replacement analysis revised

8/4/2019 Replacement Analysis Revised

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ReplacementAnalysis


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Why replacement studies?

The basic replacement study is designedto determine if a currently used assetshould be replaced.

Whether unplanned or anticipated,replacement is considered for one or moreof several reasons:

Reduced performance

Altered requirements

Obsolescence


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Reduced Performance

Because of the physical deterioration ofparts, the ability to perform at an expectedlevel of reliability(being available and

performing correctly when needed) orproductivity(performing at a given level ofquality and quantity) is not present.

This results in increased cost of operation,higher scrap and rework costs, lost sales, andlarger maintenance expenses.


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Altered Requirements

New requirements of accuracy, speed orother specifications may have beenestablished.

These new requirements cannot be met byexisting equipment or system.

Often, the choice is between completereplacement or enhancement throughretrofitting or augmentation.


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Obsolescence

International competition and the rapidlychanging technology of automation,computers, and communications make

currently used systems and assetsperform acceptably but less productivelythan equipment coming onto the market.

A replacement analysis will guideengineers in deciding if an asset needsreplacement due to obsolescence.


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Why replacement studies?

A replacement study is usually designed tofirst make the economic decision to retainor replace an asset now.

If the decision is to replace, the study iscomplete.

If the decision is to retain, the costestimates and decision will be revisitedeach year to ensure that the decision toretain is still economically correct.


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Basic Concepts

Defender This is the asset currently owned (or in place).

Challenger

This is one of the alternatives beingconsidered.

Outsider or Consultant Perspective

We assume that we currently own or useneither asset (defender nor challenger) andwe must select between the challenger andthe in-place defender.


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Basic Concepts

Outsider Perspective

In order to acquire the defender, we must

invest in the prevailing market value of the

defender. This estimated market or trade-in value

becomes the first cost of the defender.


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Basic Concepts

Outsider Perspective There will be new estimates for the

remaining economic life, annual operating

cost, and salvage value for the defender. All these values will likely differ from the

original estimates.

Because of the outsider perspective, all

estimates made and used previously shouldbe neglected in the replacement analysis.


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Sunk Cost

Since the past is common to alternatives,past costs are considered irrelevant in areplacement analysis.

This includes a sunk cost, which is anamount of money invested earlier thatcannot be recovered now or in the future.

This may occur due to changed economic,technological, or other conditions or ill-advised business decisions.


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Sunk Cost

In industry, a sunk cost occurs when anasset is considered for replacement, andthe actual market or trade-in value is less

than that predicted by the depreciationmodel used to write off the original capitalinvestment, or is less than the estimated

salvage value.


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Sunk Cost

Sunk Cost = Present BV Present MV

If the resulting difference is a negativenumber, then there is no sunk cost

involved.


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Sunk Cost

For example, an asset purchased for$100,000 five years ago now has adepreciated book value of $50,000. A

replacement study is being conducted andonly $20,000 is offered as the trade-inamount. Hence,

Sunk cost = $50,000 - $20,000 = $30,000


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Sunk Cost

In replacement studies, the sunk costshould not be included in the economicanalysis.

It is incorrect to recover the sunk cost of

the defender by adding it to the first cost ofthe challenger.

This penalizes the challenger, making its firstcost appear higher and thereby biasing thedecision.


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Replacement Analysis

To perform a replacement analysis, theevaluator takes the perspective of aconsultant to the company.

If choosing between two replacementalternatives, then it is assumed (foranalysis purposes) that

neither asset is currently owned, and

the outcome is either to acquire the used

asset or to acquire a new asset


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There are two equivalent approaches thatmay be taken in a replacement studywhen determining the first cost for the

alternatives and when performing theanalysis:

Cash-flow approach

Opportunity-cost approach


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Cash-flow Approach

When a challenger is selected, thedefenders market value is a cash inflow to

each challenger alternative.

When the defender is selected, there is noactual capital outlay.


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Cash-flow Approach

Recognize that there is an actual cash-flow advantage to the challenger if thedefender is traded.

For the analysis, use the following:

Defender: First cost amount is zero

Challenger: First cost is the actual cost minus

the quoted trade-in value of thedefender


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Cash-flow Approach

Important Note:

The estimated lives of the defender (thatis, its remaining life) and of the challengermust be equal to use this method.

The comparison must be made over thesame study period.


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Opportunity-Cost Approach

Also called the conventional approach,the opportunity-cost approach usesthe defenders trade-in or current

market value as the first cost of thedefender, and the initial cost of thereplacement as the challengers first

cost.


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If the defender is chosen, the owner forgoesan amount of capital equivalent to the trade-in value, hence an opportunity cost.

i.e., the opportunity is lost, hence the cost.

This approach is cumbersome when thereare multiple challengers each quoting a

different trade-in value for the defender,because it requires a different defender firstcost for comparison with each challenger.


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Assume the defender trade-in amount isforgone and the defender service can beacquired as a used asset at its trade-in

value.

Defender: First cost is the trade-in value

Challenger: First cost is its actual cost


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For convenience and uniformity, we willconsistently use the opportunity-cost

approach in solving problems onreplacement analysis.

S th t h ld 10 ld


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Suppose that you have an old car 10 years oldthat could be sold to a dealer for $400 cash.Assume its MV two years from now is zero. The

annual maintenance expenses will average$800 into the future, and the car averages only10 miles per gallon. Gasoline costs $1.50 per

gallon, and you average 15,000 miles per year.You now have an opportunity to replace the oldcar with a better one that costs $8,000.

Because of a two-year warranty, themaintenance expenses for the new car arenegligible. The new car averages 30 miles pergallon. (continued at next slide)

Sample Problem 1


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Use the IRR method to determinewhich alternative you shouldselect. Utilize a two-yearanalysis period and assume

that the new car can be sold for$5,000 at the end of year two.Ignore the effects of income

taxes and let your MARR be15% per year.

Sample Problem 1


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An industrial lift truck has been in service forseveral years and management is

contemplating on replacing it. A planninghorizon of five years is to be used in thereplacement study. The old lift truck has a

current market value of $1,500. If thedefender is retained, it is anticipated tohave annual expenses of $7,300. It will

have a zero market value at the end of fiveadditional years of service.

(continued at next slide)

Sample Problem 2


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The new lift truck (challenger) willcost $10,000 and will have annualexpenses of $5,100. It will have amarket value of $2,500 at the end

of the planning horizon. Determinethe preferred alternative, using apresent worth comparison and an

MARR (before tax) of 20% peryear.

Sample Problem 2


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After-Tax Investment Valueof the Defender

If the defender were sold, then

EOY BTCF d T.I. ITCF ATCF

0 MV0 none MV0-BV0 -t(MV0-BV0) MV0-t(MV0-BV0)

If the defender were NOT sold, then

EOY BTCF d T.I. ITCF ATCF

0 -MV0 none -(MV0-BV0) t(MV0-BV0) -MV0 + t(MV0-BV0)

The Midcontinent Po er A thorit p rchased


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The Midcontinent Power Authority purchasednew coal extraction equipment 3 years ago for$600,000. Management has discovered it is

technologically outdated now. A newequipment alternative (the challenger) hasbeen identified and costs $1,000,000. If a

generous trade-in of $400,000 is offered forthe current equipment, perform AW analysesusing a 7% per year after-tax MARR and an

effective tax rate of 34%. Apply straight-linedepreciation with zero salvage value for bothalternatives.

(continued at next slide) Sample Problem 3


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The following additional data are applicable:

Defender ChallengerAnnual cost $100,000 $15,000

Recovery period

(years) 8 (originally) 5

Sample Problem 3


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AW and Study Period

A replacement study is an application of theAW method of comparing unequal lifealternatives.

In a replacement study with no specifiedstudy period, the AW values are determinedby a technique of cost evaluation called theeconomic service life (ESL) analysis.

If a study period is specified, the procedure isdifferent from that used when no study periodis set.


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AW and Study Period

The AW of a primarily cost cash flowpattern is sometimes called theequivalent uniform annual cost (EUAC).

Because this term is commonly used inthe definition of the economic life of anasset, the EUAC will be used in

replacement studies when referring to theAW.


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Study Period inReplacement Analysis

The study period or planning horizon is thenumber of years selected for the economicanalysis to compare the defender and

challenger. There are two typical situations:

The anticipated remaining life of the defender

equals the life of the challenger The anticipated remaining life of the defender

is shorter than the life of the challenger


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When a defender may be replaced with achallenger having an estimated lifedifferent from that of the defenders

remaining life, the length of the studyperiod must be determined.

It is common practice to use a study period

equal to the life of the longer-lived asset. Then, the AW value of the shorter lived asset

will apply throughout the entire study period.


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Sometimes, an abbreviated study period isimposed in economic evaluations.

This approach will force the recovery ofthe initial investment and the requiredMARR over a shortened period of time

compared to what my be longer lives ofalternatives.


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When the defenders remaining life and the

challengers life are unequal, it is

necessary to preselect a study period forthe analysis.


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Useful Life < Study Period

The annual worth is commonly assumedto continue at the same calculated amountfor an alternative with an N value less

than the study period. If this assumption is inappropriate, perform

the analysis using new estimates for the

defender, the challenger, or both.


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Useful Life > Study Period

If the study period is abbreviated to be lessthan one or both of the alternatives life

estimates, it is necessary to recover the first

cost and the required return at the MARR inless time than normally expected.

This will artificially increase the AW values.

It is usually a management decision onwhether to use an abbreviated study period.


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Economic Service Life (ESL)

The minimum number of years that anasset should be retained in service tominimize its total cost, considering the

time value of money, capital investmentrecovery, and annual operating andmaintenance costs, is called the economic

service life (ESL).


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With each passing year of asset use, thefollowing trends are usually observable:

The equivalent annual worth of the annual

operating cost (AOC) or operating andmaintenance (O&M) cost increases.

The equivalent AW of the assets initial

investment or first cost decreases. The actual trade-in amount or salvage value

decreases relative to the first cost.

S S


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Economic Service Life

The aforementioned factors cause theassets total AW curve to decrease for

some years and to increase thereafter.

The minimum total AW value indicates theN value for the economic service life.

That is, the N value when replacement is

most economical. This should be the estimated asset life used

in engineering economic analysis, if onlyeconomics is considered.


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Economic Service Life

Where P = initial investment or first cost

SVk = salvage value after retainingasset for k years

AOC = annual operating cost for year j

(j = 1, 2, , k)

)k%,i,P/A()j%,i,F/P(AOC

)k%,i,F/A(SV)k%,i,P/A(PAW

kj

1j j

kk


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Robert Roe has just purchased a four-year-old used car, paying $3,000 for it. A friend

suggested that he determine in advancehow long he should keep the car so as toensure the greatest overall economy.

Robert has decided that, because of stylechanges, he would not want to keep thecar longer than four years, and he hasestimated the annual expenses andmarket values for years 1 to 4 as follows:

(continued at next slide)

Sample Problem 4


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Annual Expenses Market ValueYear 1 $950 $2,250

Year 2 $1,050 $1,800

Year 3 $1,100 $1,450

Year 4 $1,550 $1,160

If Roberts capital is worth 12% per year, at

the end of which year should he dispose ofthe car?

Sample Problem 4

An existing robot is used in a commercial material


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An existing robot is used in a commercial materiallaboratory to handle ceramic samples in thehigh-temperature environment that is part of

several test procedures. Due to changingcustomer needs, the robot will not meet futureservice requirements unless it is upgraded at a

cost of $2,000. Because of this, a newadvanced technology robot has been selectedfor potential replacement. Assume that a robot

will be needed for an indefinite period of time.The firms before-tax MARR is 25%. Should theexisting robot be replaced?

(continued at next slide)Sample Problem 5

Defender Challenger


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Sample Problem 5

Defender Challenger

Current market

value

$38,200 $51,00

(purchase price)Upgrade cost $2,000 $0

Annualexpenses

$1,400 in yearone, andincreasing by8% per year

$1,000 in yearone, andincreasing by$150 per year

Useful life 6 years 10 years

Market value atend of useful life

-$1,500 $7,000

Four years ago, the Attaboy Lawn Mower


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Four years ago, the Attaboy Lawn MowerCompany purchased a piece of equipment.Because of increasing maintenance costs for

this equipment, a new piece of machinery isbeing considered for the assembly line. Thecost characteristics of the defender and the

challenger are shown at the next slide.Suppose a $6,000 market value is availablenow for the defender. With an after-tax MARR

of 10%, determine which alternative to select.The effective tax rate is 40%. Straight-linedepreciation with zero salvage value is applied.

Sample Problem 6

Defender Challenger


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Sample Problem 6

Defender Challenger

Original Cost $9,000 $11,000

(purchase price)Maintenance $500 in year one

(four years ago)increasing by a

uniform gradientof $100 per yearthereafter

$150

Originalestimated life 9 years 5 years

Market value atend of useful life

$0 $3,000

Use the PW method to select the better of the


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Use the PW method to select the better of thefollowing alternatives:

Sample Problem 7

Annualexpenses

Defender:Alternative A

Challenger:Alternative B

Labor $300,000 $250,000

Material $250,000 $100,000

Insurance andproperty taxes

4% of initialcapital

None

Maintenance $8,000 None

Leasing cost None $100,000

Continued at next slide

Assume the defender was installed five years ago


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y gand that its MACRS (GDS) property class isseven years. The after-tax MARR is 10% peryear, and the effective income tax rate is 40%.

Definition of alternatives:

A: Retain an already owned machine (defender) inservice for eight more years

B: Sell the defender and lease a new one(challenger) for eight years

Alternative A (additional information):

Cost of defender five years ago = $500,000BV now = $111,550

Estimated MV eight years from now = $50,000

Present MV = $150,000 Sample Problem 7

A company is considering replacing a machine


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p y g p gthat was bought six years ago for $50,000.However, the machine can be repaired and its

life extended by five more years. If the currentmachine is replaced, the new machine will cost$44,000 and will reduce operating expenses by$6,000 per year. The seller of the new machinehas offered a trade-in allowance of $15,000 forthe old machine. If the MARR is 12% per yearbefore taxes, how much can the company

spend to repair the existing machine?(a) $22,371 (b) $50,628

(c) $7,371 (d) -$1,000

Sample Problem 8

Machine A was purchased three years ago for


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Machine A was purchased three years ago for$10,000 and had an estimated market value of$1,000 at the end of its 10-year life. Annual

operating costs are $1,000. The machine willperform satisfactorily for the next seven years.A salesman for another company is offeringmachine B for $50,000 with a market value of$5,000 after 10 years. Annual operating costswill be $600. Machine A could be sold now for$7,000. The MARR is 12% per year.

(a) What is the equivalent uniform annual cost(EUAC) of continuing to use Machine A?

(b) What is the EUAC of buying Machine B?

replacement analysis revised

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