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Contemporary Engineering Economics, 4th edition, © 2007
Replacement Decisions
Lecture No. 58Chapter 14Contemporary Engineering EconomicsCopyright © 2006
Contemporary Engineering Economics, 4th edition, © 2007
Required Assumptions and Decision Frameworks Planning Horizon (Study Period)
Infinite planning horizon Finite planning horizon
Technology 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
1. 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.
2. 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.
Replacement Strategies under the Infinite Planning Horizon
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 AE
N AE
N AE
N AE
N AE
1 15%) 130
2 15%) 116
3 15%) 500
4 15%) 961
5 15%) 434
: ( $5,
: ( $5,
: ( $5,
: ( $5,
: ( $6,N years
AED
D
*
* $5,
2
116
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)
N = 1 year: AE(15%) = $7,500N = 2 years: AE(15%) = $6,151N = 3 years: AE(15%) = $5,847N = 4 years: AE(15%) = $5,826N = 5 years: AE(15%) = $5,897
NC*=4 years
AEC*=$5,826
Investment cost = $10,000 Salvage value
N = 1: $6,000N > 1: decreases at a 15% over previous year
Operating costN = 1: $2,000N > 1: increases at a15%
annual rate
Contemporary Engineering Economics, 4th edition, © 2007
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
*
*
2 years
$5,116D
D
N
AEC
Contemporary Engineering Economics, 4th edition, © 2007
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
Marginal Analysis to Determine when the Defender should be Replaced
Contemporary Engineering Economics, 4th edition, © 2007
23
$3000
$2000
$5000
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 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.
$6,450
2 3
Contemporary Engineering Economics, 4th edition, © 2007
Example 14.5 Replacement Analysis under the Finite Planning Horizon
N
Annual Equivalent Cost ($)
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 patternsunder 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
Option 3 (j0, 2), (j, 4), (j, 2)
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
Option 4 (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
Example 14.5 continued
Contemporary Engineering Economics, 4th edition, © 2007
Example 14.5 continued
Option 5: (j0, 3), (j, 4), (j, 1)
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
Option 6: (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
