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Unformatted text preview: IEOR 162, Fall 2011 Suggested Solution to Homework 01 Problem 1 (Modified from Problem 3.2.3) (a) Let x 1 = hours of running process 1 and x 2 = hours of running process 2. Then the problem can be formulated as min 4 x 1 + x 2 s.t. 3 x 1 + x 2 ≥ 10 (Requirement of chemical A) x 1 + x 2 ≥ 5 (Requirement of chemical B) x 1 ≥ 3 (Requirement of chemical C) x 1 ≥ (This is not required since we already have x 1 ≥ 3) x 2 ≥ . (b) The graphical solution of this problem is shown in Figure 1. The feasible region is the shaded zone. The dotted line is the isocost line with ( 4 , 1) as the improving direction. The optimal solution is point A, which is (3 , 2). The minimized objective value is 4 × 3 + 2 = 14. Therefore, Leary Chemical should run 3 hours of process 1 and 2 hours of process 2. The company will have $14 as its cost. x + x = 5 3 x + x = 1 0 A x x x = 3 1 2 1 2 1 1 2 (  4 ,  1 ) Figure 1: Graphical solution for Problem 1 5 x + 2 x = 1 2 8 x + 2 x = 1 6 2 1 1 2 x x 1 2 A ( 4 , 1 )...
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This note was uploaded on 10/24/2011 for the course IEOR 162 taught by Professor Zhang during the Fall '07 term at Berkeley.
 Fall '07
 Zhang

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