11. 81% C 3000 0 5600 7 9. 33% Take in order of increasing first cost: A > 15% Ro. RB-A (IRR ‘(-1000 0 0 – 1000 2850)) 9. 8% => B > A Ro. RC-B (IRR – 1000 – 150 -150 – 150 -2850 5600)) returns 6. 75% => B is best. 10/20/2021 rd 5

Incremental Ro. R Analysis A B C D 5 -year life First cost 100 130 200 330 Annual income 100 90. 78 160 164. 55 Annual 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 10/20/2021 rd 6

Problem 8 -27 A (-1300 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 10/20/2021 rd 7

Problem 8 -2 X First Cost UAB Life (years) Ro. R -100 31. 5 4 9. 93% Which is better if a) MARR = 6%? b) MARR = 9% c) MARR = 10% d) MARR = 14% 10/20/2021 Y -50 16. 5 4 12. 11% X-Y -50 15 4 7. 71% X Y Y Do Nothing rd 8

Problem 8 -3 First Cost UAB Life (years) IRR (%) A -100 30 5 15. 24 B -150 43 5 13. 34 B-A -50 13 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 – A Which is better if a) MARR = 6%? B b) MARR = 8%? B c) MARR = 10% A 10/20/2021 d) MARR = 16% Do Nothing rd 9

Example 8 -6 MARR = 6% First Cost UAB Life A 4 K 639 20 B 2 K 410 20 C 6 K 761 20 D 1 K 117 20 E 9 K 785 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 10/20/2021 rd 10

Problem 8 -8 Mutually Exclusive Neutralization Precpitation First cost $700 K $500 K 40 K 110 K 175 K 125 K 5 5 Annual chemical cost Salvage value Life, years (UIRR 200 70 5 50) 26. 05% => Select Neutralization 10/20/2021 rd 11

Problem 8 -? MARR = 6% A B C 20 -year analysis First cost $10 K $15 K $20 K UAB 1625 1890 Life 10 20 20 (UIRR 10 e 3 1625 10) 9. 96% A (UIRR 15 e 3 1625 20) 8. 84% B (UIRR 20 e 3 1890 20) 7. 01% C (UIRR 5 e 3 0 10 10 e 3) 7. 18% B – A 10 -year (UIRR 5 e 3 265 20 0) 0. 56% C – B 20 -year 10/20/2021 rd 12

Problem 8 -14 MARR = 8% A First cost 1000 UAB 150 Salvage 1000 Life (yrs) 5 Ro. R 15% B 2000 150 2700 6 11. 83% C no replacement 3000 0 5600 7 13. 3% (IRR '(-1000 0 0 -1000 2850)) 9. 8% B > A (IRR '(-1000 -150 -150 -2850 5600)) 6. 75% for C – B < 8% => Reject C; Conclude B is best. 10/20/2021 rd 13

Problem 8 -15 Year 0 1 IRR (%) X -10 15 50 Y -20 28 40 Y- X -10 13 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%. 10/20/2021 rd 14

Problem 8 -19 Replace B and C when needed. Use MARR = 8% A B First cost $100 $150 UAB 10 17. 62 Life (years) ∞ 20 C $200 55. 48 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 10/20/2021 rd 15

Problem 8 -21 MARR = 12% n A B C 0 $-20 K 1 10 K 5 K (IRR '(-20 10 5 10 6)) (IRR '(-20 10 10 10 0)) (IRR '(-20 5 5 5 15)) (IRR '(-5 0 6)) (IRR '(-5 -5 -5 15)) Choose B 10/20/2021 2 5 K 10 K 5 K 21. 35% 23. 38% 14. 98% 9. 54% 0. 0% rd 3 10 K 5 K 4 6 K 0 15 K A B C A – B Reject A C – B Reject C 16

Problem 8 -25 A B C D First Cost 100 K 130 K 200 K 330 K UAB 26. 38 K 38. 78 K 47. 48 K 91. 55 K Life 5 5 Ro. R 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. Ro. RD-B = 9. 84% > 8% MARR => Select D. 10/20/2021 rd 17

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 4315. 2 4315. 2) 0. 8) 8. 836% MARR < 8. 836% Choose B 8. 836 < MARR < 14. 9% Choose C at 9% 14. 9% < MARR < i% Choose A 30976(A/P, i%, 20) 10/20/2021 rd 18

Problem 8 -27 MARR = 6% First Cost, $ Ann-Benefits Salvage Life Ro. R A 2 K 800 2 K 5 40% (UIRR 2 0. 8 5 2) 40%; (UIRR 4 0. 4 7 1. 4) 0. 98%; B 5 K 500 1. 5 K 6 -2. 4% C D 4 K 3 K 400 1300 1. 4 K 3 K 7 4 1% 43. 3% Reject B & C (UIRR 5 0. 5 1. 5 6 1. 5) -2. 38% (UIRR 3 1. 3 4 3) 43. 33% (IRR '(-1000 500 500 3500 -2800)) 51. 9% D - A (cum-add '(-1000 500 500 3500 -2800)) (-1000 -500 0 500 4000 1200) => unique positive Ro. R 2800(P/F, 6%, 1) = $2641. 51 (IRR '(-1000 500 500 858. 49 0)) 41. 1% 10/20/2021 rd 19

Problem 8 -29 MARR = 8% Atlas Zippy First cost $6700 $16, 900 AO&M cost 1500 1200 UAB 4000 4500 Salvage 1000 3500 Life (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 -3200 2500 3500 Zippy cf: -16900 3300 3300 6800 (IRR '(-10200 800 6500 800 3300)) 6. 802% Select Atlas 10/20/2021 rd 20

Problem 8 -33 A B B-A First Cost $100 K $300 K 200 K Annual Benefit 30 K 66 K 33 K Profit Rate (%) 30% 22% 18% MARR = 20% thus eliminating C. C $500 K 80 K 16% B – A cash flow is -200 K 36 K returns a profit rate of 18%. Thus best to choose A as the $200 K difference can be making MARR money at 20%. 10/20/2021 rd 21

Problem 8 -34 First cost AB AOC A B C $10 K $18 K $25 K 4 K 6 K 7. 5 K 2 K 3 K 3 K D $30 K 9 K 4 K $30 K Budget MARR = 15% A earns 2 K + 0. 15 * 20 K = $5 K / year B earns 3 K + 0. 15 * 12 K = $4800 / year C earns 4. 5 K + 0. 15 * 5 K = $5200 / year *** D earns 5 K + 0. 15 * 0 = $5 K Choose C 10/20/2021 rd 22

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) = 26. 78215 < 1% 0. 94% => 11. 28% APR => Lease at rate above 11. 28%. 10/20/2021 rd 23

ME; MARR = 9%; Life 10 years First cost UAB 1. 2. 3. 4. 5. A B C D E $4 K $5 K $2 K $3 K $6 K $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 (UIRR 2 K 266 10 0) 5. 52% => Reject C; A is best. 10/20/2021 rd 24