1. 2/20/2016rd1 Engineering Economic Analysis Chapter 8  Incremental Analysis

2. 2/20/2016rd2 X Y $15 $10 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% Benefit-Cost Graph Rejection $20

3. 2/20/2016rd3 MARR = 6% A B C 20-years First cost $2000$4000$5000 UAB 410 639 700 PWBenefits 4703 7329 8029 Benefit-Cost Graph 4703 7329 8029 A B C 2000 4000 5000 i=6% NPW=0

4. Incremental Analysis ABCMARR = 10% First cost$18K$25K$15KLife 25 years UAB105521251020 Salvage ~ 0 IRR (%) 7 9 8 IRR A-C = (UIRR 3000 35 25 0)  -7.86% C > A IRR B-C = (UIRR 10E3 1105 25 0)  10.04% B > C 2/20/2016rd4

5. 2/20/2016rd5 Incremental Analysis MARR = 8% A B C First Cost100020003000 UA Benefits 150 150 0 Salvage value100027005600 Life 5 6 7 IRR15%11.81%9.33% Take in order of increasing first cost: A > 15% RoR B-A (IRR ‘(-1000 0 0 0 0 –1000 2850))  9.8% => B > A RoR C-B (IRR –1000 –150 -150 -150 –150 -150 -2850 5600)) returns 6.75% => B is best.

6. Incremental RoR Analysis ABCD5-year life First cost100 130200330 Annual income10090.78160164.55 Annual cost73.6252.00112.52 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 2/20/2016rd6

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 2/20/2016rd7

8. 2/20/2016rd8 Problem 8-2 X Y X - Y First Cost-100-50 -50 UAB 31.516.5 15 Life (years) 4 4 4 RoR9.93%12.11% 7.71% Which is better if a) MARR = 6%?X b) MARR = 9%Y c) MARR = 10%Y d) MARR = 14%Do Nothing

9. 2/20/2016rd9 Problem 8-3 A B B-A First Cost-100-150 -50 UAB 30 43 13 Life (years) 5 5 IRR (%) 15.2413.349.43 (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%? Bb) MARR = 8%? B c) MARR = 10% Ad) MARR = 16% Do Nothing

10. 2/20/2016rd10 Example 8-6 MARR = 6%ABCDE First Cost4K2K6K1K9K UAB639410761117785 Life2020202020 (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

11. Problem 8-8 2/20/2016rd11 Mutually ExclusiveNeutralizationPrecpitation First cost$700K$500K Annual chemical cost 40K 110K Salvage value 175K 125K Life, years 5 5 (UIRR 200 70 5 50)  26.05% => Select Neutralization

12. 2/20/2016rd12 Problem 8-? MARR = 6% A B C20-year analysis First cost$10K$15K$20K UAB1625 1625 1890 Life 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

13. 2/20/2016rd13 Problem 8-14 MARR = 8%ABC no replacement First cost100020003000 UAB 150 150 0 Salvage 1000 2700 5600 Life (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 Reject C; Conclude B is best.

14. 2/20/2016rd14 Problem 8-15 Year X YY- X 0-10-20 -10 1 15 28 13 IRR (%) 50 40 30 Over what range of MARR is Y preferred over X? Y is better for MARR 50%.

15. 2/20/2016rd15 Problem 8-19 Replace B and C when needed. Use MARR = 8% A B C First cost$100$150$200 UAB 10 17.62 55.48 Life (years) ∞ 20 5 Capitalized Costs Analysis NPW A = 10/0.08 – 100 = $25 NAW B = 17.62 -150(A/P, 8%, 20) = $2.34 perpetuity NPW B = 2.34/0.08 = $29.28 NAW C = 55.48 -200(A/P, 8%, 5) = $5.39 or perpetuity NPW C = 5.39/0.08 = $67.36 *** C

16. 2/20/2016rd16 Problem 8-21 MARR = 12% n 0 1 2 34 A$-20K10K 5K10K6K B -20K10K10K10K0 C -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 C Choose B

17. 2/20/2016rd17 Problem 8-25 A B C D First Cost100K130K200K330K UAB26.38K38.78K47.48K 91.55K Life 5 5 5 5 RoR10%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.  RoR D-B = 9.84% > 8% MARR => Select D.

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 B 8.836 < MARR < 14.9% Choose C at 9% 14.9% < MARR < i% Choose A30976(A/P, i%, 20) 2/20/2016rd18

19. 2/20/2016rd19 Problem 8-27 MARR = 6% A B C D First Cost, $ 2K 5K 4K 3K Ann-Benefits800 500 400 1300 Salvage 2K1.5K 1.4K 3K Life 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%

20. 2/20/2016rd20 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 2500 -3200 2500 2500 3500 Zippy cf: -16900 3300 3300 3300 3300 3300 6800 (IRR '(-10200 800 800 6500 800 800 3300))  6.802% Select Atlas

21. 2/20/2016rd21 Problem 8-33 A B B-A C First Cost$100K $300K 200K$500K Annual Benefit 30K 66K 33K 80K Profit 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%.

22. Problem 8-34 ABCD $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 2/20/2016rd22

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 11.28% APR => Lease at rate above 11.28%. 2/20/2016rd23

24. ME; MARR = 9%; Life 10 years ABCDE First cost$4K$5K$2K$3K$6K UAB$797$885$259$447$1063 1.(UIRR 2000 259 10 0)  5% Reject 2.(UIRR 3000 447 10 0)  8% Reject D 3.(UIRR 4000 797 10 0)  15% Accept A 4.(UIRR 1000 88 10 0)  -% => Reject B 5. (UIRR 2K 266 10 0)  5.52% => Reject C; A is best. 2/20/2016rd24