This document discusses replacement analysis and economic service life. It defines key replacement terminology and outlines two approaches - the cash flow approach and opportunity cost approach. It explains how to determine the economic service life by minimizing the total annual equivalent cost. When the required service period is long, the strategies for replacement analysis under an infinite planning horizon are discussed. An example problem compares retaining an old machine versus buying a new one, calculating the economic life for each and determining the optimal replacement time.

1. 1
REPLACEMENT DECISIONS
CHAPTER 12

2. 2
Replacement Decisions
 Replacement Analysis Fundamentals
 Economic Service Life
 Replacement Analysis When a Required Service is Long

3. 3
Defender:
an old machine
Challenger:
a new machine
Current market value:
selling price of the defender
in the market place
Sunk cost:
any past cost unaffected
by any future investment
decisions.
Trade-in allowance:
value offered by the
vendor to reduce the price
of a new equipment
Replacement Terminology

4. Replacement Analysis Fundamentals
4
 Replacement projects are decision problems that involve the
replacement of existing out of date or worn-out assets.
 When existing equipment should be replaced with more efficient
equipment?
Examine three aspects of the replacement problem
1) Approaches for comparing defender and challenger
2) Determination of economic service life
3) Replacement analysis when the required service period is long

5. Two basic approaches to analyze replacement problems
5
Cash Flow Approach
 Treat the proceeds (amount of
money received) from sale of the
old machine as down payment
toward purchasing the new
machine. (Use it to bye the new
one)
 This approach is meaningful when
both the defender and challenger
have the same service life.
 Use PW or AE values in the
analysis
Opportunity Cost Approach
 Treat the proceeds (amount of
money received) from sale of the
old machine as the investment
required to keep the old machine.
(Do not use the money to bye the
new one).
 This approach is more commonly
practiced in replacement analysis.

6. 6

7. 7

8. 8

9. 9

10. 10

11. 11

12. Economic Service Life
 Economic service life is the
remaining useful life of an asset
that results in the minimum annual
equivalent cost.
 We should use the respective
economic service lives of the
defender and the challenger when
conducting a replacement analysis.
12
Capital (Ownership)
cost
Operating
cost
+
Minimize
Annual
Equivalent
Cost

13. Economic Service Life Continue….
13
 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
(potential benefit that is given up as you seek an alternative
action) 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.

14. Mathematical Relationship
Objective: Find n* that minimizes total AEC
14
AEC CR i OC i
 
( ) ( )
AE of Capital Cost:
AE of Operating Cost:
Total AE Cost:

15. 15

16.  Here are some special cases that the economic
service life can be determined easily:
 If the salvage value is constant and equal to the
initial cost, and the annual operating cost increases
with time, AEC is an increasing function of N and
attains its minimum at N = 1. In this case, we should
try to replace the asset as soon as possible.
 If the annual operating cost is constant and the
salvage value is less than the initial cost and
decreases with time, AEC is a decreasing function
of N. In this case, we would try to delay
replacement of the asset as long as possible.
 If the salvage value is constant and equal to the
initial cost and the annual operating costs are
constant, AEC will also be constant. In this case,
the time at which the asset is replaced does not
16

17. Example 12.4 Economic Service Life for a Lift Truck
17
A company considers buying a new electric forklift truck that
would cost $20,000, have operating cost of $6,000 in the first year,
and have a salvage value of $14,000 at the end of the first year. For
the remaining years, operating costs increase each year by 25%
over the previous year’s operating costs. Similarly, the salvage
value declines each year by 30% from the previous year’s salvage
value. The truck has a maximum life of eight years, without any
major engine overhaul. The firm’s required rate of return is 12%.
Find the economic service life of this new machine.

18. 18

19. 19
CR (12%)1 = I (A/P, 12%, 1) – S1 (A/F, 12%, 1)
CR (12%)1 = $20,000 (1.12) – $14,000 (1) = $8,400
OC (12%)1 =Σ [OC1 (P/F, 12%, 1)] (A/P, 12%, 1)
OC (12%)1 =Σ [$6,000 (0.8929)] (1.12) = $6,000
AEC1 = CR 1 + OC1 = $8,400 + $6,000 AEC1 = $14,400
N = 1

20. 20
CR (12%)2 = I (A/P, 12%, 2) – S2 (A/F, 12%, 2)
CR (12%)2 = $20,000 (0.5917) – $9,800 (0.4717) = 11,834 – 4,623 = $7,211
OC (12%)2 =Σ [OC1 (P/F, 12%, 1) + OC2 (P/F, 12%, 2)] x (A/P, 12%, 2)
OC (12%)2 =Σ [$6,000 (0.8929) + $7500 (0.7972)] x (0.5917) = $6,708
AEC2 = CR (12%)2 + OC2 = $7,211 + $6,708 AEC2 = $13,919
N =
2

21. 21

22. 22
CR (12%)3 = I (A/P, 12%, 3) – S3 (A/F, 12%, 3)
CR (12%)3 = $20,000 (0.4163) – $6,860 (0.2963) = 8,326 – 2,032 = $6,294
OC (12%)3 =Σ [ OC1 (P/F, 12%, 1) + OC2 (P/F, 12%, 2) + OC3 (P/F, 12%, 3)] (A/P, 12%, 3)
OC (12%)3 =Σ [$6,000 (0.8929) + $7,500 (0.7972) + $9,375 (0.7118)] (0.4163) = $7,498
AE3 = CR (12%)3 + OC3 AEC3 = $6,294 + $7,498 = $13,792
N = 3

23. 23
N = 4
CR (12%)4 = I (A/P, 12%, 4) – S4 (A/F, 12%, 4)
CR (12%)4 = $20,000 (0.3292) – $4,802 (0.2092) = $6,584 – $1,004 = $5,580
OC (12%)4 =Σ [OC1 (P/F, 12%, 1) + OC2 (P/F, 12%, 2) + OC3 (P/F, 12%, 3)
+ OC4 (P/F, 12%, 4)] (A/P, 12%, 4)
OC (12%)4 =Σ [ $6,000 (0.8929) + $7,500 (0.7972) + $9,375 (0.7118)
+ $11,719 (0.6355)] (0.3292) = $8,381
AE4 = CR (15%)4 + OC4 AEC4 = $5,580 + $8,381 = $13,961

24. AEC if the Asset were Kept N Years
24
N = 1, AEC1 = $14,400
N = 2, AEC2 = $13,919
N = 3, AEC3 = $13,792
N = 4, AEC4 = $13,961
N = 5, AEC5 = $14,387
N = 6, AEC6 = $15,044
N = 7, AEC7 = $15,920
N = 8, AEC8 = $17,008
Minimum cost
If you purchase the asset, it is
most economical to replace
the asset every 3 years
Economic Service Life

25. 25

26. Replacement Analysis
When the Required Service Period is Long
26
 We know 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 meant the service period required by the
defender and future challengers.
 The infinite planning horizon is used when we are unable to predict
when the activity under consideration will be terminated.
 In other situations, the project will have a definite and predictable
duration. In these cases, replacement policy should be formulated
based on a finite planning horizon.

27. Decision Frameworks continue…….
27
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. When the planning horizon is finite, the PW
method is convenient to be used.
 Although the economic service life of the defender is defined as the
number of years of service that minimizes the annual equivalent
cost (or maximizes the annual equivalent revenue), the end of the
economic life is not necessarily the optimum time to replace the
defender.

28. 28
Handling Unequal Service Life Problems in Replacement Analysis
 Let us consider a situation where you are comparing a defender
(D) with an economic service life of three years and a challenger
(C) with an economic service life of six years.
 This means that if we decide to keep the old machine (D) for
three years, we will replace it at time 3 by an asset similar to the
new machine. This asset will in turn be replaced six years later, at
time 9, by another asset C.
 There are two implied infinite sequences in this scenario:
 Keep defender (D), buy a challenger (C) at time 3, buy another
challenger (C) at time 9, buy another challenger (C) at time 15, etc.
 Buy challenger (C) at time 0, buy a challenger (C) at time 6, buy a
challenger (C) at time 12, and so on

29. 29
Handling Unequal Service Life Problems in Replacement Analysis
 It is clear that the AE cost approach for either sequence of assets
is the same after the remaining life of the defender. Therefore, we
can directly compare the AEC for the remaining life of the
defender with the AEC for the challenger over its economic
service life.

30. 30
Replacement Strategies
under the Infinite Planning Horizon
 Compute the economic lives of both defender and challenger.
Let’s use ND* and NC* to represent economic lives of the
defender and the challenger.
 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*, it is more
costly to keep the defender than to replace it with the challenger.
Thus, the challenger should replace the defender now.

31. 31
Replacement Strategies
under the Infinite Planning Horizon
 If the defender is not to be replaced now, when should it be
replaced?
 First, we need to continue to use it until its economic life is over.
Then, calculate the cost of running the defender for one more
year. If this cost is greater than AEC*, the defender should be
replaced at the end of its economic life.
 This process should be continued until you find the optimal
replacement time. This approach is called marginal analysis.

32. 32
Replacement Strategies
under the Infinite Planning Horizon

33. EXAMPLE 12.5 Replacement Analysis under the Infinite
Planning Horizon
General Engineering Company is considering replacing an old vertical
cylinder honing (polishing) machine. They are considering two options:
Option 1:
Retain or keep the old machine. If it is kept, the old machine can be
used for another six years with proper maintenance. The market value
of the machine is expected to decline 25% annually over the previous
years. The operating costs are estimated at $3,500 during the first year
and are expected to increase by $1,000 per year thereafter.
33

34. EXAMPLE 12.5 continue..........
Option 2:
Alternatively, the firm can sell the machine to another firm in the
industry now for $4,000 and buy new honing machine. The new
machine costs $12,000 and will have operating costs of $2,300 in the
first year, increasing by 20% per year thereafter. The expected
salvage value is $8,000 after one year and will decline 30% each
year thereafter. The company requires rate of return of 12%.
Find the economic life for each option, and determine when the
defender should be replaced.
34

35. EXAMPLE 12.5
Replacement Analysis under the Infinite Planning Horizon
35

36. SOLUTION: Economic Service Life for Defender
36
N = 1
CR (12%)N = I (A/P, 12%, N) – SN (A/F, 12%, N) and
AEOC (12%)N =Σ [OCn (P/F, 12%, N)] (A/P, 12%, N)
CR (12%)1 = $4,000 (1.12) – 3,000 (1.0) = 4,480 – 3,000 = $1,480
AEOC1 = 3,500 (0.8929) (1.12) = $3,500
Σ AEC1 = CR (15%)1 + AEOC1 = 1,480 + 3,500 = $4,980

37. SOLUTION - Economic Service Life for Defender
37
n
Forecasted
Operating
Cost
Market Value if
Disposed of
0
$4,000
1
$3,500 $3,000
2
$4,500 $2,250
3
$5,500 $1,688
4
$6,500 $1,266
5
$7,500 $949
6
$8,500 $712

38. Economic Service Life Calculation For Defender
38

39. SOLUTION - Economic Service Life for Challenger
39
N = 1
CR (12%)N = I (A/P, 12%, N) – SN (A/F, 12%, N)
AEOC =Σ [OCn (P/F, 12%, N)] (A/P, 12%, N)
CR (12%)1 = $12,000 (1.12) – 8,000 (1.0) = 13,440 – 8,000 = $5,440
AEOC1 = 2,300 (0.8929) (1.12) = $2,300
Σ AEC1 = CR (15%)1 + AEOC1 = 5,440 + 2,300 = $7,740

40. Economic Service Life Calculation for Challenger
40
n Market
Value
O & M
Cost
CR
(12%)
OC
(12%)
AEC(12%
)
0 $12,000
1 $8,000 $2,300 $5,440 $2,300 $7,740
2 $5,600 $2,760 $4,459 $2,517 $6,976
3 $3,920 $3,312 $3,834 $2,753 $6,587
4 $2,774 $3,974 $3,377 $3,008 $6,385
5 $1,921 $4,769 $3,027 $3,285 $6,312
6 $1,345 $5,723 $2,753 $3,586 $6,339
7 $941 $6,868 $2,536 $3,911 $6,447
8 $659 $8,241 $2,362 $4,263 $6,625
Annual changes in
Market Value
- 30%
Annual increases in
O & M
20%
Interest rate 12%

41. Replacement Decisions
NC* = 5 years
AECC* = $6,312
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.
ND* = 1 year
AECD* = $4,980

42. Marginal Analysis – When to Replace the Defender
42
Question:
What is the additional (incremental) cost for keeping the defender
one more year from the end of its economic service life, from Year
1 to Year 2?
Financial Data:
Opportunity cost at the end of year 1: $3,000 (market value of the
defender at the end year 1)
Operating cost for the 2nd year: $4,500
Salvage value of the defender at the end of year 2: $2,250
n
Forecasted
Operating
Cost
Market Value
if Disposed of
0
$4,000
1
$3,500 $3,000
2
$4,500 $2,250
3
$5,500 $1,688
4
$6,500 $1,266
5
$7,500 $949
6
$8,500 $712

43. Step 1:
Calculate the equivalent cost of retaining the defender
one more year from the end of its economic
service life, say 1 to 2.
$3,000 (F/P,12%,1) + $4,500 - $2,250
= $5,610
Step 2:
Compare this cost with
AECC* = $6,312 of the challenger.
Conclusion:
Since keeping the defender for the 2nd year is less
expensive than replacing it with the challenger,
keep the defender beyond its economic service life.
43
1
2
$3000
$2250
$4,500
$5,610
1 2

44. Step 1: Calculate the equivalent cost of
retaining the defender one more year from
year 2 to 3.
$2,250 (F/P,12%,1) + $5,500 - $1,688
$2,250 x 1.12 + $5,500 - $1,688
= $6,332
Step 2: Compare this cost with
AECC* = $6,312 of the challenger.
Conclusion:
For year three, keeping the defender is more
expensive. This means that we should
replace the defender at the end of year two.
44
2
3
$2250
$1688
$5,500
$6,332
2 3

45. 45
SOLVED PRACTICE PROBLEMS

46. 46
12.1
NEWNAN Furniture owns and operates an industrial lift truck in their warehousing
operation. The record indicates that the lift truck was purchased four years ago at
$15,000. The estimated salvage value is $4,000 after four years of operation. First-
year O&M expenses were $2,000, but the O&M expenses have increased by $400
each year for the first four years of operation. Using i = 10% compute the annual
equivalent costs of the lift truck for four years.
12.1 SOLUTION

47. 47
12.5
A firm is considering the replacement of a 2,000 kg capacity pressing machine.
The machine was purchased five years ago at a cost of $22,000. The machine
was originally expected to have a useful life of 10 years and a $2,000 estimated
salvage value at the end of that period. However, the machine has not been
dependable and is frequently out of service while awaiting repairs. The
maintenance expenses of the pressing machine have been rising steadily and
currently amount about $5,000 per year. The machine could be sold now for
$6,000. If it is retained or kept the machine will require an immediate $2,500
overhaul to keep it in operable condition. This overhaul will neither extend the
originally estimated service life nor increase the value of the machine. The
updated annual operating costs, engine overhaul cost, and market values over the
next five years are estimated as follows:

48. 48
12.5
N Engine Market
O & M
Overhau
l Value
-5 22,000
-4
-3
-2
-1
0 $2,500 $6,000
1 $5,000 $4,500
2 $5,500 $3,500
3 $6,000 $3,000
4 $6,500 $3,000 $2,500
5 $7,500 $2,000

49. 49
12.5
A drastic increase in operating costs during the fourth year is expected as a result
of another overhaul, that is about $3,000 which will be required in order to keep
the machine in operating condition. The firm’s MARR is 15%.
a) If the machine is to be sold now, what will be its sunk cost?
b) What is the opportunity cost of not replacing the machine now?
c) What is the equivalent annual cost of owning and operating the machine for
two more years?
d) What is the equivalent annual cost of owning and operating the machine for
five more years?

50. 12.5) SOLUTION
(a) Purchase cost = $22,000, market value = $6,000,
sunk cost = $22,000 - $6,000 = $16,000
(b) Opportunity cost = $6,000
(c)
(d)
50
PW(15%) $6,000 $2,500 $5,000( / ,15%,1)
($5,500 $3,500)( / ,15%,2)
$14,360.2
AEC(15%) $14,360.2( / ,15%,2)
$8,832.96
P F
P F
A P
   
 
 


PW(15%) $8,500 $5,000( / ,15%,1) $5,500( / ,15%,2)
$6,000( / ,15%,3) $9,500( / ,15%,4)
($7,500 $2,000)( / ,15%,5)
$29,117.84
AEC(15%) $29,117.84( / ,15%,5)
$8,686.30
P F P F
P F P F
P F
A P
   
 
 
 


N Engine Market
O & M Overhaul Value
-5 22,000
-4
-3
-2
-1
0 $2,500 $6,000
1 $5,000 $4,500
2 $5,500 $3,500
3 $6,000 $3,000
4 $6,500 $3,000 $2,500
5 $7,500 $2,000

51. 51
12.11
A special-purpose machine is to be purchased at a cost of $30,000. The
following table shows the expected annual operating and maintenance cost
and the salvage value for each year of service: If the interest rate is 12%,
what is the economic service life for this machine?
Year of
Service
O & M
Cost
Market
Value
0 $30,000
1 $5,000 $25,800
2 $6,500 $16,000
3 $10,000 $10,000
4 $12,500 $5,000
5 $14,800 $0

52. 52
12.11) SOLUTION
At i = 12%, the economic service life is 1 year.
Interest rate 12%
n Market Value O&M Costs CR(12%) OC(12%) AEC(12%)
0 $30,000
1 $25,800 $5,000 $7,800 $5,000 $12,800
2 $16,000 $6,500 $10,204 $5,708 $15,911
3 $10,000 $10,000 $9,527 $6,980 $16,507
4 $5,000 $12,500 $8,831 $8,135 $16,966
5 $0 $14,800 $8,322 $9,184 $17,506

53. 53
12.15
Advanced Electrical Insulator Company is considering replacing a broken inspection
machine, which has been used to test the mechanical strength of electrical
insulators, with a newer and more efficient one. If repaired, the old machine can be
used for another five years, although the firm does not expect to realize any salvage
value from scrapping it in five years.
Alternatively, the firm can sell the machine to another firm in the industry now for
$5,000. If the machine is kept, it will require an immediate $1,200 overhaul to
restore it to operable condition. The overhaul will neither extend the service life
originally estimated nor increase the value of the inspection machine. The operating
costs are estimated at $2,000 during the first year and are expected to increase by
$1,500 per year thereafter. Future market values are expected to decline by $1,000
per year.
The new machine costs $10,000 and will have operating costs of $2,000 in the first
year, increasing by $800 per year thereafter. The expected salvage value is $6,000
after one year and will decline 15% each year. The company requires a rate of
return of 15%.
Find the economic life for each option, and determine when the defender should be
replaced.

54. 54
Interest rate 15%
n Market Value O&M Costs CR(15%) OC(15%) AEC(15%)
0 $5,000 $1,200
1 $4,000 $2,000 $1,750 $3,380 $5,130
2 $3,000 $3,500 $1,680 $3,436 $5,116
3 $2,000 $5,000 $1,614 $3,886 $5,500
4 $1,000 $6,500 $1,551 $4,410 $5,961
5 $0 $8,000 $1,492 $4,942 $6,434
12.15) Defender: Economic service year is 2 years

55. 55
Interest rate 15%
n Market Value O&M Costs CR(15%) OC(15%) AEC(15%)
0 $5,000 $1,200
1 $4,000 $2,000 $1,750 $3,380 $5,130
2 $3,000 $3,500 $1,680 $3,436 $5,116
3 $2,000 $5,000 $1,614 $3,886 $5,500
4 $1,000 $6,500 $1,551 $4,410 $5,961
5 $0 $8,000 $1,492 $4,942 $6,434
12.15) Defender: Economic service year is 2 years
Marginal analysis:
1. Opportunity cost at the end of year two,
which is equal to the market value then,
or $3,000
2. Operating cost for the third year: $5,000
3. Salvage value of the defender at the end
of year three: $2,000
The cost of using the defender for one more
year from the end of its economic service life is
3 $3,000( / ,15%,1) $5,000 $2,000
$6, 450
F F P
  

Compare this cost with the challenger.
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.