Replacement Decisions
Lecture No. 58
Chapter 14
Contemporary Engineering Economics
Copyright © 2006
Contemporary Engineering Economics, 4th
edition, © 2007
Required Assumptions and Decision
Frameworks

Planning Horizon (Study Period)



Technology




Infinite planning horizon
Finite planning horizon
No technology improvement is anticipated over
the planning horizon
Technology improvement cannot be ruled out
Revenue Cash Flow Information
Decision Frameworks
Contemporary Engineering Economics, 4th
edition, © 2007
Replacement Strategies under the Infinite
Planning Horizon
1.
2.
Replace the defender now: The cash flows of the
challenger estimated today will be used. An identical
challenger will be used thereafter if replacement becomes
necessary again in the future. This stream of cash flows is
equivalent to a cash flow of AECC* each year for an infinite
number of years.
Replace the defender, say, x years later: The cash flows of
the defender will be used in the first x years. Starting in
year x+1,the cash flows of the challenger will be used
indefinitely thereafter.
Contemporary Engineering Economics, 4th
edition, © 2007
Types of Typical Replacement
Decision Frameworks
Contemporary Engineering Economics, 4th
edition, © 2007
Example 14.4 Relevant Cash Flow
Information (Defender)
Cash flow diagram for defender
when N = 4 years
Contemporary Engineering Economics, 4th
edition, © 2007
Economic Service Life for the
Defender (Example 14.4)

Defender: Find the
remaining useful
(economic) service life.
N  1: A E (15% )  $5,130
N  2 : A E (15% )  $5,116
N  3: A E (15% )  $5,500
N  4 : A E (15% )  $5, 961
N D *  2 years
N  5: A E (15% )  $6, 434
A E D *  $ 5,1 1 6
Contemporary Engineering Economics, 4th
edition, © 2007
AEC as a Function of the Life of the
Defender (Example 14.4)
Contemporary Engineering Economics, 4th
edition, © 2007
Economic Service Life for the
Challenger (Example 14.3)
 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 at a15%
annual rate
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
Contemporary Engineering Economics, 4th
edition, © 2007
Replacement Decisions
N D *  2 years
A E C D *  $ 5,1 1 6
NC*= 4 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.
AEC*=$5,826
Contemporary Engineering Economics, 4th
edition, © 2007
Marginal Analysis to Determine when
the Defender should be Replaced
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
Contemporary Engineering Economics, 4th
edition, © 2007

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
$3000
- $2,000 = $6,450


Step 2: Compare this cost
with AECC = $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.
Contemporary Engineering Economics, 4th
edition, © 2007
2
$5000
3
$6,450
Example 14.5 Replacement Analysis under
the Finite Planning Horizon
Annual Equivalent Cost ($)
N
Defender
Challenger
1
5,130
7,500
2
5,116
6,151
3
5,500
5,857
4
5,961
5,826
5
6,434
5,897
Some likely replacement patterns
under a 8-year finite planning horizon
Contemporary Engineering Economics, 4th
edition, © 2007
Replacement Analysis under the Finite Planning Horizon
(Example 14.5)

Option 1:
(j0, 0), (j, 4), (j, 4)
PW(15%)1=$5,826(P/A, 15%, 8)
=$26,143

Option 2:
(j0, 1), (j, 4), (j, 3)
PW(15%)2=$5,130(P/F, 15%, 1)
+$5,826(P/A, 15%, 4)(P/F, 15%, 1)
+$5,857(P/A, 15%, 3)(P/F, 15%, 5)
=$25,573
Contemporary Engineering Economics, 4th
edition, © 2007
Example 14.5 continued


Option 3
(j0, 2), (j, 4), (j, 2)
Option 4
PW(15%)3=$5,116(P/A, 15%, 4)(P/F, 15%, 2)
+$5,826(P/A, 15%, 4)(P/F, 15%, 2)
+$6,151(P/A, 15%, 2)(P/F, 15%, 6)
= $25,217
minimum cost
(j0, 3), (j, 5)
PW(15%)4= $5,500(P/A, 15%, 3)
+$5,897(P/A, 15%, 5)(P/F, 15%, 3)
=$25,555
Contemporary Engineering Economics, 4th
edition, © 2007
Example 14.5 continued


Option 5:
(j0, 3), (j, 4), (j, 1)
Option 6:
PW(15%)5= $5,500(P/A, 15%, 3)
+ $5,826(P/A, 15%, 4)(P/F, 15%, 3)
+ $7,500(P/F, 15%, 8)
= $25,946
(j0, 4), (j, 4)
PW(15%)6= $5,826(P/A, 15%, 4)(P/F, 15%, 4)
+ $5,826(P/A, 15%, 4)(P/F, 15%, 4)
= $26,529
Contemporary Engineering Economics, 4th
edition, © 2007
Graphical Representation of Replacement
Strategies under a Finite Planning Horizon
Contemporary Engineering Economics, 4th
edition, © 2007