Chapter 15 Replacement Decisions

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Chapter 15Replacement Decisions

• Replacement Analysis Fundamentals
• Economic Service Life
• Replacement Analysis When Required Service is
  Long
• Replacement Analysis with Tax Consideration

2
Replacement Terminology

• 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

• Defender an old machine
• Challenger new machine
• Current market value selling price of the
  defender in the market place

3
Sunk Cost associated with an Assets Disposal
Original investment
20,000
Lost investment (economic depreciation)
Market value
Repair cost
10,000
5000
10,000
Sunk costs 15,000
0 5000 10,000 15,000
20,000 25,000 30,000
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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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Replacement Analysis Cash Flow Approach
Sales proceeds from defender
10,000
5500
2500
0 1 2 3
0 1 2 3
6000
8000
(a) Defender
(b) Challenger
15,000
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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
• PW(12)C 5,500 (P/F, 12, 3) - 5,000
• - 6,000 (P/A, 12, 3)
• -15,495.90
• AE(12)C PW(12)C(A/P, 12, 3)
• -6,451.79

Replace the defender now!
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Opportunity Cost Approach
Challenger
5500
Defender
2500
0 1 2 3
0 1 2 3
6000
8000
10,000
Proceeds from sale viewed as an opportunity cost
of keeping the asset
15,000
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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) -10,615.33
Replace the defender now!
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Economic Service Life

• DefEconomic 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
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Mathematical Relationship

• Capital Recov. Cost.
• Operating Cost
• Total Cost
• Objective Find n that minimizes AEC

AEC
OC(i)
CR(i)
n
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Economic Service Life for a Lift Truck
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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 N 7,
AEC7 6,394
Minimum cost
Economic Service Life
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Required Assumptions and Decision Frameworks

• Planning horizon (study period)
• Technology
• Relevant cash flow information
• Decision Frameworks

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Replacement Strategies under the Infinite
Planning Horizon

• 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 x1,the cash
  flows of the challenger will be used
  indefinitely.

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Example 15.5

• Defender Find the remaining useful (economic)
  service life.

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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

NC4 years AEC5,826
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Replacement Decisions

• Should replace the defender now? No, because AED
  lt AEC
• If not, when is the best time to replace the
  defender? Need to conduct marginal analysis.

• NC4 years
• AEC5,826

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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.
• 3,000(F/P,15,1) 5,000
• - 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.

2000
2
3
3000
5000
2
3
6,450
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Replacement Analysis under the Finite Planning
Horizon
Some likely replacement patterns under a finite
planning horizon of 8 years
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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)15,826(P/A, 15, 8)
• 26,143
• Option 2 (j0, 1), (j, 4), (j, 3)
• PW(15)25,130(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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Example 15.6 continued

• Option 3 (j0, 2), (j, 4), (j, 2)
• PW(15)35,116(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
• Option 4 (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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Example 15.6 continued

• Option 5 (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
• Option 6 (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), (j0, 1), (j, 4), (j,
3), (j0, 2), (j, 4), (j, 2), (j0, 3),
(j, 5), (j0, 3), (j, 4), (j, 1), (j0,
4), (j, 4),
Option 1 Option2 Option 3 Option 4 Option
5 Option 6
0 1 2 3 4 5 6
7 8
Years in service
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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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Depreciation basis
20,000
20,000
Total depreciation
Book value
14,693
5307
Book loss
Market value
10,000
4693
Loss tax credit
Market value
10,000
Net proceeds from disposal (11,877)
0 4000 8000 12,000
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 defenders economic analysis. Sunk
  costspast costs that cannot be changed by any
  future investment decisionshould not be
  considered in a defenders economic analysis.

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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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• 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.