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Unformatted text preview: Homework 3 Solutions Solution to Exercise 4, page 101 : (a) Since we want to maximize x 1 , we would like it to be as large as possible while feasible. From the given system of equations, we see that the maximum feasible value of x 1 is the minimum of 4 / 5 , 8 / 6 , and 1 (obtained by setting the other variables to zero). Thus, the maximum value for x 1 is 1. To justify the preceding in terms of the basic variables, we note that from the given equations we can immediately find a basic feasible solution x 2 = 4 , x 3 = 8 , x 4 = 3, whose cost is zero. By doing a ratio test, we find that increasing x 1 = 1, will cause x 4 to become zero, and x 4 will leave the basis, which does not affect the cost. Hence, the new basic feasible solution will have x 1 = 1 and it will be optimal. (b) To maximize x 1 , we would like it to be as small as possible while feasible. From the given system of equations, we see that the minimum feasible value of x 1 is 0, and the current basic feasible solution x 2 = 4 , x 3 = 8 , x 4 = 3 is actually optimal. Solution to Exercise 5, page 101: We can think of a slack variable x 6 introduced to the inequality, so that we have maximize z = 5 x 1 6 x 2 + 3 x 3 5 x 4 + 12 x 5 subject to x 1 + 3 x 2 + 5 x 3 + 6 x 4 + 3 x 5 + x 6 = 90 x 1 , x 2 , x 3 , x 4 , x 5 , x 6 . There are 6 variables and only 1 equation, so that every basic feasible solution consists of one variable only. The equation as given can be interpreted as: if x 6 is basic, then x 6 = 90 and its objective value is 0. If we choose x 1 to be basic, then from the equation we have x 1 = 90 and its objective value is z = 5 90 = 450. Choosing x 2 to be basic, we get x 2 = 90 / 3 = 30 with z = 6 30 = 180 . Choosing x 3 as basic, we get x 3 = 90 / 5 = 18 with z = 3 18 = 180 . Choosing x 4 as basic, we get...
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This note was uploaded on 08/31/2010 for the course IESE GE 330 taught by Professor Nedich during the Spring '09 term at University of Illinois at Urbana–Champaign.
 Spring '09
 Nedich

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