engineering economics in canada chapter 7 replacement decisions

Engineering Economics in Canada

Chapter 7

Replacement Decisions

Copyright © 2006 Pearson Education Canada Inc. 7-2

Introduction

• Long-lived assets require regular evaluation as they age.

• When they are evaluated, one of four choices can be made:

1. Keep the asset in use without major change (do nothing)

2. Overhaul the asset to improve its performance

3. Remove it from use without replacement

4. Replace the current asset with another asset


Introduction (con’t)

• Chapter 7 is concerned with replacement decisions for long-lived assets.

• Why a special chapter?– The relevant costs for making such

decisions are not always obvious.– The service live has been assumed in

earlier chapters – Chapter 7 shows how the service life of an asset is calculated.

– Assumptions about how an asset may be replaced in the future can affect a current decision.


• Recall from Chapter 6 – Assets lose value, or depreciate, for a variety of reasons:

– Use related physical loss: Measured as a function of units of production, miles driven, or other measures of use

– Time related physical loss: Measured in units of time

– Functional loss: Measured in terms of the function lost

Reasons for Depreciation


7.2 A Replacement Example

• How long should you keep a car before replacing it?

• Assume that you have an ongoing need for a car.– The equivalent annual capital costs of

replacing the car every N years can be found from the capital recovery formula

EAC(capital) = A = (P – S)(A/P, i, N) + Si• The longer the time between replacements, the

lower the EAC(capital).


7.2 A Replacement Example

equivalent annual capital costs

EAC(capital) = A = (P – S)(A/P, i, N) + Si

EAC: Annualized Capital Cost

P: Purchase Value

S: Salvage Value

i: Interest Rate

N: N-year replacement Period


7.2 A Replacement Example (con’t)

• The replacement decision also depends on operating and maintenance costs– These costs tend to increase the longer you

keep a car– The EAC of operating and maintaining a car

can be computed assuming the car is kept for N years, N=1,2,…

• The economic life is the number of years that minimizes:

EAC(total) = EAC(capital) + EAC(op. & maint.)


A Replacement Example

0

2,000

4,000

6,000

8,000

10,000

12,000

14,000

16,000

18,000

20,000

0 2 4 6 8 10

Years between replacement, N

EA

C

EAC(capital)

EAC(op. and maint.)

EAC(Total)

Economic Life = 5 yrs


7.3 Reasons for Replacement or Retirement

• If there is an ongoing need for the service an asset provides…

• Replacement becomes necessary if there is a less expensive means of getting the service it provides, or if the service provided is no longer adequate.

• If there is no longer a need for the service an asset provides, an asset may be retired – removed from service without replacement.


• What are the relevant costs associated with owning and assets?– Purchasing an asset is a decision to acquire

capacity, or the ability to produce a good or service. Relevant costs are:

1. Capital costs: – The difference between the price paid and

what it can be sold for some time after purchase, usually expressed as EAC.

– The largest portion of capital costs of an asset typically occurs early in its life.

7.4 Capital Costs and Other Costs


2. Installation costs:– Installation costs occur at the beginning of the

life of new assets and are not reversible once the capacity is put in place

– E.g. lost production time, lost output due to learning a new system

3. Operating and maintenance costs– the cost of using the asset to produce goods

or services• Most replacement decisions are made on the

basis of total Equivalent Annual Costs (EAC)

Capital Costs and Other Costs (con’t)


Equivalent Annual Costs (EAC)

P = first cost I = installation cost

S = salvage value N = number of years asset kept

EAC(capital) = (P + I – S)(A/P, i, N) + S i

EAC(op & maint):

(1)Treat as an arithmetic or gradient series if applicable, or

(2) Convert each cash flow to PW, then convert to annuity over N periods

EAC(total) = EAC(capital) + EAC(op & maint)


Replacement Scenarios• Defender – an existing asset• Challenger – a potential replacement • We will look at three situations:

1. The Defender and Challenger are identical and the need for the asset continues indefinitely.

2. The Defender is different from the Challenger; the challenger repeats indefinitely.

3. The Defender and Challenger are different, and successive Challengers are not identical.


7.5 Defender and Challenger are Identical

• All long-lived assets will need to be replaced eventually.

• We consider the situation where there is an ongoing need for an asset and the technology is not changing rapidly.– We model the replacement decision as if it

were to take place an indefinitely large number of times i.e. cyclic replacement


• Cyclic replacement: When technology, prices and interest rates are not changing quickly, an assumption can be made that an asset will be replaced with the same type of asset over and over

• EAC(capital) usually decrease over time• EAC(op&maint) usually increase over time

• Hence, there usually is a lifetime that will minimize EAC(total): this is the economic life of the asset

Economic Life of an Asset


Economic Life Illustrated

0

2,000

4,000

6,000

8,000

10,000

12,000

14,000

16,000

18,000

20,000

0 2 4 6 8 10

Years between replacement, N

EA

C

EAC(capital)

EAC(op. and maint.)

EAC(Total)

Economic Life = 5 yrs


• A new bottle capping machine costs $40 000, plus $5000 for installation.

• The machine is expected to have a useful life of 8 years with no salvage value at that time (assume straight line depreciation).

• Operating and maintenance costs are expected to be $3000 for the first year, increasing by $1000 each year thereafter.

• Interest is 12%. What is the economic life of a bottle capper?

Example 7-1(very important)


Replacement Analysis

• The minimum Total EAC occurs when replacement occurs every six years.

• The economic life of a bottle capper is 6 years

Years Salvage O & M EAC EAC EACkept Value Costs Capital O & M Total

0 $40,0001 $35,000 $3,000 $15,400 $3,000 $18,4002 $30,000 $4,000 $12,475 $3,472 $15,9473 $25,000 $5,000 $11,327 $3,925 $15,2524 $20,000 $6,000 $10,631 $4,359 $14,9905 $15,000 $7,000 $10,122 $4,775 $14,8976 $10,000 $8,000 $9,713 $5,172 $14,8857 $5,000 $9,000 $9,365 $5,551 $14,9168 $0 $10,000 $9,059 $5,913 $14,972


Decision Rule:

1. Determine the Economic Life of the challenger and its associated EAC

2. Determine the remaining Economic Life of the defender and its associated EAC

3. If the EAC(defender) > EAC(challenger), replace now. Otherwise, do not replace now.

7.6 Challenger is Different from Defender; Challenger repeats


• Brockville Brackets (BB) has a 3 year old robot and its Challenger is an upgraded used robot.

Defender Challenger

Price (when new) $300 000 $175 000

Installation $50 000 $10 000

Useful life 12 years 9 years

Annual depr. rate 20% 20%

• The annual maintenance costs for the upgraded robot are $40 000 the first year and will increase by 10% per year thereafter.

• The current robot's maintenance costs are expected to be $50 000 next year, also increasing at 10% per year.

• MARR is 15%. Should BB replace the current robot? If yes, when?

Example 7-2


Replacement Analysis (a)

• The Economic life of challenger robots is 9 years. The EAC of this replacement interval is $92 440 per year.

(a) Challenger - Used RobotYears EAC EAC EACKept Salvage Maint Capital Maint Total

0 175,0001 140,000 40,000 72,750 40,000 112,7502 112,000 44,000 61,703 41,860 103,5643 89,600 48,400 55,223 43,744 98,9674 71,680 53,240 50,444 45,645 96,0905 57,344 58,564 46,683 47,561 94,2456 45,875 64,420 43,643 49,487 93,1317 36,700 70,862 41,150 51,419 92,5698 29,360 77,949 39,088 53,352 92,4409 23,488 85,744 37,372 55,281 92,653


Replacement Analysis (b)

• The remaining economic life of the defender is 5 years. The EAC of keeping the defender this additional time is $97 808

• The EAC(defender) > EAC(challenger) replace now.

(b) Defender - Current RobotYears EAC EAC EACKept Salvage Maint Capital Maint Total

0 153,6001 122,880 50,000 53,760 50,000 103,7602 98,304 55,000 48,759 52,326 101,0853 78,643 60,500 44,626 54,680 99,3054 62,915 66,550 41,201 57,057 98,2585 50,332 73,205 38,356 59,452 97,8086 40,265 80,526 35,987 61,859 97,8467 32,212 88,578 34,009 64,274 98,2828 25,770 97,436 32,352 66,689 99,0429 20,616 107,179 30,962 69,102 100,064


In some cases, the computations for the Defender can be simplified

For assets that have been in place several years, the annual capital costs will often be low in comparison to operating and maintenance costs and the EAC(total) will be increasing in N

One year principle: if the cost of keeping the defender one more year exceeds the EAC of the challenger at its economic life, then the defender should be replaced immediately.

One Year Principle


Cost Components for Certain Older Assets


• Normally we expect the future challengers to be better than the current challenger– Do we skip over the current challenger and wait for

the next “new and improved” challenger?

• We must enumerate all possible combinations of replacement options and evaluate all to make a choice– EAC for each combination must be calculated– Number of combinations increases quickly– Typically, very little information is available on the

costs and benefits of future challengers

7.7 Challenger is Different from Defender;Challenger Does not Repeat


• In example 7-2, Brockville Brackets (BB) may use the current robot for its remaining life of 9 years, and replace it with a stream of new robots.

• BB may replace the current robot with an upgraded used robot, with a remaining life also of 9 years, followed by a stream of new robots.

… and so on

• BB can list all the possible options as Mutually Exclusive projects and select the best based on PW, AW (EAC) or IRR.

Example 7-2 (revisited)


Project

#

Keep Current

Robot for

Keep Upgrade

Robot for

Start stream of

New Robots in

1 0 years 0 years 0 years

2 1 year 0 years 1 year


… … … …


11 0 years 1 year 1 year

12 1 year 1 year 2 years

13 2 years 1 year 3 years

… … … …

19 8 years 1 year 9 years

… … … …



Summary

• Reasons for Replacement or Retirement of an asset

• Costs relevant to replacement decisions• Replacement analysis

1. Defender and Challenger are Identical• Economic life; cyclic replacement.

2. Challenger is different from Defender; Challenger repeats Indefinitely

3. Challenger is different from Defender; Challenger does not repeat

engineering economics in canada chapter 7 replacement decisions

Documents

asset slide

pearson education canada

costs cont slide

replacement decisions

cyclic replacement slide

installation costs

relevant costs

maintenance costs