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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?
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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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11.1 Trade-In Issues
Investment in the challenger
PC – (TIV – MVD)
Cash Price for the challenger less:
(Trade-in Value – Market Value of the Defender)
This represents the true investment in thechallenger to the firm!
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11.1 Trade-In Issues
Investment in the challenger
PC – (TIV – MVD)
The Cash pricefor
The challenger with
No trade-in
The Opportunity CostGiven up by not
Selling theDefender outright!
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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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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.
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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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11.2 Example: Find the ESL
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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11.2 Example: Find the ESL
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%)
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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!
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11 2 S i C ll F l
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11.2 Specimen Cell Formulas:AW(AOC)
(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 -$6,580
Note: Application of the PMT financial function incombination with the NPV function for AW of AOC’s!
=IF($B18>$C$14,"",-PMT($C$12,$B18,NPV($C$12,$D$17:$D18)+0))
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11.2
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Specimen Cell Formulas: TotalAW(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 -$6,580
PMT and NPV function to compute the total AW year byyear.
=IF($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
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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!
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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.
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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.
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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
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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.
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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.
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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).
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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,$13,000}
Assume the interest rate is set at
10%/year.
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11 3 D f d A l i
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11.3 Defender AnalysisInput Parameters
Interrest Rate (%) 10.00%
Investment Cost ($) 15,000.00$
No of Years to Study 3
(1) (2)
Year Mkt. Value AOC/Yr
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
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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
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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)
Year Mkt. Value AOC/Yr
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
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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
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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.
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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)
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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!
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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
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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.
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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!
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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!
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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)
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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 ……
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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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11.5 Specified Study Period
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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11.5 Specified Study Period
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?
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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