(c) 2001 Contemporary Engineering Economics 1 Chapter 15 Replacement Decisions Replacement Analysis...

(c) 2001 Contemporary Engineering Economics

1

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 Asset’s

Disposal

$0 $5000 $10,000 $15,000 $20,000 $25,000 $30,000

Original investment

$10,000 $5000

Market value

$10,000

Lost investment(economic depreciation) Repair cost

$20,000

Sunk costs = $15,000


4

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.


5

Replacement Analysis – Cash Flow Approach

0 1 2 3 0 1 2 3

$8000

$2500

$15,000

$6000

$5500

(a) Defender (b) Challenger

$10,000

Sales proceeds from defender


6

Annual Equivalent Cost - Cash Flow Approach

Defender: PW(12%)D = $2,500 (P/F, 12%, 3) - $8,000 (P/A, 12%, 3)

= - $17,434.90AE(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) = -$15,495.90AE(12%)C = PW(12%)C(A/P, 12%, 3) = -$6,451.79

Replace the

defender now!


7

Opportunity Cost Approach

0 1 2 3 0 1 2 3

$8000

$2500

$15,000

$6000

$5500Defender Challenger

$10,000Proceeds from sale viewed asan opportunity cost of keepingthe asset


8

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

Opportunity Cost Approach

Replace the defender now!


9

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.

Ownership (Capital)Cost (init.+salvg.)

Operating cost

+

Minimize

CR i I A P i N S A F i NN( ) ( / , , ) ( / , , )


10

Mathematical Relationship

• Capital Recov. Cost.

• Operating Cost:

• Total Cost:

• Objective: Find n* that minimizes AEC

CR i I A P i N S A F i NN( ) ( / , , ) ( / , , )

OC i OC P F i n A P i Nnn

N

( ) ( / , , ) ( / , , )

1

AEC CR i OC i ( ) ( )

CR(i)

OC(i)

AEC

n*


11

Economic Service Life for a Lift Truck


12

Economic Service Life Calculation (Example 15.4)

• N = 1

AEC1 = $18,000(A/P, 15%, 1) + $1,000 - $10,000

= $11,700


13

• N = 2

AEC2 = [$18,000 + $1,000(P/A,15%, 15%, 2)](A/P, 15%, 2)

- $7,500 (A/F, 15%, 2)

= $8,653


14

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


15

Required Assumptions and Decision Frameworks

• Planning horizon (study period)

• Technology

• Relevant cash flow information

• Decision Frameworks


16

Replacement Strategies under the Infinite Planning Horizon

1. 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.

2. 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.


17

Example 15.5

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

N AE

N AE

N AE

N AE

N AE

1 15%) 130

2 15%) 116

3 15%) 500

4 15%) 961

5 15%) 434

: ( $5,

: ( $5,

: ( $5,

: ( $5,

: ( $6,N years

AED

D

*

* $5,

2

116


18

• Challenger: find the economic service life.

N = 1 year: AE(15%) = $7,500N = 2 years: AE(15%) = $6,151N = 3 years: AE(15%) = $5,847N = 4 years: AE(15%) = $5,826N = 5 years: AE(15%) = $5,897

NC*=4 years

AEC*=$5,826


19

Replacement Decisions

• 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.

NC*=4 years

AEC*=$5,826

N

AED

D

*

* $5,

2

116

years


20

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


21

23

$3000

$2000

$5000

• 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.

$6,450

2 3


22

Replacement Analysis under the Finite Planning Horizon

Annual Equivalent Cost ($)

N 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 5,897Some likely replacement patternsunder a finite planning horizon of8 years


23

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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• Option 3 (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

• 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

Example 15.6 continued


25

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


26

(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

Planning horizon = 8 years


27

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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$0 $4000 $8000 $12,000 $16,000 $20,000

$10,000

$10,000

$14,693

$20,000$20,000

$4693

Depreciation basis

Book valueTotaldepreciation

$5307

Market value

Market valueLosstax credit

Book loss

Net proceeds from disposal ($11,877)

$4693 $1877 40%


29

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.


30

• 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.

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(c) 2001 Contemporary Engineering Economics 1 Chapter 15 Replacement Decisions Replacement Analysis...

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