IEOR162_hw04_sol

IEOR162_hw04_sol - IEOR 162 Spring 2012 Suggested Solution...

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Unformatted text preview: IEOR 162, Spring 2012 Suggested Solution to Homework 04 Problem 1 (a) The manufacturer’s problem is max 0 . 4( p 1- 10 q 1 ) + 0 . 6( p 2- 10 q 2 ) s.t. 20 q 1- p 1 ≥ (IR-1) 15 q 2- p 2 ≥ (IR-2) 20 q 1- p 1 ≥ 20 q 2- p 2 (IC-1) 15 q 2- p 2 ≥ 15 q 1- p 1 . (IC-2) (b) Adding (IC-1) and (IC-2), we get 20 q 1 + 15 q 2 ≥ 20 q 2- 15 q 1 ⇒ 5 q 1 ≥ 5 q 2 ⇒ q 1 ≥ q 2 . (c) To show that (IR-1) is redundant, we apply (IC-1) and (IR-2) as follows: 20 q 1- p 1 ≥ 20 q 2- p 2 (IC-1) ≥ 15 q 2- p 2 (Because 20 > 15) ≥ . (IR-2) (d) To show that (IC-1) is binding at any optimal solution, consider the manufacturer’s problem without (IR-1). Now, to maximize her expected profit, the manufacturer would like to make p 1 as large as possible. If there is no (IR-1), then (IC-1) is the only constraint that has p 1 at the LHS. The manu- facturer will keep increasing p 1 until (IC-1) is binding (i.e., when the LHS hits the RHS). Therefore, (IC-1) must be binding at any optimal solution. Note that when the manufacturer increases p 1 , (IC-2) will be relaxed as the RHS becomes smaller. We thus do not need to worry about (IC-2)....
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This note was uploaded on 03/14/2012 for the course IEOR 162 taught by Professor Zhang during the Spring '07 term at Berkeley.

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IEOR162_hw04_sol - IEOR 162 Spring 2012 Suggested Solution...

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