1/9/20161 engineering economic analysis chapter 9 other analysis techniques

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Engineering Economic AnalysisEngineering Economic Analysis

Chapter 9 Other Analysis Techniques

Public Sector ProjectsPublic 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

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Criticisms of B/CCriticisms 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

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Benefit/Cost RatiosBenefit/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

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Benefits. Costs, DisbenefitsBenefits. 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 IRRMIRR 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

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Incremental AnalysisIncremental 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.

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Benefits - DisbenefitsBenefits - 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.

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Social Discount RateB/C and Modified B/C ratios Benefits (B) Disbenefits (D) Costs (C) Present Worth (B, D, C) Disbenefits SUBTRACTED from BenefitsAlternatives 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

B/C & Modified B/C RatiosB/C & Modified B/C Ratios

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Benefits, Disbenefits, or CostsBenefits, 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.

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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 C 10% MARR 15% Choose A

15% MARR Choose Do Nothing

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Incremental B/C 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 68B/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 YB/C Y-X = 10/25(A/P,15%,5)= 7.46 = 1.34 => Y is betterB/C U-Y = 16/50(A/P,15%,5)=14.92 = 1.073 => U is best.

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Example 9-6Example 9-6

Payback PeriodYear A B0 -1000 -27831 200 12002 200 12003 1200 12004 1200 12005 1200 1200Payback for A (cum-add -1000 200 200 1200 1200 1200)) (-1000 -800 -600 600 1800 3000) => 2.5 yearsPayback for B = 2783/1200 = 2.32 years

Problem 9-8Problem 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

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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)


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.

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n A B C D E F MARR = 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.


ME A B

First cost $500 $800

UA Cost 200 150

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 = 0.827 => Select A

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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

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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


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

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BreakevenBreakeven

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.

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BreakevenBreakeven

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)

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SensitivitySensitivity

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.

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Sensitivity AnalysisSensitivity Analysis

Strategy First Cost Salvage AOC Life

P $20K 0 $11K 3A ML 20K 0 9K 5 O 20K 0 5K 8

P 15K 500 4K 2B ML 15K 1K 3.5K 4 O 15K 2K 2K 7

P 30K 3K 8K 3C ML 30K 3K 7K 7 O 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

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MARR = 12%

A BridgeA 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

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Future Worth AnalysisFuture 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

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BreakevenBreakeven

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) 500 c) 5000 d)50,000

4.5X + 10K = 1.5X + 25K

3X = 15K

X = 5K

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BreakevenBreakeven

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

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Multiple RatesMultiple 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

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$100K Budget at 10% Rate$100K Budget at 10% Rate

Plan Investment Annual Profit

I $20K $5K

II $100K $16.5K

III $70K $10K

IV $50K $8.5K

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Plan Annual Return

I 5K + (100 – 20)K * 0.10 = $13K

II 16.5K + (100 – 100)K * 0.1 = $16.5K Best

III 10K + (100 – 70)K * 0.1 = $13K

IV 8.5K + (100 – 50)k * 0.1 = $13.5K

Mutually Exclusively; No Replacement; 8%Mutually Exclusively; No Replacement; 8%

A B B- A

First cost $10K $15K

Annual maintenance $1K $500

Uniform annual benefit $3000 $4500

Salvage value $1K $2K

Life (years) 8 5

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(IRR '(-5000 2000 2000 2000 2000 2000 2000 -2000 -2000 -1000)) 25.17%

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Future Worth AnalysisFuture 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

1/9/20161 engineering economic analysis chapter 9 other analysis techniques

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