1. Chapter 15 Replacement Decisions
Replacement Analysis Fundamentals
Economic Service Life
Replacement Analysis When Required Service is Long
Replacement Analysis with Tax Consideration
(c) 2001 Contemporary Engineering Economics

2. 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 fundamentals
Approaches for comparing defender and challenger
Determination of economic service life
Replacement analysis when the required service period is long
(c) 2001 Contemporary Engineering Economics

3. Replacement Terminology
Defender: an old machine
Challenger: new machine
Current market value: selling price of the defender in the market place
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
(c) 2001 Contemporary Engineering Economics

4. Sunk Cost associated with an Asset’s Disposal (example 15.1)
Original investment
$20,000
Lost investment
(economic depreciation)
Market value
Repair cost
$10,000
$10,000
$5000
Sunk costs = $15,000
$ $ $10, $15, $20, $25, $30,000
(c) 2001 Contemporary Engineering Economics

5. (c) 2001 Contemporary Engineering Economics
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.
(c) 2001 Contemporary Engineering Economics

6. Cash Flow Approach Example 15.2 Replacement Analysis Sales proceeds
from defender
$10,000
$5500
$2500
$6000
$8000
(b) Replace the defender
With Challenger
(a) Keep the Defender
$15,000
(c) 2001 Contemporary Engineering Economics

7. Annual Equivalent Cost
Cash Flow Approach
 Defender:
PW(12%)D = $2,500 (P/F, 12%, 3) - $8,000 (P/A, 12%, 3)
= - $17,434.90
AE(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)
= -$15,495.90
AE(12%)C = PW(12%)C(A/P, 12%, 3)
= -$6,451.79
Annual difference is $807.31
Replace the defender now!
(c) 2001 Contemporary Engineering Economics

8. Example 15.3 Opportunity Cost Approach
Defender
Challenger
$2500
$5500
$8000
$6000
$10,000
Proceeds from sale viewed as
an opportunity cost of keeping
the asset
$15,000
(c) 2001 Contemporary Engineering Economics

9. Opportunity Cost Approach
 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 $ in favor of the challenger.
Replace the
defender now!
(c) 2001 Contemporary Engineering Economics

10. (c) 2001 Contemporary Engineering Economics
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.
Minimize
Ownership (Capital)
cost +
Operating
cost
(c) 2001 Contemporary Engineering Economics

11. 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.
(c) 2001 Contemporary Engineering Economics

12. Mathematical Relationship
AE of Capital Cost:
AE of Operating Cost:
Total AE Cost:
Objective: Find n* that minimizes total AEC
AEC
OC(i)
CR(i) n*
(c) 2001 Contemporary Engineering Economics

13. Economic Service Life for a Lift Truck Example 15.4
(c) 2001 Contemporary Engineering Economics

14. 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
(c) 2001 Contemporary Engineering Economics

15. (c) 2001 Contemporary Engineering Economics
(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
(c) 2001 Contemporary Engineering Economics

16. (c) 2001 Contemporary Engineering Economics
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
(c) 2001 Contemporary Engineering Economics

17. Required Assumptions and Decision Frameworks
Now we understand how the economic service life of an asset is
determined, 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.
(c) 2001 Contemporary Engineering Economics

18. 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.
(c) 2001 Contemporary Engineering Economics

19. Replacement Strategies under the Infinite Planning Horizon
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*.
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.
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.
(c) 2001 Contemporary Engineering Economics

20. Replacement Analysis under the Infinite Planning Horizon Example 15.5
Step 1: Find the remaining useful (economic) service life of the defender.
(c) 2001 Contemporary Engineering Economics

21. (c) 2001 Contemporary Engineering Economics
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%) = $5,897
NC*=4 years
AEC*=$5,826
(c) 2001 Contemporary Engineering Economics

22. 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
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23. (c) 2001 Contemporary Engineering Economics
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
(c) 2001 Contemporary Engineering Economics

24. (c) 2001 Contemporary Engineering Economics
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.
$2000 2 3
$3000
$5000 2 3
$6,450
(c) 2001 Contemporary Engineering Economics