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Engineering Economic Analysis

This chapter explores various analysis techniques for evaluating larger public sector projects with longer lifespans. It covers benefit/cost ratios, criticisms, modified B/C ratios, MIRR, incremental analysis, and more.

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Engineering Economic Analysis

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  1. Engineering Economic Analysis Chapter 9  Other Analysis Techniques

  2. Public Sector Projects • Larger investment • Longer life 30 to 50+ years (capitalized costs) • (A/P, 7% , 30) = 0.0805864; (AGP, 7%, 50) = 0.0724598 • No profit and Publicly owned • Benefit citizenry (undesirable consequences) • Lower MARR exempt from taxes called discount rate • $ from taxes, bonds, and fees (toll roads), philanthropy • Multiple criteria vice RoR • Politically inclined vice economic • Examples: Hospitals, parks and recreation, sports arenas emergency relief, airports, courts, utilities, schools

  3. Criticisms of B/C • Used after the fact for justification rather than evaluation • Serious distribution inequities; one group reaps the benefits while the other group incurs the costs • The benefits to whosoever they accrue • Qualitative information is largely ignored. • Public policy generally should favor t the disadvantaged • NOT a scientifically unbiased method of analysis

  4. Benefit/Cost Ratios B/C = PW of benefits = AW of benefits = FW of benefits PW of costs AW of costs FW of costs If B/C  1.0, accept project; else reject. Conventional B/C ratio = benefits – disbenefits = B - D costs C Modified B/C = benefits – disbenefits – M&O costs initial investment Salvage value is included in the denominator as a negative cost for the modified ratio. Ratios can change, but not the decision. (B D C) = (10 8 8)  B – C = 10 – 8 – 8 = -6

  5. Benefits. Costs, Disbenefits Build new Convention Center Benefits – improve downtown image, attract conferences, sports franchises, revenue from rental facilities, increased revenue for downtown merchants Costs – Architectural design, construction, design and construction of parking garage, facilities and maintenance, facility insurance Disbenefits -- loss of bike trail, park, nature trail, and pond loss of wildlife habitat in urban area

  6. MIRR for Indeterminate IRR • n 0 1 2 3cf $5K -7K 2K 2K • (cubic 5 -7 2 2) -0.375 and 2 complex roots • Real root -1  Indeterminate IRR not positive • By using a reinvestment rate of 15% and borrowing rate of 12% per year the MIRR gives • (MIRR '(5 -7 2 2) 12 15)  23.95% > 15% • => good investment. PW(24%) = $2.81K

  7. Incremental Analysis • Consider projects 20-year life with MARR at 6% • A B C D E F1st cost $4K 2K 6K 1K 9K 10KPW benefit 7330 4700 8730 1340 9K 9500B/C ratio 1.83 2.35 1.46 1.34 1.0 0.95 • B-D = 3360/1K = 3.36 => B > DA-B = 2630/2K = 1.32 => A > B • C-A = 1400/2K = 0.70 => A > C • E-A = 1670/5K = 0.33 => A > E and A is best.Notice that B has highest B/C ratio, but not chosen.

  8. Benefits - Disbenefits • Benefits of airport runway expansion – reduced delays, taxing time, fuel costs, landing and departure fees, lost earnings from public point of view. • Return to the moon? • Disbenefits -- Undesirable consequences from a viewpoint • May be included or disregarded in the analysis, but can make a distinctive difference in the analysis.

  9. B/C & Modified B/C Ratios Social Discount Rate B/C and Modified B/C ratios Benefits (B) Disbenefits (D) Costs (C) Present Worth (B, D, C) Disbenefits SUBTRACTED from Benefits Alternatives B D C (B – D)/C B/C B C B/C X 10 3 5 (10 – 3)/5 = 1.4 9–7 = 2 6-5=1 2/1 = 2 Y 13 4 6 (13 – 4)/6 = 1.5 Disbenefits ADDED to costs B/C B /C B C B/C X  10/8 = 1.25 Y 13/10 = 1.3 3 2 1.5

  10. Benefits, Disbenefits, or Costs • Expenditure of $30M to pave highway • $100K loss of revenue to farmers because of highway right-of-way • $300K annual income to local businesses due to tourism • $500K per year upkeep for causeway • Fewer highway accidents due to improved lighting.

  11. 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 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  10% Choose C10%  MARR  15% Choose A 15%  MARR Choose Do Nothing

  12. Incremental B/C • Use benefit/cost ratio to select the best of the 5 mutually exclusive alternatives at 15%. • Year U V W X Y Z • 0 -200 -125 -100 -125 -150 -225 • 1-5 68 40 25 42 52 68 • B/C 1.14 ** 0.84 1.13 1.16 ** • By inspection U > Z, X > V, W < 1. Choose from U, X and Y. • Start with X challenged by Y • B/C Y-X = 10/25(A/P,15%,5)= 7.46 = 1.34 => Y is better • B/C U-Y = 16/50(A/P,15%,5)=14.92 = 1.073 => U is best.

  13. Example 9-6 • Payback Period • Year A B0 -1000 -27831 200 12002 200 12003 1200 12004 1200 12005 1200 1200 • Payback for A(cum-add -1000 200 200 1200 1200 1200))  (-1000 -800 -600 600 1800 3000) => 2.5 years • Payback for B = 2783/1200 = 2.32 years

  14. Problem 9-8 • Year 0 is akin to year 20 • $100 on year 21 with $100 gradient from year 22 to 55. • Compute future worth on 65th birthday. • (FGP (FGP (PGG 100 100 12 35) 12 35) 12 10)  • $1,160,715.44

  15. Problem 9-17 • Geometric gradient, A1 = 100, g = 100%, i = 10% n = 10 F10 = ? • (FGP (PGGG 100 100 10 10) 10 10) $113,489.59 • (FGP $43755.14 10 10)

  16. Problem 9-30 • Use benefit-cost analysis for the 5-year cash flows below at a MARR of 10%. • A B C • Cost $600 $500 $200 • UAB 158.3 138.7 58.3 • B/CC (10%) = 58.3/(200*A/P,10%,5) = 1.105 • B/CB-C(10%) = 80.4/300(A/P, 10%, 5) = 1.016 • B/CA-B(10%) = 19.6/100(A/P, 10%, 5) = 9.743 • Conclude: Select B.

  17. Problem 9-36 • n A B C D E FMARR = 15%0 -200 -125 -100 -125 -150 -2251-5 68 40 25 42 52 68 • By inspection: A > F; D > B;B/C ratios: A = 1.14; C = 0.83; D = 1.21; E = 1.25Incremental analysis: • B/CE-D = (52-42)(P/A, 15%, 5) / (150 – 125) = 1.34 => E > D • B/CA-E = (68-52)(P/A, 15%, 5) / (200 – 150) = 1.15 => A > E • A is best.

  18. Problem 9-46 • ME A B • First cost $500 $800 • UA Cost 200 150 • Life (years) 8 8 • Compute payback if B is picked. PB = 300/50 = 6 yrs. • Use 12% MARR and B/C to select. • 50(P/A, 12%, 8)/300 = 248/300 = 0.827 => Select A

  19. Problem 9-49 • n 0 1 2 3 4 MARR = 10%X -100 25 25 25 25Y -50 16 16 16 16Z -50 21 21 21 21 • B/CX < 1 by inspection and has zero rate of return • B/CY is dominated by Z which is best, showing a positive rate of return ($16.67) at 10%. • PWZ(10%) = -50 + 21(P/A, 10%, 4) = -50 + 66.57 = $16.67

  20. Problem 9-57 • A B MARR = 12%First cost $500 $800UAB 200 150Life 8 8 • Incremental payback = (800 – 500)/50 = 6 years. • By inspection both A and B have B/C ratios > 1. • B/CB-A = 50 / 300(A/P,12%,8) = 50 / 60.4 < 1 => A is better than B

  21. Problem 9-67 • A BFirst cost $55K $75KAnnual expenses Operations 9500 7200 Maintenance 5000 3000 • Taxes/Insurance 1700 2250 • Find useful life for equivalency at 10% MARR • 55K + 16.2K(P/A,10%,n) = 75K + 12.45K(P/A,10%,n) • 20 = 3.75(P/A,10%, n) • (P/A,10%, n) = 5.333 => 8 years • At 0% MARR, (P/A, 0%, n) = 5.333 => n = 5.333 years

  22. Breakeven • 1. A firm can buy parts from a supplier for $96 delivered or can make for $46. A fixed cost of $8000 per year is needed to make the parts. Find the number of units per year for which the cost of the two alternatives will break even. •  Cost to make = Cost to buy => 8000 + 46x = 96x => x = 160 parts. • 2. A firm can least a fleet of cars for its sales personnel for $15 per day plus $0.09 per mile for each car. Alternatively the firm can pay each salesperson $0.18 per mile to use his or her own car. How many miles per day must a salesperson drive to break even?  • 15 + 0.09x = 0.18x => x = 167 miles.

  23. Breakeven • Initial cost is $200M with 20% salvage value within 5 years. Cost to produce car is $21K but expected to sell for $33K. Production capacity for year 1 is 4000 cars. Rate is 12%. • Find uniform amount of production increase to recoup investment in 3 years? • $200M = (33 – 21)K(P/A, 12%, 3) * (X + 4000)

  24. Sensitivity • 3. A B C MARR = 6% • First Cost $2K $4K $5K • UAB 410 639 700 • Life 20 20 20 • NPW(6%) $2703 3329 3029 • Find the maximum first cost of B to become non-preferable. • 639(P/A, 6%, 20) - FCB = 3029 => FCB = $4300.28. B is preferred if its initial cost does not exceed $4300.

  25. Sensitivity Analysis MARR = 12% • Strategy First Cost Salvage AOC Life • P $20K 0 $11K 3A ML 20K 0 9K 5O 20K 0 5K 8 • P 15K 500 4K 2B ML 15K 1K 3.5K 4O 15K 2K 2K 7 • P 30K 3K 8K 3C ML 30K 3K 7K 7O 30K 3K 3.5K 9 AP = 20K(A/P, 12%, 3) + 11K = $19327; BP = $12,640; CP = $19,601AML = $14,548 BML = 8229***; CML = 13,276AO = 9026 BO = 5089; CO = 8927

  26. A Bridge • Life 30 years Bid A ($M) Bid B MARR = 5% • Material costs 4.2 6.2Labor costs 0.6 0.7Overhead costs 0.35 0.03Admin & legal 0.60 0.01Annual OC 0.05 0.075Annual revenue ? 0.40 • Annual benefits 0.22 0.25 • Find annual benefits for Bid A to be equivalent to Bid B. • 374,045.75 + 50,000 – 220,000 + X = 1 • 451,457 + 75,000 - 250,000 - 400,000 ans. 0.76759975

  27. Future Worth Analysis • The firms has retained earnings of $1.2M, $1M and $950K respectively 3, 2, and 1 year ago. This year the firm has • $1.8M to invest. The MARR is 18%. Find the value of a project that can undertaken at 25% down payment. • Present value of funds is • $4,485,038.40 + 1.8E6 = $6,285,038.40

  28. Breakeven • Process A has fixed costs of $10K and unit costs of $4.50 each. Process B has fixed costs of $25K and unit costs of $1.50 each. At what level of annual production are the costs of the two processes the same? • 50 b) 500 c) 5000 d)50,000 • 4.5X + 10K = 1.5X + 25K • 3X = 15K • X = 5K

  29. Breakeven • Trailers, Inc makes 30 per month but fixed costs are $750K/month with variable costs of $35K and production has dropped to 25 units per month in the last 5 months. • Revenue generated is $75K per unit. a) How does 25 compare with BE? b) Calculate current profit. • a) 75K * Q = 750K + 35K * Q => QBE = 18.75 units => • 25 units is still above breakeven. • b) 25(75 – 35)K – 750K = $250K/month

  30. Multiple Rates • An investment pays interest at the rate of 10%, 12%, 11%, 8%, and 9% for 5 years. The equivalent uniform rate of interest is ____. If you invested $10K on this deal, how much would you have at the end of 5 years? • (1 + i)5 = 1.10 * 1.12 * 1.11 * 1.08 * 1.09 • = (* 1.1 1.12 1.11 1.08 1.09) • = 1.609845 • i = 1.6098451/5 – 1 = 10%. • F = (FGP 10E3 10 5) = $16,105.10

  31. $100K Budget at 10% Rate

  32. Mutually Exclusively; No Replacement; 8% (IRR '(-5000 2000 2000 2000 2000 2000 2000 -2000 -2000 -1000))  25.17%

  33. Future Worth Analysis • 9-7 Find the future worth of $50K in 5 years if invested at 16% with 48 compounding periods per year. • F = 50K(1 + 0.00333)240 = $111,129.10. • 9-10 Salary is now $32K with $600 increase each year for 30 years of which 10% of yearly salary earns at 7% per year. Find future worth. • F = 3200(F/A, 7, 30) + 60(P/G, 7%, 30)(F/P, 7%, 30) • = $357,526.62

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