Engineering Economic Analysis Chapter 5 Present Worth Analysis

Engineering Economic Analysis Chapter 5 Present Worth Analysis 2/24/2021 rd 1

Proposals • Independent • Mutually Exclusive • Contingent • Essential 2/24/2021 rd 2

Minimum Attractive Rate of Return MARR – desired interest rate that a business wishes to earn on its investment the rate at which one can always invest money in an investment pool Also known as the required rate of return Do Nothing - an alternative in all decisions which is assumed to earn at the MARR Net Present Worth: PW of income – PW of outgo 2/24/2021 rd 3

Independent Proposals Alternatives A 0 A 1 A 2 A 3 2/24/2021 Decision P 1 P 2 0 1 0 0 1 1 rd Do Nothing Accept P 1 Accept P 2 Accept P 1 & P 2 4

Four Mutually Exclusive Proposals MARR = 5%; 7 -year life A B C D 1 st Cost 575 1200 1700 2300 Net Annual Benefit 140 350 425 700 0 0 340 200 Salvage Value NPWA(5%) = NPWB(5%) = NPWC(5%) = NPWD(5%) = 2/24/2021 -575 + 140(P/A, 5%, 7) = 235. 09 -1200 + 350(P/A, 5%, 7) = 825. 23 -1700 + 425(P/A, 5%, 7) + 340(P/F, 5%, 7) = 1000. 84 -2300 + 700(P/A, 5%, 7) + 200(P/F, 5%, 7) = 1892. 60 *** rd 5

Mutually Exclusive Present Worth MARR = 15% A First Cost $100 K Ann-Benefit 15. 2 K Salvage value 10 K Life (years) 10 B $152 K 31. 9 K 0 10 C $184 K 35. 9 K 15 K 10 D $220 K 41. 5 K 20 K 10 PWA = -100 K + 15. 2 K(P/A, 15%, 10) + 10 K(1. 1)-10 = -$21243 PWB = -152 K + 31. 9 K(P/A, 15%, 10) = $8099 *** PWC = -184 K + 35. 9 K(P/A, 15%, 10) + 15 K(1. 1)-10 = -$118. 44 PWD = -220 K + 41. 5 K(P/A, 15%, 10) + 20 K(1. 1)-10 = -$6777 2/24/2021 rd 6

Present Worth & Salvage Value A new stamping machine cost $100 K and is used for 5 years generating revenues of $25 K per year. Find the salvage value to get a 12% per year rate of return. -100 K + 25 K(P/A, 12%, 5) + S(P/F, 12%, 5) = 0 -100 K + 90119. 41 + 0. 5674 S = 0 S = $17, 413. 80 Check: (UIRR 1 e 5 25 e 3 5 17413. 80) 12% 2/24/2021 rd 7

PW Cash Flow Find the present worth of the following cash flow with the MARR set at 10%: n 0 1 2 3 4 cf -100 -200 300 400 500 PW(10%) = -100 -200(1. 1)-1 + 300(1. 1)-2 +400(1. 1)-3 + 500(1. 1)-4 = $608. 14 (List-pgf '(-100 -200 300 400 500) 10) $608. 14 2/24/2021 rd 8

Least Common Multiple (Life) MARR = 7% n A B 0 1 -1000 400 -1200 340 2 400 340 3 400 340 4 5 6 340 340 Method I NPWA(7%) = -1000 + 400(P/A, 7%, 6) – 1000(P/F, 7%, 3) = $90. 32 NPWB(7%) = -1200 + 340(P/A, 7%, 6) = $420. 62. *** Method II NPWA(7%) = -1000 + 400(P/A, 7%, 3) = $49. 73 (one cycle) For 2 cycles: NPWA(7%) = 49. 73[1 + (P/F, 7%, 3)] = 49. 23(1 + 0. 81629787689) = $90. 32 2/24/2021 rd 9

Problem 5 -50 First cost UAB Life (yrs) A $5300 $1800 4 B $10, 700 $2100 8 MARR = 10% PWA(10%) = -5300 + 1800(P/A, 10%, 8) – 5300(P/F, 10%, 4) = $682. 90 *** (* (+ -5300 (P/A 1800 10 4))(PW-cycle 4 8 10)) 682. 90 PWB(10%) = -10, 700 + 2100(P/A, 10%, 8) = $503. 35 2/24/2021 rd 10

Problem 5 -54 Model A B First Cost $50 K $80 K AOC UAB S Life $2 K $9 K $10 K 10 $1 K $12 K $30 K 10 9% PWA(9%) = -50 K + 7 K(P/A, 9%, 10) + 10 K(P/F, 9%, 10) = $-852. 29 PWB(9%) = -80 K +11 K(P/A, 9%, 10) + 30 K(P/F, 9%, 10) = $3266. 56 ** If no B's available, Do Nothing, or seek other alternatives 2/24/2021 rd 11

Incremental Analysis MARR = 10% n A B B-A 0 1 PW -1000 2000 818. 18 -5000 7000 1363. 64 -4000 545. 45 IRR 100% 40% 25% > 10% Observe that using the higher rate of return criterion is flawed for selection of mutually exclusive alternatives. 2/24/2021 rd 12

Incremental Analysis MARR = 10% n A B A-B 0 -100 -50 1 35 16. 5 18. 5 2 35 16. 5 18. 5 3 35 16. 5 18. 5 4 35 16. 5 18. 5 Ro. R 14. 96 * 12. 11 17. 76 A (+ -100 (PGA 35 10 4)) 10. 94 B (+ -50 (PGA 18. 5 10 4)) 8. 64 A-B (+ -50 (PGA 16. 5 10 4)) 2. 30 2/24/2021 rd 13

Incremental Analysis Unknown MARR n A B C A-B C-A C-B 0 -2000 -1000 -3000 -1000 -2000 1 1500 800 1500 700 2 1000 500 2000 500 1000 1500 0 < MARR < 8. 08% 8. 08 < MARR < 27. 61 < MARR < 40. 76% < MARR 2/24/2021 3 800 500 1000 300 200 500 IRR 34. 37% 40. 76% 24. 81% 27. 61% 8. 08% 17. 20% choose C; choose A choose B; choose Do Nothing rd 14

MARR = 10% 1 st Cost Annual Benefits Salvage Value Life (years) PW (1 -cycle) Cycle factor A 10 K 6 K 1 K 2 $1239. 67 3. 92598 B 15 K 10 K -2 K 3 $8365. 89 C 12 K 5 K 3 K 4 $5898. 37 2. 739886 2. 149521 PWA(10%) = -10 K + 6 K(P/A, 10%, 12) +1 K(P/F, 10%, 12)9 K[(P/F, 10%, 2) + (P/F, 10%, 4) +(P/F, 10%, 6) + (P/F, 10%, 8) + (P/F, 10%, 10)] = $4, 866. 92. (* (+ -10 e 3 (PGA 6 e 3 10 2) (P/F 1 e 3 10 2))(PW-cycle 2 12 10)) 4866. 93 PWB(10%) = -15 K + 10 K(P/A, 10%, 12) – 2 K(P/F, 10%, 12) – 17 K[(P/F, 10%, 3) + (P/F, 10%, 6) + (P/F, 10%, 9)] = $22, 921. 59 *** PWC(10%) = -12 K + 5 K(P/A, 10%, 12) + 3 K(P/F, 10%, 12) – 9 K[(P/F, 10%, 4) + P/F, 10%, 8)] = $12, 678. 66 2/24/2021 rd 15

Annual Worth Analysis MARR = 8% 1 st Cost Annual Maintenance Life (years) Salvage A $5 K 0 5 0 B $8 K 150 12 $2 K AWA(8%) = -5 K(A/P, 8%, 5) = -$1252. 28 AWB(8%) = -8 K(A/P, 8%, 12) - 150 + 2 K(A/F, 8%, 12) = -$1106. 17 2/24/2021 rd 16

AW Analysis period MARR = 10% for 7 -year study 1 st Cost UAB 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 ** 2/24/2021 rd 17

Internal Rate of Return (Ro. R) Equivalent definitions of rate of return: Interest rate earned on unrecovered balance PW(benefits) = PW(costs) NPW = 0 EUAB – EUAC = 0 Annual Benefits = Annual Costs 2/24/2021 rd 18

Rate of Return Table Lookup n CF 0 -2 K 1 1 K 2 1 K 3 1 K 2000 = 1000(P/A, i%, 3) => (P/A, i%, 3) = 2 => i = 23. 375% at 20%, n = 3, P/A = 2. 106/154 = -0. 6883 * 5 = 3. 44 at 25%, n = 3, P/A = 1. 952 Linear Interpolation yields 23. 44% (UIRR 2000 1000 3 0) 23. 375188% 2/24/2021 rd exact answer 19

Exact Ro. R Calculation n CF 0 -1292. 52 1 500 2 900 -1292. 52(1 + i)2 + 500(1 + i) + 900 = 0; let x = (1 + i) 1292. 52 X 2 + 500 X + 900 = 0 (quadratic -1292. 52 500 900) 1. 0499 0; -0. 6631 => i = 5% Let x = (1 + I) and resolve at year 2. 2/24/2021 rd 20

Newton’s Method tan = f(x) / (x 0 – x 1) = f ’(x) f(x) root x 1 2/24/2021 x 0 rd 21

Ro. R or IRR n cash flow 0 1 2 3 -1000 4100 -5580 2520 3 sign changes => 3 real roots or 1 real and 2 complex -1000 x 3 + 4100 x 2 – 5580 x + 2520 = 0 Find the roots. (irr '(-1000 4100 -5580 2520) 0. 25) 50 (irr '(-1000 4100 -5580 2520) 0. 95 ) 20 (irr '(-1000 4100 -5580 2520) 0. 75 10 ) 40 (cubic -1000 4100 -5580 2520) (1. 5 1. 4 1. 2) 2/24/2021 rd 22

Incremental Analysis IRR MARR = 10% n A B A–B 0 1 -100 35. 0 -50 16. 5 -50 18. 5 2 35. 0 16. 5 18. 5 3 35. 0 16. 5 18. 5 4 35. 0 16. 5 18. 5 Ro. R 14. 96% 12. 11 17. 76% (UIRR 50 18. 5 4 0) 17. 76% > 10% Conclude A earns at B’s rate of 12. 11% for the first $50 and at 17. 76% for the next $50. Choose A. 2/24/2021 rd 23

PW & AE & Ro. R Analyses MARR = 8% n A B 0 1 2 3 4 5 -2500 746 746 746 -6000 1664 1664 Ro. R 15% 12% Solve by PW, AE and Ro. R analyses. PWA(8%) = -2500 + 746(P/A, 8%, 5) = $478. 56 PWB(8%) = -6000 + 1664(P/A, 8%, 5) = $643. 87** AEA(8%) = 746 – 2500(A/P, 8%, 5) = $119. 86 AEB(8%) = 1664 – 6000(A/P, 8%, 5) = $161. 26 ** B – A => 3500 = 918(P/A, i%, 5) => (P/A, i%, 5) = 3. 8126 => i = 9. 78% > MARR => Select B which earns at 12%. (UIRR 6000 1664 5 0) 11. 988878 2/24/2021 (UIRR 3500 918 5) 9. 78% rd 24

External Interest Rate External rate = 6% n cf 0 19 1 10 2 -50 3 -50 4 20 5 60 Move 19 and 10 to year 2 to modify cash flow. 19(F/P, 6%, 2) + 10(F/P, 6%, 1) = 32 New CF -18 -50 20 60 from which the Ro. R is (IRR ‘(– 18 – 50 20 60)) 8. 4% 2/24/2021 rd 25

Incremental Ro. R Analysis n A B B-A 0 -10 -20 -10 1 15 28 13 Ro. R 50% 40% 30% If MARR <= 30%, choose B If 30% < MARR < 50% choose A If MARR > 50%, choose Do Nothing 2/24/2021 rd 26

Incremental Ro. R Analysis n A B C 0 1 2 3 4 5 -200 59. 7 -300 77. 1 -600 165. 2 Ro. R 15. 03% 8. 99% 11. 69% B-A C-B C-A -100 17. 4 -300 88. 1 -400 105. 5 -4. 47% 14. 34% 9. 99% 0% < MARR < 10% < MARR < 11. 7% < MARR < 15% MARR > 15% 2/24/2021 Choose C; Choose A; Choose Do Nothing rd 27

Capitalized Costs P = A[(1 + i)n - 1] / i(1 + i)n and as n ∞ P=A/i Cash requirements for an endowment run $100 K to establish and $30 K per year indefinitely and $20 K at the end of every 4 th year. Interest rate at 8%. CC = [-100 K(A/P, 8%, ∞) -30 K -20 K(A/F, 8%, 4)]/0. 08 = [(-8 K -30 K – 4438. 42)]/0. 08 = $530, 480. 20 2/24/2021 rd 28

Problem 5 -40 How much was deposited 50 years ago at 8% to provide perpetual payments of $10 K per year? Needed is 10 K/0. 08 = $125 K (P/F 125 K, 8%, 50) $2665. 15 2/24/2021 rd 29

Problem 5 -50 Use an 8 -year analysis period at 10% for the two mutually exclusive proposals. A B First Cost $5300 $10, 700 UAB 1800 2100 Life (years) 4 8 NPWA(10%) = -5300 + 1800(P/A, 10%, 8) – 5300(P/F, 10%, 4) = $682. 90 ** NPWB(10%) = -10, 700 + 2100(P/A, 10%, 8) = $503. 35 2/24/2021 rd 30

Problem 5 -62 Interest rate at 10% for 10 -year life, no salvage value. Mutually exclusive proposals First Cost UAB 1 -5 UAB 6 -10 A $600 100 50 B $600 100 C $600 110 D $600 150 0 E $600 150 50 C > B > A and E > D. Choose between C and E. PWC(10%) = -600 + 100(P/A, 10%, 5) + 110(P/A, 10%, 5)*1. 1 -5 = $38 PWE(10%) = -600 + 150(P/A, 10%, 5) + 50(P/A, 10%, 5)*1. 1 -5 = $86. 31 ** 2/24/2021 rd 31

Net Cash Flow Two mutually exclusive service projects are given below. MARR = 12% n 0 1 2 3 6 -year study A -$1 K -400 + 200 (200 is salvage value) B -$800 -200 + 0 a) Project B saves you $344 in PW. True NPWA(12%) = -1 K -400(P/A, 12%, 6) – 800(P/F, 12, 3) + 200(P/F, 12%, 6) = $3112. 66 NPWB(12%) = -800 -200(P/A, 12%, 6) -800[(1. 12)-2 + (1. 12)-4] = $2768. 45 b) Project A cost $1, 818 in first cycle. True, but not for 2 nd cycle NPWA(12%) = -1000 -400(P/A, 12%, 3) + 200(P/F, 12%, 3) = $1818. 38 NPWA(12%) = -800 -400(P/A, 12%, 3) + 200(P/F, 12%, 3) c) Project B's NPW exceeds A's by $680 False d) None of the above NPWB = [-800 -200(P/A, 12%, 2)][1 + (P/F, 12%, 4) + (P/F, 12%, 6)] = $2768. 45 2/24/2021 rd 32

Problem 5 -71 Use a 12 -year analysis and i = 10% to select the best of the 3 mutually exclusive alternatives A B C First cost $10 K $15 K $12 K Annual benefit 6 K 10 K 5 K Salvage value 1 K -2 K 3 K Life (years) 2 3 4 Find A for one cycle: PWA = -10 K + 6 K(P/A, 10%, 2) + 1 K(P/F, 10%, 2) = $1239. 67 (* 1239. 67 (+ 1 (PGF 1 10 2) (PGF 1 10 4) (PGF 1 10 6) (PGF 1 10 8) (PGF 1 10 10))) $4866. 93 PWB = $8364. 78 * 2. 739886 = $22, 918. 53 *** $1000 $600 PWC = 5898. 37 * 2. 149521 = $12, 678. 67 $10 K 2/24/2021 rd $900 33

Bond You bought a 5 -year, $10 K par value bond paying a coupon rate of 6% compounded semiannually and wish to earn an 8% effective return. How much should you pay for the bond? PW = 300(P/A, 3. 923%, 10) + 10 K(P/F, 3. 923%, 10) = $9, 248. 48 Note: Effective rate of 8% => APR = 7. 85% => effective semiannual rate is 3. 923%. 2/24/2021 rd 34

Breakeven Lease You can buy a piece of equipment for $50, 000 or lease it for $6000 for 10 years. The lease is preferred rate for an interest rate exceeding a) 3% b) 3. 33% c) 3. 4% d) 3. 5% => 30 K = 6000(P/A, i%, 10) => (P/A, i%, 10) = 8. 3333 => i = 3. 5% 2/24/2021 rd 35