Chapter 15
Replacement Decisions
• Replacement Analysis
Fundamentals
• Economic Service Life
• Replacement Analysis
When Required
Service is Long
• Replacement Analysis
with Tax
Consideration
(c) 2001 Contemporary Engineering
Economics
1
Replacement Terminology
• Defender: an old
machine
• Challenger: new
machine
• Current market value:
selling price of the
defender in the market
place
• Sunk cost: any past
cost unaffected by any
future decisions
• Trade-in allowance:
value offered by the
vendor to reduce the
price of a new
equipment
• Operating Cost
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Economics
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Sunk Cost associated with an Asset’s
Disposal
Original investment
$20,000
Market value
Lost investment
(economic depreciation)
$10,000
Repair cost
$5000
$10,000
Sunk costs = $15,000
$0
$5000
$10,000
$15,000
$20,000
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Economics
$25,000
$30,000
3
Replacement Decisions
•
Cash Flow Approach
–
–
–
Treat the proceeds from
sale of the old machine as
down payment toward
purchasing the new
machine.
Can be used in the
analysis period is same for
all alternatives.
Use NPW or AE analysis
to decide
• Opportunity Cost Approach
– Treat the proceeds from sale
of the old machine as the
investment required to keep
the old machine.
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Economics
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Replacement Analysis – Cash Flow
Approach
Sales proceeds
from defender
$10,000
$5500
$2500
0
1
2
3
0
2
3
$6000
$8000
(a) Defender
1
$15,000
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Economics
(b) Challenger
5
Annual Equivalent Cost - Cash Flow
Approach
 Defender:
PW(12%)D = $2,500 (P/F, 12%, 3) - $8,000 (P/A, 12%, 3)
= - $17,434.90
AE(12%)D = PW(12%)D(A/P, 12%, 3)
= -$7,259.10
 Challenger:
Replace
PW(12%)C = $5,500 (P/F, 12%, 3) - $5,000
the
- $6,000 (P/A, 12%, 3)
defender
= -$15,495.90
now!
AE(12%)C = PW(12%)C(A/P, 12%, 3)
= -$6,451.79
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Economics
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Opportunity Cost Approach
0
Defender
$2500
1
3
2
Challenger
0
1
2
$5500
3
$6000
$8000
$10,000
Proceeds from sale viewed as
an opportunity cost of keeping
the asset
$15,000
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Economics
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Opportunity Cost Approach
 Defender:
PW(12%)D = -$10,000 - $8,000(P/A, 12%, 3) + $2,500(P/F, 12%, 3)
= -$27,434.90
AE(12%)D = PW(12%)D(A/P, 12%, 3)
= -$11,422.64
 Challenger:
PW(12%)C = -$15,000 - $6,000(P/A, 12%, 3) + $5,500(P/F, 12%, 3)
= -$25,495.90
AE(12%)C = PW(12%)C(A/P, 12%, 3)
Replace the
defender now!
= -$10,615.33
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Economics
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Economic Service Life
• Def:Economic service life
is the useful life of a
defender, or a challenger,
that results in the
minimum equivalent
annual cost
• Why do we need it?: We
should use the respective
economic service lives of
the defender and the
challenger when
conducting a replacement
analysis.
Minimize
Ownership (Capital)
Cost (init.+salvg.)
+
Operating
cost
CR(i )  I ( A / P, i , N )  S N ( A / F , i , N )
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Mathematical Relationship
• Capital Recov. Cost.
CR(i )  I ( A / P, i , N )  S N ( A / F , i , N )
AEC
• Operating Cost:
OC(i)
N
OC(i )   OCn ( P / F , i , n) ( A / P, i , N )
n 1
CR(i)
• Total Cost:
AEC  CR(i)  OC(i)
• Objective: Find n*
that minimizes AEC
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Economics
n*
10
Economic Service Life for a Lift Truck
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Economics
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Economic Service Life Calculation (Example 15.4)
• N=1
AEC1 = $18,000(A/P, 15%, 1) + $1,000 - $10,000
= $11,700
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• N=2
AEC2 = [$18,000 + $1,000(P/A,15%, 15%, 2)](A/P, 15%, 2)
- $7,500 (A/F, 15%, 2)
= $8,653
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N = 3, AEC3 = $7,406
N = 4, AEC4 = $6,678
N = 5, AEC5 = $6,642
N = 6, AEC6 = $6,258
Minimum cost
N = 7, AEC7 = $6,394
Economic Service Life
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Economics
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Required Assumptions and Decision
Frameworks
•
•
•
•
Planning horizon (study period)
Technology
Relevant cash flow information
Decision Frameworks
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Economics
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Replacement Strategies under the
Infinite Planning Horizon
1.
2.
Replace the defender now: The cash flows of the
challenger will be used from today and will be repeated
because an identical challenger will be used if
replacement becomes necessary again in the future. This
stream of cash flows is equivalent to a cash flow of AEC*
each year for an infinite number of years.
Replace the defender, say, x years later: The cash flows
of the defender will be used in the first x years. Starting
in year x+1,the cash flows of the challenger will be used
indefinitely.
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Economics
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Example 15.5
• Defender: Find the
remaining useful
(economic) service
life.
N  1: AE (15%)  $5,130
N  2: AE (15%)  $5,116
N  3: AE (15%)  $5,500
N  4: AE (15%)  $5,961
N D*  2 years
AE D*  $5,116
N  5: AE (15%)  $6,434
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• Challenger: find the
economic service life.
N = 1 year: AE(15%) = $7,500
N = 2 years: AE(15%) = $6,151
N = 3 years: AE(15%) = $5,847
N = 4 years: AE(15%) = $5,826
N = 5 years: AE(15%) = $5,897
NC*=4 years
AEC*=$5,826
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Replacement Decisions
N D*  2 years
AED*  $5,116
NC*=4 years
AEC*=$5,826
• Should replace the
defender now? No,
because AED < AEC
• If not, when is the best
time to replace the
defender? Need to
conduct marginal
analysis.
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Economics
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Marginal Analysis
Question: What is the additional (incremental)
cost for keeping the defender one more year
from the end of its economic service life, from
Year 2 to Year 3?
Financial Data:
• Opportunity cost at the end of year 2: Equal to the market
value of $3,000
• Operating cost for the 3rd year: $5,000
• Salvage value of the defender at the end of year 3: $2,000
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• Step 1: Calculate the equivalent
cost of retaining the defender
one more from the end of its
economic service life, say 2 to
3.
$2000
2
3
$3,000(F/P,15%,1) + $5,000
$3000
- $2,000 = $6,450
• Step 2: Compare this cost with
AEC = $5,826 of the challenger.
• Conclusion: Since keeping the
defender for the 3rd year is more
expensive than replacing it with
the challenger, DO NOT keep
the defender beyond its
economic service life.
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Economics
2
$5000
3
$6,450
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Replacement Analysis under the Finite
Planning Horizon
N
Annual Equivalent Cost
($)
Defender
Challenger
1
5,130
7,500
2
5,116
6,151
3
5,500
5,857
4
5,961
5,826
5
6,434
5,897
Some likely replacement patterns
under a finite planning horizon of
8 years
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Economics
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Example 15.6 Replacement Analysis under the
Finite Planning Horizon (PW Approach)
•
Option 1: (j0, 0), (j, 4), (j, 4)
PW(15%)1=$5,826(P/A, 15%, 8)
=$26,143
• Option 2: (j0, 1), (j, 4), (j, 3)
PW(15%)2=$5,130(P/F, 15%, 1)
+$5,826(P/A, 15%, 4)(P/F, 15%, 1)
+$5,857(P/A, 15%, 3)(P/F, 15%, 5)
=$25,573
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Economics
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Example 15.6 continued
• Option 3
• Option 4
(j0, 2), (j, 4), (j, 2)
PW(15%)3=$5,116(P/A, 15%, 4)(P/F, 15%, 2)
+$5,826(P/A, 15%, 4)(P/F, 15%, 2)
+$6,151(P/A, 15%, 2)(P/F, 15%, 6)
= $25,217
minimum cost
(j0, 3), (j, 5)
PW(15%)4= $5,500(P/A, 15%, 3)
+$5,897(P/A, 15%, 5)(P/F, 15%, 3)
=$25,555
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Economics
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Example 15.6 continued
• Option 5:
• Option 6:
(j0, 3), (j, 4), (j, 1)
PW(15%)5= $5,500(P/A, 15%, 3)
+ $5,826(P/A, 15%, 4)(P/F, 15%, 3)
+ $7,500(P/F, 15%, 8)
= $25,946
(j0, 4), (j, 4)
PW(15%)6= $5,826(P/A, 15%, 4)(P/F, 15%, 4)
+ $5,826(P/A, 15%, 4)(P/F, 15%, 4)
= $26,529
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Planning horizon = 8 years
(j0, 0), (j, 4), (j, 4),
Option 1
(j0, 1), (j, 4), (j, 3),
Option2
(j0, 2), (j, 4), (j, 2),
Option 3
(j0, 3), (j, 5),
Option 4
(j0, 3), (j, 4), (j, 1),
Option 5
(j0, 4), (j, 4),
Option 6
0
1
2
3
4
5
6
7
8
Years in service
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Economics
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Replacement Analysis with Tax
Consideration
• Whenever possible, replacement decisions should
be based on the cash flows after taxes. (Example
15.8)
• When computing the net proceeds from sale of the
old asset, any gains or losses must be identified to
determine the correct amount of the opportunity
cost. (Example 15.7)
• All basic replacement decision rules including the
way of computing economic service life remain
unchanged. (Example 15.10)
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Economics
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Depreciation basis
$20,000
$20,000
Total
depreciation
Book value
$14,693
$5307
Market value
Book loss
$10,000
$4693
Loss
tax credit
Market value
$4693  40%  $1877
$10,000
Net proceeds from disposal ($11,877)
$0
$4000
$8000
$12,000
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Economics
$16,000
$20,000
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Summary
• In replacement analysis, the defender is an
existing asset; the challenger is the best available
replacement candidate.
• The current market value is the value to use in
preparing a defender’s economic analysis. Sunk
costs—past costs that cannot be changed by any
future investment decision—should not be
considered in a defender’s economic analysis.
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Economics
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• Two basic approaches to analyzing replacement
problems are the cash flow approach and the
opportunity cost approach.
– The cash flow approach explicitly considers the
actual cash flow consequences for each
replacement alternative as they occur.
– The opportunity cost approach views the net
proceeds from sale of the defender as an
opportunity cost of keeping the defender.
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Economics
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• Economic service life is the remaining useful life of a
defender, or a challenger, that results in the minimum
equivalent annual cost or maximum annual equivalent
revenue. We should use the respective economic service
lives of the defender and the challenger when conducting a
replacement analysis.
• Ultimately, in replacement analysis, the question is not
whether to replace the defender, but when to do so.
• The AE method provides a marginal basis on which to
make a year-by-year decision about the best time to replace
the defender.
• As a general decision criterion, the PW method provides a
more direct solution to a variety of replacement problems,
with either an infinite or a finite planning horizon, or a
technological change in a future challenger.
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• The role of technological change in asset improvement
should be weighed in making long-term replacement plans
• Whenever possible, all replacement decisions should be
based on the cash flows after taxes.
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