HW13sol - ORIE 3300/5300 2010 ASSIGNMENT 13 SOLUTION Fall...

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ORIE 3300/5300 ASSIGNMENT 13 SOLUTION Fall 2010 Problem 1 After introducing a slack variable and solving the linear programming relaxation using the simplex method, we arrive at the following optimal tableau: z + 1 5 x 1 + 3 5 x 3 = 27 5 2 5 x 1 + x 2 + 1 5 x 3 = 9 5 . We can see the optimal solution is ( x 1 , x 2 ) = (0 , 9 / 5), with optimal value z = 27 / 5. Now we will construct the branch-and-bound tree. Since x 2 has a fractional value now, we branch on x 2 and consider two subproblems with extra con- straint x 2 2 or x 2 1. It is easy to see the subproblem with constraint x 2 2 is infeasible since 2 x 1 + 5 x 2 0 + 5 × 2 = 10 > 9, but in a large instance we will have to determine infeasibility as we reoptimize using the dual simplex method. We will show how to do that below. For the subproblem with constraint x 2 2, after introducing an additional surplus variable for this constraint, and adding the resulting equation to the optimal tableau above, we obtain: z + 1 5 x 1 + 3 5 x 3 = 27 5 2 5 x 1 + x 2 + 1 5 x 3 = 9 5 x 2 - x 4 = 2 . x 4 will be a basic variable, but we want to eliminate the basic variable x 2 from this last row, so we multiply the last equation by -1 and add the second equation to it to obtain the dual-feasible tableau: z + 1 5 x 1 + 3 5 x 3 = 27 5 2 5 x 1 + x 2 + 1 5 x 3 = 9 5 2 5 x 1 + 1 5 x 3 + x
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HW13sol - ORIE 3300/5300 2010 ASSIGNMENT 13 SOLUTION Fall...

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