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Engineering Economic Analysis

Engineering Economic Analysis. Chapter 7  Rate of Return Analysis. Rate of Return. If the MARR is not given, Present Worth and Annual Equivalent cannot be used. Need more computational effort for RoR analysis. RoR is immediately meaningful to most lay people.

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Engineering Economic Analysis

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  1. Engineering Economic Analysis Chapter 7  Rate of Return Analysis rd

  2. Rate of Return If the MARR is not given, Present Worth and Annual Equivalent cannot be used. Need more computational effort for RoR analysis. RoR is immediately meaningful to most lay people. Defines a project as a measure of efficiency and not necessarily its monetary value. rd

  3. Internal Rate of Return (IRR) Interest Rate equating benefits to costs Net Present Worth at the IRR = $0 NPW(IRR) = 0 Net Annual Worth at the IRR = $0 NAW(IRR) = 0 Net Future Worth at the IRR = $0 NFW(IRR) = 0 Return on the unpaid balance of an amortized loan. PW of Benefits - PW of Costs = 0 PW of Benefits = 1PW of Costs rd

  4. Rate of Return Rate of return is the interest rate earned on the unpaid balance of an amortized loan. You borrow $10K from a bank and agree to paid $4,021 at the end of each year for 3 years at 10%/yr. Year Balance (B) Return on B Payment B at end 0 -$10K 0 0 -$10K 1 -10K -1000 4021 -6979 2 -6979 -698 4021 -3656 3 -3696 -336 4021 0 rd

  5. Simple vs. Non-simple Investments Simple ~ initial investment is negative and there is only one sign change in the cash flow. If first sign is positive, cash flow is borrowing, if negative, cash flow is investing. Non-simple investment has more than one sign change. n 0 1 2 3 4A -1K -500 800 1500 2000 SimpleB -1K 3900 -5031 2145 Non-simpleC 1K -450 -450 -450 Simple borrowing rd

  6. Example 1 Compute the IRR on the following cash flow: n 0 1 2 cf -1340.73 500 1000 PW(i*) = PWCash Flow – PWCash outflows = 0 Solve equation 13.4073x2 -5x - 10 = 0; where x = (1 + i). (quadratic 13.4073 -5 -10) (1.07 -0.70) Accept root 1.07 and i = 7%. (IRR ‘(-13.4073 5 10)) 7% rd

  7. Example 2 Compute the rate of return on the following cash flow: n 0 1 2 3 4 cf -700 100 175 250 325 (quartic -700 100 175 250 325) 1.0691 or 6.91%. (IRR ‘(-700 100 175 250 325))  6.91%. rd

  8. Example 3 n 0 1 2 3 4 i = 6.92%cf -700 100 175 250 325 Owe Pay PReduction Interest 700.00 100 51.56 48.44 = 0.0692 * 700.00 648.44 175 130.13 44.87 = 0.0692 * 648.44518.31 250 214.13 35.87 = 0.0692 * 518.31304.18 325 303.95 21.05 = 0.0692 * 304.18 Your un-recovered investment is earning at the internal rate of return 6.92%. rd

  9. Bond Rate of Return A 10-year $1000 bond is bought at face value and sold after one year for $950. The bond pays $40 interest each 6 months. a) Find the rate of return on the bond for the seller, and b) for the buyer if kept until maturity. a) (IRR ‘(-1000 40 990)) 1.52 => APR = 3.04 => ieff = 3.06%.(Quadratic -100 4 99) (1.015188 -0.975188) b) (UIRR 950 40 18 1000) 4.41% semi-annual or APR = 8.82% and effective rate of 9.01% for the buyer. rd

  10. Bond Maturity You buy a 12-year, $1000 par value, 15% semiannual coupon rate bond for $1,298.68. Compute your yield to maturity. 1298.68 = 75(P/A, i%, 24) + 1000(P/F, i%, 24) (UIRR 1298.68 75 24 1000)  5.28% => 10.55% APR or an effective yield of 10.83%. rd

  11. Mutually Exclusive IRRs Year A B B - A 0 -$10 -$20 -$10 1 +15 +28 +13IRR 50% 40% 30% For a MARR below 30%, B is preferred over A; 30% to 50% A is preferred, above 50%, Do Nothing is preferred. NPWA(35%) = -10 + 11.11 = $1.11NPWB(35%) = -20 + 20.74 = $0.74 rd

  12. Use rate of return (RoR) analysis for the following 3 mutually exclusive alternatives in reference to an unknown MARR. A B CFirst Cost $200 $300 $600Uniform annual benefits 59.7 77.1 165.2Useful life (years) 5 5 5End salvage value 0 0 0Computed RoR 15% 9% 11.7% Incremental RoR B - A => 100 = 17.4(P/A, i%, 5) => i = -4.47% C - A => 400 = 105.5(P/A, i%, 5) => i = 10% C - B => 300 = 88.1(P/A, i%, 5) => i = 14.3% Conclude: if MARR  9% Choose C Reject B 9%  MARR  10% Choose C Reject A 10%  MARR  11.7% Choose A Reject C 11.7%  MARR  15% Choose A MARR  15% Do Nothing rd

  13. Multiple Rates of Return n A B C D0 -100 -100 0 -1001 -200 50 -50 502 200 -100 115 03 200 60 -66 2004 200 -100 (cum-add '(-100 -200 200 200 200)) (-100 -300 -100 100 300) Note that a unique RoR exists. (cum-add '(-100 50 -100 60 -100))  (-100 -50 -150 -90 -190) => no rate of return Project fails the test and (-100 -50 -50 150) => unique RoR rd

  14. Multiple RoRs n 0 1 2 3 cash flow $-1000 4100 -5580 2520 PW(20%) = -1000 +4100(1.2)-1 –5580(1.2)-2 +2520(1.2)-3 = -1000 + 3416.67 – 3875 + 1458.33 = 0PW(40%) = -1000 +4100(1.4)-1 –5580(1.4)-2 +2520(1.4)-3 = 0 PW(50%) = -1000 +4100(1.5)-1 –5580(1.5)-2 +2520(1.5)-3 = 0.(cubic -100 410 -558 252) (1.5 1.4 1.2)(IRR '(-100 410 -558 252) 50)  50%(IRR '(-100 410 -558 252) 70)  40%(IRR '(-100 410 -558 252) 90)  20% rd

  15. Problem 7A-1 n 0 1 2 3 4 cf -15K 10K -8K 11K 13K Cum+ -15 -5 -13 -2 11 and one sign change => unique RoR (IRR '(-15 10 -8 11 13))  21.22% (quartic -15 10 -8 11 13) (1.2122 -0.586) rd

  16. Polynomial Equations Fundamental Theorem of Algebra – Every polynomial of degree n has n roots in the complex field. (make-poly '(1 12 7 9)) (1 -29 283 -1011 756) X4 -29X3 + 283X2 -1011X + 756 = 0 has the 4 roots 1, 7, 12 and 9 (quartic 1 -29 283 -1011 756) (12 9 7 1) rd

  17. Problem 7-21 Reduce to one cycle to get 80 80 80 120 200 80(F/A, i%, 2) = 200 => (F/A, i%, 2) = 2.5 => i = 50% 80x – 120 = 0 => x = 3/2 => i = 0.5 or 50%. rd

  18. Problem 7-22 500 250 -3810(1 + i)2 + 500(1 + i) + 250 = -3810 (quadratic -3810 500 4060) roots 1.099987 -0.968754 => I = 10% 3810 rd

  19. Problem 7-23 412 + 5000(P/F, i%, 10) = [1000/i](P/F, i%, 10) 412(1 + i)10 + 5000 = 1000/i ; Resolving at year 10 i = 15% by guess and test. Must be < 20%. A = 1000 10 412 5000 rd

  20. Problem 7-6 n 0 1 2 3 4cf -500 200 150 100 50 Observe 500 = Σ(200 150 100 50) => i = 0%. (IRR '(-500 200 150 100 50)) 0% (quartic -500 200 150 100 50)  1 etal rd

  21. Problem 7-14 A person pays $119.67 a month for 30 months after borrowing $3000. Calculate the APR and the APY. 3000 = 119.67(P/A, i%, 30) (UIRR 3000 119.67 30 0) 1.20 per month => => 14.4 APR => 15.39% APY. rd

  22. Problem 7-16 Find RoR for cash flow n 0 1 2 3 4 cf -500 0 0 0 4500 4500 = 500(F/P, i%, 4) => (F/P, i%, 4) = 9 (can't use tables) (1 + i)4 = 9 => i = 91/4 - 1 = 73.205% rd

  23. Problem 7A-22 Find RoR. n 0 1 2 3 4 5 6 7 8cf -200 100 100 100 -300 100 200 200 -124.5 4 sign changes (cum-add '(-200 100 100 100 -300 100 200 200 -124.5)) (-200 -100 0 100 -200 -100 100 300 175.5) one sign change (IRR '(-200 100 100 100 -300 100 200 200 -124.5)) 20% Unique IRR rd

  24. Problem 7-20 Buy home for $9,000 and pay $80 per year for 4 years and sell for $15,000 to yield RoR of ________. (UIRR 9000 -80 4 15000) 12.88% rd

  25. Problem 7A-25 Find the rate of return for the following cash flow: $400 1 2 -$200 -$100 (Quadratic -2 4 -1)  (1.707107 0.292893) => 70.7% rd

  26. Problem 7-28 Buy a $1000 face value bond for $925 which pays 4% coupon rate per year for 10 years. Find RoR. (UIRR 925 40 10 1000) 4.97% rd

  27. Problem 7-38 Buy at $140K and rent 4 units at $500/month for 5 years. Maintenance etc. is $8K annually. Can sell at $160K. Find RoR. (UIRR 140E3 (- (* 12 500 4) 8E3) 5 160E3)  13.61% Yes, since 13.61% > 12%. rd

  28. Problem 7-40 A 30 year loan at 6% for a home costing $120K and you've saved $12K for a 10% down payment. Lender charges "points" of 2% of the loan value added to the initial balance. Find monthly payment and effective annual interest. 120K – 12K = 108K, which at 2% ($2160) is $110,160. (AGP 110160 0.005 360) $660.47 monthly payment 108K = 660.47(P/A, i%, 360) => (P/A, i%, 360) = 163.52 Between ¼ and ½ about 0.52% per month or 6.42% ieff rd

  29. Problem 7-46 MARR = 6% Gas Station Ice Cream n = 20 years First cost $80K $120KAnnual Taxes 3K 5KAnnual Income 11K 16KIRR 7.75% 6.63% Check the increment Ice Cream - Gas Station RoR: (UIRR 40 3 20 0) 4.21% < 6% => go with gas station. rd

  30. Problem 7-51 n 0 1 2 3 4 MARR = 8%. A -5K -3K 4K 4K 4KB -5K 2K 2K 2K 2K (A-B)cf 0K -5K 2K 2K 2K (IRR ‘(-5 2 2 2))  9.7% > 8% => A > B. NPWA(8%) = (list-pgf '(-5 -3 4 4 4) 8) $1.7670K NPWB(8%) = (list-pgf '(-5 2 2 2 2) 8)  $1.6245K NPWA-B(8%) = (list-pgf '(0 -5 2 2 2) 8) $142.77 > 0 rd

  31. Problem 7-52MARR = 8% n 0 1 2 3 4 5 A -2500 746 746 746 746 746 B -6000 1664 1664 1664 1664 1664 B-A -3500 918 918 918 918 918 (UIRR 3500 918 5 0)  9.78% > 8% => B > AA ~ 2500/746 = 3.3512; B ~ 6000/1664 = 3.606 B – A ~ 3.81 => B > A at 8% MARR. NPWB-A(8%) = -3500 + 918(P/A, 8%, 5) = $165.30 > 0 rd

  32. 7-58 MARR = 10% A BB not replaced at n = 10 First Cost 150 100 UAB 25 22.25 Salvage 20 0 Life 15 10 Regard the increment B - A: 50 = 2.75(P/A, i%, 10) + 25(P/A, i%, 5) + 20(P/F, i%, 15)(IRR '(-50 2.75 2.75 2.75 2.75 2.75 2.75 2.75 2.75 2.75 2.75 25 25 25 25 45)) 11.60% Choose A if the MARR is 10%. rd

  33. Non-Simple Project Given the following cash flow, compute the IRRs and make an accept or reject decision if the MARR is 15%. n 0 1 2cf -1e6 2.3e6 -1.32e6(quadratic -1 2.3 -1.32)  (1.1999 1.10000) => 10% and 20% (MIRR '(-1e6 2.3e6 -1.32e6) 15 15)  15.054382 > 15% Accept Confirm: PW(15%) = -1 + 2.3 * 1.15-1 -1.32 * 1.15-2 = 0.00189 > 0 => Accept rd

  34. Mutually Exclusive Alternatives n 0 1 IRR $PW(10%) A -1000 2000 100% 818.18B -5000 7000 40% 1363.64B – A -4000 5000 25% 545.45Select B PW is an absolute measure while RoR is a relative measure. B = A + (B – A) => B earns at A's 100% up to $1000, then at 25%. rd

  35. Incremental Cost Only MARR = 15% A B Invest $4.5E6 $12.5E6AOC 7.41292E6 5.5041E6Salvage 5e5 1e6Life 6 6 B – A (UIRR 8 1.90882 6 0.5) 12.43% < 15% => A AEA (15%) = (+ (CRWR 4.5E6 5E5 15 6) 7.41292E6) $8,544,867.84 AEB (15%) = (+ (CRWR 12.5E6 1E6 15 6) 5.4041E6)  $8,592,825 A is cheaper. rd

  36. Computing the MIRR • Compute the MIRR for the following cash flow using 6% for the borrowing rate and 12% for the investing rate. • n 0 1 2 3 cf -1000 500 900 -200 • (mirr '(-1000 500 900 -200) 6 12)  11.87% • (list-pgf '(1000 0 0 200) 6)  $1167.92 • (list-fgp '(0 500 900 0) 12)  $1635.20 • (igpfn 1167.92384 1635.20 3)  11.87% rd 1/2/2020 rd

  37. Problem You buy a new car for $32K and put $8K down and borrow the rest from the dealer. The dealer charges monthly payments of $900 for 4 years. Find the dealer's ROR. Ans. 37.86% APY rd

  38. Compute the MIRR Find the MIRR for the following cash flow by using 5% for borrowing rate and 9% for the investment rate. n 0 1 2 3 4 5 cf -20 70 -15 30 -10 -20 Ans. 18.52% rd

  39. Rate of Return An investment of $350K is made followed by income of $200K per year for 3 years. The IRR of the investment is nearest to: • 15.3% b) 32.7% c) 41.7% d) 57.1% (UIRR 350 200 3) 32.68% (cubic -350 200 200 200)  1.326752 => 32.675% rd

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