Lec%2018_Full

Lec%2018_Full - IE 343 Engineering Economics Lecture 18:...

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IE 343 Engineering Economics Lecture 18: Chapter 5 – Evaluating a Single Project Instructor: Tian Ni Oct.5, 2011 IE 343 Fall 2011 1
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Methods to evaluate profitability Present Worth (PW) Future Worth (FW) Annual Worth (AW) IE 343 Fall 2011 Last Time … 2
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Rule : If PW(MARR) 0, the project is profitable. (Returns can recover all my initial investment under MARR. ) If PW(MARR) < 0, the project is not profitable. (Returns can not recover all my initial investment under MARR) IE 343 Fall 2011 Present Worth (PW) method 3
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Rule : If FW(MARR) 0, the project is profitable. (Returns will at least satisfy my expected rate of return MARR) If FW(MARR) < 0, the project is not profitable. (Returns can not satisfy my expected rate of return MARR) IE 343 Fall 2011 Future Worth (FW) method 4
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The AW of a project is an equal annual series of dollar amounts , for a stated study period, that is equivalent to the cash inflows and outflows at an interest rate that is generally the MARR. You can calculate AW as annual equivalent value However, here we introduce a new concept: AW(i%)=R – E – CR(i%) R: Annual equivalent revenues E: Annual equivalent expenses CR: Annual equivalent Capital Recovery cost IE 343 Fall 2011 Annual Worth (AW) method 5
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CR includes the loss in value of asset and interest on invested capital . The CR can be computed by: CR(i%)=I(A/P,i%,N)-S(A/F,i%,N) I: Initial investment S: Salvage value. IE 343 Fall 2011 Annual Worth (AW) method 6
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IE 343 Fall 2011 Example 5.1 - Cash Flow Diagram 0 1 2 3 4 $100,000 5 MARR = 20% $40,000 $5,000 S=$20,000 7 CR(i%)=I(A/P,i%,N)-S(A/F,i%,N) I: Initial investment S: Salvage value
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IE 343 Fall 2011 Example 5.1 - Cash Flow Diagram 0 1 2 3 4 $100,000 5 MARR = 20% $40,000 $5,000 S=$20,000 8 CR(i%)=I(A/P,i%,N)-S(A/F,i%,N) I: Initial investment S: Salvage value
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Back to Example 5.1: I = $100,000 R = $40,000 E = $5,000 S = $20,000 MARR = 20%. N: 5 years.
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Lec%2018_Full - IE 343 Engineering Economics Lecture 18:...

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