Session 29 Economic Life of Defender and Challenger
Problem 9-5 (Economic Life of Challenger) •
In a replacement analysis for a vacuum seal on a spacecraft, the following data are known about the challenger: •
•
•
•
•
The initial investment is $12,000 There is no annual maintenance cost for the first three years, however, it will be $2,000 in years four and five, and then $4,500 in the sixth year and increasing by $2,500 each year thereafter. The salvage value is $0 at all times The MARR is 10% per year.
What is the economic life of the challenger?
Problem 9-5 (Economic Life of Challenger) EOY Cash Flow 0 -12000 1 0 2 0 3 0 4 -2000 5 -2000 6 -4500 7 -7000
PW EUAC -12000 0 ($13,200.00) 0 ($6,914.29) 0 ($4,825.38) ($1,366.03) ($4,216.59) ($1,241.84) ($3,853.52) ($2,540.13) ($3,937.31) ($3,592.11) ($4,260.13)
Economic Life •
•
Proper analysis requires knowing the economic life (minimum EUAC) of the alternatives. The EUAC of a new asset can be computed if the capital investment, annual expenses, and year-by-year market values are known or can be estimated. •
•
The difficulties in estimating these values are encountered in most engineering economy studies, and can be overcome in most cases.
In this course, we will focus on understanding the concepts of economic life (the calculation becomes extremely messy and timeconsuming).
Economic Life (minimum EUAC) •
•
Finding the EUAC of the challenger requires finding the total marginal cost of the challenger, for each year. The minimum such value identifies the economic life. The equation below represents the present worth, through year k , of total costs. (Although the sign is positive, it is a cost. Eq. 9-1.)
Total Marginal Cost •
•
The total marginal cost is the equivalent worth, at the end of year k , of the increase in PW of total cost from year k-1 to year k .
This can be simplified to (eq. 9-2):
Economic Life of Challenger (example)
Economic Life of Challenger (example)
Economic Life of Defender •
•
•
If a major overhaul is needed, the life yielding the minimum EUAC is likely the time to the next major overhaul. If the MV is zero (and will be so later), and operating expenses are expected to increase, the economic life will be the one year. The defender should be kept as long as its marginal cost is less than the minimum EUAC of the best challenger.
Replacement Cautions •
•
•
•
In general, if a defender is kept beyond where the TC exceeds the minimum EUAC for the challenger, the replacement becomes more urgent. Rapidly changing technology, bringing about significant improvement in performance, can lead to postponing replacement decisions. When the defender and challenger have different useful lives, often the analysis is really to determine if now is the time to replace the defender. Repeatability or cotermination can be used where appropriate.
After-Tax Replacement •
Taxes can affect replacement decisions.
•
Most replacement analyses should consider taxes.
•
•
Taxes must be considered not only for each year of operation of an asset, but also in relation to the sale of an asset. Since depreciation amounts generally change each year, spreadsheets are an especially important tool to use.
Replacement (example) •
•
Given the following information, what is the most economical time to replace the defender? Challenger:
Defender:
N
EUAC
N
TC
1
$20,000
1
$17,000
2
$19,000
2
$17,300
3
$18,000
3
$17,600
Replacement (example) •
The minimum EUAC of the Challenger is never less than the TC of the Defender, so we don’t replace!
•
Challenger:
Defender:
N
EUAC
N
TC
1
$20,000
1
$17,000
2
$19,000
2
$17,300
3
$18,000
3
$17,600
Replacement (example) •
•
Given the following information, what is the most economical time to replace the defender? Challenger:
Defender:
N
EUAC
N
TC
1
$18,800
1
$17,000
2
$18,000
2
$17,300
3
$17,200
3
$17,600
Replacement (example) •
•
Since the minimum EUAC of the Challenger is less than the TC of the Defender by Year 2, we replace the Defender in Year 2. Challenger:
Defender:
N
EUAC
N
TC
1
$18,800
1
$17,000
2
$18,000
2
$17,300
3
$17,200
3
$17,600
