Lec%2024_Full

# Lec%2024_Full - IE 343 Engineering Economics Lecture 24...

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IE 343 Engineering Economics Lecture 24: Chapter 6 - Comparison and Selection among Alternatives Instructor: Tian Ni Oct.24, 2011 IE 343 Fall 2011 1

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• Quiz 6 will be given at the beginning of the class 4:30PM – 4:45PM on Wednesday Oct.26 • Quiz 6 will have 2 problems totally 10 points • You will have 15 minutes to finish quiz 6 • Please bring a CALCULATOR to the quiz • All quiz 6 problems will be selected from the following three sources. You will see the same problems but maybe with numbers changed IE 343 Fall 2011 Tips for Quiz 6
• Three sources: 1. Homework 6 Problem 1-7 (solutions for Homework 6 has been posted on Blackboard Vista) 2. Examples on Lecture Notes (please refer to the full version lecture notes for the example solutions) Lec 20_Full: Example 5.1,5.2 (simple payback, discounted payback) Lec 21_Full: Example 6.1,6.2,6.3 3. Examples on Textbook Chapter 5 section 5.8 and Chapter 6 Example 6.1-6.6 IE 343 Fall 2011 Tips for Quiz 6

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Basic Idea: When the alternatives have different useful lives, we need to make some modifications to compare the alternatives over the same period of time , which is needed for a fair comparison. IE 343 Fall 2011 Study period ≠ useful life (of at least one alternative) 4
Case 1:Repeatability Assumption(AW method) IE 343 Fall 2011 5

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Case 1: Let’s assume that we can repeat the same cash flow of each alternative during its initial useful life span in its succeeding life spans. (Repeatability assumption) This “ repeatability ” assumption is useful for situations where – the study period is very long (can be assumed infinity) – the study period is equal to a common multiple of the alternatives’ lives. IE 343 Fall 2011 Case 1:Repeatability Assumption 6
MARR=10%. Assume Repeatability holds: This can be done by identical replacements of alternatives at the end of their useful life spans. IE 343 Fall 2011 Example 6.7 – Case 1:Repeatability Assumption A B Investment 5,250 7,500 Annual revenues 1,882.5 2,220 Useful life(years) 4 6 7

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Based on this assumption, we can use a study period of 12 years (the least common multiple (LCM) of 4 and 6) where Alternative A repeats itself 3 times and Alternative B repeats itself 2 times during the entire period . IE 343 Fall 2011 Example 6.7 – Case 1 – Solution A A A B B 0 4 6 8 12 8
IE 343 Fall 2011 Example 6.7 – Case 1 – Solution 0 5,250 A=1,882.5 A A 1 2 3 4 5 6 7 8 9 10 11 12 5,250 5,250 A A …………………………………. . Alternative A with 3 identical replacement 0 7,500 A=2,220 A A 1 2 3 4 5 6 7 8 9 10 11 12 7,500 A A …………………………………. . Alternative B with 2 identical replacement 9

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Let’s use the PW method to find the best alternative: (study period=12 years) PW(10%) A = -5,250- 5,250[(P/F,10%,4)+(P/F,10%,8)]+1,882.5(P/A,10%,12) = \$1,542.
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Lec%2024_Full - IE 343 Engineering Economics Lecture 24...

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