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# hw4_sol - Homework 4 Solution Solution to Exercise 3 page...

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Homework 4 Solution February 17, 2009 Solution to Exercise 3, page 111 By the two-phase 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 0 . 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 0 . 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 0 . 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 0 0 0 0 - 1 - 1 0 R 1 1 1 1 0 1 0 7 R 2 2 - 5 1 - 1 0 1 10 1

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Next, we need to transform the table so that the reduced costs of R 1 and R - 2 are zero. By adding the R 1 -row to the v -row, and by adding the R 2 -row to the v -row, we obtain: Basic x 1 x 2 x 3 s 1 R 1 R 2 Solution v 3 4 2 - 1 0 0 17 R 1 1 1 1 0 1 0 7 R 2 2 - 5 1 - 1 0 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 0 7 / 2 1 / 2 1 / 2 0 - 3 / 2 2 R 1 0 7 / 2 1 / 2 1 / 2 1 - 1 / 2 2 x 1 1 - 5 / 2 1 / 2 - 1 / 2 0 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 0 0 0 0 - 1 - 1 0 x 2 0 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. The initial table for the solution is: Basic x 1 x 2 x 3 s 1 Solution z - 2 - 3 5 0 0 x 2 0 1 1 / 7 1 / 7 4 / 7 x 1 1 0 6 / 7 - 1 / 7 45 / 7 We transform the table to get the reduced costs of the basic variables to be zero, and the table is: 2
Basic x 1 x 2 x 3 s 1 Solution z 0 0 50 / 7 1 / 7 102 / 7 x 2 0 1 1 / 7 1 / 7 4 /

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hw4_sol - Homework 4 Solution Solution to Exercise 3 page...

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