Chapter 11 Replacement Decisions

1
Chapter 11Replacement 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
Sunk Cost associated with an Assets Disposal
(Example 11.1 Macintosh Printing Inc.)
Original investment (printing machine)
20,000
Lost investment (economic depreciation)
Market value
Repair cost
10,000
5000
10,000
Sunk costs 15,000
0 5000 10,000 15,000
20,000 25,000 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
• Market price 10,000
• Remaining useful life 3 years
• Salvage value 2,500
• OM cost 8,000

• Challenger
• Cost 15,000
• Useful life 3 years
• Salvage value 5,500
• OM cost 6,000

7
Replacement Analysis Cash Flow Approach
Sales proceeds from defender
10,000
5500
2500
0 1 2 3
0 1 2 3
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
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 P
W(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

• 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 (OM) costs, labor costs,
  material costs and energy consumption costs.

13
Mathematical RelationshipObjective 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. Lets 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)
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)
2000 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

• N 1
• AE (15)1 6,200 (A/P, 15, 1) 2000 1,500
  (A/G, 15, 1)
• 1,000 (5 1) (A/F, 15, 1)
• AE (15)1 7,130 2,000 0 4,000 5,130
• OR
• CR (15)N I (A/P, 15, N) SN (A/F, 15, N)
• AEOC S 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) 1739.2 x (1.15)
  2,000
• S 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
• AE (15)2 6,200 (A/P, 15, 2) 2000 1,500
  (A/G, 15, 2)
• 1,000 (5 2) (A/F, 15,
  2)
• AE (15)2 3,813.62 2,000 697.65 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 S 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
• S AE2 CR (15)2 AEOC2 2,418.32 2,697.55
  5,116

27

• N 3
• AE (15)3 6,200 (A/P, 15, 3) 2000 1,500
  (A/G, 15, 3)
• 1,000 (5 3) (A/F, 15, 3)
• AE (15)3 2,715.6 2,000 1,360.65 576
  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 S 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
• S AE3 CR (15)3 AEOC3 2,139.6 3,360.7
  5,500

28

• N 4
• AE (15)4 6,200 (A/P, 15, 4) 2000 1,500
  (A/G, 15, 4)
• 1,000 (5 4) (A/F, 15,
  4)
• AE (15)4 3,813.62 2,000 697.65 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 S 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
• S AE4 CR (15)4 AEOC4 1,971.56 3,989.75
  5,961

29
ECONOMIC SERVICE LIFE OF DEFENDER

• N 5
• AE (15) 5 6,200 (A/P, 15, 5) 2000 1,500
  (A/G, 15, 5)
• 1,000(5 5) (A/F, 15, 3)
• AE (15)5 1,849.46 2,000 2,584.2 0
  6,434
• OR
• CR (15)5 I (A/P, 15, 5) S5 (A/F, 15, 5)
• CR (15)5 6,200 (0. 2983) 0 (0.1483)
  1,850
• AEOC5 S 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) 8,000 (0.4972)
  x (0.2983)
• AEOC5 4,584
• S AE5 CR (15)5 AEOC 5 1,850 4,584
  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 defenders economic life is two years.

31
AEC as a Function of the Life of the Defender
(Example 11.5)
32
ECONOMIC SERVICE LIFE OF CHALLENGER (Example 11.5)

• Investment cost 10,000
• Salvage value
• N 1 6,000
• N gt 1 decreases at a 15 over previous year
• Operating cost
• N 1 2,000
• N gt 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) 2000 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) 2000 800
  (A/G, 15, 1)
• 6,000(0.85)1-1 (A/F, 15, 1)
• AE (15)1 11500 2,000 0 6,000 7,500
• OR
• CR (15)N I (A/P, 15, N) SN (A/F, 15, N)
• AEOC S 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) 1739.2 x (1.15)
  2,000
• S AE1 CR (15)1 AEOC1 3,130 2,000
  7,500

34

• N 2
• AE (15)2 10,000(A/P, 15, 2) 2000 800
  (A/G, 15, 2)
• 6,000(0.85)2-1 (A/F, 15, 2)
• AE (15)1 6151 2,000 372.08 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 S 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
• S AE2 CR (15)2 AEOC2 2,418.32 2,697.55
  6,151

35

• N 3
• AE (15)3 10,000(A/P, 15, 3) 2000 800
  (A/G, 15, 3)
• 6,000(0.85)3-1 (A/F, 15, 3)
• AE (15)3 4380 2,000 625.68 1,248.48
  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 S 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
• S AE3 CR (15)3 AEOC3 2,139.6 3,360.7
  5,857

36

• N 4
• AE (15)4 10,000(A/P, 15, 4) 2000 800
  (A/G, 15, 4)
• 6,000(0.85)4-1 (A/F, 15, 4)
• AE (15)4 3,503 2,000 1,061.04 738
  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 S 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
• S AE4 CR (15)4 AEOC4 2,765 3,061
  5,826

37

• N 5
• AE (15)5 10,000(A/P, 15, 5) 2000 800
  (A/G, 15, 5)
• 6,000(0.85)5-1 (A/F, 15, 5)
• AE (15)5 2,983 2,000 1378.24 464.48
  5,897
• OR
• CR (15)5 I (A/P, 15, 5) S5 (A/F, 15, 5)
• CR (15)5 10,000 (0. 2983) 3,132 (0.1483)
  2,519
• AEOC5 S 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) 5,200 (0.4972) x
  (0.2983)
• AEOC5 3,378
• S AE5 CR (15)5 AEOC 5 2,519 3,378
  5,897

38

• 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.

NC4 years AEC5,826
39
Replacement Decisions

• Should replace the defender now? No, because AECD
  lt AECC
• If not, when is the best time to replace the
  defender? Need to conduct the marginal analysis.

• NC 4 years
• AEC5,826

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