Lec%2038 - IE 343 Engineering Economics Lecture 38: Chapter...

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Unformatted text preview: IE 343 Engineering Economics Lecture 38: Chapter 9 – Replacement Analysis Instructor: Tian Ni Dec.2, 2011 IE 343 Fall 2011 1 Retirement without Replacement (Abandonment) IE 343 Fall 2011 2 IE 343 Fall 2011 When to retire an asset ¡ Consider a project for which the period of required service is finite and that has positive net cash flows following an initial capital investment . ¡ Market values , or abandonment values , are estimated for the end of each remaining year in the project’s year ¡ In view of an opportunity cost (MARR) of i% per year, should the project be undertaken? In other words, what is the economic life of the project? 3 IE 343 Fall 2011 Retirement without Replacement ¡ Two Key Assumptions: 1. Once a capital investment has been made, the firm desires to postpone the decision to abandon a project as long as its present equivalent value(PW) is not decreasing 2. The existing project will be terminated at the best abandonment time and will not be replaced by the firm 4 IE 343 Fall 2011 Retirement without Replacement ¡ Rule of finding abandonment time: Abandonment time is year of max PW 5 IE 343 Fall 2011 Example 9.9 ¡ An $80,000 baling machine for recycled paper was purchased by the XYZ company two years ago. The current MV of the machine is $50,000, and it can be kept in service for seven more years. MARR is 12% per year and the projected net annual receipts (revenues less expenses) and end-of-year market values for the machine are shown on the table. When is the best time for the company to abandon this project? 6 IE 343 Fall 2011 Example 9.9 End of Year 1 2 3 4 5 6 7 Net Annual Receipts 20,000 20,000 18,000 15,000 12,000 6,000 3,000 Market Value 40,000 32,000 25,000 20,000 15,000 10,000 5,000 7 IE 343 Fall 2011 Example 9.9 – Solution ¡ Use the PWs that result from deciding to keep the machine exactly one, two, three, four, five, six, and seven years: ¡ Keep for one year: ¡ PW 1 (12%) = -$50,000 + ($20,000 + $40,000)(P/F, 12%, 1) = $3,571 ¡ Keep for two year: ¡ PW 2 (12%) = -$50,000 + $20,000(P/F, 12%, 1) + ($20,000 + $32,000)(P/F, 12%, 2) = $3,571 8 IE 343 Fall 2011 Example 9.9 – Solution ¡ Keep for three year: PW 3 (12%) = $14,408 ¡ Keep for four year: PW 4 (12%) = $18,856 ¡ Keep for five year: PW 5 (12%) = $21,466 ¡ Keep for six year: PW 6 (12%) = $21,061 ¡ Keep for seven year: PW 7 (12%) = $19,614 ¡ Solution : FIVE years maximizes the PW, so keep the machine for FIVE years. 9 After-Tax Replacement Studies IE 343 Fall 2011 10 IE 343 Fall 2011 After-Tax Replacement Studies ¡ Income taxes associated with a proposed project may represent a major cash outflow for a firm ¡ Taxes have significant effects on cash flows ¡ In replacement studies, the replacement of an asset often results in gains or losses from the sale of the existing asset(Defender)....
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This note was uploaded on 01/28/2012 for the course IE 343 taught by Professor Vincent,g during the Winter '08 term at Purdue University.

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Lec%2038 - IE 343 Engineering Economics Lecture 38: Chapter...

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