www.izmirekonomi.edu.tr asst. prof. dr. mahmut ali gÖkÇe, izmir university of economics spring,...

(c) 2001 Contemporary Engineering Economics

1www.izmirekonomi.edu.tr

Asst. Asst. Prof. Dr. Prof. Dr. Mahmut Ali Mahmut Ali GÖKÇE, Izmir University of EconomicsGÖKÇE, Izmir University of Economics

SpringSpring, 2007, 2007

Replacement Decision Models

(c) 2001 Contemporary Engineering Economics www.izmirekonomi.edu.tr



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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

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.




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Example 11.5 Replacement Analysis under the Infinite Planning Horizon Step 1:Find the

remaining useful (economic) service life of the defender.

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




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Step 2: find the economic service life of the challenger.

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%) = $5,897NC*=4 years

AEC*=$5,826




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Step 3: Replacement Decisions

Should we replace the defender now? No, because AED < AEC

If not, when is the best time to replace the defender? Need to conduct a marginal analysis.

NC*=4 years

AEC*=$5,826

N

AED

D

*

* $5,

2

116

years




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Marginal Analysis – When to Replace the DefenderQuestion: 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: $3,000 (market value of the defender at the end year 2) • 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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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




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Replacement Analysis with Tax Consideration Whenever possible, replacement decisions should

be based on the cash flows after taxes. 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.

All basic replacement decision rules including the way of computing economic service life remain unchanged.




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

$10,000

$10,000

$14,693

$20,000$20,000

$4693

Cost basis

Book valueTotal depreciation

$5307

Market value

Market valueLosstax credit

Book loss

Net proceeds from disposal ($11,877)

$4693 $1877 40%

Determining the Opportunity Cost with Tax Consideration




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Example 11.6 Replacement Analysis under an Infinite Planning Horizon Economic service life calculation with tax

consideration for Defender (Table 11.1) Economic service life calculation with tax

consideration for Challenger (Table 11.2) Replacement Decision under the Infinite

Planning Horizon




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Current After-tax

n Market Value Salvage Value

0 $5,000 $5,000(1 – 0.40) = $3,000

1 4,000 4,000(1 – 0.40) = 2,400

2 3,000 3,000(1 – 0.40) = 1,800

3 2,000 2,000(1 – 0.40) = 1,200

4 1,000 1,000(1 – 0.40) = 600

5 0 0

6 0 0

Forecasted After-tax

n Overhaul O&M Cost O&M Cost

0 $1,200 $1,200(1 – 0.40) = $720 1 0 $2,000 2,000(1 – 0.40) = 1,200

2 0 3,500 3,500(1 – 0.40) = 2,100

3 0 5,000 5,000(1 – 0.40) = 3,000

4 0 6,500 6,500(1 – 0.40) = 3,900

5 0 8,000 8,000(1 – 0.40) = 4,800




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D E F G H I J K L

Example 11.6 : Remaing Useful Life for DefenderEquivalent Annual Cost if the

Defender is Kept for N More Years

Net Investment Net A/T

A/T Depreciation Operating and Cash Capital Operating Total

Depreciation O&M Credit Cost Net Salvage Flow Cost Cost Cost

(720)$ (720)$ (3,000)$ (3,720)$

(1,200)$ (1,200)$ 2,400$ 1,200$ (1,050)$ (2,028)$ (3,078)$

(720)$ (720)$ (3,000)$ (3,720)$

(1,200)$ (1,200)$ (1,200)$

(2,100)$ (2,100)$ 1,800$ (300)$ (1,008)$ (2,061)$ (3,070)$

(720)$ (720)$ (3,000)$ (3,720)$

(1,200)$ (1,200)$ (1,200)$

(2,100)$ (2,100)$ (2,100)$

(3,000)$ (3,000)$ 1,200$ (1,800)$ (968)$ (2,332)$ (3,300)$

(720)$ (720)$ (3,000)$ (3,720)$

(1,200)$ (1,200)$ (1,200)$

(2,100)$ (2,100)$ (2,100)$

(3,000)$ (3,000)$ (3,000)$

(3,900)$ (3,900)$ 600$ (3,300)$ (931)$ (2,646)$ (3,576)$

(720)$ (720)$ (3,000)$ (3,720)$

(1,200)$ (1,200)$ (1,200)$

(2,100)$ (2,100)$ (2,100)$

(3,000)$ (3,000)$ (3,000)$

(3,900)$ (3,900)$ (3,900)$

(4,800)$ (4,800)$ -$ (4,800)$ (895)$ (2,965)$ (3,860)$




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A B C D E F G H I J K L M N O

Permitted Annual Depreciation Amounts over the Expected Net A/T

Holding Holding Period Total Book Market Taxable Gains Salvage

Period O&M 1 2 3 4 5 6 7 Depreciation Value Value Gains Tax Value

1 2,000$ 2,000$ 2,000$ 8,000$ 6,000$ (2,000)$ (800)$ 6,800$

2 3,000$ 2,000$ 1,600$ 3,600$ 6,400$ 5,100$ (1,300)$ (520)$ 5,620$

3 4,000$ 2,000$ 3,200$ 960$ 6,160$ 3,840$ 4,335$ 495$ 198$ 4,137$

4 5,000$ 2,000$ 3,200$ 1,920$ 576$ 7,696$ 2,304$ 3,685$ 1,381$ 552$ 3,133$

5 6,000$ 2,000$ 3,200$ 1,920$ 1,152$ 576$ 8,848$ 1,152$ 3,132$ 1,980$ 792$ 2,340$

6 7,000$ 2,000$ 3,200$ 1,920$ 1,152$ 1,152$ 576$ -$ 10,000$ -$ 2,662$ 2,662$ 1,065$ 1,597$

7 8,000$ 2,000$ 3,200$ 1,920$ 1,152$ 1,152$ 576$ -$ 10,000$ -$ 2,263$ 2,263$ 905$ 1,358$

Note: Asset price of $10,000, depreciated under MACRS for 5-year property with the half-year convention

Table 11.2(a) Forecasted O&M Costs and Net Proceeds from Sale as a Function of Holding Period (Example 11.6)




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101112131415161718192021222324252627

A B C D E F G H I J K L M N(1) (2) (3) (4) (5) (6) (7) (8) (9) (10) (11) (12)

Before-Tax After-Tax Cash Flow If the Asset Equivalent Annual Cost if the

Holding Period Operating Expenses is Kept for N More Years Challenger is Kept for N More Years

Net Investment Net A/T

A/T Depreciation Operating and Cash Capital Operating Total

N n O&M Depreciation O&M Credit Cost Net Salvage Flow Cost Cost Cost

0 (10,000)$ (10,000)$

1 1 2,000$ 2,000$ (1,200)$ 800$ (400)$ 6,800$ 6,400$ (4,700)$ (400)$ (5,100)$

0 (10,000)$ (10,000)$

1 2,000$ 2,000$ (1,200)$ 800$ (400)$ (400)$

2 2 3,000$ 1,600$ (1,800)$ 640$ (1,160)$ 5,620$ 4,460$ (3,536)$ (753)$ (4,290)$

0 (10,000)$ (10,000)$

1 2,000$ 2,000$ (1,200)$ 800$ (400)$ (400)$

2 3,000$ 3,200$ (1,800)$ 1,280$ (520)$ (520)$

3 3 4,000$ 960$ (2,400)$ 384$ (2,016)$ 4,137$ 2,121$ (3,188)$ (905)$ (4,094)$

0 (10,000)$ (10,000)$

1 2,000$ 2,000$ (1,200)$ 800$ (400)$ (400)$

2 3,000$ 3,200$ (1,800)$ 1,280$ (520)$ (520)$

3 4,000$ 1,920$ (2,400)$ 768$ (1,632)$ (1,632)$

4 4 5,000$ 576$ (3,000)$ 230$ (2,770)$ 3,133$ 363$ (2,875)$ (1,190)$ (4,065)$

0 (10,000)$ (10,000)$

1 2,000$ 2,000$ (1,200)$ 800$ (400)$ (400)$ Economic Service Life2 3,000$ 3,200$ (1,800)$ 1,280$ (520)$ (520)$

3 4,000$ 1,920$ (2,400)$ 768$ (1,632)$ (1,632)$

4 5,000$ 1,152$ (3,000)$ 461$ (2,539)$ (2,539)$

5 5 6,000$ 576$ (3,600)$ 230$ (3,370)$ 2,340$ (1,030)$ (2,636)$ (1,474)$ (4,110)$




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Annual Equivalent Cost ($)

n Defender Challenger

1

2

3

4

5

6

7

3,078

3,070

3,300

3,576

3,860

5,100

4,290

4,094

4,065

4,110

4,189

4,287

N*D = 2 years N*C = 4 years




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When to Replace the Defender?n PW(15%)

CostCalculation

0

1

2

3

4

5

27,100

26,242

25,482

25,353

25,704

26,413

PW P A P Fn

15% = $3,078 , 15%, 1 $27,100 , 15%, 10 =1b g b g b g

PW P A P Fn

15% = $3,070 , 15%, 2 $27,100 , 15%, 20 =2b g b g b g

PW P A P Fn

15% = $3,300 , 15%, 3 $27,100 , 15%, 30 =3b g b g b g

PW P A P Fn

15% = $3,576 , 15%, 4 $27,100 , 15%, 40 =4b g b g b g

PW P A P Fn

15% = $3,860 , 15%, 5 $27,100 , 15%, 50 =5b g b g b g

PWn

15% = 1/ 0.15 $4,0650 =0b g b gb g




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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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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 of owning and operating the asset.

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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