Lec%2014_Full - IE 343 Engineering Economics Lecture 14:...

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IE 343 Engineering Economics Lecture 14: Chapter 5 – Evaluating a Single Project Instructor: Tian Ni Sep.23, 2011 IE 343 Fall 2011 1
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New Plan: 5 th Week Sep 19 Sep 21 Sep 23 (Quiz 3&Chap4, 5) (Hw4 due &Chap 5) 6 th Week Sep 26 Sep 28 Sep 30 (Table&Excel&VBA (Quiz 4& (Exam 1 Review 1) 7 th Week Oct 3 Oct 5 Oct 7 8 th Week Oct 10 Oct 12 Oct 14 (Oct Break) (HW5 due & HW6 Out) IE 343 Fall 2011 Exam 1 Date Change! 2
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No homework next week. Homework 5 will be assigned next Friday Sep. 30 after Exam 1. Homework 5 is due Oct. 14 Exam 1 will have two parts: (Chapter 1 – 4) Part 1 : Old problems randomly picked from Homework, Lecture Notes and Textbook. (Just like Quizzes) Part 2 : New problems to test your understanding Details will be announced next Monday You can start to review Homework, Lecture Notes and Textbook now! IE 343 Fall 2011 Announcement 3
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Textbook Tables (Appendix C) Excel VBA Solve some Practice Problems in Chapter 4 Note : Monday’s class is part of review 1. I will teach how to use textbook tables (appendix C), Excel, and VBA to simply the computations, which is very useful in the Exam 1. IE 343 Fall 2011 Next Class … Monday (Sep. 26) 4
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Question: What is the corresponding present equivalent value of the cash flow diagram below under a nominal interest rate r = 12% compounded annually, monthly and continuously? Is it a good investment? IE 343 Fall 2011 *Example 4.22* 5
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IE 343 Fall 2011 *Example 4.22* 0 1 13 14 15 A = 3,000 ….. 22 23 24 ……. .…. .. 12 $19,000 2 3 4 5 A A 1.06A A 06 . 1 2 A 06 . 1 9 A 06 . 1 10 A 06 . 1 11 EOY 6
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IE 343 Fall 2011 *Example 4.22* - Decomposition 0 1 $19,000 2 3 23 24 0 1 A = 3,000 ………. .. 2 3 4 5 12 A Single Cash Flow Deferred Annuities Start at EOY 3 23 24 P S P D 7
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IE 343 Fall 2011 *Example 4.22* - Decomposition Geometric Gradient Series Start at the end of year 13 13 14 15 ….. 22 23 24 A 1.06A A 06 . 1 2 A 06 . 1 9 A 06 . 1 10 A 06 . 1 11 0 1 P G 12 ……. . 8
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Basic Idea: Evaluate the present equivalent value of the three decomposed cash flow diagrams separately and then sum them up. Notation: P S : Present Value for the single cash flow P D : Present Value for the Deferred Annuities P G : Present Value for the Geometric Gradient Series i A : Effective interest rate with annual compounding i M : Effective interest rate with monthly compounding i C : Effective interest rate with continuous compounding IE 343 Fall 2011 *Example 4.22* – Solution 9
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General Solution: P S = -19,000 P D = 3,000(P/A, i%, 10)(P/F, i%, 2) f = 6%, r = 12%, f ≠ r P G = (P/F, i%, 12) P = P S + P D + P G IE 343 Fall 2011 *Example 4.22* – Solution 06 . 0 % )] 12 , 06 . 0 , / )( 12 %, , / ( 1 [ 000 , 3 i P F i F P 10
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Solution: Annual Compounding i A = r = 12% per year P S = -19,000 P D = 3,000(P/A, 12%, 10)(P/F, 12%, 2) = 13,513 f = 6%, i = 12%, f ≠ i
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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-West Lafayette.

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Lec%2014_Full - IE 343 Engineering Economics Lecture 14:...

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