This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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 )...
View Full
Document
 Fall '07
 Zhang

Click to edit the document details