1/9/20161 Engineering Economic Analysis Chapter 9  Other Analysis Techniques

Public Sector Projects Larger investment Longer life 30 to 50+ years (capitalized costs) (A/P, 7%, 30) = ; (AGP, 7%, 50) = 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 1/9/20162

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 1/9/20163

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

1/9/20165 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

MIRR for Indeterminate IRR n cf$5K-7K2K2K (cubic )  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 '( ) 12 15)  23.95% > 15% => good investment. PW(24%) = $2.81K 1/9/20166

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8 Incremental Analysis Consider projects 20-year life with MARR at 6% A B C D E F 1 st cost $4K 2K 6K 1K 9K 10K PW benefit K 9500 B/C ratio B-D = 3360/1K = 3.36 => B > D A-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.

1/9/20169 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.

1/9/ 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 = 1.4 9–7 = 2 6-5=1 2/1 = 2 Y (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 = B/C & Modified B/C Ratios

1/9/ 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.

1/9/ Use rate of return (RoR) analysis for the following 3 mutually exclusive alternatives in reference to an unknown MARR. A B C First Cost$200$300$600 Uniform annual benefits Useful life (years) End salvage Computed 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 C 10%  MARR  15% Choose A 15%  MARR Choose Do Nothing

1/9/ 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 B/C 1.14 ** ** 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 = => U is best.

1/9/ Example 9-6 Payback Period Year A B Payback for A (cum-add ))  ( ) => 2.5 years Payback for B = 2783/1200 = 2.32 years

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 65 th birthday. (FGP (FGP (PGG ) 12 35) 12 10)  $1,160, /9/201615

1/9/ Problem 9-17 Geometric gradient, A 1 = 100, g = 100%, i = 10% n = 10 F 10 = ? (FGP (PGGG ) 10 10)  $113, (FGP $ )

Problem 9-30 Use benefit-cost analysis for the 5-year cash flows below at a MARR of 10%. ABC Cost$600$500$200 UAB B/C C (10%) = 58.3/(200*A/P,10%,5) = B/C B-C (10%) = 80.4/300(A/P, 10%, 5) = B/C A-B (10%) = 19.6/100(A/P, 10%, 5) = Conclude: Select B. 1/9/201617

1/9/ Problem 9-36 n A B C D E F MARR = 15% By inspection: A > F; D > B; B/C ratios: A = 1.14; C = 0.83; D = 1.21; E = 1.25 Incremental analysis: B/C E-D = (52-42)(P/A, 15%, 5) / (150 – 125) = 1.34 => E > D B/C A-E = (68-52)(P/A, 15%, 5) / (200 – 150) = 1.15 => A > E A is best.

Problem 9-46 ME A B First cost $500$800 UA Cost Life (years) 8 8 a)Compute payback if B is picked.  PB = 300/50 = 6 yrs. b)Use 12% MARR and B/C to select. 50(P/A, 12%, 8)/300 = 248/300 = => Select A 1/9/201619

1/9/ Problem 9-49 n MARR = 10% X Y Z B/C X < 1 by inspection and has zero rate of return B/C Y is dominated by Z which is best, showing a positive rate of return ($16.67) at 10%. PW Z (10%) = (P/A, 10%, 4) = = $16.67

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

Problem 9-67 AB First cost$55K$75K Annual expenses Operations Maintenance Taxes/Insurance Find useful life for equivalency at 10% MARR 55K K(P/A,10%,n) = 75K K(P/A,10%,n) 20 = 3.75(P/A,10%, n) (P/A,10%, n) = => 8 years At 0% MARR, (P/A, 0%, n) = => n = years 1/9/201622

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 => x = 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? x = 0.18x => x = 167 miles. 1/9/201623

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 ) 1/9/201624

Sensitivity 3.ABC MARR = 6% First Cost$2K$4K$5K UAB Life NPW(6%)$ Find the maximum first cost of B to become non-preferable. 639(P/A, 6%, 20) - FC B = 3029 => FC B = $ B is preferred if its initial cost does not exceed $ /9/201625

Sensitivity Analysis StrategyFirst CostSalvageAOC Life P $20K 0$11K 3 A ML 20K 0 9K 5 O 20K 0 5K 8 P 15K 500 4K 2 B ML 15K 1K 3.5K 4 O 15K 2K 2K 7 P 30K 3K 8K 3 C ML 30K 3K 7K 7 O 30K 3K 3.5K 9 A P = 20K(A/P, 12%, 3) + 11K = $19327; B P = $12,640; C P = $19,601 A ML = $14,548 B ML = 8229***; C ML = 13,276 A O = 9026B O = 5089; C O = /9/ MARR = 12%

A Bridge Life 30 yearsBid A ($M)Bid B MARR = 5% Material costs Labor costs Overhead costs Admin & legal Annual OC Annual revenue ?0.40 Annual benefits Find annual benefits for Bid A to be equivalent to Bid B. 374, ,000 – 220,000 + X = 1 451, , , ,000 ans /9/201627

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, E6 = $6,285, /9/201628

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? a)50 b) 500c) 5000 d)50, X + 10K = 1.5X + 25K 3X = 15K X = 5K 1/9/201629

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 => Q BE = units => 25 units is still above breakeven. b) 25(75 – 35)K – 750K = $250K/month 1/9/201630

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 = (* ) = i = /5 – 1 = 10%. F = (FGP 10E3 10 5) = $16, /9/201631

$100K Budget at 10% Rate PlanInvestmentAnnual Profit I$20K$5K II$100K$16.5K III$70K$10K IV$50K$8.5K 1/9/ PlanAnnual Return I5K + (100 – 20)K * 0.10 = $13K II16.5K + (100 – 100)K * 0.1 = $16.5K Best III10K + (100 – 70)K * 0.1 = $13K IV8.5K + (100 – 50)k * 0.1 = $13.5K

Mutually Exclusively; No Replacement; 8% ABB- A First cost$10K$15K Annual maintenance $1K$500 Uniform annual benefit$3000$4500 Salvage value$1K$2K Life (years)85 1/9/ (IRR '( ))  25.17%

1/9/ 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( ) 240 = $111, 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