Engineering Economy (17th Edition)
17th Edition
ISBN: 9780134870069
Author: William G. Sullivan, Elin M. Wicks, C. Patrick Koelling
Publisher: PEARSON
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Textbook Question
Chapter 9, Problem 13P
Use the PW method to select the better of the following alternatives:
Assume that the defender was installed five years ago. The MARR is 10% per year.
Definition of alternatives:
A: Retain an already owned machine (defender) in service for eight more years.
B: Sell the defender and lease a new one (challenger) for eight years.
Alternative A (additional information):
Cost of defender five years ago = $500,000
BV now = $111,550
Estimated MV eight years from now = $50,000
Present MV = $150,000
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The AW values for retaining a presently owned machine for additional years are shown in the table. Note that the values represent the AW amount for each of the n years that the asset is kept, that is, if it is kept 5 more years, the annual worth is $95,000 for each of the 5 years. Assume that future costs remain as estimated for the replacement study and that used machines like the one presently owned will always be available. (a) What is the ESL and associated AW of the defender at a MARR of 12% per year? (b) A challenger with an ESL of 7 years and an AWC = $ −89,500 per year has been identified. Which AW will be less for the respective ESL periods? Retention Period, Years AW Value, $ per Year 1 −92,000 2 −88,000 3 −85,000 4 −89,000 5 −95,000
Give typing answer with explanation and conclusion
A piece of equipment has a first cost of $175,000, a maximum useful life of 7 years, and a market (salvage) value described by the relation S = 120,000 – 11,000k, where k is the number of years since it was purchased. The salvage value cannot go below zero. The AOC series is estimated using AOC = 60,000 + 14,000k. The interest rate is 12% per year. Determine the economic service life and the respective AW.
The AW at the Economic Service Life, (in $)
In a replacement analysis problem, the following information is available
Initial cost : 12.000 YTL
Annual maintenance cost: None in the first three years
2.000 YTL at the end of the 4th year
2.000 YTL at the end of the 5th year
2,500 YTL per year after the 5th year
will increase smoothly (at the end of the 6th year, 4.500 YTL
7.000 YTL at the end of the 7th year, etc.
The scrap value in any given year is zero. Take the discount rate of 10% and ignore income taxes. Calculate the most economical life for this candidate equipment.
A) 6
B) 2
C) 5
D) 8
E) 4
Chapter 9 Solutions
Engineering Economy (17th Edition)
Ch. 9 - Prob. 1PCh. 9 - Prob. 2PCh. 9 - Prob. 3PCh. 9 - Prob. 4PCh. 9 - Prob. 5PCh. 9 - Prob. 6PCh. 9 - Prob. 7PCh. 9 - A city water and waste-water department has a...Ch. 9 - Prob. 9PCh. 9 - Prob. 10P
Ch. 9 - Prob. 11PCh. 9 - Prob. 12PCh. 9 - Use the PW method to select the better of the...Ch. 9 - Prob. 14PCh. 9 - Prob. 15PCh. 9 - Prob. 16PCh. 9 - Prob. 17PCh. 9 - Prob. 18PCh. 9 - Prob. 19PCh. 9 - Prob. 20PCh. 9 - Prob. 21PCh. 9 - Prob. 22PCh. 9 - Prob. 23PCh. 9 - Prob. 24PCh. 9 - Prob. 25PCh. 9 - Prob. 26PCh. 9 - Prob. 27SECh. 9 - Prob. 28SECh. 9 - Prob. 29CSCh. 9 - Prob. 30CSCh. 9 - Prob. 31CSCh. 9 - Prob. 32FECh. 9 - Prob. 33FECh. 9 - Prob. 34FECh. 9 - Prob. 35FECh. 9 - Prob. 36FE
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