## Presentation on theme: "Engineering Economics in Canada"— Presentation transcript:

Chapter 7 Replacement Decisions

Introduction Long-lived assets require regular evaluation as they age. When they are evaluated, one of four choices can be made: Keep the asset in use without major change (do nothing) Overhaul the asset to improve its performance Remove it from use without replacement Replace the current asset with another asset Copyright © 2006 Pearson Education Canada Inc.

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. Copyright © 2006 Pearson Education Canada Inc.

Reasons for Depreciation
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 Copyright © 2006 Pearson Education Canada Inc.

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). Copyright © 2006 Pearson Education Canada Inc.

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 Copyright © 2006 Pearson Education Canada Inc.

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.) Copyright © 2006 Pearson Education Canada Inc.

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. Copyright © 2006 Pearson Education Canada Inc.

7.4 Capital Costs and Other Costs
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. Copyright © 2006 Pearson Education Canada Inc.

Capital Costs and Other Costs (con’t)
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) Copyright © 2006 Pearson Education Canada Inc.

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): Treat as an arithmetic or gradient series if applicable, or Convert each cash flow to PW, then convert to annuity over N periods EAC(total) = EAC(capital) + EAC(op & maint) Copyright © 2006 Pearson Education Canada Inc.

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. Copyright © 2006 Pearson Education Canada Inc.

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 Copyright © 2006 Pearson Education Canada Inc.

Economic Life of an Asset
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 Copyright © 2006 Pearson Education Canada Inc.

Economic Life Illustrated

Example 7-1(very important)
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? Copyright © 2006 Pearson Education Canada Inc.

Replacement Analysis The minimum Total EAC occurs when replacement occurs every six years. The economic life of a bottle capper is 6 years Copyright © 2006 Pearson Education Canada Inc.

7.6 Challenger is Different from Defender; Challenger repeats
Decision Rule: Determine the Economic Life of the challenger and its associated EAC Determine the remaining Economic Life of the defender and its associated EAC If the EAC(defender) > EAC(challenger), replace now. Otherwise, do not replace now. Copyright © 2006 Pearson Education Canada Inc.

Example 7-2 Brockville Brackets (BB) has a 3 year old robot and its Challenger is an upgraded used robot. Defender Challenger Price (when new) \$ \$ Installation \$ \$10 000 Useful life 12 years 9 years Annual depr. rate 20% 20% The annual maintenance costs for the upgraded robot are \$ the first year and will increase by 10% per year thereafter. The current robot's maintenance costs are expected to be \$ next year, also increasing at 10% per year. MARR is 15%. Should BB replace the current robot? If yes, when? Copyright © 2006 Pearson Education Canada Inc.

Replacement Analysis (a)
The Economic life of challenger robots is 9 years. The EAC of this replacement interval is \$ per year. Copyright © 2006 Pearson Education Canada Inc.

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. Copyright © 2006 Pearson Education Canada Inc.

One Year Principle 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. Copyright © 2006 Pearson Education Canada Inc.

Cost Components for Certain Older Assets

7.7 Challenger is Different from Defender; Challenger Does not Repeat
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 Copyright © 2006 Pearson Education Canada Inc.

Example 7-2 (revisited) 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. Copyright © 2006 Pearson Education Canada Inc.