Engineering Economic Analysis Chapter 6 Annual Cash Flow

Engineering Economic Analysis Chapter 6 Annual Cash Flow Analysis 12/4/2020 rd 1

Annual Worth Analysis MARR = 8% First Cost Annual Maintenance Life (years) Salvage Value A $5000 0 5 0 B $8000 150 12 $2000 AWA(8%) = 5000(A/P, 8%, 5) = $1252. 28 AWB(8%) = 8000(A/P, 8%, 12) + 150 – 2 K(A/F, 8%, 12) = $1106. 17 *** Choose B, Less Cost 12/4/2020 rd 2

Capital Recovery Factor CR = (P – S)(A/P, i%, n) + Si [cost] (A/F, i, n) = (A/P, i, n) – i n 1 10% F/P P/F A/P F/A P/A A/G 1. 1000 0. 9091 1. 0000 1. 1000 1. 0000 0. 9091 0. 00 P/G 0. 00 F=S 0 1 2 3. . . n– 1 n A P 12/4/2020 rd 3

Annual Equivalent in Cycles Find the annual equivalent of the following cash flow at 12%: n 0 1 2 3 4 cf -$1000 $800 700 600 500 PW = -1000 + 800(P/A 12%, 4) -100(P/G, 12% 4) = $1017. 15 AE = 1017. 15(A/P, 12%, 4) = $334. 88 12/4/2020 rd 4

AW (6 -32) MARR = 10% First Cost Uniform Annual Benefit Useful life (years) A B $100 $150 16 24 20 AWA(10%) = 16 – 100(A/P, 10%, ) = 16 – 100(0. 1) = $6. 00 AWB(10%) = 24 – 150(A/P, 10%, 20) = 24 – 17. 62 = $6. 38* 12/4/2020 rd 5

Mutually Exclusive Find the annual equivalent savings for A and B at 15%. n A B 0 1 2 3 4 -$12 K 9120 6840 4560 2280 -$12 K 6350 6360 6350 AEB = 6350 – 12 K(A/P, 15%, 4) = $2146. 82 PWA = (list-PGF '(-12000 9120 6840 4560 2280) 15) $5404. 33 AEA = 5404. 33(A/P, 15%, 4) = $1892. 95 => Choose B. 12/4/2020 rd 6

Annual Equivalent MARR = 15%. Choose best of A, B and C using annual equivalent n 0 1 2 3 4 A -$2. 5 K 1000 1800 1000 400 B -$4 K 1600 1500 C -$5 K 1800 2000 PWA = (list-PGF '(-2500 1000 1800 1000 400) 15) = $616. 84 AEA = 616. 84(A/P, 15%, 4) = $216. 06 PWB = (list-PGF '(-4000 1600 1500) 15) = $369. 42 AEB = 369. 42(A/P, 15%, 4) = $129. 40 => Choose A over B PWC = (list-PGF '(-5000 1800 2000) 15) = $384. 82 AEC = 384. 82(A/P, 15%, 4) = $134. 79 => C over B and Select C 12/4/2020 rd 7

Unknown Life A B First Cost $800 $1000 UA Benefit 230 Life (years) 5 n Use a MARR of 12% to find n for equally desirable alternatives. AEA = 230 – 800(A/P, 12%, 5) = 8. 07 = 230 – 1000(A/P, 12%, n) = 0. 22193 => n = 6. 87 years 12/4/2020 rd 8

Unknown First Cost Find B's first cost to equate A and B at 12%. A B First Cost $150 $FC UAB 40 65 Salvage 100 200 Life (years) 6 6 CRFA = (150 -100)(A/P, 12%, 6) + 100 * 0. 12 = -$24. 16 AEA = 40 – 24. 16 = $15. 83 AEB = -[(FC -200)(A/P, 12%, 6) + 200 * 0. 12] + 65 = $15. 83 => FC = -$303. 44 12/4/2020 rd 9

Investment Alternatives MARR = 5% P 1 P 2 P 3 Investment $800 K $600 K $400 K Annual Revenue 450 K 400 K 300 K Annual Cost 200 K 180 K 150 K Salvage Value 100 K 80 K 60 K Proposals 1 & 2 are mutually exclusive, and 3 is contingent on 2. Budget limit is $1, 200, 000 with 8 -year lives. P 1 P 2 P 3 12/4/2020 a 0 0 0 ok b 1 0 0 ok c d e f 0 0 1 1 1 0 0 1 x x g 0 1 1 ok rd h 1 1 1 x 10

Time Value of Money at i = 10% n A B 0 1 -100 40 -100 20 2 40 20 3 40 60 4 40 60 PWA(10%) = -100 + 40(P/A, 10%, 4) = $26. 79 PWB(10%) = -100 + 20(P/A, 10%, 2) + 60(P/A, 10%, 2)(P/F, 10%, 2) = $20. 77 Get return ASAP for maximum benefit. 12/4/2020 rd 11

AW Analysis period MARR = 10% for 7 -year study First Cost Uniform Annual Benefit Life (years) A 100 55 3 B 150 61 4 AEA(10%) = 55 – [100 + 100(P/F, 10%, 3) + 100(P/F, 10%, 6)](A/P, 10%, 7) = $7. 43 AEB(10%) = 61 – [150 + 150(P/F, 10%, 4)](A/P, 10%, 7)] = $9. 14 ** 12/4/2020 rd 12

Problem 6 -20 An engineer wants to have $5 M by the time he retires in 40 years. Assuming a 15% rate of interest compounded continuously, find the annual sum. A = Fi / [(1 + i)n - 1] A = 5 M(e 0. 15 -1)/(e 0. 15*40) = $2005. 73 12/4/2020 rd 13

Problem 6 -45 First cost Annual Maint Salvage value Life (years) Masonite $250 20 10 4 Brick $1000 10 100 20 Find present worth of Masonite for 20 years MARR = 6%. One cycle 250 + 20(P/A, 6%, 4) – 10(A/F, 6%, 4) = $317. 02[1 + (P/F, 6%, 4) + (P/F, 6%, 8) + (P/F, 6%, 12) + (P/F, 6%, 16)] = 317. 02(3. 310123) = $1049. 36 (PW-cycle n m i) ~ (PW-cycle 4 20 6) 3. 310123 PWbrick = 1 K + 10(P/A, 6%, 20) – 100(P/F, 6%, 20) = $1083. 52 12/4/2020 rd 14

Problem 6 -37 Use annual worth to choose the better machine at 8%. A B First cost $5000 $8000 Life (yrs) 5 12 Salvage value 0 $2000 Annual maintenance 0 150 AEA(8%) = -5000(A/P, 8%, 5) = -$1252. 28 AEB(8%) = (8000 - 2000)(A/P, 8%, 12) + 2000 * 0. 08 150 + 150 = $1106. 17 12/4/2020 rd 15

Problem 6 -42 Use 7 -year analysis study at 10%. A B Cost $100 $150 UAB 55 61 Life (yrs) 3 4 AEA(10%) = 55 – [100 + 100(P/F, 10%, 3 + 100(P/F, 10%, 6)] (A/P, 10%, 7) = $7. 55 AEB(10%) = 61 – [150 + (P/F, 10%, 4)](A/P, 10%, 7) = $8. 91 ** 12/4/2020 rd 16

Mutually Exclusive n 0 1 2 3 4 5 MARR = 12% A -2500 750 750 2750 B -4000 1200 3200 a. What is the AW of A? b. 1. 371 2. -162 3. 135 4. 1338 5. 1590 b. What is the IRR for A? c. 1. 12% 2. 31% 3. 16% 4. 28% 5 25% c. What is the IRR of the increment? 1. 18% 2. 21% 3. 12% 4. 24% 5. 15% d. Which should be selected? B 12/4/2020 rd 17

Best Rate Which of the following options is best at the end of 5 years if you invest $1000 now? a) b) c) d) e) 12% compounded annually 11. 75% compounded semiannually 11. 5% compounded quarterly 11. 25% compounded monthly 11. 2% compounded continuously 12/4/2020 rd 18

Capitalized Equivalent Find the capitalized equivalent amount at 10%/yr for a series of annual receipts of $400 for the last 10 years which will increase to $500/yr after 10 years and remain constant thereafter. … 1 10 a)$4, 621 b) $4, 386 c) $4, 452 d) $9, 854 b) CC = 400/0. 1 + 100/0. 1(P/F, 10) 12/4/2020 rd 19