chapter 15 replacement decisions

(c) 2001 Contemporary Engineering Economics

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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


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Replacement Analysis Fundamentals

• Replacement projects are decision problems involve the replacement of existing obsolete or worn-out assets.

• When existing equipment should be replaced with more efficient equipment.

Examine replacement analysis fundamentals1) Approaches for comparing defender and challenger2) Determination of economic service life3) Replacement analysis when the required service period

is long


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Replacement Terminology

• Sunk cost: any past cost that is unaffected by any future decisions

• Trade-in allowance: value offered by the vendor to reduce the price of a new equipment

• Defender: an old machine

• Challenger: new machine

• Current market value: selling price of the defender in the market place


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Sunk Cost associated with an Asset’s

Disposal (example 15.1)

$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


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Replacement Decisions Two basic approaches to analyze replacement problems.

• Cash Flow Approach

- Treat the proceeds from sale of the old machine as down payment toward purchasing the new machine.

• 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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Example 15.2 Replacement Analysis Cash Flow Approach

0 1 2 3 0 1 2 3

$8000

$2500

$15,000

$6000

$5500

(a) Keep the Defender (b) Replace the defender

With Challenger

$10,000

Sales proceeds from defender


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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)

= - $17,434.90AE(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

Annual difference is $807.31

Replace the

defender now!


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Example 15.3 Opportunity Cost Approach

0 1 2 3 0 1 2 3

$8000

$2500

$15,000

$6000

$5500

Defender Challenger

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


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Defender: Charge $ 10,000 as an opportunity cost or incurred cost.

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

Because of the annual difference of $807.31 in favor of the challenger.

Opportunity Cost Approach

Replace the defender now!


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Economic Service Life• Definition: The economic

service of an asset is defined to be the period of useful life that minimizes the annual equivalent cost of owning and operating asset.

• We should use the economic service lives of the defender and the challenger when conducting a replacement analysis.

Ownership (Capital)cost

Operating cost

+

Minimize


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Economic Service Life Continue….

• Capital cost have two components: Initial investment and the salvage value at the time of disposal.

• The initial investment for the challenger is its purchase price. For the defender, we should treat the opportunity cost as its initial investment.

• Use N to represent the length of time in years the asset will be kept; I is the initial investment, and SN is the salvage value at the end of the ownership period of N years.

• The operating costs of an asset include operating and maintenance (O&M) costs, labor costs, material costs and energy consumption costs.


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Mathematical Relationship• AE of Capital Cost:

• AE of Operating Cost:

• Total AE Cost:

• Objective: Find n* that minimizes total 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*


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Economic Service Life for a Lift Truck Example 15.4


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Economic Service Life Calculation (Example 15.4)

• N = 1 (if you replace the asset every year)

AEC1 = $18,000(A/P, 15%, 1) + $1,000 - $10,000 = $11,700


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• N = 2 (Truck will be used for two years and disposed of at the end of year 2)

AEC2 = [$18,000 + $1,000(P/A1, 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

Now we understand how the economic service life of an asset isdetermined, the next question is to decide whether now is the time to replace the defender.

Consider the following factors:

Planning horizon (study period)• The project will have a exact a predictable duration. Therefore, the

replacement policy should be formulated based on a finite planning horizon.


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Decision Frameworks continue…….

Technology• Predictions of technological patterns over the planning horizon refer to the

development of types of challengers that may replace those under study. • If we assume that all future machines will be same as those now in service,

there is no technological progress in the area will occur. • In other cases, we may recognize the possibility of machines become available, that

will be more efficient, reliable, productive than those in the current market.

Relevant cash flow information• Many varieties of predictions can be used to estimate the pattern of revenue, cost and

salvage value over the life of an asset.

Decision Criterion• The AE method provides a more direct solution when the planning horizon is infinite

(endless). When the planning horizon is finite (fixed), the PW method is convenient to be used.


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Replacement Strategies under the Infinite Planning Horizon

1. Compute the economic lives of both defender and challenger. Let’s use ND* and NC* to indicate the economic lives of the defender and the challenger, respectively. The annual equivalent cost for the defender and the challenger at their respective economic lives are indicated by AED* and AEC*.

2. Compare AED* and AEC*. If AED* is bigger than AEC*, we know that it is more costly to keep the defender than to replace it with the challenger. Thus, the challenger should replace the defender now.

3. If the defender should not be replaced now, when should it be replaced? First, we need to continue to use until its economic life is over. Then, we should calculate the cost of running the defender for one more year after its economic life. If this cost is greater than AEC*, the defender should be replaced at the end of is economic life. This process should be continued until you find the optimal replacement time. This approach is called marginal analysis, that is, to calculate the incremental cost of operating the defender for just one more year.


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Replacement Analysis under the Infinite Planning Horizon Example 15.5

• 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%) = $7,500N = 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


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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


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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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$3000

$2000

$5000

• Step 1: Calculate the equivalent cost of retaining the defender one more year from the end of its economic service life, say 2 to 3.

AED = $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

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chapter 15 replacement decisions

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