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Unformatted text preview: Homework 4 Solution February 17, 2009 Solution to Exercise 3, page 111 By the twophase method, we need to maximize z = 2 x 1 + 3 x 2 5 x 3 subject to x 1 + x 2 + x 3 = 7, 2 x 1 5 x 2 + x 3 10, and x i 0 for all i . We first transform the problem to standard form: maximize z = 2 x 1 + 3 x 2 5 x 3 subject to x 1 + x 2 + x 3 = 7 2 x 1 5 x 2 + x 3 s 1 = 10 x 1 ,x 2 ,x 3 ,s 1 . In each equality , we introduce an artificial variable. Thus, the problem in standard form and with artificial variables is given by: maximize z = 2 x 1 + 3 x 2 5 x 3 subject to x 1 + x 2 + x 3 + R 1 = 7 2 x 1 5 x 2 + x 3 s 1 + R 2 = 10 x 1 ,x 2 ,x 3 ,s 1 ,R 1 ,R 2 . In Phase I, we always solve the feasibility problem, i.e., we minimize the sum of the artificial variables to determine a basic feasible solution. Thus, in this case we solve (by the standard simplex method) the following problem: mimimize v = R 1 + R 2 subject to x 1 + x 2 + x 3 + R 1 = 7 2 x 1 5 x 2 + x 3 s 1 + R 2 = 10 x 1 ,x 2 ,x 3 ,s 1 ,R 1 ,R 2 . We start the initial simplex table for this problem by choosing R 1 and R 2 as basic variables. The initial table is: Basic x 1 x 2 x 3 s 1 R 1 R 2 Solution v 1 1 R 1 1 1 1 1 7 R 2 2 5 1 1 1 10 1 Next, we need to transform the table so that the reduced costs of R 1 and R 2 are zero. By adding the R 1row to the vrow, and by adding the R 2row to the vrow, we obtain: Basic x 1 x 2 x 3 s 1 R 1 R 2 Solution v 3 4 2 1 17 R 1 1 1 1 1 7 R 2 2 5 1 1 1 10 This is the first table ready for simplex implementation. Since we are minimizing, the candidate variables that can improve the cost are those with positive reduced cost values, i.e., x 1 and x 3 . We choose x 1 for example, and by ratio test we see that x 1 = min { 7 , 5 } = 5. Thus, R 2 leaves the basis, as x 1 enters; and the new table is given by: Basic x 1 x 2 x 3 s 1 R 1 R 2 Solution v 7 / 2 1 / 2 1 / 2 3 / 2 2 R 1 7 / 2 1 / 2 1 / 2 1 1 / 2 2 x 1 1 5 / 2 1 / 2 1 / 2 1 / 2 5 At this point, the candidates to enter the basis are x 2 , x 3 , and s 1 . Suppose we choose x 2 to enter the basis. The ratio test gives x 2 = 4 / 7 and R 1 leaves the basis, resulting in the following table: Basic x 1 x 2 x 3 s 1 R 1 R 2 Solution v 1 1 x 2 1 1 / 7 1 / 7 2 / 7 1 / 7 4 / 7 x 1 1 0 6 / 7 1 / 7 5 / 7 1 / 7 45 / 7 This is optimal table for Phase I. The basic feasible solution is x 1 = 45 / 7, x 2 = 4 / 7 and x 3 = 0. We use this solution to proceed with phase II of the method, solving the original problem....
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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
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