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Module 7: Incremental Method

SI-4251 Ekonomi Teknik

Muhamad Abduh, Ph.D.

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Outline Module 7

MARR

Incremental Analysis

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The Minimum Attractive Rate of Return

The Minimum Attractive Rate of Return(MARR) is the rate at which an entity can

always invest. MARR is set as the result of apolicy decision by the entity, whichrepresents the entitys profit objective.

MARR is set a based on entitys view of futureopportunities along its financial situation: MARR too low may allow proposal that is

marginally productive or result in a loss.

MARR too high may result in rejectinginvestment that would have good returns.

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Establishing MARR

An entity (corporation) accumulates funds(capital) by means of two sources: debt

financing, equity financing, or the mix of the two

Debt financing refers to capital borrowed fromother party that will be paid back at stated

interest by a specific date. No direct risk involving the lender on repayment offunds and interest, or profits resulting from thefunds

(short, medium, long) terms loans, bonds, mortgage

Capital financing represents capital owner by thecorporation used to generate revenue. Sales of common or preferred stocks for public

corporations

Own money for private companies

Retained earning

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Establishing MARR For capital budgeting and alternative evaluation MARR (the

cost of capital) is set calculated independently for eachtype of financing

The interest rate paid for (cost capital) for mixed financing iscalculated from weighted proportion of source of financing

Weighted Average Cost of Capital:

WACC = ( ) (equity fraction x cost of equity) + ( ) (capital debt fraction x cost of

)debt capital:xampleA company is deciding to increase its capital in order to

. - -finance an alternative investment With a 40 60 D E mix with

. % %, .debt costing 8 5 and equity costing 10 calculate WACC

ACC = ( %)( . %) + ( %)( %) = . %40 8 5 60 10 9 4

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Establishing MARR

MARR is then set based on that cost, whichreflects the view and/or preference of the

entity (corporation) toward alternatives ofinvestment

The MARR varies from one alternative toanother, because of:

Project risk which should return higher thatMARR

Sensitivity of project area lowering MARR inone area may provide incentive to encourageinvestment in other area

Tax structure tax adds to the reduction of netincome

Capital-financing method demand supply for

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Incremental Analysis

With respects to MARR, where unlimitedinvestment opportunities yielding return at theMARR is extended into the future, it can beassumed that the proceeds produced by thecurrent investments can be invested at the

minimum attractive rate of return.The decision for selection of alternatives is based

on the analysis of the difference betweenmutually exclusive alternatives.

The incremental investment analysis considers allfeasible alternatives (that is yielding return >MARR), starting from the least cost investment.

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Incremental AnalysisFund of $ 1,500,000 is available for investment. MARR is set

at 15%

Alternative P: investment $ 1,000,000 @ 21% return

Alternative Q: investment $ 1,400,000 @ 18% return

Alternative R: investment $ 1,250,000 @ 20% return

Alternative

Investmen( )t $ K

Yield RemainingFund

($)K

Yield Total Return

Rate(%)

Return( )$ K

Rate(%)

Return( )$ K

Yield( )$ K

RoR(%)P ,1 000 21 .210 0 500 15 .75 0 .285 0 .19 0

Q ,1 400 18 .252 0 100 15 .15 0 .267 0 .17 8R ,1 250 20 .250 0 250 15 .37 5 .287 5 .9 2

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Incremental ROR Net Cash Flow Tabulation

Rate of return can be calculated from cash flowtabulation of individual alternative.

Selection of alternatives is done by sequentialcomparison of two alternatives, starting fromthe lowest to the next higher initial investment.

For positive cash flow, start with do nothingalternative

Net cash flow (difference between two cash flow)is to be used to calculate incremental ROR

=Net cash flow cash flow B - cash flow = 0 i A B

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Incremental Investment:Net Cash Flow Tabulation

ash Flow A ash Flow B ( - )ash Flow B AInitial cost - , ,125 000 000 - , ,157 750 000 - , ,32 750 000

End of year 1 - , ,9 800 000 + , ,2 800 000 , ,12 600 000

End of year 2 + , ,21 750 000 + , ,11 000 000 - . ,10 750 000

End of year 3 + , ,45 900 000 + , ,65 500 000 , ,19 600 000End of year 3 + , ,88 750 000 + , ,82 750 000 - , ,6 000 000

Salvage value + , ,75 000 000 + , ,95 000 000 , ,20 000 000

/ %PW C F @ 10

Higher initial cost

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Example

Three alternatives investment are beingconsidered at MARR 12%

X Y Z

Initial cost - 650,000,000 -540,000,000 -720,000,000

Yearly expenses - 135,000,000 -123,500,000 -130,000,000

Yearly revenues 330,000,000 321,000,000 357,500,000

Salvage value 45,000,000 52,000,000 202,000,000

period 5 5 5

(yearly cash flow) 195,000,000 197,500,000 227,500,000

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Solution

PW- /Net CF = - + ( / , *, ) + ( / , *, ) =

P A P A i 5 SV P F i 5 0

= % ( / , , ) = .For i 12 P A 12 5 3 6048 ( / , , ) = .P F 12 5 0 5674

= % ( / , , ) = .For i 10 P A 10 5 3 7908 ( / , , ) = .P F 10 5 0 6209

= % ( / , , ) = .For i 15 P A 15 5 3 3522 ( / , , ) = .P F 15 5 0 4972

Y X Z

comparison do nothing to Y

Incremental cost, P -540,000,000

Incremental C/F, A 195,000,000

IncrementalSV, SV 45,000,000

Present Worth C/F @ MARR

Incremental i*

Decision

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Solution

PW- /Net CF = - + ( / , *, ) + ( / , *, ) =

P A P A i 5 SV P F i 5 0

= % ( / , , ) = .For i 12 P A 12 5 3 6048 ( / , , ) = .P F 12 5 0 5674

= % ( / , , ) = .For i 10 P A 10 5 3 7908 ( / , , ) = .P F 10 5 0 6209

= % ( / , , ) = .For i 15 P A 15 5 3 3522 ( / , , ) = .P F 15 5 0 4972

Y X Z

comparison do nothing to Y Y to X Y to Z

Incremental cost, P -540,000,000 -110,000,000 -180,000,000

Incremental C/F, A 195,000,000 2,500,000 32,500,000

IncrementalSV, SV 45,000,000 7,000,000 157,000,000

Present Worth C/F @ MARR ? ? ?

Incremental i* > 12% < 12% > 12%

Decision Select Y Retain Y Select Z

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Homework #7

A ready-mix concrete producer is considering to install a new

mixer system:

a) develop net cash flow tabulation b) if the company has set MARR at 12%, which system

should be installed?

c) if all alternatives are to use MARR, will yourecommend otherwise?

Operating characteristics System A System B System C

( )Installed cost $ , ,2 250 000 , ,2 950 000 , ,2 750 000

( )Annual Operating cost $ ,320 000 ,495 000 ,401 500

( )Annual production cm ,10 500 ,21 200 ,19 900

( / )Unit price $ cm .122 50 .122 50 .122 50( / )Overhaul cost $ 2 years ,220 000 ,245 000 ,295 000

( )Salvage value $ ,221 500 ,308 000 ,367 500

( )Useful life year 4 6 6