Chapter 11
Replacement Decisions

Replacement Analysis
Fundamentals

Economic Service Life

Replacement Analysis
When a Required
Service is Long
1
Replacement Terminology

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
decisions

Trade-in allowance: value
offered by the vendor to
reduce the price of a new
equipment
2
Sunk Cost associated with an Asset’s Disposal
(Example 11.1
Macintosh Printing Inc.)
Original investment (printing machine)
$20,000
Market value
Lost investment
(economic depreciation)
$10,000
Repair cost
$5000
$10,000
Sunk costs = $15,000
$0
$5000
$10,000
$15,000
$20,000
$25,000
$30,000
3
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
1)
2)
3)
Approaches for comparing defender and challenger
Determination of economic service life
Replacement analysis when the required service period is
long
4
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.
5
Example 11.2

Defender




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
6
Replacement Analysis – Cash Flow Approach
Sales proceeds
from defender
$10,000
$5500
$2500
0
1
2
3
0
2
3
$6000
$8000
(a) Defender
1
$15,000
(b) Challenger
7
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!
8
Example 11.3
Comparison of Defender and Challenger Based on
Opportunity Cost Approach
9
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)
Replace the
defender now!
= $10,615.33
10
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
Annual Equivalent Cost

Ownership (Capital)
cost
+
Operating
cost
11
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.
12
Mathematical Relationship
Objective: Find n* that minimizes total AEC
AE of Capital Cost:
CR (i )  I ( A / P , i , N )  S N ( A / F , i , N )
AE of Operating Cost:
OC (i ) 
N
 OC
n
( P / F , i , n) ( A / P , i , N )
n 1
Total AE Cost:
AEC  CR(i)  OC(i)
13
AEC
OC (i)
CR(i)
n*
14
Example 11.4 Economic Service Life for a
Lift Truck
15
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
16

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
17
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
Economic Service Life
Minimum cost
If you purchase the asset,
it is most economical to replace
the asset for every 6 years
18
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.
19
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.
20
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.
21
Example 11.5 Relevant Cash Flow
Information (Defender)
22
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
23
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 =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) = 1739.2 x (1.15) = $2,000
Σ AE1 = CR (15%)1 + AEOC1 = 3,1,30 + 2,000 = $5,130
24
Objective is : Find n* that minimizes total AEC
AE of Capital Cost:
CR (i )  I ( A / P , i , N )  S N ( A / F , i , N )
AE of Operating Cost:
OC (i ) 
N
 OC
n
( P / F , i , n) ( A / P , i , N )
n 1
Total AE Cost:
AEC  CR(i)  OC(i)
25
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 =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 = 2,418.32 + 2,697.55 = $5,116
26
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 =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 = 2,139.6 + 3,360.7 = $5,500
27
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 =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 = 1,971.56 + 3,989.75 = $5,961
28
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 =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) +
+ 8,000 (0.4972)] x (0.2983)
AEOC5 = $4,584
Σ AE5 = CR (15%)5 + AEOC 5 = 1,850 + 4,584 = $6,434
29
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.
AEC as a Function of the Life of the Defender
(Example 11.5)
31
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%
32
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 =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) = 1739.2 x (1.15) = $2,000
Σ AE1 = CR (15%)1 + AEOC1 = 3,130 + 2,000 = $7,500
33
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 =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 = 2,418.32 + 2,697.55 = $6,151
34
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 =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 = 2,139.6 + 3,360.7 = $5,857
35
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 =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 = 2,765 + 3,061 = $5,826
36
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 =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) +
5,200 (0.4972)] x (0.2983)
AEOC5 = $3,378
Σ AE5 = CR (15%)5 + AEOC 5 = 2,519 + 3,378 = $5,897
37
N = 1 year: AE(15%) = $7,500
The economic service life
of the challenger is four
years.
N = 2 years: AE(15%) = $6,151
N = 3 years: AE(15%) = $5,857
N = 4 years: AE(15%) = $5,826
NC*=4 years
N = 5 years: AE(15%) = $5,897
AEC*=$5,826
38
Replacement Decisions
N D*  2 years

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.
AECD*  $5,116
NC*= 4 years
AEC*=$5,826
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
40

Step 1: Calculate the equivalent
cost of retaining the defender one
more from the end of its economic
service life, say 2 to 3.
$2000
2
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.
$3000
2
$5000
3
$6,450
41