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Authored by Don Smith, Texas A&M University 2004

CHAPTER 11

Replacement and RetentionDecisions

McGrawHill

ENGINEERING ECONOMY SixthEdition

Blank and

Tarquin

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Chapter 11 Learning Objectives

1. Basics of Replacement Study;2. Economic Service Life;

3. Performing a Replacement Study;

4. Additional Considerations in aReplacement Study;

5. Replacement Study over aSpecified Study Period;

6. Chapter Summary

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CHAPTER 11

Section 11.1 Basics of Replacement Study

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11.1 Why Replace Assets

Reduced Performance: Wear and Tear;

Decreasing reliability and Productivity;

Increasing operating and maintenance costs.

Altered Requirements: New production needs, accuracy, speed, etc.

Obsolescence:

Current assets may be less productive;

Not state of the art – meet competition.

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11.1 Terminology

Defender Asset:

Current installed asset;

Challenger Asset:

The potential replacement or “challenging” asset;

Under consideration to replace the defender asset.

Together, the Defender and Challenger:

Constitute mutually exclusive alternatives;

Select one and reject the other.

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11.1 Annual Worth Values

Analysis Approach forReplacement:

Annual Worth Approach;

EUAC – since costs tend todominate the study (-) cash

flows;

Salvage values – if any – are alsopart of the analysis (+) value.

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11.1 Economic Service Life

Economic Service Life (ESL) Number of years for an alternative for

which the AW or EUAC is Minimum;

Implies that a period by period analysis

is performed;

Computing the AW for 1 year; then 2

years; … until a minimum cost time

period is found; Performed manually or by spreadsheet.

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11.1 Investment Concerns-Defender

For a replacement analysis twoinvestment costs are critical:

1. The proper investment cost to

apply to keeping the defenderin service;

2. The proper investment cost to

apply to any challenger asset

that might replace the current

defender asset.

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11.1 Investment Concerns-Defender

While it may seem strange tocharge an investment cost for

keeping one’s own asset (the

defender) this is what must occur.

Keeping the defender is not free!

Why? Because the firm is giving up the

opportunity to receive a possible cashflow from selling the current defender!

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11.1 Investment Concerns

One must assign an investmentcost to KEEPING the defender

asset!

The appropriate investment costto assign to the defender asset is:

The current fair market value of

the defender at the time the

replacement decision is being

examined.

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11.1 Defender First Cost

Defender First Cost: Initial investment in the defender asset

back in time;

This investment (cost) is considered

“sunk ” for analysis purposes;

A past cost that cannot be changed or

altered;

The issue of the relevance of theinvestment cost in the analysis will be

addressed soon.

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11.1 Challenger First Cost

This is the total investment (Pchallenger )required in a new (challenger) asset

that will possibly replace the current

defender.

In a replacement study this investment

is know with a fair amount of certainty.

What IF a trade in value is offered for

the defender to apply to thechallenger?

C i ht © Th M G Hill C i I P i i i d f d ti di l

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11.1 Trade In Concerns

Often a trade in value is offered bya vendor to take in the defender

with a credit on the purchase

towards the challenger.

Be careful how this is handled!

Points to focus upon….


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11.1 Basic Principles

The past investment in the defender is“sunk ” and not totally relevant to the

analysis.

Only the Fair Market Value (FMV) of the

defender is relevant. FMV of the defender is the net economic

worth of the current defender;

Sale or disposal price less any costs

associated with removing the defender.


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11.1 Basic Principles – ImportantQuestion

At times, a “high” trade-in valuemay be offered for the defender

compared to its current fair market

value.

If this is the case: What should be the investment cost in

the challenger for a replacement study

analysis if a trade-in value is offered?

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11.1 Trade-In Issues

For a Trade-In, the correct investment cost to assign

to the challenger is:

Investment in the Challenger:

PC – (TIV – MVD)

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Investment in the challenger


Cash Price for the challenger less:

(Trade-in Value – Market Value of the Defender)

This represents the true investment in thechallenger to the firm!

Copyright © The McGraw-Hill Companies Inc Permission required for reproduction or display

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Investment in the challenger


The Cash pricefor

The challenger with

No trade-in

The Opportunity CostGiven up by not

Selling theDefender outright!


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Copyright © The McGraw Hill Companies, Inc. Permission required for reproduction or display.


11.1 Trade-In Issues - Example

Bought a system 3 years ago for$120,000. (Defender);

A fair market value of the current

defender is $70,000 right now;A challenger can be purchased for

cash for $100,000 now!

The vendor selling the challengeroffers a trade-in of $80,000 on the

current defender.


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11.1 Trade-In Issues - Example

What should be the properinvestment cost for the challenger to

the firm if the defender is traded?

PC = $100,000;

TIV = $80,000

FMVD = $70,000

InvestmentChallenger now = $100,000 –

($80,000 - $70,000) = $90,000.

This represents the “true” investment in

the challenger with the trade.


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11.1 Other Issues…

Investment in the challenger assetmust include:

Actual cash price for purchase;

Transportation costs;

Installing/make-ready for use costs;

Other one-time costs at time t = 0

associated with placing the challenger

in-service.


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11.1 Warning! What Not to Do!

At times a decision maker might dothe following:

Take the investment cost in the

challenger;

Then add the remaining book value of the

defender to that investment;

This is wrong!

Overly penalizes the challenger with a

sunk cost associated with the defender

asset!

Do not do this!


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py g p , q p p y


11.1 Sunk Costs

A “sunk” cost is any cost that hasoccurred in the past and cannot be

changed or altered by a current

decision.

The past investment in any

defender or its remaining book

value is not relevant! Unless, an after-tax replacement

analysis is being conducted!

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11.1 Outsider’s or Consultant’sView

One assumes that you own neither of the assets in question;

The service provided by the defender

can be “purchased” with an investment

of the firm’s money equal to the currentfair market value of the defender. Buy your own asset!

Hard to comprehend? Perhaps…but this is an

objective approach to costing the defender!


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11.1 Costing the Defender

The consultant’s view assumes: If the defender is retained in service,

the firm is giving up (forgone

opportunity) a potential cash inflow – IF

NOT REPLACED now!

This view attempts to minimize any bias

towards either the defender or the

challenger!


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11.1 Replacement Approach

The traditional approach toconducting a replacement

analysis is:

The Annual Cost or AnnualWorth approach! With an assumed interest rate;

Assumed lives for each

alternative.


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11.1 Assumptions

The traditional approach toconducting a replacement

analysis is:

The Annual Cost or AnnualWorth approach! With an assumed interest rate;

Assumed lives for each

alternative.

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11.1 Study Period forReplacement

If Infinite then: The required services needed are

needed indefinitely;

The challenger is the bestavailable and if selected will have

the same repeated cycles of costs

forever!

Cost estimates for every future

cycle will be the same!


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11.1 Study Period forReplacement

Study Period Finite: The previous assumptions do not

hold!

See Section 11.5 for a fixed studyperiod analysis.


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Section 11.2Economic Service Life (ESL)

The best value for “n” is notknown in this type of problem.

The ESL for a given asset is:

The number of years where the AW of the future costs is minimum;

Using the cost estimates of all possible

years that the asset may provide a

needed service! Termed, “The minimum cost life”

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11.2 ESL – General Format

Compute: AW(i%)t =:

-Capital Recovery

- AW of operating costs

Salvage values may be incorporated into

the capital recovery term.

Do this for n = 1 then n = 2, then n = …

and observe the min cost “n” value.

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11.2 Components of ESL

Capital Recovery Costs (CRC) CRS’s generally decrease with each year of

operation;

The longer one uses an asset the costs

associated with owning the asset are spread

out over more and more time periods.

/ / / /0 1 2 . . . n-1 n

Investment at

t = 0 (P)

Sn

Diagram for Capital Recovery


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11.2 Capital Recovery Formula

CRC Setup

/ / / /0 1 2 . . . n-1 n

$P

Sn

CRC(i%) = -P(A/P,i%,n) + S(A/F,i%,n)

CRC(i%) is the annual cost of “owing” anasset over “n” time periods


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11.2 Annual Operating CostComponent

Annual Operating Costs (AOC);End-of-year estimated costs of

operating the asset in question.

AOC’s tend to increase over time;One wants to distribute the AOC

over a range of assumed number of

years;“n” = {1, then 2, then 3, …. }


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11.2 Plotting ESL

The ESL can be visualized by plottingthree curve forms: 1. Plot the CRC’s over assumed values of “n”;

2. Plot the AOC’s over the same assumed values

of “n”; Plot the sum of the CRC and AOC over the same

assumed values of “n” (Total AW of AOC’s)

Examine the AW plot to observe the minimum

cost life of the respective asset.


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11.2 Typical ESL Plot

Min. Total AW of costs life


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11.2 AW over “k” Years

Notation: P = initial investment in the asset;

Sk = estimated salvage value after

“k” years; AOC j = annual operating costs for

year j (j = 1 to k)

“k” the number of years for theanalysis.


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11.2 Closed Form of AWk

Analytical Form for Total AWk:

k

k

j

j=1

AW ( / , , ) ( / , , )

AOC ( / , , ) ( / , , )

k Total P A P i k S A F i k

P F i j A P i k

= − + −

∑

Procedure: Year-by-year analysis for “k” years – where“k” is given or assumed.


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11.2 Example 11.2 - Overview

Defender Asset;3 years old now;

Market value now: $13,000;

5-year study period assumed;Require Estimates of the future

salvage values and annual

operating costs for the 5-yearperiod.


11 2 E l F t M k t

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11.2 Example: Future MarketValues

Estimated Future Market Valuesand AOC’s:

MktVt

AOCt

t = 1: $9000 $-2500 t = 2: $8000 -2700

t = 3: $6000 -3000

t = 4: $2000 -3500

t = 5: $0 -4500

Mkt. Values aredecreasing:

AOC’s areincreasing:

Assume the interest rate is 10% per year.


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11.2 Example: Find the ESL

Period – by – period analysisFor “k” = 1 year:

0 1

P=$13,000

S1 =

$9000

AOC1 = -2500

AW(10%)1 = (-$13,000)(A/P,10%,1) + $9000(A/F,10%,1) -2500

= -$7800 ( for one year!)


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Period – by – period analysisFor “k” = 2 years:

0 1 2

P=$13,000

S2 = $8000

AOC1 = -2500

AW(10%)2 = (-13,000)(A/P,10%,2) + 8000(A/F,10%,2)

-[2500(P/F,10%,1) + 2700(P/F,10%,2)](A/P,10%,2)

= -$6276/yr for 2 years.

AOC2 = -$2700


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Period – by – period analysisFor “k” = 3 years:

0 1 2 3

P=$13,000

S3 = $6000

AOC1 = -2500

AW(10%)3 = (-13,000)(A/P,10%,3) +6000(A/F,10%,3)

-[2500(P/F,10%,1) + 2700(P/F,10%,2) + 3000(P/F,10%,3](A/P,10%,3) =

-$6132/yr for 3 years.

AOC2 = -$2700AOC3 = -$3000


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11.2 Example - continued

A similar analysis for k = 4 and 5 isconducted;

The AW(10)k , K = {1,2,3,4,5} are

tabulated as:

Total AWk

k=1: -7800

k=2: -6276

k=3: -6132k=4: -6556

k=5: -6579

Min. Cost Year = 3 years


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11.2 Spreadsheet Format

Input Parameters

1 Interrest Rate (%) 10.00%2 Investment Cost ($) 13,000.00$

3 No of Years to Study 5

(1) (2)

Year Mkt. Value AOC/Yr

1 $9,000.00 -$2,500.002 $8,000.00 -$2,700.00

3 $6,000.00 -$3,000.00

4 $2,000.00 -$3,500.00

5 $0.00 -$4,500.00

(3) (4) (5) Min LifeCap. Rec. Costs AW of AOC's Year Total AW(i%) ID

-$5,300 -$2,500 1 -$7,800

-$3,681 -$2,595 2 -$6,276

-$3,415 -$2,718 3 -$6,132 Min Life

-$3,670 -$2,886 4 -$6,556

-$3,429 -$3,150 5 -$6,580

Base Input

Parameters

Schedule of Est. Mkt. Values

And AOC’s/year

Tabulation of CRS’s, AOC’s and

Total AW(i%)


11 2 Specimen Cell Formulas:

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11.2 Specimen Cell Formulas:CRCk

(3) (4) (5) Min Life

Cap. Rec. Costs AW of AOC's Year Total AW(i%) ID

-$5,300 -$2,500 1 -$7,800

-$3,681 -$2,595 2 -$6,276

-$3,415 -$2,718 3 -$6,132 Min Life

-$3,670 -$2,886 4 -$6,556

-$3,429 -$3,150 5 -$6,580

=IF($B18>$C$14,"",PMT($C$12,$B18,$C$13,-$C18))

Note: Application of the PMT financial function!


11 2 S i C ll F l

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11.2 Specimen Cell Formulas:AW(AOC)

(3) (4) (5) Min Life


-$5,300 -$2,500 1 -$7,800

-$3,681 -$2,595 2 -$6,276

-$3,415 -$2,718 3 -$6,132 Min Life

-$3,670 -$2,886 4 -$6,556

-$3,429 -$3,150 5 -$6,580

Note: Application of the PMT financial function incombination with the NPV function for AW of AOC’s!

=IF($B18>$C$14,"",-PMT($C$12,$B18,NPV($C$12,$D$17:$D18)+0))


11.2

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Specimen Cell Formulas: TotalAW(3) (4) (5) Min Life


-$5,300 -$2,500 1 -$7,800

-$3,681 -$2,595 2 -$6,276

-$3,415 -$2,718 3 -$6,132 Min Life

-$3,670 -$2,886 4 -$6,556

-$3,429 -$3,150 5 -$6,580

PMT and NPV function to compute the total AW year byyear.

=IF($B18>$C$14,"",-PMT($C$12,$B18,NPV($C$12,$D$17:$D18)+0))


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11.2 Building a Plotting Table

year CRC AOC AW

1 $5,300.00 $2,500 $7,800

2 $3,680.95 $0 $3,681

3 $3,414.80 $2,718 $6,132

4 $3,670.18 $2,886 $6,556

5 $3,429.37 $3,150 $6,580

Table below is used for Plotting Purposes

NO Data Entr re uired!

Summary Table from the spreadsheet to assist in

plotting the curve forms


11 2 Plot of Capital Recovery

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11.2 Plot of Capital RecoveryCosts

ESL

$0.00

$1,000.00

$2,000.00

$3,000.00

$4,000.00

$5,000.00

$6,000.00

0 1 2 3 4 5 6

Years

$ A W

CRC


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11.2 Plot of the AOC’s

ESL

$0

$500

$1,000

$1,500

$2,000

$2,500

$3,000

$3,500

0 1 2 3 4 5 6

Years

$ A

AOC


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11.2 Combined Plots for Example

ESL

$0.00

$1,000.00

$2,000.00

$3,000.00

$4,000.00

$5,000.00

$6,000.00

$7,000.00

$8,000.00

$9,000.00

0 1 2 3 4 5 6

Years

$ A W

CRC AOC AW

Min AW Year!


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11.2 ESL vs. AW Analysis

Traditional AW Analysis: “n” is fixed or assumed;

First cost at t = 0;

Est. salvage value at t = “n”;

ESL Analysis: “n” varies from t = 1 to t = “k:

Year-by year analysis using AW(i%)

Table of possible future salvage (market)values for the asset in question.

Table of future AOC’s, year by year.


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11.2 ESL vs. AW

AW analysis is for a fixed time periodwith a table of AOC’s and one estimated

future salvage value out at t = “n”.

ESL is a period-by-period variant of AW

where a table of estimated future salvagevalues may be provided and a tabulation

of a set of AW’s for each time period

evaluated.

Seeking the min. AW life in an ESL analysis.


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11.2 Important Conclusions

If “n” fixed (known) thendetermine the AW(i%) for “n” time

periods.

This is the ESL given “n”.

This is the correct value to use in a

replacement study!


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11.2 “n” Not Known

For “n” of a defender or challengerthat is not known or assumed:

Compute the ESL for a range of

“n” values where n = {1, then 2, …,

then K};

From this period-by-period

analysis, determine the min cost

life and the associated AW for that

life.


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11.2 “n” Not Known

Will need a table of future, estimatedmarket (salvage) values for the asset

in question and a table of the future

estimated annual operating costs.

For estimating future market values

can: Apply a sequence of % loss of value;

See Chapter 15 for other cost estimatingtechniques.


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11.2 Marginal Cost Approach

Marginal Costs are year-by-yearestimates of the costs to:

Own the asset and,

Operate the asset,

For the current year in question.

Three Components of Marginal Costs:

1. Cost of ownership (loss in Mkt. Value/yr;

2. Foregone interest of Mkt. Value at beg. Of

the year; 3. AOC for each year.


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11.2 Marginal Cost Analysis

Compute the marginal costs per year;Find their equivalent annual worth;

AW of marginal costs = total AW of

costs;

Can perform either a ESL analysis or a

Marginal Cost analysis when yearly Mkt.

Values are estimated;

Same result!


11 2 Marginal Cost Format: Example

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11.2 Marginal Cost Format: Example11.2

Loss in MV Lost Interest Est. Year MV for Year on MV for Year AOC/Year

1 $9,000 -$4,000.00 -$1,300.00 -$2,500.00

2 $8,000 -$1,000.00 -$900.00 -$2,700.00

3 $6,000 -$2,000.00 -$800.00 -$3,000.00

4 $2,000 -$4,000.00 -$600.00 -$3,500.00

5 $0 -$2,000.00 -$200.00 -$4,500.00

Marginal Cost AW of

time For the Year Marginal Costs

1 -$7,800.00 -$7,800

2 -$4,600.00 -$6,276

3 -$5,800.00 -$6,132

4 -$8,100.00 -$6,556

5 -$6,700.00 -$6,580

Min. Cost Life:At t = 3.

Same Result as

The ESL analysis.


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11.2 Specific Conclusions:

1. To perform an ESL analysis year-by-year market value estimates

are made relative to the specific

asset;

Determine the “n” value with the

lowest AW of costs;

The “n” value and the associated AWn

are then used in the replacementanalysis.


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11.2 Specific Conclusions:

2. If no estimates of future marketvalues are made and “n” is fixed

then: Fix “n”;

Estimate Sn;

Calculate AW over “n” years given P

and S;

Use the “n” and AW given “n” in thereplacement analysis for the specific

asset.


11 2 Setting Up for a Replacement

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11.2 Setting Up for a ReplacementAnalysis

Conduct the ESL for the defenderand the challenger(s);

Form the following alternatives:

Challenger Alternative (C): AWCfor n

C

yrs;

Defender Alternative: (D): AWDfor n

D

yrs.


i h l i

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11.2 Begin the Evaluation

The next section (11.3) illustrates

how to conduct a before-tax

replacement study.


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CHAPTER 11

Section 11.3

Performing a

Replacement Study

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11.3 Overview of the two

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11.3 Overview of the twoMethods


11 3 N R l t St d

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11.3 New Replacement Study

Given: {(C) or (D)};Apply:

AWDvs. AW

C

Select the best alternativeStay with the Defender for nD years

or,

Go with the challenger for nCyears.


11 3 O Y L t A l i

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11.3 One-Year Later Analysis

Validate all cost and market valueestimates;

Is the current year nD?

If “YES”, replace the defender with (C); If “NO”, retain defender for one more

year and re-evaluate then.

If cost estimates have changed then: Update all estimates

Calculate AWC and AWD

Initiate a new replacement study.


11.3 Case Problem ( Ex. 11.4) -

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11.3 Case Problem ( Ex. 11.4) Defender

Defender Data Current Market Value: $15,000;

Future Mkt. Values will decrease by 20%/yr;

Keep for no more than 3 years;

AOC’s: {$4,000,$8,000,$12,000}

Retrofit next year = $16,000;

AOC’sD:= {$20,000, $8,000,$12,000}

(costs).


11 3 C P bl Ch ll

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11.3 Case Problem: Challenger

First Cost: $50,000Future Mkt. Values decreasing by

20%/year;

Retain for no more than 5 years;AOC’s

C:=

{$5,000,$7,000,$9,000,$11,000,$13,000}

Assume the interest rate is set at

10%/year.


11 3 D f d A l i

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11.3 Defender AnalysisInput Parameters

Interrest Rate (%) 10.00%

Investment Cost ($) 15,000.00$

No of Years to Study 3

(1) (2)


1 $12,000.00 -$20,000.00, . - , .

3 $7,680.00 -$12,000.00

(3) (4) (5) Min Life

Cap. Rec. Costs AW of AOC's Year Total AW(i%) ID-$4,500 -$20,000 1 -$24,500- , - , - ,

-$3,711 -$13,595 3 -$17,307 Min Life

ESL(Defender) = 3 yrs: AW = -$17,307/year


11 3 Cost Plots for Defender

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11.3 Cost Plots for Defender

ESL

$0.00

$5,000.00

$10,000.00

$15,000.00

$20,000.00

$25,000.00

$30,000.00

0 1 2 3

Years

$ A W

CRC AOC AW

Min AW Cost Life = 3 years


11 3 Example 11 4: Challenger

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11.3 Example 11.4: ChallengerInput Parameters

Interrest Rate (%) 10.00%

Investment Cost ($) 50,000.00$

No of Years to Study 5

(1) (2)


1 $40,000.00 -$5,000.00

2 $32,000.00 -$7,000.00

3 $25,600.00 -$9,000.00

4 $20,480.00 -$11,000.00

5 $16,384.00 -$13,000.00

(3) (4) (5)

Cap. Rec. Costs AW of AOC's Year Total AW(i%)-$15,000 -$5,000 1 -$20,000

-$13,571 -$5,952 2 -$19,524

-$12,372 -$6,873 3 -$19,245

-$11,361 -$7,762 4 -$19,123

-$10,506 -$8,620 5 -$19,126

Min CostLifeN = 4 Years

AWC = -19,123/year


11 3 Challenger Plots

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11.3 Challenger Plots

$19,000

$19,100

$19,200

$19,300

$19,400

$19,500

$19,600

$19,700

$19,800

$19,900

$20,000

$20,100

0 1 2 3 4 5 6

Years

$ A W

AW

ESL: Challenger

Min Cost Life = 4 yearsAt -$19,123/year


11 3 Case Problem: Summary

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11.3 Case Problem: Summary

AWD = -$17,307/year;n

D= 3 years;

AWC= -$19,123/year;

nC = 4 years;

Conclusion:

Stay with the Defender for at least one moreyear – lowest AW(10%) cost: -$17,307/yr vs. -$19,123/yr.


11 3 Market Value of Defender

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11.3 Market Value of Defender

What minimum market value of thedefender will make the current

challenger economically attractive?

If a high enough market value (trade-

in) is possible for the defender asset,one should do so and go with the

Challenger immediately!

Break-even or replacement value (RV)


11.3 Replacement Value

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p(Defender)

Economic Fact:

If the actual market value (trade-in)

exceed the breakeven replacement value,the challenger is the better alternative.

If this is the case, replace now with the

challenger!


11 3 RV Analysis

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11.3 RV Analysis

If a spreadsheet analysis has beenconducted, one may use: Goal Seek to find the RV or,

SOLVER provided the cells are properly

linked.

Build you own spreadsheet to gain the

experiences needed – replacement

problems are difficult when one appliesmanual calculations.


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CHAPTER 11

Section 11.4

Additional Considerations

in a Replacement Study


11.4 Three Additional

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Considerations

1. Future-year ReplacementDecisions;

2. Opportunity-cost vs. Cash Flow

approaches;

3. Anticipation of improved future

challengers.


11.4 Future Replacement

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Decisions

1. Future-year Replacement Decisions: Replace now? One year from now? Two

years from now?

The procedure just presented does

assist with answering this question

provided:

The cost estimates for (C) and (D) do

not change!


11.4 Future Replacement

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Decisions

1. Future-year ReplacementDecisions:

If cost estimates change for either

(D) or (C) then the analysis should

be initiated again within areasonable time frame.

With changing estimates the

decision to replace may be altered!

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11 4 Cash Flow Approach

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11.4 Cash Flow Approach

2. Cash Flow Approach:Assumes that when the challenger

is selected and a cash inflow for the

defender is received then: The investment in the challenger is

immediately reduced.

Discourage this approach;

Works only if the lives of the (C) and

(D) are the same!


11.4 Anticipation of Future

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Challengers

3. Assumption: At some time in the

future a worthy challenger will

appear and replace the defender. Always study future trends of

challengers; May be best to augment the defender

until such time a more worthy challenger

becomes available.

Tax considerations should always beinvolved (see Chapter 17)


Section 11.5

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Section 11.5Replacement Study over a

Specified Study Period

At times, a fixed or “impressed”

study period will apply to both the

challenger and the defender.

Such a study may not be based

upon the ESL approach.

If a fixed study period then ……


11 5 Specified Study Period

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11.5 Specified Study Period

At times, a fixed or “impressed”study period will apply to both the

challenger and the defender.

Such a study may not be based

upon the ESL approach.

If a fixed study period then ……



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Determine the AW for both C and Dover the prescribed study period;

No need to perform an ESL

analysis;

Assumption is that the services of

C and D are not needed beyond the

study period.

Use the AW approach.



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Focus on the most accurateestimates available over the fixed

study period.

Assume “equal service” for both D

and C over the study period.

What happens IF the defender’s

remaining life is shorter than the

study period?


11.5 Defender’s Remaining Life

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11.5 Defender s Remaining Life

If the defender’s remaining life isshorter than the study period then:

Focus on upgrading the defender and obtaincost estimates in order to extend the

defender out to the study period.

These costs become part of retaining the

defender out to the prescribed study period.

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Chapter 11 Summary cont.

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Chapter 11 Summary cont.

For ESL, need estimates of futuremarket values and annual

operating costs for both C and D.

Annual Worth is the best method

especially if the lives of each

alternative are different.

For a fixed study period for both D

and C, apply the traditional AWmethod.

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ENGINEERING ECONOMY Sixth

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CHAPTER 5

End of Slide Set

Mc

GrawHill

EditionBlank andTarquin