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Dealing with Uncertainty Uncertainty Decisions under Certainty vs. Decisions under Uncertainty (risk) Sources of Uncertainty – possible inaccuracy of the estimates – the type of business involved in relation to the future health of the economy – the type of physical plant and equipment involved – length of the assumed study period HKUST, IEEM, Dr. C. J. Su Nonprobabilistic Methods for Dealing with Uncertainty Breakeven analysis Sensitivity Analysis HKUST, IEEM, Dr. C. J. Su Breakeven Analysis When the selection among alternatives is heavily dependent on a single factor, such as capacity utilization, energy utilization, etc. that is uncertain. A breakeven point for the factor is determined such that two alternatives are equally desirable. Estimating the most likely value of the uncertain factor then choose between the alternatives by and comparing this estimate to the breakeven value. HKUST, IEEM, Dr. C. J. Su Breakeven Analysis EWA (y) = EWB (y) where EWA and EWB = an equivalent worth value for the net cash flow of Alternatives A & B . y = a common factor of interest affecting the equivalent worth of Alternative A and B . The common factor y can be : – – – – Annual revenue and expenses Rate of return Market (or salvage) value Equipment life HKUST, IEEM, Dr. C. J. Su Example Two alternative electric motors that provide 100 hp output are under consideration. An Alpha motor can be purchased for $12,500 and has an efficiency of 74%, an estimated useful life of 10 years, and maintenance expenses of $500 /year. A Beta motor will cost $16,000 and has an efficiency of 92%, a life of 10 years, and annual maintenance expenses of $250. Annual taxes and insurance expenses on either motor will be 1.5 % of the investment. If the MARR is 15%, how many hours per year would the motors have to be operated at full load for the annual expenses to be equal? Assume that market values for both are negligible and that electricity costs $0.05 per kilowatt-hour HKUST, IEEM, Dr. C. J. Su Alpha Machine 1 hp = 0.746 kW. If X = # of hours of operation /year Capital recovery amount (depreciation and minimum profit): -12,500(A/P,15%,10) = -12,500(0.1993) = -2,490 / year Operating expense for power: - (100)(0.746) X ($0.05) /0.74 = - 5.04X per year Maintenance expense: - 500 /year Taxes and insurance: -12,500(0.015) = -187 /year Beta Machine Capital recovery amount: - 16,000(A/P,15%,10) = - 16,000(0.1993) = - 3,190 /year Operating expense for power: - (100) (0.746) X($0.05) /0.92 = - 4.05X /year Maintenance expense: - 250 /year Taxes and insurance: -16,000 (0.015) = - 240 /year At the breakeven point, AW (X) = AW (X) -2,490 - 5.04X - 500 - 187 = - 3,190 - 4.05X- 250 - 240 -5.04X - 3,177 = - 4.05X- 3,680, X = 508 hours/year Graphical Solution At the breakeven point, AW (X) = AW (X) -2,490 - 5.04X - 500 - 187 = - 3,190 - 4.05X- 250 - 240 AW (X) = -5.04X - 3,177 & AW (X) = - 4.05X- 3,680 HKUST, IEEM, Dr. C. J. Su Example The Universal Postal Service is considering the possibility of putting wind deflectors on their long-haul tractors. Three types of deflectors, with the following characteristics, are being considered (MARR = 10% per year): Windshear Blowbv Air-vantage Capital investment - 1,000 - 400 - 1,200 Drag reduction 20% 10% 25% Maintenance/year - 10 -5 -5 Useful life 10 years 10 years 5 years If 5% in drag reduction means 2% in fuel savings per mile, how many miles do tractors have to be driven per year before the Windshear deflector is favored over the other deflectors? Over what range of miles driven per year is Air-vantage the best choice? (Fuel cost is $1.00 per gallon and average fuel consumption is 5 miles per gallon without the deflectors). State any assumptions you make HKUST, IEEM, Dr. C. J. Su Assumption: Repeatability (the deflector will continuously) Let X be the mileage driven per year then be Windshear: Total 8% reduction in fuel, operating cost/year : - (X/5) (0.92)(1.00) = -0.184 X /year AW = - 1,000(A/P,10%,10) - 10 - 0.184 X = - 172.75 - 0.184 X Blowby: Total 4% reduction in fuel, operating cost/year : - (X/5) (0.96)(1.00) = -0.192 X /year AW = - 400(A/P,10%,10) - 5 - 0.192 X = - 70.1 - 0.192 X Air-vantage: Total 10% reduction in fuel, operating cost/year : - (X/5) (0.90)(1.00) = -0.180 X /year AW = -1200(A/P, 10%, 5) - 5 - 0.18 X = -321.56 - 0.18 X used Breakeven points: Windshear vs. Blowby deflector - 172.75 - 0.184X = - 70.1 - 0.192X, X = 12,831 miles/year Windshear vs. Air-vantage deflector - 172.75 - 0.184X = -321.56 - 0.18 X, X = 37,203 miles/year Blowby vs. Air-vantage deflector - 70.1 - 0.192X = -321.56 - 0.18 X, X = 20,955 miles/year Graphically ... HKUST, IEEM, Dr. C. J. Su Example In planning 2-story office building, the architect has submitted two designs. The first provides foundation and structural details so that two additional stories can be added to the required initial two stories at a later date and without modifications to the original structure. This building would cost $1,400,000. The 2nd design, without such provisions, would cost only $1,250,000. If the first plan is adopted, an additional two stories could be added at a later date at a cost of $850,000. If the second plan is adopted, however, considerable strengthening and reconstruction would be required, which would add $300,000 to the cost of a 2-story addition. Assuming that the building is expected to be needed for 75 years, by what time would the additional two stories have to be built to make the adoption of the first design justified? (The MARR is 10% per year.) Let T be the time when the additional 2-stories is needed 1st Design 2nd Design PW cost: First unit - 1,400,000 Second unit - 850,000(P/F, 10%, T) PW(1st design) = PW(2nd Design) - 1,400,000 - 850,000 (P/F, 10%. T) = - 1,250,000 - 1,150,000 (P/F, 10%,T) => (P/F, 10%. T) = 0.5 => T = 7 years - 1,250,000 - 1,150,000(P/F, 10%, T) If the additional space will be required < 7 years then choose the 1st design (with the foundation); Else choose the 2nd design (without the foundation) PW - 2,250,000 1st Design - 2,400,000 2nd Design 7 years T HKUST, IEEM, Dr. C. J. Su Sensitivity Analysis Sensitivity - the relative magnitude of change in the measure of merit (such as PW) caused by one or more changes in estimated study factor values. provide information about the potential impact of uncertainty in selected factor estimates. Its routine use is fundamental to achieving sound results useful in the decision process. HKUST, IEEM, Dr. C. J. Su Example A machine for which cash flow estimates are given in the following list is being considered for immediate installation. Because of the new technology built into this machine, it is desired to investigate its PW over a range of 40% in (a) capital investment, (b) annual net cash flow, (c) market value, and (d) useful life. Based on these estimates, how much can the initial investment increase without making the machine an unattractive venture? MARR = 10% Capital investment, I -$11,500 Revenues/yr 5,000 Expenses/yr -2,000 Market value, MV 1,000 Useful life, N 6 years HKUST, IEEM, Dr. C. J. Su Sensitivity to Capital Investment (I) PW(10%) = - 11,500 + 3,000(P/A, 10%,6) + 1,000(P/F, 10%,6) = 2,130 (a) When the capital investment varies by p% the PW(10%) = - (1 p%)(11,500) + 3,000(P/A,10%,6) + 1,000(P/F,10%,6) If p% = 0 => PW(10%) = 2,130 If p% = 10% PW(10%) = - ( 1.1 )( 11,500 ) + 3,000(P/A,10%,6) + 1,000(P/F,10%,6) = - (1.1 )(11,500) + 3,000 (4.3553) + 1,000( 0.5645) = 1060.4 HKUST, IEEM, Dr. C. J. Su PW (10%) I 4,000 3,000 2,130 2,000 1,060.4 1,000 -40 -30 -20 -10 % Change in Parameter 0 10 -1,000 -2,000 -3,000 20 30 40 Sensitivity to Net Annual CF (A) PW(10%) = - 11,500 + 3,000(P/A, 10%,6) + 1,000(P/F, 10%,6) = 2,130 (a) When the net annual CF varies by a% the PW(10%) = - 11,500 + (1 a%) 3,000(P/A,10%,6) + 1,000(P/F,10%,6) If a% = 0 => PW(10%) = 2,130 If a% = 10% PW(10%) = - 11,500 + ( 1.1 ) 3,000(P/A,10%,6) + 1,000(P/F,10%,6) = - 11,500 + ( 1.1 ) 3,000 (4.3553) + 1,000( 0.5645) = 3886.99 HKUST, IEEM, Dr. C. J. Su PW (10%) A 3886.99 4,000 3,000 2,130 2,000 1,000 -40 -30 -20 -10 % Change in Parameter 0 10 -1,000 -2,000 -3,000 20 30 40 Sensitivity to Market Value (MV) PW(10%) = - 11,500 + 3,000(P/A, 10%,6) + 1,000(P/F, 10%,6) = 2,130 (a) When the Market Value (MV) varies by s % the PW(10%) = - 11,500 + 3,000(P/A,10%,6) + (1 s%) 1,000(P/F,10%,6) If s% = 0 => PW(10%) = 2,130 If s% = 10% PW(10%) = - 11,500 + 3,000(P/A,10%,6) + ( 1.1 ) 1,000(P/F,10%,6) = - 11,500 + 3,000 (4.3553) + ( 1.1 ) 1,000( 0.5645) = 2,186.85 HKUST, IEEM, Dr. C. J. Su PW (10%) 4,000 3,000 2,130 MV Low Slope => Insensitive 2,000 2,186.85 1,000 -40 -30 -20 -10 % Change in Parameter 0 10 -1,000 -2,000 -3,000 20 30 40 Sensitivity to Useful Life (n) PW(10%) = - 11,500 + 3,000(P/A, 10%,6) + 1,000(P/F, 10%,6) = 2,130 (a) When the Useful Life (n) varies by n% the PW(10%) = - 11,500 + 3,000(P/A,10%, 6 n% ) + 1,000 (P/F,10%, 6 n%) If n % = 0 => PW(10%) = 2,130 If n % = 10% PW(10%) = - 11,500 + 3,000(P/A,10%, 6.1) + ( 1.1 ) 1,000(P/F,10%,6.1 ) = - 11,500 + 3,000 (4.3553) + ( 1.1 ) 1,000( 0.5645) = 2,560.35 HKUST, IEEM, Dr. C. J. Su PW (10%) 4,000 N 3,000 2,560.35 2,000 1,000 -40 -30 -20 -10 % Change in Parameter 0 10 -1,000 -2,000 -3,000 20 30 40 Which factor is the most sensitive to PW ? I PW (10%) A 4,000 N MV 3,000 2,000 1,000 -40 -30 -20 -10 % Change in Parameter 0 10 -1,000 -2,000 -3,000 20 30 40 END HKUST, IEEM, Dr. C. J. Su