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Lecture 4

# Lecture 4 - ISE 460 Engineering Economy University of...

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ISE 460: Engineering Economy University of Southern California Summer 2010 Joseph Chow Lecture 4

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A Note on the Textbook I checked with the department – yes, we had recalled the textbook from the bookstore because they stocked upon Dr. Bottlik’s book which is the wrong one They tell me that the textbook should hopefully arrive by next week. Sorry for the confusion. If the textbooks don’t arrive by next Tuesday I will postpone the HW #1 deadline until Tuesday 9/14/10 Those of you who don’t have a textbook, please try your best to keep up with the lecture slides so that you can quickly review through the textbook once you get it 10/12/10 2
A Note on Air Conditioning I sent a note to the USC Facilities so hopefully it’s more comfortable in the classroom now Is it? 10/12/10 3

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Review Role of Engineering Economics Accounting Basics Time Value of Money Compound Interest Economic Equivalence Today… Some more economic equivalence Nominal and effective interest rates 10/12/10 4
More Economic Equivalence What can we do with the ability to move cash flows back and forth along the time axis? 10/12/10 5

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Refresher Example 10/12/10 6 Compute the value of P in the accompanying cash flow diagram, assuming that i = 9%.
Deferral Strategy (or, Shifting Payments Forward in Time) E.g. deferring your loan, deferring an investment decision Suppose we want to push off payments for k periods how much would we need to pay, A’, to make the present value equivalent to the original payments A? 10/12/10 7 N 1 2 … N-1 P 0 A N+k 1 k+1 …N+k-1 P 0 A’

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Deferral Strategy (or, Shifting Payments over Time) Find A’ given P, i, and N, and k period delay -- A’ = V k (A/P,i,N), where V k is the value of P at the end of period k, V k = P(F/P,i,k) Combining the two: A’ = P(F/P,i,k)(A/P,i,N) 10/12/10 8 N 1 2 … N-1 P 0 A A’ Why defer? N+k 1 k+1 …N+k-1 P 0 A’
Composite Cash Flows Suppose we have multiple cash flows comprised of A 1 on the first two periods, followed by a larger A 2 on the following three periods. This can be decomposed: 10/12/10 9 0 5 1 2 3 4 A 2 A 1 0 5 1 2 3 4 A 1 0 5 1 2 3 4 A 2 – A 1

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Linear Gradient Series Present-Worth Factor: find P given G, N, I P = 0 + G(1+i) -2 + 2G(1+i) -3 + … + (N-1)G(1+i) -N = {this is an arithmetic-geometric series with finite sum} = 10/12/10 10 0 5 1 2 3 4 4 G G 2 G 3 G = - + - N n n i G n 1
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Lecture 4 - ISE 460 Engineering Economy University of...

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