Advantages and Uses of Annual WorthII)

8/8/2019 Advantages and Uses of Annual WorthII)

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Chapter 6Chapter 6Annual Worth AnalysisAnnual Worth Analysis


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6.1 Advantages and Uses of Annual6.1 Advantages and Uses of Annual

WorthWorth Popular Analysis TechniquePopular Analysis Technique

Easily understoodEasily understood results are reported inresults are reported in

dollars per time perioddollars per time period Eliminates the LCM computation problemEliminates the LCM computation problem

associated with the present worth methodassociated with the present worth method

Only have to evaluate one life cycle of a projectOnly have to evaluate one life cycle of a project


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Ch. 6: AgendaCh. 6: Agenda

AW over one project life cycleAW over one project life cycle

AW calculationsAW calculations

Capital Recovery analysisCapital Recovery analysisTwo approachesTwo approaches

Selecting Alternatives by Annual WorthSelecting Alternatives by Annual Worth

AW analysis for a permanent investmentAW analysis for a permanent investment


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66 -- Learning Objectives: TopicsLearning Objectives: Topics

AW over one project life cycleAW over one project life cycle

AW calculationsAW calculations

Capital Recovery analysisCapital Recovery analysisTwo approachesTwo approaches

SelectingAlternatives by Annual WorthSelectingAlternatives by Annual Worth

AW analysis for a permanent investmentAW analysis for a permanent investment


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6.1 Advantages and Uses of Annual6.1 Advantages and Uses of Annual

WorthWorth Popular Analysis TechniquePopular Analysis Technique

Easily understoodEasily understood results are reportedresults are reported

in dollars per time periodin dollars per time period Eliminates the LCM computationEliminates the LCM computation

problem associated with the presentproblem associated with the presentworth methodworth method

Only have to evaluate one life cycle of aOnly have to evaluate one life cycle of aprojectproject


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6.1 AWCalculations6.1 AWCalculations

General in nature such that:General in nature such that:

AW = PW(A/P,i%,n)AW = PW(A/P,i%,n) AW = FW(A/F,i%,n)AW = FW(A/F,i%,n)

Convert all cash flows to their equivalentConvert all cash flows to their equivalent

endend--ofof--period amountsperiod amounts


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6.16.1 EEquivalentquivalent AAnnualnnual CCostost

CashCash--Flow analysis approach where the cashFlow analysis approach where the cashflows are converted to their respective equal,flows are converted to their respective equal,endend--ofof--period amounts.period amounts.

The result is reported in terms of $/periodThe result is reported in terms of $/period Variant of the present worth approachVariant of the present worth approach

PopularPopular with some managers who tend to thinkwith some managers who tend to thinkin terms of $/year, $/months, etc.in terms of $/year, $/months, etc.


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6.1 AW and Repeatability Assumption6.1 AW and Repeatability Assumption

If two or more alternatives possessIf two or more alternatives possess

unequal lives, then one need only evaluateunequal lives, then one need only evaluate

theAW for any given cycletheAW for any given cycle

The annual worth of one cycle is theThe annual worth of one cycle is the

same as the annual worth of the othersame as the annual worth of the other

cycles (by assumption)cycles (by assumption)


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6.1 Repeatability Assumption6.1 Repeatability Assumption

Given alternatives with unequal livesGiven alternatives with unequal lives

The assumptions areThe assumptions are::

1.1. The services so provided are needed foreverThe services so provided are needed forever

2.2. The first cycle of cash flows is repeated forThe first cycle of cash flows is repeated forsuccessive cyclessuccessive cycles

3.3. All cash flows will have the same estimatedAll cash flows will have the same estimatedvalues in every life cyclevalues in every life cycle


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6. 1. One or More Cycles6. 1. One or More Cycles

Cycle 1 Cycle 2 Cycle K

Annualize any one of the cycles

AW assumes repeatability of CFs

Find the annualworth of any given

cycle ($/period)


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6.1 66.1 6--year & 9year & 9--year Problem (Ex. 6.1)year Problem (Ex. 6.1)

See Example 6.1See Example 6.1

Present Worth would mandate a 18Present Worth would mandate a 18--yearyear

study periodstudy period 3 Cycles of the 63 Cycles of the 6--year projectyear project

2 cycles of the 92 cycles of the 9--year projectyear project

6 year Project

9 year Project


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6.1 66.1 6--year & 9year & 9--year Problem (Ex. 6.1)year Problem (Ex. 6.1)

Need an 18Need an 18--year study period for bothyear study period for both

Present Worth would mandate a 18Present Worth would mandate a 18--year studyyear studyperiodperiod

3 Cycles of the 63 Cycles of the 6--year projectyear project 2 cycles of the 92 cycles of the 9--year projectyear project

Means a lot of calculation time!Means a lot of calculation time!

6-year Project 6-year Project 6-year Project

9-year Project 9-year Project


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6.1 Example 6.16.1 Example 6.1 ContinuedContinued

WithAW analysis if one assumes the cashWithAW analysis if one assumes the cash

flow patterns remain the same for the 6flow patterns remain the same for the 6--

and 9and 9--year projects, then all one has to doyear projects, then all one has to do

is:is:

6-year Project

9-year Project

Find the AW of any 6 year cycle

Find the annual worth ofany 9-year cycle

And then compare the AW6 yr to AW9 yr.


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6.1 Multiple cycle..same result!6.1 Multiple cycle..same result!

For two cyclesFor two cycles

EAC

= 3000 + 5000 (1+(P

|F, .10, 5))(A|P, .10, 10)EA

C= 3000 + 5000 (1+(

P|F, .10, 5))(A|

P, .10, 10)

= 3000 + 1319 == 3000 + 1319 = $4319/yr$4319/yr

0 1 2 3 4 5 6 7 8 9 10

$5,000

$5,000

A 1-10 = $3,000


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6.1 Advantages of AW6.1 Advantages of AW

Applicable to a variety of engineeringApplicable to a variety of engineering

economy studieseconomy studies

Asset ReplacementAsset Replacement

BreakevenAnalysisBreakevenAnalysis

Make or Buy DecisionsMake or Buy Decisions

Studies dealing with mfg. CostsStudies dealing with mfg. Costs


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6.1 AW Requirements6.1 AW Requirements

Similar to the Present Worth Method,AWSimilar to the Present Worth Method,AW

analysis requires:analysis requires:

Adiscount rateAdiscount rate beforebefore the analysis is startedthe analysis is started

Estimates of the future cash flowsEstimates of the future cash flows

Estimate of the time period(s) involvedEstimate of the time period(s) involved


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6.2 Capital Recovery and AW Values6.2 Capital Recovery and AW Values

Assume the potential purchase of anyAssume the potential purchase of anyproductive assetproductive asset

One needs to know or estimate:One needs to know or estimate: Initial InvestmentInitial Investment -- PP

Estimated Future Salvage ValueEstimated Future Salvage Value -- SS

Estimated life of the assetEstimated life of the asset -- NN

Estimated operating costs and timingEstimated operating costs and timing AOCAOC(Asset Operation Cost)(Asset Operation Cost)

Operative interest rateOperative interest rate i%i%


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6.2 Capital RecoveryCost6.2 Capital RecoveryCost

Management is concerned about the equivalentManagement is concerned about the equivalentannual cost ofannual cost of owningowning a productive asset.a productive asset.

This cost is termed Capital Recovery CostThis cost is termed Capital Recovery Cost

CR is a function of {P, S, i%, and n }CR is a function of {P, S, i%, and n }


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6.2 Capital RecoveryCost (CR)6.2 Capital RecoveryCost (CR)

CR = the equivalent annual worth of theasset given:

Capital Recovery (CR) is the annualizedCapital Recovery (CR) is the annualized

equivalent of the initial investment Pequivalent of the initial investment P00 and theand the

annualized amount of the future salvage value Fannualized amount of the future salvage value Fnn

.

P0

FN

0 1 2 3 N-1 N

S


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6.2 Capital RecoveryCost6.2 Capital RecoveryCost CRCR

Given:

Convert to:

.0 1 2 3 N-1 N

P0

FN

$A per year (CRC)

P0

FN

0 1 2 3 N-1 N.

S


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6.2 Capital RecoveryCost6.2 Capital RecoveryCost

COMPUTING CR FORINVESTMENTSCOMPUTING CR FORINVESTMENTSWITH SALVAGE VALUES:WITH SALVAGE VALUES:

Method 1:Method 1:

CR =CR = --[P(A|P, i, n)[P(A|P, i, n) -- S(A|F, i, n)]S(A|F, i, n)]

P

.. .

Invest P 0 $

N

Salvage for FN $ at t = N

S


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6.2 More Traditional CR Approach6.2 More Traditional CR Approach

Method IIMethod II Subtract the salvage value from theSubtract the salvage value from theoriginal cost and compute the annual cost of theoriginal cost and compute the annual cost of the

difference.difference. Add to that the interest that the salvageAdd to that the interest that the salvagevalue would return each year, SV (i).value would return each year, SV (i).

CR(i%)=CR(i%)= --[([(PP -- SS) (A|P,) (A|P, ii,, nn) +) + SS((ii)])]


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6.2 CR6.2 CR -- ExplainedExplained

A firms management invests the ownersA firms management invests the owners

money in productive assets.money in productive assets.

Not managements moneyNot managements money belongs to thebelongs to theowners of the firm.owners of the firm.

The owners expect a return on their fundsThe owners expect a return on their funds

being invested.being invested.


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If the firm invests in a productive asset, thenIf the firm invests in a productive asset, then

the annual cost of the investment at time t =the annual cost of the investment at time t =0 and0 and

The estimated future salvage value n timeThe estimated future salvage value n timeperiods henceperiods hence

Must be evaluated using an appropriateMust be evaluated using an appropriatediscount rate because.discount rate because.


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The owners funds are tied up in theThe owners funds are tied up in theinvestment cost at time t = 0 and could haveinvestment cost at time t = 0 and could have

been invested elsewhere.been invested elsewhere.

Thus, the annualized cost of the time t = 0Thus, the annualized cost of the time t = 0investment is determined to reflect theinvestment is determined to reflect the

commitment of funds to the asset.commitment of funds to the asset.


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6.3 Alternatives by Annual Worth6.3 Alternatives by Annual Worth

Given a discount rate (in advance)Given a discount rate (in advance)

AW is perhaps the easiest method to apply forAW is perhaps the easiest method to apply for

analysis of alternativesanalysis of alternatives

Mutually ExclusiveAnalysisMutually ExclusiveAnalysis

Select the one best alternativeSelect the one best alternative

SingleAlternativeSingleAlternative

Accept ifAW positive, else rejectAccept ifAW positive, else reject


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6.3 AW6.3 AW Mutually ExclusiveMutually Exclusive

Given a set of two or more alternatives,Given a set of two or more alternatives,

determine the AW(i%), thendetermine the AW(i%), then

Select the alternative with the lowest annual cost orSelect the alternative with the lowest annual cost or

the highest annual net cash flowthe highest annual net cash flow

If pure cost situationIf pure cost situation select min. cost alternativeselect min. cost alternative

If mixed costs and revenuesIf mixed costs and revenues select the max.AWselect the max. AW

(i%) alternative(i%) alternative


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Example 6.3Example 6.3

In this example a pizza delivery service isIn this example a pizza delivery service isconsidering the purchase of 5 in car mappingconsidering the purchase of 5 in car mappingsystems to improve delivery. Each costs $4600systems to improve delivery. Each costs $4600

and has a 5 yr useful life. Salvage is $300 forand has a 5 yr useful life. Salvage is $300 for

each unit. Operating costs for 5 units is $650 theeach unit. Operating costs for 5 units is $650 thefirst year, increasing by $50 per year thereafter.first year, increasing by $50 per year thereafter.

The MARR is 10%.The MARR is 10%.

a. What new income is required to justify thea. What new income is required to justify theinvestment?investment?


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6.3 Example 6.36.3 Example 6.3

Cash Flow Diagram is:Cash Flow Diagram is:

1 2 3 4 5

P=-23,000

S = +$1,500

-$650

-$700

-$750

-$800

-$850


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The Capital Recovery component is:The Capital Recovery component is:

1 2 3 4 5

P=-23,000

S =+$1,500

CR(10%) = -23,000(A/P,10%,5) +

1,500(A/F,10%,5) = -$5,822


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RevenueRevenue Op Costs are:Op Costs are:

1 2 3 4 5

$650

$700

$750$800

$850


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6.3 Example 6.36.3 Example 6.3 Cost Revenue component is seen to equal (costsCost Revenue component is seen to equal (costs

treated as positive values):treated as positive values):

==--650650 -- 50(A/G,10%,5)50(A/G,10%,5)

==--650650 90.5090.50

= $= $--740.50

740.50

1.8101


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TotalAnnual worth (CR + Cost)TotalAnnual worth (CR + Cost)

CR(10%) =CR(10%) = --$5,822$5,822

Revenue/Cost Annual amount: $Revenue/Cost Annual amount: $--740.50740.50

AW(10%) =AW(10%) = --$5,822+$$5,822+$--740.50740.50

AW(10%) =AW(10%) = $$--6,562.506,562.50

Thisamount would be required to recover the investmentThisamountwouldbe requiredtorecoverthe investmentandoperatingcostsatthe 10% rate onaperandoperatingcostsatthe 10% rate onaper--yearbasisyearbasis


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6.4 AW of a Perpetual Investment6.4 AW of a Perpetual Investment

EAC of a perpetual investmentEAC of a perpetual investment

If an investment has no finite cycle, it is called aIf an investment has no finite cycle, it is called a

perpetual investmentperpetual investment. If P is the present worth. If P is the present worthof the cost of that investment, then EAC is P timesof the cost of that investment, then EAC is P timesi, the interest P would have earned each year.i, the interest P would have earned each year.

EAC=A =EAC=A = P* iP* i

Remember: P = A/i

From the previous chapter


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6.4 Example:P

erpetual Investment6.4 Example:P

erpetual Investment EXAMPLEEXAMPLE

Two alternatives are considered for covering aTwo alternatives are considered for covering a

football field.football field.The first is toThe first is to plant natural grassplant natural grass and the second isand the second istoto install AstroTurfinstall AstroTurf. Interest rate is 10%/year.. Interest rate is 10%/year.

Assume the field is to last a long time.Assume the field is to last a long time.

Use AW to evaluate the alternatives.Use AW to evaluate the alternatives.


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6.4 Example6.4 Example ContinuedContinued

Alternative A:Alternative A:

Natural GrassNatural Grass Replanting will be required each 10Replanting will be required each 10years at a cost of $10,000. Annual cost foryears at a cost of $10,000. Annual cost formaintenance is $5,000. Equipment must bemaintenance is $5,000. Equipment must bepurchased for $50,000, which will be replaced afterpurchased for $50,000, which will be replaced after

5 years with a salvage value of $5,0005 years with a salvage value of $5,000


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6.4 Example6.4 Example ContinuedContinued

Alternative B:Alternative B:

Astro TurfAstro Turf This option has a first cost ofThis option has a first cost of

$150,000, and an annual maintenance cost of$150,000, and an annual maintenance cost of$5,000.$5,000.


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6.46.4 Example:Natural GrassExample:Natural Grass

Alternative A:Alternative A:

0 1 2 3 4 5 6 7 8 9 10

$10,000

A = $5,000

F5 = $5,000

P = $50,000+ $10,000

F5 = $5,000

F5=$50,000


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6.4 Example: Natural Grass6.4 Example: Natural Grass AnalysisAnalysis

((--) $60,000 (A/P,10%,10)) $60,000 (A/P,10%,10)

((--) $5,000 (already an annual cost)) $5,000 (already an annual cost)

((--) $50,000 (P/F,10%,5)(A/P,10%,10)) $50,000 (P/F,10%,5)(A/P,10%,10) (+) $5,000 (P/F,10%,5)(A/P,10%,10)(+) $5,000 (P/F,10%,5)(A/P,10%,10)

((--) $10,000 (A/F,10%,10)) $10,000 (A/F,10%,10)

(+) $5,000 (A/F,10%,10)(+) $5,000 (A/F,10%,10) == $$--19,046/year19,046/year


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6.4 Example: Natural Grass6.4 Example: Natural Grass AnalysisAnalysis

((--) $10,000 (A/P,10%,10)) $10,000 (A/P,10%,10)

((--) $5,000 (already an annual cost)) $5,000 (already an annual cost)

((--) $50,000 (A/P,10%,5)) $50,000 (A/P,10%,5)

(+) $5,000 (A/F,10%,5)(+) $5,000 (A/F,10%,5)

== $$--18,998/year18,998/year


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6.4 Example: Artificial Carpet (Surface)6.4 Example: Artificial Carpet (Surface)

A = P(i) for a perpetual life projectA = P(i) for a perpetual life project

Annual Cost of Installation:Annual Cost of Installation:

=$=$--150,000 (.10) = $150,000 (.10) = $--15,000/ year15,000/ year

Annual Maintenance = $Annual Maintenance = $--5,000/year5,000/year

Total: $Total: $--15,00015,000 -- $5,000 =$5,000 = $$--20,000/Yr20,000/Yr Choose A, cost less per year!Choose A, cost less per year!


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Problem 6Problem 6--1010

The machines shown are underThe machines shown are underconsideration for anconsideration for animprovement to an automatedimprovement to an automatedcandy bar wrapping process.candy bar wrapping process.

Determine which should beDetermine which should beselected on the basis of anselected on the basis of anannual worth analysis using anannual worth analysis using aninterest rate of 15% per year.interest rate of 15% per year.

MachineMachineCC

MachineMachineDD

First CostFirst Cost $$--40,00040,000 $$--65,00065,000

AnnualAnnualCostCost

--10,00010,000 --12,00012,000

SalvageSalvage 12,00012,000 25,00025,000

Life, yrsLife, yrs 33 66


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Advantages and Uses of Annual WorthII)

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