1. Chapter 11 Replacement Decisions
Replacement Analysis Fundamentals
Economic Service Life
Replacement Analysis When a Required Service is Long

2. Replacement Terminology
Sunk cost: any past cost 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: a new machine
Current market value: selling price of the defender in the market place

3. (economic depreciation)
Sunk Cost associated with an Asset’s Disposal (Example Macintosh Printing Inc.)
Original investment (printing machine)
$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

4. 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

5. Replacement Decisions
Cash Flow Approach
Treat the proceeds from sale of the old machine as down payment toward purchasing the new machine.
This approach is meaningful when both the defender and challenger have the same service life.
Opportunity Cost Approach
Treat the proceeds from sale of the old machine as the investment required to keep the old machine.
This approach is more commonly practiced in replacement analysis.

6. Example 11.2 Defender Challenger Market price: $10,000
Remaining useful life: 3 years
Salvage value: $2,500
O&M cost: $8,000
Challenger
Cost: $15,000
Useful life: 3 years
Salvage value: $5,500
O&M cost: $6,000

7. Replacement Analysis – Cash Flow Approach
Sales proceeds
from defender
$10,000
$5500
$2500
$6000
$8000
(a) Defender
(b) Challenger
$15,000

8. Annual Equivalent Cost - Cash Flow Approach
Defender:
PW(12%)D = $8,000 (P/A, 12%, 3) -$2,500 (P/F, 12%, 3)
= $17,434.90
AEC(12%)D = PW(12%)D(A/P, 12%, 3)
= $7,259.10
Challenger:
PW(12%)C = $5,000 + $6,000 (P/A, 12%, 3)
- $5,500 (P/F, 12%, 3)
= $15,495.90
AEC(12%)C = PW(12%)C(A/P, 12%, 3)
= $6,451.79
Replace the defender now!

9. Example 11.3 Comparison of Defender and Challenger Based on Opportunity Cost Approach
Figure: 14-02EXM

10. Annual Equivalent Cost - Opportunity Cost Approach
Defender:
PW(12%)D = -$10,000 - $8,000(P/A, 12%, 3) + $2,500(P/F, 12%, 3)
= -$27,434.90
AEC(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
AEC(12%)C = -PW(12%)C(A/P, 12%, 3)
= $10,615.33
Replace the
defender now!

11. Economic Service Life + Minimize Ownership (Capital) cost Operating
Definition: Economic service life is the remaining useful life of an asset that results in the minimum annual equivalent cost.
Why do we need it?: We should use the respective economic service lives of the defender and the challenger when conducting a replacement analysis.
Minimize
Ownership (Capital)
cost
Annual Equivalent Cost +
Operating
cost

12. Economic Service Life Continue….
Capital cost have two components: Initial investment (I) and the salvage value (S) 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.

13. Mathematical Relationship Objective: Find n* that minimizes total AEC
AE of Capital Cost:
AE of Operating Cost:
Total AE Cost:

14. AEC
OC (i)
CR(i) n*

15. Example 11.4 Economic Service Life for a Lift Truck

16. Steps to Determine an Economic Service Life
N = 1 (if you replace the asset every year)
AEC1 = $18,000(A/P, 15%, 1) + $1,000 - $10,000
= $11,700

17. N = 2 (if you replace the asset every other year)
AEC2 = [$18,000 + $1,000(P/A, 15%, 15%, 2)](A/P, 15%, 2)
- $7,500 (A/F, 15%, 2)
= $8,653

18. AEC if the Asset were Kept N Years
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
If you purchase the asset,
it is most economical to replace
the asset for every 6 years
Economic Service Life

19. 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)
By planning horizon, it is mean that the service period required by the defender and a sequence of future challengers. The infinite planning horizon is used when we are unable to predict when the activity under consideration will be terminated. In other situation, the project will have a definite and predictable duration. In these cases, replacement policy should be formulated based on a finite planning horizon.

20. 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.
A number of possibilities exist in predicting purchase cost, salvage value, and operating cost as dictated by the efficiency of the machine over the life of an asset.
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 explicitly recognize the possibility of future machines that will be significantly more efficient, reliable, or productive than those currently on the market. (such as personal computers)
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.

21. 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.

22. Example 11.5 Relevant Cash Flow Information (Defender)
Figure: 14-07EXM

23. ECONOMIC SERVICE LIFE OF DEFENDER
General equation for AE calculation for the defender is as follows:
AE (15%) = $6,200(A/P, 15%, N) $1,500 (A/G, 15%, N)
– 1,000 (5 – N) (A/F, 15%, N) for N = 1,2,3,4, and 5
Cash flow diagram for defender When N = 4 years

24. ECONOMIC SERVICE LIFE OF DEFENDER
AE (15%)1 = $6,200 (A/P, 15%, 1) $1,500 (A/G, 15%, 1)
– 1,000 (5 – 1) (A/F, 15%, 1)
AE (15%)1 = 7, , – 4,000 = $5,130 OR
CR (15%)N = I (A/P, 15%, N) – SN (A/F, 15%, N)
AEOC =Σ [OCn (P/F, 15%, N)] (A/P, 15%, N)
CR (15%)1 = 6,200 (1.15) – 4,000 (1.0) = 7,130 – 4,000 = $3,130
AEOC1 = 2,000 (0.8696) (1.15) = x (1.15) = $2,000
Σ AE1 = CR (15%)1 + AEOC1 = 3,1,30 + 2,000 = $5,130

25. Objective is : Find n* that minimizes total AEC
AE of Capital Cost:
AE of Operating Cost:
Total AE Cost:

26. N = 2 Σ AE2 = CR (15%)2 + AEOC2 = 2,418.32 + 2,697.55 = $5,116
AE (15%)2 = $6,200 (A/P, 15%, 2) $1,500 (A/G, 15%, 2)
– 1,000 (5 – 2) (A/F, 15%, 2)
AE (15%)2 = 3, , – 1,395.3 = $5,116 OR
CR (15%)2 = I (A/P, 15%, 2) – S2 (A/F, 15%, 2)
CR (15%)2 = 6,200 (0.6151) – 3,000 (0.4651) = $2,418.32
AEOC2 =Σ [ OC1 (P/F, 15%, 1) + OC2 (P/F, 15%, 2) ] (A/P, 15%, 2)
AEOC2 = [2,000 (0.8696) + 3,500 (0.7561)] (0.6151) = $2,697.55
Σ AE2 = CR (15%)2 + AEOC2 = 2, , = $5,116

27. Σ AE3 = CR (15%)3 + AEOC3 = 2,139.6 + 3,360.7 = $5,500 N = 3
AE (15%)3 = $6,200 (A/P, 15%, 3) $1,500 (A/G, 15%, 3)
– 1,000 (5 – 3) (A/F, 15%, 3)
AE (15%)3 = 2, , , – 576 = $5,500 OR
CR (15%)3 = I (A/P, 15%, 3) – S3 (A/F, 15%, 3)
CR (15%)3 = 6,200 (0.4380) – 2,000 (0.2880) = $2,139.6
AEOC3 =Σ [ OC1 (P/F, 15%, 1) + OC2 (P/F, 15%, 2) + OC3 (P/F, 15%, 3) ] x (A/P, 15%, 3)
AEOC3 = [2,000 (0.8696) + 3,500 (0.7561) + 5,000 (0.6575)] (0.6151)
AEOC3 = $3,360.7
Σ AE3 = CR (15%)3 + AEOC3 = 2, ,360.7 = $5,500

28. Σ AE4 = CR (15%)4 + AEOC4 = 1,971.56 + 3,989.75 = $5,961 N = 4
AE (15%)4 = $6,200 (A/P, 15%, 4) $1,500 (A/G, 15%, 4)
– 1,000 (5 – 4) (A/F, 15%, 4)
AE (15%)4 = 3, , – 1,395.3 = $5,961 OR
CR (15%)4 = I (A/P, 15%, 4) – S4 (A/F, 15%, 4)
CR (15%)4 = 6,200 (0.3503) – 1,000 (0.2003) = $1,971.56
AEOC4 =Σ [ OC1 (P/F, 15%, 1) + OC2 (P/F, 15%, 2) + OC3 (P/F, 15%, 3) OC4 (P/F, 15%, 4)] x (A/P, 15%, 4)
AEOC4 = [2,000 (0.8696) + 3,500 (0.7561) + 5,000 (0.6575) + 6,500 (0.5718) ] x (0.6151)
AEOC4 = $3,989.75
Σ AE4 = CR (15%)4 + AEOC4 = 1, , = $5,961

29. ECONOMIC SERVICE LIFE OF DEFENDER
AE (15%) 5 = $6,200 (A/P, 15%, 5) $1,500 (A/G, 15%, 5)
– 1,000(5 – 5) (A/F, 15%, 3)
AE (15%)5 = 1, , , = $6,434 OR
CR (15%)5 = I (A/P, 15%, 5) – S5 (A/F, 15%, 5)
CR (15%)5 = 6,200 ( ) – 0 (0.1483) = $1,850
AEOC5 =Σ [ OC1 (P/F, 15%, 1) + OC2 (P/F, 15%, 2) + OC3 (P/F, 15%, 3) OC4 (P/F, 15%, 4) + OC5 (P/F, 15%, 5)] x (A/P, 15%, 5)
AEOC5 = [2,000 (0.8696) + 3,500 (0.7561) + 5,000 (0.6575) + 6,500 (0.5718) ,000 (0.4972)] x (0.2983)
AEOC5 = $4,584
Σ AE5 = CR (15%)5 + AEOC 5 = 1, ,584 = $6,434

30. For N = 1 to 5, the results are as follows:
N = 1: AE (15%) = $5,130
N = 2: AE (15%) = $5,116
N = 3: AE (15%) = $5,500
N = 4: AE (15%) = $5,961
N = 5: AE (15%) = $6,434
When N = 2 years, we get the lowest AE value. Thus the defender’s economic life is two years.

31. AEC as a Function of the Life of the Defender (Example 11.5)
Figure: 14-08

32. ECONOMIC SERVICE LIFE OF CHALLENGER (Example 11.5)
Investment cost = $10,000
Salvage value
N = 1: $6,000
N > 1: decreases at a 15% over previous year
Operating cost
N = 1: $2,000
N > 1: increases by $800 per year (G = $800)
Rate of return 15%

33. ECONOMIC SERVICE LIFE OF CHALLENGER
General equation for AE calculation for the challenger is as follows:
AE (15%)N = $10,000(A/P, 15%, N) $800 (A/G, 15%, N)
– $6,000(1 – 15%)N-1 (A/F, 15%, N) for N = 1,2,3,4, and 5
 N =1
AE (15%)1 = $10,000(A/P, 15%, 1) $800 (A/G, 15%, 1)
– $6,000(0.85)1-1 (A/F, 15%, 1)
AE (15%)1 = , – 6,000 = $7,500 OR
CR (15%)N = I (A/P, 15%, N) – SN (A/F, 15%, N)
AEOC =Σ [OCn (P/F, 15%, N)] (A/P, 15%, N)
CR (15%)1 = 10,000 (1.15) – 6,000 (1.0) = 11,500 – 6,000 = $5,500
AEOC1 = 2,000 (0.8696) (1.15) = x (1.15) = $2,000
Σ AE1 = CR (15%)1 + AEOC1 = 3, ,000 = $7,500

34. N =2 Σ AE2 = CR (15%)2 + AEOC2 = 2,418.32 + 2,697.55 = $6,151
AE (15%)2 = $10,000(A/P, 15%, 2) $800 (A/G, 15%, 2)
– $6,000(0.85)2-1 (A/F, 15%, 2)
AE (15%)1 = , – 2,372 = $6,151 OR
CR (15%)2 = I (A/P, 15%, 2) – S2 (A/F, 15%, 2)
CR (15%)2 = 10,000 (0.6151) – 5,100 (0.4651) = $3,779
AEOC2 =Σ [ OC1 (P/F, 15%, 1) + OC2 (P/F, 15%, 2) ] (A/P, 15%, 2)
AEOC2 = [2,000 (0.8696) + 2,800 (0.7561)] (0.6151) = $2,372
Σ AE2 = CR (15%)2 + AEOC2 = 2, , = $6,151

35. Σ AE3 = CR (15%)3 + AEOC3 = 2,139.6 + 3,360.7 = $5,857 N =3
AE (15%)3 = $10,000(A/P, 15%, 3) $800 (A/G, 15%, 3)
– $6,000(0.85)3-1 (A/F, 15%, 3)
AE (15%)3 = , – 1, = $5,857 OR
CR (15%)3 = I (A/P, 15%, 3) – S3 (A/F, 15%, 3)
CR (15%)3 = 10,000 (0.4380) – 4335 (0.2880) = $3,132
AEOC3 =Σ [ OC1 (P/F, 15%, 1) + OC2 (P/F, 15%, 2) + OC3 (P/F, 15%, 3) ] x (A/P, 15%, 3)
AEOC3 = [2,000 (0.8696) + 2,800 (0.7561) + 3,600 (0.6575)] (0.4380)
AEOC3 = $2,725
Σ AE3 = CR (15%)3 + AEOC3 = 2, ,360.7 = $5,857

36. Σ AE4 = CR (15%)4 + AEOC4 = 2,765 + 3,061 = $5,826 N =4
AE (15%)4 = $10,000(A/P, 15%, 4) $800 (A/G, 15%, 4)
– $6,000(0.85)4-1 (A/F, 15%, 4)
AE (15%)4 = 3, , , – 738 = $5,826 OR
CR (15%)4 = I (A/P, 15%, 4) – S4 (A/F, 15%, 4)
CR (15%)4 = 10,000 (0.3503) – 3,685 (0.2003) = $2,765
AEOC4 =Σ [ OC1 (P/F, 15%, 1) + OC2 (P/F, 15%, 2) + OC3 (P/F, 15%, 3) OC4 (P/F, 15%, 4)] x (A/P, 15%, 4)
AEOC4 = [2,000 (0.8696) + 2,800 (0.7561) + 3,600 (0.6575) + 4,400 (0.5718) ] x (0.3503)
AEOC4 = $3,061
Σ AE4 = CR (15%)4 + AEOC4 = 2, ,061 = $5,826

37. Σ AE5 = CR (15%)5 + AEOC 5 = 2,519 + 3,378 = $5,897 N =5
AE (15%)5 = $10,000(A/P, 15%, 5) $800 (A/G, 15%, 5)
– $6,000(0.85)5-1 (A/F, 15%, 5)
AE (15%)5 = 2, , – = $5,897 OR
CR (15%)5 = I (A/P, 15%, 5) – S5 (A/F, 15%, 5)
CR (15%)5 = 10,000 ( ) – 3,132 (0.1483) = $2,519
AEOC5 =Σ [ OC1 (P/F, 15%, 1) + OC2 (P/F, 15%, 2) + OC3 (P/F, 15%, 3) OC4 (P/F, 15%, 4) + OC5 (P/F, 15%, 5)] x (A/P, 15%, 5)
AEOC5 = [2,000 (0.8696) + 2,800 (0.7561) + 3,600 (0.6575) + 4,400 (0.5718) ,200 (0.4972)] x (0.2983)
AEOC5 = $3,378
Σ AE5 = CR (15%)5 + AEOC 5 = 2, ,378 = $5,897

38. The economic service life of the challenger is four years.
N = 1 year: AE(15%) = $7,500 N = 2 years: AE(15%) = $6,151 N = 3 years: AE(15%) = $5,857 N = 4 years: AE(15%) = $5,826 N = 5 years: AE(15%) = $5,897
The economic service life of the challenger is four years.
NC*=4 years
AEC*=$5,826

39. Replacement Decisions
Should replace the defender now? No, because AECD < AECC
If not, when is the best time to replace the defender? Need to conduct the marginal analysis.
NC*= 4 years AEC*=$5,826

40. Marginal Analysis – When to Replace the Defender
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: $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

41. 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.
$2000 2 3
$3000
$5000 2 3
$6,450