dr. c.j. su ieem dept. hkust definitions finvestment proposal: a single undertaking or project being...

Dr. C.J. Su IEEM Dept. HKUST

DefinitionsDefinitions Investment Proposal: a single undertaking Investment Proposal: a single undertaking

or project being considered as an or project being considered as an investment possibility.investment possibility.– Independent ProposalIndependent Proposal: the acceptance from a : the acceptance from a

set of alternatives has no effect on the set of alternatives has no effect on the acceptance of any other proposals in the set.acceptance of any other proposals in the set.

– Dependent Proposals 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.


ExamplesExamplesIndependent Proposals - the purchase of a Independent Proposals - the purchase of a

CNC milling machine, a security system, CNC milling machine, a security system, office furniture, and fork lift trucks.office furniture, and fork lift trucks.

Dependent Proposals Dependent Proposals – Mutually exclusive proposal : Select a course Mutually exclusive proposal : Select a course

from a set of courses that have the same time from a set of courses that have the same time slot. Select different brand of equipment that slot. Select different brand of equipment that perform the same functions.perform the same functions.

– Contingent proposal: the purchase of software Contingent proposal: the purchase of software is contingent on the purchase of hardware. is contingent on the purchase of hardware. The construction of the 3rd floor is contingent The construction of the 3rd floor is contingent on the construction of 1st & 2nd floors.on the construction of 1st & 2nd floors.


Comparing Alternatives Comparing Alternatives with Equal Planning Horizonwith Equal Planning HorizonA & B Are Mutually ExclusiveA & B Are Mutually ExclusiveAt i%= MARRAt i%= MARRIf PW(A) > PW(B) => Accept AIf PW(A) > PW(B) => Accept A

Else Accept BElse Accept BIf FW(A) > FW(B) => Accept AIf FW(A) > FW(B) => Accept A

Else Accept BElse Accept BIf AW(A) > AW(B) => Accept AIf AW(A) > AW(B) => Accept A

Else Accept BElse Accept B


ExampleExample Three mutually exclusive investment alternatives Three mutually exclusive investment alternatives

for implementing an office automation plan in a for implementing an office automation plan in a firm are being considered. The study period is firm are being considered. The study period is 1010 years, and the useful lives of all three alternatives years, and the useful lives of all three alternatives are also 10 years. Market values of all alternatives are also 10 years. Market values of all alternatives are zero at the end of their useful lives. If the firm's are zero at the end of their useful lives. If the firm's MARR is 10% /year, which alternative should be MARR is 10% /year, which alternative should be selected ?selected ?

A B C

Alternative

InvestmentNet Revenue

-390,000 -920,000 -660,000 69,000 167,000 133,500


SolutionSolution PW(10%)A = -$390,000 + $69,000(P/A, PW(10%)A = -$390,000 + $69,000(P/A, 10%, 10) 10%, 10)

= $33,977= $33,977

PW(10%)B = -$920,000 + $167,000PW(10%)B = -$920,000 + $167,000(P/A, (P/A, 10%, 10) 10%, 10)

= $106,148= $106,148

PW(10%)C = -$660,000 + $133,500PW(10%)C = -$660,000 + $133,500(P/A, (P/A, 10%,10) 10%,10)

= $160,304= $160,304 C > B > A, means C is preferred to B and B is C > B > A, means C is preferred to B and B is

preferred to A. preferred to A.


IRR MethodIRR Method

If IRR of If IRR of (A - B) > MARR => the (A - B) > MARR => the incremental investment is justified; incremental investment is justified; therefore proposal A should be selectedtherefore proposal A should be selected

If A & B are mutually exclusive If A & B are mutually exclusive alternativesalternatives

If If IRR(A - B) > MARR => Accept A, Reject IRR(A - B) > MARR => Accept A, Reject BB

Else Accept B and Reject Else Accept B and Reject AA


ExampleExampleAlternativeAlternative

A BA B (B -A)(B -A)

Capital investmentCapital investment - 60,000 - 60,000 - 73,000 -13,000- 73,000 -13,000Net Annual revenues 22,000Net Annual revenues 22,000 26,225 4,225 26,225 4,225

N = 4 years, MARR = 10%N = 4 years, MARR = 10%

Alternative IRR PW (10%)

A 17.3% 9,738 B 16.3% 10,131


Incremental Analysis Incremental Analysis ProcedureProcedure Arrange the alternatives on the order of Arrange the alternatives on the order of

increasing capital investmentincreasing capital investmentCase 1: For Investment Alternatives:Case 1: For Investment Alternatives: Compute the IRR for each alternativesCompute the IRR for each alternatives– If all IRR < MARR => Do nothingIf all IRR < MARR => Do nothing– If exactly one alternative’s IRR > MARR => If exactly one alternative’s IRR > MARR =>

Select this alternativeSelect this alternative– If more than one alternative’s IRR > MARR, If more than one alternative’s IRR > MARR,

use incremental criterion to select the best use incremental criterion to select the best alternative.alternative.

Case 2: For Cost Alternatives:Case 2: For Cost Alternatives:Use incremental criterion to select the best Use incremental criterion to select the best

alternativealternative


Investment Proposal Investment Proposal ExampleExample Suppose that we are analyzing the following six Suppose that we are analyzing the following six

mutually exclusive alternatives for a project (arranged mutually exclusive alternatives for a project (arranged in ascending order of initial investment) using the IRR in ascending order of initial investment) using the IRR method. The useful life of each alternative is method. The useful life of each alternative is 10 years10 years, , and the MARR is and the MARR is 10%10% per year. Also, net annual per year. Also, net annual revenues less expenses vary among all alternatives. If revenues less expenses vary among all alternatives. If the study period is 10 years, and the salvage (market) the study period is 10 years, and the salvage (market) values are 0, which alternative should be chosen?values are 0, which alternative should be chosen?

A B C D E FA B C D E FCapital investment - 900 - 1,500 - 2,500 - 4,000 - 5,000 -7,000Capital investment - 900 - 1,500 - 2,500 - 4,000 - 5,000 -7,000Annual revenuesAnnual revenues less expensesless expenses 150 276 400 925 1,125 150 276 400 925 1,125

1,4251,425


SolutionSolution A B C D E FA B C D E FIRR 10.6% 13.0% IRR 10.6% 13.0% 9.6%9.6% 19.1% 18.3% 15.6% 19.1% 18.3% 15.6%Alternative C is unacceptable IRR(C) < MARRSelect A as the base for comparison


Cost Proposal ExampleCost Proposal ExampleDesign AlternativeDesign Alternative

D1 D2 D3 D1 D2 D3 D4 D4

Capital investmentCapital investment -100,000 -140,600 -148,200 -122,000

Annual expenses - 29,000 - 16,900 - 14,800Annual expenses - 29,000 - 16,900 - 14,800 - 22,100 - 22,100

Useful life (years) 5 5 5 Useful life (years) 5 5 5 5 5

Market value 10,000Market value 10,000 14,000 14,000 25,60025,600 14,000 14,000


ERR ExampleERR Example The analysis period is six years, and the MARR for capital The analysis period is six years, and the MARR for capital

investments at the plant is 20% per year before taxes. Using investments at the plant is 20% per year before taxes. Using the ERR method, which alternative should be selected? (the ERR method, which alternative should be selected? ( = = MARR.) MARR.)

| - 640,000|(F/P, i '%,6)= 262,000(F/P,20%,5) +... + 260,000 = 2,853,535| - 640,000|(F/P, i '%,6)= 262,000(F/P,20%,5) +... + 260,000 = 2,853,535


Parking Lot Construction Parking Lot Construction ExampleExample

Capital Net AnnualCapital Net Annual Investment IncomeInvestment Income

P. Keep existing parking P. Keep existing parking -- 200,000 22,000 200,000 22,000lot, but improvelot, but improve

B1. Construct one-story - 4,000,000 600,000B1. Construct one-story - 4,000,000 600,000buildingbuilding

B2. Construct two-story - 5,550,000 720,000B2. Construct two-story - 5,550,000 720,000buildingbuilding

B3. Construct three-story - 7,500,000 960,000B3. Construct three-story - 7,500,000 960,000buildingbuilding

MARR = 10%, N= 15 years, Salvage = Initial InvestmentMARR = 10%, N= 15 years, Salvage = Initial Investment


Selection based on PW MethodSelection based on PW Method

PW(10%)p = -$200,000 + $22,000(P/A,10%,15) + PW(10%)p = -$200,000 + $22,000(P/A,10%,15) +

$200,000(P/F,10%,15) = $200,000(P/F,10%,15) = $15,214$15,214

PW(10%)PW(10%)B1B1 = - 4,000,000 + $600,000(P/A,10%.15) + = - 4,000,000 + $600,000(P/A,10%.15) +

$4,000,000(P/F,10%,15) = $4,000,000(P/F,10%,15) = $1,521,260$1,521,260

PW(10%)PW(10%)B2B2 = - $5,550,000 + $720,000(P/A,10%,15) = - $5,550,000 + $720,000(P/A,10%,15)

+ $5,550,000(P/F,10%,15) = + $5,550,000(P/F,10%,15) = $1,255,062$1,255,062

PW(10%)PW(10%)B3B3 = - $7,500,000 + $960,000(P/A,10%,15) = - $7,500,000 + $960,000(P/A,10%,15)

+ $7,500,000(P/F,10%,15) = + $7,500,000(P/F,10%,15) = $1,597,356$1,597,356

From PW => Select B3From PW => Select B3


Selection based on IRRSelection based on IRRMutually Exclusive AlternativesMutually Exclusive Alternatives

PP B1 - P B2 - B1 B3 - B1 B1 - P B2 - B1 B3 - B1

Capital - 200,000 - 4,000,000 - 5,550,000 - Capital - 200,000 - 4,000,000 - 5,550,000 - 7,500,0007,500,000

investmentinvestment

Net annual 22,000 600,000 720,000 Net annual 22,000 600,000 720,000 960,000960,000

income income

Residual Residual 200,000 4,000,000 5,550,000 200,000 4,000,000 5,550,000 7,500,0007,500,000

valuevalue

IRRIRR 11%11% 15% 15% 13% 13% 12.8% 12.8%


ConclusionConclusion

PW and IRR Incremental PW and IRR Incremental Analysis methods reach Analysis methods reach consistent selection for consistent selection for mutually exclusive alternatives.mutually exclusive alternatives.

Whenever possible try Whenever possible try NotNot to to use IRR to compare use IRR to compare alternatives. Use PW, FW, or alternatives. Use PW, FW, or AW.AW.


Comparing Alternatives Comparing Alternatives with Unequal Liveswith Unequal LivesWhen comparing alternatives When comparing alternatives

with unequal lives, the principle with unequal lives, the principle that that all alternatives under all alternatives under consideration must be consideration must be compared over the same time compared over the same time spanspan is basic to sound decision is basic to sound decision making.making.


Method 1 Method 1 Estimation of Required Cash Estimation of Required Cash FlowFlowWhen the required cash flow (salvage When the required cash flow (salvage

value) can be estimated, this method value) can be estimated, this method can be applied.can be applied.

Case 1: Alternative’s Useful life > study Case 1: Alternative’s Useful life > study periodperiodthe salvage value for the alternative the salvage value for the alternative

extendingextending

beyond the study period must be directlybeyond the study period must be directly

estimated.estimated.


ExampleExample

AlternativesAlternatives

EOYEOY A A B B

00 -15,000-15,000 - 20,000- 20,000

11 - 6,000- 6,000 - 2,000- 2,000

22 - 6,000 - 6,000 - 2,000 - 2,000

33 - 6,000 - 6,000 - 2,000 - 2,000

44 - 6,000 - 6,000 ____ ____

55 - 6,000 + 3,000 - 6,000 + 3,000 ____ ____


Suppose study period (planning Suppose study period (planning horizon) = 3horizon) = 3

0 1 2 3 4 5

- 15,000- 6,000 /year

0 1 2 3

- 20,000- 2,000 /year

4,000 Estimated

A

B

3,000


If MARR = 20%, Alternatives A & B If MARR = 20%, Alternatives A & B “coterminated” at year 3“coterminated” at year 3

AW (A) = -15,000 (A/P, 20%, 3) - 6,000 + AW (A) = -15,000 (A/P, 20%, 3) - 6,000 +

4,000(A/F, 20%, 3) = 4,000(A/F, 20%, 3) = -12,021-12,021

AW(B) = -20,000(A/P, 20%, 3) - 2,000 = AW(B) = -20,000(A/P, 20%, 3) - 2,000 = --11,49411,494

=> B > A=> B > A

Using PW or FW will yield the same result.Using PW or FW will yield the same result.


Case 2: Alternative’s Useful life < study Case 2: Alternative’s Useful life < study periodperiod

the operational cost and/or revenue the operational cost and/or revenue value for the alternative extending value for the alternative extending beyond the study period must be beyond the study period must be directly estimated.directly estimated.

Assuming that the study period = 5 Assuming that the study period = 5 and at year 4 & 5 will require costs and at year 4 & 5 will require costs $3,000 per year for the last two $3,000 per year for the last two years of alternative B’s life. years of alternative B’s life.


0 1 2 3 4 5

- 15,000- 6,000 /year

0 1 2 3

- 20,000- 2,000 /year

A

B

3,000

- 3,000 - 3,000

PW (A) = -15,000 (A/P, 20%,5) - 6,000 + 3,000(A/F, 20%, 5) = -10,613

Estimated PW (B) = - 20,000(A/P, 20%, 5) - 2,000 - 1,000(F/A, 20%, 2) (A/F, 20%, 5) = -8,984=> B > A


Method 2Method 2For alternatives that are repeatable For alternatives that are repeatable

(long term planning horizon). For (long term planning horizon). For example, public service facility.example, public service facility.

The repeatability assumption The repeatability assumption assuming that the alternative will assuming that the alternative will repeat identical cash flow pattern repeat identical cash flow pattern until the common study period is until the common study period is reached.reached.


ExampleExample Two mutually exclusive investment alternatives, A and B, Two mutually exclusive investment alternatives, A and B,

associated with a small engineering project for which associated with a small engineering project for which revenues as well as expenses are involved. They have revenues as well as expenses are involved. They have useful lives of useful lives of 4 4 and and 6 6 years, respectively. If the MARR = years, respectively. If the MARR = 10%10% per year, show which feasible alternative is more per year, show which feasible alternative is more desirable by using equivalent worth methods. Use the desirable by using equivalent worth methods. Use the repeatability assumption.repeatability assumption.

AA BBCapital investmentCapital investment - $3,500 - $3,500 - $5,000 - $5,000Annual revenueAnnual revenue 1,900 1,900 2,500 2,500Annual expenses - 645Annual expenses - 645 -1,020 -1,020Useful life (years)Useful life (years) 4 4 6 6Market value at end of useful lifeMarket value at end of useful life 0 0 0 0


AW(10%)A = -3,500(A/P,10%,4) + (1,900 - 645) = 151AW(10%)A = -3,500(A/P,10%,4) + (1,900 - 645) = 151

AW(10%)B = -5,000(A/P,10%,6) + (2,500 - 1,020) = 332AW(10%)B = -5,000(A/P,10%,6) + (2,500 - 1,020) = 332 B > AB > A


Consistency in AW, PW, and Consistency in AW, PW, and FWFWPW(10%)A = - 3,500 - 3,500[(P/F,10%,4) + (P/F,10%,8)] + PW(10%)A = - 3,500 - 3,500[(P/F,10%,4) + (P/F,10%,8)] +

(1,900 - 645)(P/A,10%,12) = 1,028(1,900 - 645)(P/A,10%,12) = 1,028

PW(10%)B = - 5,000 - 5,000(P/F,10%,6) + (2,500 - 1,020)PW(10%)B = - 5,000 - 5,000(P/F,10%,6) + (2,500 - 1,020)

(P/A,10%,12) = 2,262(P/A,10%,12) = 2,262

B > AB > A

FW(10%)A = [- 3,500(F/P,10%,4) + FW(10%)A = [- 3,500(F/P,10%,4) +

(1,900 - 645)(F/A,10%,4)](F/P,10%,2) = 847(1,900 - 645)(F/A,10%,4)](F/P,10%,2) = 847

FW(10%)B = - 5,000(F/P,10%,6) + FW(10%)B = - 5,000(F/P,10%,6) +

(2,500 - 1,020) (F/A,10%,6) = 2,561(2,500 - 1,020) (F/A,10%,6) = 2,561

B > AB > A


Pump ModelSP240 HEPS9

Capital investmentCapital investment - 33,200- 33,200 - 47,600 - 47,600Annual expenses:Annual expenses: Electrical energyElectrical energy - 2,165 - 2,165 - 1,720- 1,720 Maintenance - 1,100 in year 1, - 500 in year 4, and increasing Maintenance - 1,100 in year 1, - 500 in year 4, and increasing

and increasing - 100/yr thereafterand increasing - 100/yr thereafter -500/yr thereafter -500/yr thereafter

Useful life (years)Useful life (years) 5 5 9 9Salvage valueSalvage value 0 0 5,0005,000

The new processing facility is needed by your firm at least as far into the The new processing facility is needed by your firm at least as far into the future as the strategic plan forecasts operating requirements. The future as the strategic plan forecasts operating requirements. The MARR, before taxes, is 20% per year. Based on this information, which MARR, before taxes, is 20% per year. Based on this information, which model slurry pump should you select?model slurry pump should you select?


With the repeatability assumptionWith the repeatability assumption

AW(20%)AW(20%)Sp240Sp240 = - 33,200(A/P,20%,5) - 2,165 = - 33,200(A/P,20%,5) - 2,165

- [1,100 + 500(A/G,20%,5)] = - [1,100 + 500(A/G,20%,5)] = -15,187-15,187

AW(20%)AW(20%)HEPS9HEPS9 = - 47,600(A/P,20%,9) + 5,000(A/F,20%,9) = - 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 - 1,720 - [500(P/A,20%,6) + 100(P/G,20%,6)] x

(P/F,20%,3) x (A/P,20%,9) = (P/F,20%,3) x (A/P,20%,9) = - 13,622- 13,622

HEPS9 > SP240HEPS9 > SP240


Suppose that the estimated market value of pump model Suppose that the estimated market value of pump model HEPS9 in five years is $15,000, and the firm's MARR HEPS9 in five years is $15,000, and the firm's MARR remains 20% per year. Which pump model should be remains 20% per year. Which pump model should be selected for this replacement action?selected for this replacement action?

AW(20%)AW(20%)HEPS9 HEPS9 = - 47,600(A/P,20%,5) + = - 47,600(A/P,20%,5) +

15,000(A/F,20%,5) - 1,72015,000(A/F,20%,5) - 1,720

- [$500(P/F,20%,4) + $600(P/F,20%,5)] x - [$500(P/F,20%,4) + $600(P/F,20%,5)] x

(A/P,20%,5) = - 15,783(A/P,20%,5) = - 15,783

AW(20%)AW(20%)SP240SP240 = - 15,187 (from previous example) = - 15,187 (from previous example)

SP240 > HEPS9SP240 > HEPS9


Imputed (Implied) Market Value TechniqueWhen study period T < Useful LifeWhen study period T < Useful Life

This technique estimates the value of This technique estimates the value of the remaining life for an assetthe remaining life for an asset

The market value of an asset at time T, The market value of an asset at time T, MVMVTT

MVMVT T = = [EW at end of year T of remaining capital [EW at end of year T of remaining capital recovery amounts] + [EW at end of year T of original recovery amounts] + [EW at end of year T of original market value at end of useful life]market value at end of useful life]

where EW means equivalent worth at i = MARRwhere EW means equivalent worth at i = MARR..


ExampleExampleSuppose that the pump example is modified Suppose that the pump example is modified

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


EWEWCRCR = [47,600(A/P,20%,9) - 5,000(A/F,20%,9)] = [47,600(A/P,20%,9) - 5,000(A/F,20%,9)] x (P/A,20%,4) = 29,949x (P/A,20%,4) = 29,949

Compute the EW at end of year five, based on Compute the EW at end of year five, based on the original MV at end of useful life:the original MV at end of useful life:

EWEWMVMV = 5,000(P/F,20%,4) = 2,412 = 5,000(P/F,20%,4) = 2,412Then, the new market value estimate at the Then, the new market value estimate at the

end of year five is as follows:end of year five is as follows:

MVs = EWMVs = EWCRCR + EW + EWMVMV = 29,949 + 2,412 = = 29,949 + 2,412 = 32,36132,361

AW(20%)AW(20%)HEPS9HEPS9 = - 47,600(A/P,20%,5) + 32,361 = - 47,600(A/P,20%,5) + 32,361 (A/F,20%,5) - 1,720 - [$500(P/F,20%,4) + (A/F,20%,5) - 1,720 - [$500(P/F,20%,4) + 600(P/F,20%,5)]x (A/P,20%,5) = 600(P/F,20%,5)]x (A/P,20%,5) = -13,449-13,449

AW(20%)AW(20%)SP240SP240 = = -15,187,-15,187, => pump model => pump model HEPS9 > SP240HEPS9 > SP240


Capitalized Worth (CW) Capitalized Worth (CW) MethodMethodCW 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.).

Suppose the end of period uniform payment = A CW = PWN-> = A(A/P, i%, N)

= A {limN-> [(1+ i)N - 1]/[i [(1+ i)N]} = A/i

=> P * i = A = P(A/P, i, N) => i = (A/P, i, N)


ExampleExampleA firm wishes to endow an advanced manufacturing processes A firm wishes to endow an advanced manufacturing processes laboratory at a university. The endowment principal will earn laboratory at a university. The endowment principal will earn interest that averages interest that averages 8% 8% per year, which will be sufficient to per year, which will be sufficient to cover all expenditures incurred in the establishment and cover all expenditures incurred in the establishment and maintenance of the laboratory for an indefinitely long period maintenance of the laboratory for an indefinitely long period of time (forever). Cash requirements of the laboratory are of time (forever). Cash requirements of the laboratory are estimated to be $100,000 now (to establish it), $30,000 per estimated to be $100,000 now (to establish it), $30,000 per year indefinitely, and $20,000 at the end of every fourth year year indefinitely, and $20,000 at the end of every fourth year (forever) for equipment replacement.(forever) for equipment replacement.

(a) For this type of problem, what study period (a) For this type of problem, what study period (N) is, (N) is, practically practically speaking, defined to be "forever"?speaking, defined to be "forever"?

(b) What amount of endowment principal is required to establish (b) What amount of endowment principal is required to establish the laboratory and then earn enough interest to support the the laboratory and then earn enough interest to support the remaining cash requirements of this laboratory forever?remaining cash requirements of this laboratory forever?


(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


A scholarship offers a student A scholarship offers a student $15,000 a month. What’s the deposit $15,000 a month. What’s the deposit required if the bank’s annual interest required if the bank’s annual interest rate is 12% nominal.rate is 12% nominal.

i= 12% / 12 = 1%i= 12% / 12 = 1%

CW = A / i = 15,000 / 0.01 = CW = A / i = 15,000 / 0.01 = 1,500,0001,500,000


ExampleExampleA selection is to be made between two A selection is to be made between two structural designs. Because revenues do not structural designs. Because revenues do not exist (or can be assumed to be equal), only exist (or can be assumed to be equal), only negative cash flow amounts (costs) and the negative cash flow amounts (costs) and the market value at the end of useful life are market value at the end of useful life are estimated, as follows:estimated, as follows:

Structure MStructure M Structure NStructure NCapital investmentCapital investment - $12,000- $12,000 - $40,000- $40,000Market valueMarket value 0 0 10,000 10,000Annual expensesAnnual expenses - 2,200 - 2,200 - - 1,0001,000Useful life (years)Useful life (years) 1010 25 25

Using the repeatability assumption and the CW method Using the repeatability assumption and the CW method of analysis, determine which structure is better if the of analysis, determine which structure is better if the MARR is 15% per year.MARR is 15% per year.


AW(15%)AW(15%)MM = -12,000(A/P,15%,10) - 2,200 = -12,000(A/P,15%,10) - 2,200 = - 4,592= - 4,592

AW(15%)AW(15%)NN = - 40,000(A/P,15%,25) = - 40,000(A/P,15%,25)

+ 10,000(A/F,15%,25)+ 10,000(A/F,15%,25)

- 1,000 = - 7,141- 1,000 = - 7,141

CW(15%)CW(15%)MM = AW = AWMM / i = - 4,592 / 0.15 = - 30,613 / i = - 4,592 / 0.15 = - 30,613

CW(15%)CW(15%)NN = AW = AWNN / i = -7,141 / 0.15 = - 47,607 / i = -7,141 / 0.15 = - 47,607M > NM > N


Forming Forming mutually exclusive mutually exclusive AlternativesAlternatives

– 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.


If Xj = 1 => Accept XjIf Xj = 1 => Accept Xj

Xj = 0 => Reject XjXj = 0 => Reject Xj

For three mutually exclusive projects, For three mutually exclusive projects, the alternatives are:the alternatives are:


If there are k independent proposals, If there are k independent proposals,

then there are then there are 22kk possible selections possible selections of alternatives.of alternatives.


A 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


Five proposed projects are being considered. B1 and Five proposed projects are being considered. B1 and B2 are independent of C1 and C2. Also, certain B2 are independent of C1 and C2. Also, certain projects are dependent on others that may be projects are dependent on others that may be included in the final portfolio. Using the PW method included in the final portfolio. Using the PW method and MARR = 10% per year, determine what and MARR = 10% per year, determine what combination of projects is best if the capital to be combination of projects is best if the capital to be invested is (a) unlimited, and (b) limited to invested is (a) unlimited, and (b) limited to $48,000.$48,000.

Project B1 & B2 mutually exclusive and independent Project B1 & B2 mutually exclusive and independent of of

C setC set

Project Project C1 & C2 mutually exclusiveC1 & C2 mutually exclusive and dependent and dependent ((contingent) on the acceptance of B2contingent) on the acceptance of B2

Project Project D contingent on the acceptance of C1D contingent on the acceptance of C1


(a) Alternative 6 is the best(a) Alternative 6 is the best

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

dr. c.j. su ieem dept. hkust definitions finvestment proposal: a single undertaking or project being...

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