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Dr. C.J. Su IEEM Dept. HKUST Definitions FInvestment Proposal: a single undertaking or project being considered as an investment possibility. –Independent Proposal: the acceptance from a set of alternatives has no effect on the acceptance of any other proposals in the set. –Dependent Proposals Mutually exclusive proposal : the acceptance of one proposal precludes the acceptance of any of the others. Contingent proposal: the acceptance of the proposal is dependent on the acceptance of some prerequisite proposal.

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Dr. C.J. Su IEEM Dept. HKUST Examples FIndependent Proposals - the purchase of a CNC milling machine, a security system, office furniture, and fork lift trucks. FDependent Proposals –Mutually exclusive proposal : Select a course from a set of courses that have the same time slot. Select different brand of equipment that perform the same functions. –Contingent proposal: the purchase of software is contingent on the purchase of hardware. The construction of the 3rd floor is contingent on the construction of 1st & 2nd floors.

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Dr. C.J. Su IEEM Dept. HKUST Comparing Alternatives with Equal Planning Horizon A & B Are Mutually Exclusive FAt i%= MARR FIf PW(A) > PW(B) => Accept A Else Accept B FIf FW(A) > FW(B) => Accept A Else Accept B FIf AW(A) > AW(B) => Accept A Else Accept B

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Dr. C.J. Su IEEM Dept. HKUST Example FThree mutually exclusive investment alternatives for implementing an office automation plan in a firm are being considered. The study period is 10 years, and the useful lives of all three alternatives are also 10 years. Market values of all alternatives are zero at the end of their useful lives. If the firm's MARR is 10% /year, which alternative should be selected ? A B C Alternative Investment Net Revenue -390, , ,000 69, , ,500

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Dr. C.J. Su IEEM Dept. HKUST Solution FPW(10%)A = -$390,000 + $69,000(P/A, 10%, 10) = $33,977 FPW(10%)B = -$920,000 + $167,000(P/A, 10%, 10) = $106,148 FPW(10%)C = -$660,000 + $133,500(P/A, 10%,10) = $160,304 FC > B > A, means C is preferred to B and B is preferred to A.

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Dr. C.J. Su IEEM Dept. HKUST IRR Method If IRR of (A - B) > MARR => the incremental investment is justified; therefore proposal A should be selected If A & B are mutually exclusive alternatives If IRR(A - B) > MARR => Accept A, Reject B Else Accept B and Reject A

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Dr. C.J. Su IEEM Dept. HKUST Example Alternative A B (B -A) A B (B -A) Capital investment - 60, , ,000 Net Annual revenues 22,000 26,225 4,225 N = 4 years, MARR = 10% Alternative IRR PW (10%) A 17.3% 9,738 B 16.3% 10,131

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Dr. C.J. Su IEEM Dept. HKUST Incremental Analysis Procedure FArrange the alternatives on the order of increasing capital investment Case 1: For Investment Alternatives: Compute the IRR for each alternatives Compute the IRR for each alternatives –If all IRR Do nothing –If exactly one alternative’s IRR > MARR => Select this alternative –If more than one alternative’s IRR > MARR, use incremental criterion to select the best alternative. Case 2: For Cost Alternatives: Use incremental criterion to select the best alternative

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Dr. C.J. Su IEEM Dept. HKUST Investment Proposal Example FSuppose that we are analyzing the following six mutually exclusive alternatives for a project (arranged in ascending order of initial investment) using the IRR method. The useful life of each alternative is 10 years, and the MARR is 10% per year. Also, net annual revenues less expenses vary among all alternatives. If the study period is 10 years, and the salvage (market) values are 0, which alternative should be chosen? A B C D E F A B C D E F Capital investment , , , ,000 -7,000 Annual revenues less expenses ,125 1,425 less expenses ,125 1,425

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Dr. C.J. Su IEEM Dept. HKUSTSolution A B C D E F A B C D E F IRR 10.6% 13.0% 9.6% 19.1% 18.3% 15.6% F FAlternative C is unacceptable IRR(C) < MARR F FSelect A as the base for comparison

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Dr. C.J. Su IEEM Dept. HKUST Cost Proposal Example Design Alternative D1 D2 D3 D4 D1 D2 D3 D4 Capital investment Capital investment -100, , , ,000 Annual expenses - 29, , , ,100 Useful life (years) Market value 10,000 14,000 25,600 14,000

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Dr. C.J. Su IEEM Dept. HKUST ERR Example F The analysis period is six years, and the MARR for capital investments at the plant is 20% per year before taxes. Using the ERR method, which alternative should be selected? ( = MARR.) | - 640,000|(F/P, i '%,6)= 262,000(F/P,20%,5) ,000 = 2,853,535 | - 640,000|(F/P, i '%,6)= 262,000(F/P,20%,5) ,000 = 2,853,535

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Dr. C.J. Su IEEM Dept. HKUST Parking Lot Construction Example Capital Net Annual Capital Net Annual Investment Income Investment Income P. Keep existing parking - 200,000 22,000 lot, but improve B1. Construct one-story - 4,000, ,000 building B2. Construct two-story - 5,550, ,000 building B3. Construct three-story - 7,500, ,000 building MARR = 10%, N= 15 years, Salvage = Initial Investment

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Dr. C.J. Su IEEM Dept. HKUST Selection based on PW Method PW(10%)p = -$200,000 + $22,000(P/A,10%,15) + $200,000(P/F,10%,15) = $15,214 PW(10%) B1 = - 4,000,000 + $600,000(P/A,10%.15) + $4,000,000(P/F,10%,15) = $1,521,260 PW(10%) B2 = - $5,550,000 + $720,000(P/A,10%,15) + $5,550,000(P/F,10%,15) = $1,255,062 PW(10%) B3 = - $7,500,000 + $960,000(P/A,10%,15) + $7,500,000(P/F,10%,15) = $1,597,356 From PW => Select B3

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Dr. C.J. Su IEEM Dept. HKUST FSelection based on IRR Mutually Exclusive Alternatives P B1 - P B2 - B1 B3 - B1 Capital - 200, ,000, ,550, ,500,000 investment Net annual 22, , , ,000 income Residual 200,000 4,000,000 5,550,000 7,500,000 value IRR 11% 15% 13% 12.8%

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Dr. C.J. Su IEEM Dept. HKUST

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Conclusion FPW and IRR Incremental Analysis methods reach consistent selection for mutually exclusive alternatives. FWhenever possible try Not to use IRR to compare alternatives. Use PW, FW, or AW.

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Dr. C.J. Su IEEM Dept. HKUST Comparing Alternatives with Unequal Lives FWhen comparing alternatives with unequal lives, the principle that all alternatives under consideration must be compared over the same time span is basic to sound decision making.

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Dr. C.J. Su IEEM Dept. HKUST Method 1 Estimation of Required Cash Flow FWhen the required cash flow (salvage value) can be estimated, this method can be applied. Case 1: Alternative’s Useful life > study period the salvage value for the alternative extending beyond the study period must be directly estimated.

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Dr. C.J. Su IEEM Dept. HKUST Example Alternatives EOY A B 0-15, , , , ,000- 2, ,000- 2, , , , , , , , , ,000 ____ 4 - 6,000 ____ 5 - 6, ,000 ____ 5 - 6, ,000 ____

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Dr. C.J. Su IEEM Dept. HKUST Suppose study period (planning horizon) = , ,000 /year , ,000 /year 4,000 Estimated A B 3,000

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Dr. C.J. Su IEEM Dept. HKUST FIf MARR = 20%, Alternatives A & B “coterminated” at year 3 AW (A) = -15,000 (A/P, 20%, 3) - 6, ,000(A/F, 20%, 3) = -12,021 4,000(A/F, 20%, 3) = -12,021 AW(B) = -20,000(A/P, 20%, 3) - 2,000 = -11,494 => B > A => B > A Using PW or FW will yield the same result.

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Dr. C.J. Su IEEM Dept. HKUST Case 2: Alternative’s Useful life < study period the operational cost and/or revenue value for the alternative extending beyond the study period must be directly estimated. Assuming that the study period = 5 and at year 4 & 5 will require costs $3,000 per year for the last two years of alternative B’s life.

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Dr. C.J. Su IEEM Dept. HKUST , ,000 /year , ,000 /year A B 3, ,000 PW (A) = -15,000 (A/P, 20%,5) - 6, ,000(A/F, 20%, 5) = -10,613 Estimated PW (B) = - 20,000(A/P, 20%, 5) - 2, ,000(F/A, 20%, 2) (A/F, 20%, 5) = -8,984 => B > A

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Dr. C.J. Su IEEM Dept. HKUST Method 2 FFor alternatives that are repeatable (long term planning horizon). For example, public service facility. FThe repeatability assumption assuming that the alternative will repeat identical cash flow pattern until the common study period is reached.

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Dr. C.J. Su IEEM Dept. HKUST Example Two mutually exclusive investment alternatives, A and B, associated with a small engineering project for which revenues as well as expenses are involved. They have useful lives of 4 and 6 years, respectively. If the MARR = 10% per year, show which feasible alternative is more desirable by using equivalent worth methods. Use the repeatability assumption. Two mutually exclusive investment alternatives, A and B, associated with a small engineering project for which revenues as well as expenses are involved. They have useful lives of 4 and 6 years, respectively. If the MARR = 10% per year, show which feasible alternative is more desirable by using equivalent worth methods. Use the repeatability assumption. AB Capital investment - $3,500 - $5,000 Annual revenue 1,900 2,500 Annual expenses ,020 Useful life (years) 4 6 Market value at end of useful life 0 0

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Dr. C.J. Su IEEM Dept. HKUST AW(10%)A = -3,500(A/P,10%,4) + (1, ) = 151 AW(10%)B = -5,000(A/P,10%,6) + (2, ,020) = 332 B > A B > A

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Dr. C.J. Su IEEM Dept. HKUST Consistency in AW, PW, and FW PW(10%)A = - 3, ,500[(P/F,10%,4) + (P/F,10%,8)] + (1, )(P/A,10%,12) = 1,028 PW(10%)B = - 5, ,000(P/F,10%,6) + (2, ,020)(P/A,10%,12) = 2,262 B > A FW(10%)A = [- 3,500(F/P,10%,4) + (1, )(F/A,10%,4)](F/P,10%,2) = 847 (1, )(F/A,10%,4)](F/P,10%,2) = 847 FW(10%)B = - 5,000(F/P,10%,6) + (2, ,020) (F/A,10%,6) = 2,561 (2, ,020) (F/A,10%,6) = 2,561 B > A

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Dr. C.J. Su IEEM Dept. HKUST Pump Model SP240 HEPS9 Capital investment- 33, ,600 Annual expenses: Electrical energy- 2, ,720 Electrical energy- 2, ,720 Maintenance - 1,100 in year 1, in year 4, and increasing and increasing - 100/yr thereafter Maintenance - 1,100 in year 1, in year 4, and increasing and increasing - 100/yr thereafter -500/yr thereafter -500/yr thereafter Useful life (years) 5 9 Salvage value 05,000 The new processing facility is needed by your firm at least as far into the future as the strategic plan forecasts operating requirements. The MARR, before taxes, is 20% per year. Based on this information, which model slurry pump should you select?

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Dr. C.J. Su IEEM Dept. HKUST With the repeatability assumption AW(20%) Sp240 = - 33,200(A/P,20%,5) - 2,165 - [1, (A/G,20%,5)] = -15,187 - [1, (A/G,20%,5)] = -15,187 AW(20%) HEPS9 = - 47,600(A/P,20%,9) + 5,000(A/F,20%,9) - 1,720 - [500(P/A,20%,6) + 100(P/G,20%,6)] x (P/F,20%,3) x (A/P,20%,9) = - 13, ,720 - [500(P/A,20%,6) + 100(P/G,20%,6)] x (P/F,20%,3) x (A/P,20%,9) = - 13,622 HEPS9 > SP240

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Dr. C.J. Su IEEM Dept. HKUST Suppose that the estimated market value of pump model HEPS9 in five years is $15,000, and the firm's MARR remains 20% per year. Which pump model should be selected for this replacement action? AW(20%) HEPS9 = - 47,600(A/P,20%,5) + 15,000(A/F,20%,5) - 1,720 - [$500(P/F,20%,4) + $600(P/F,20%,5)] x (A/P,20%,5) = - 15,783 AW(20%) SP240 = - 15,187 (from previous example) SP240 > HEPS9

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Dr. C.J. Su IEEM Dept. HKUST Imputed (Implied) Market Value Technique FWhen study period T < Useful Life This technique estimates the value of the remaining life for an asset The market value of an asset at time T, MV T MV T = [EW at end of year T of remaining capital recovery amounts] + [EW at end of year T of original market value at end of useful life] where EW means equivalent worth at i = MARR.

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Dr. C.J. Su IEEM Dept. HKUST Example Suppose that the pump example is modified such that another market value for pump model HEPS9, at the end of year five, is developed using the imputed market value technique. The same question is again asked, which pump model (SP240 or HEPS9) should be selected for replacement of the current pump in the catalytic system? The MARR remains 20% per year and the study period remains five years.

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Dr. C.J. Su IEEM Dept. HKUST EW CR = [47,600(A/P,20%,9) - 5,000(A/F,20%,9)] x (P/A,20%,4) = 29,949 Compute the EW at end of year five, based on the original MV at end of useful life: EW MV = 5,000(P/F,20%,4) = 2,412 Then, the new market value estimate at the end of year five is as follows: MVs = EW CR + EW MV = 29, ,412 = 32,361 AW(20%) HEPS9 = - 47,600(A/P,20%,5) + 32,361 (A/F,20%,5) - 1,720 - [$500(P/F,20%,4) + 600(P/F,20%,5)]x (A/P,20%,5) = -13,449 AW(20%) SP240 = -15,187, => pump model HEPS9 > SP240

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Dr. C.J. Su IEEM Dept. HKUST Capitalized Worth (CW) Method F FCW method involves in determining the present worth of all revenues and/or expenses over an infinite length of time (e.g., charity fund, scholarship, scientific foundation, etc.). F FSuppose the end of period uniform payment = A F FCW = PW N-> = A(A/P, i%, N) = A {lim N-> [(1+ i) N - 1]/[i [(1+ i) N ]} = A/i => P * i = A = P(A/P, i, N) => i = (A/P, i, N)

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Dr. C.J. Su IEEM Dept. HKUST Example A firm wishes to endow an advanced manufacturing processes laboratory at a university. The endowment principal will earn interest that averages 8% per year, which will be sufficient to cover all expenditures incurred in the establishment and maintenance of the laboratory for an indefinitely long period of time (forever). Cash requirements of the laboratory are estimated to be $100,000 now (to establish it), $30,000 per year indefinitely, and $20,000 at the end of every fourth year (forever) for equipment replacement. (a) For this type of problem, what study period (N) is, practically speaking, defined to be "forever"? (b) What amount of endowment principal is required to establish the laboratory and then earn enough interest to support the remaining cash requirements of this laboratory forever?

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Dr. C.J. Su IEEM Dept. HKUST (a) As N-> , i = (A/P, i, N) For i = 8%, (A/P,8%,N) = 0.08 = i when N = 100. => N = 100 is essentially forever (b) CW = - 100,000 - [30,000 + $20,000(A/F,8%,4)] / 0.08 = -$530,475

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Dr. C.J. Su IEEM Dept. HKUST FA scholarship offers a student $15,000 a month. What’s the deposit required if the bank’s annual interest rate is 12% nominal. i= 12% / 12 = 1% CW = A / i = 15,000 / 0.01 = 1,500,000

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Dr. C.J. Su IEEM Dept. HKUST Example A selection is to be made between two structural designs. Because revenues do not exist (or can be assumed to be equal), only negative cash flow amounts (costs) and the market value at the end of useful life are estimated, as follows: Structure MStructure N Capital investment- $12,000- $40,000 Market value 0 10,000 Annual expenses - 2,200- 1,000 Useful life (years)10 25 Using the repeatability assumption and the CW method of analysis, determine which structure is better if the MARR is 15% per year.

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Dr. C.J. Su IEEM Dept. HKUST AW(15%) M = -12,000(A/P,15%,10) - 2,200 = - 4,592 = - 4,592 AW(15%) N = - 40,000(A/P,15%,25) + 10,000(A/F,15%,25) + 10,000(A/F,15%,25) - 1,000 = - 7, ,000 = - 7,141 CW(15%) M = AW M / i = - 4,592 / 0.15 = - 30,613 CW(15%) N = AW N / i = -7,141 / 0.15 = - 47,607 M > N

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Dr. C.J. Su IEEM Dept. HKUST Forming mutually exclusive Alternatives – – Independent Proposal: the acceptance from a set of alternatives has no effect on the acceptance of any other proposals in the set. – – Dependent Proposals Mutually exclusive proposal : the acceptance of one proposal precludes the acceptance of any of the others. Contingent proposal: the acceptance of the proposal is dependent on the acceptance of some prerequisite proposal.

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Dr. C.J. Su IEEM Dept. HKUST If Xj = 1 => Accept Xj Xj = 0 => Reject Xj Xj = 0 => Reject Xj For three mutually exclusive projects, the alternatives are:

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Dr. C.J. Su IEEM Dept. HKUST F If there are k independent proposals, then there are 2 k possible selections of alternatives.

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Dr. C.J. Su IEEM Dept. HKUST F FA company is considering two independent sets of mutually exclusive projects. That is, projects A1 and A2 are mutually exclusive, as are B1 and B2. However, the selection of any project from the set of projects A1 and A2 is independent of the selection of any project from the set of projects B1 and B2

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Dr. C.J. Su IEEM Dept. HKUST Five proposed projects are being considered. B1 and B2 are independent of C1 and C2. Also, certain projects are dependent on others that may be included in the final portfolio. Using the PW method and MARR = 10% per year, determine what combination of projects is best if the capital to be invested is (a) unlimited, and (b) limited to $48,000. Project B1 & B2 mutually exclusive and independent of C set Project C1 & C2 mutually exclusive and dependent (contingent) on the acceptance of B2 Project D contingent on the acceptance of C1

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Dr. C.J. Su IEEM Dept. HKUST

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(a) Alternative 6 is the best (b) Alternatives 2 & 6 are excluded due to the budget limit $48,000 => alternative 5 is the best

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