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lect24 - • Solve the LP relaxation of problem P(relaxing...

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ISE 536–Fall03: Linear Programming and Extensions December 1, 2003 Lecture 24: Integer Programming, Branch and Bound Lecturer: Fernando Ord´ o˜nez 1 Branch and Bound We now present a method used to solve mixed integer programs, such as: z * = min c t x + d t y s . t . Ax + By = b x, y 0 x integer The key ideas for Branch and Bound to solve the above problem are: If we remove a constraint “ x j integer”, the modified problem is a lower bound on z * . If we fix x j to a feasible integer, or add an additional constraint on x j , the modified problem gives an upper bound on z * . Branch and Bound uses these ideas in a divide and conquer strategy. The division is done by adding constraints on the integer variables. Assume that we have problem P with some integer variables further constrained. The algorithm also uses some know upper bound on z * , U (for example the objective value of the best integer feasible solution so far). In an iteration of branch and bound the algorithm does the following
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Unformatted text preview: • Solve the LP relaxation of problem P (relaxing the integrality constraint for non-fixed integer variables). • Let x * be the optimal solution obtained, and z P the optimal objective function value. • (bounding) If z P ≥ U we delete the problem (as z P is a lower bound on all integer solutions in this branch of the tree). • (branching) Pick some x j whose optimal solution is fractional, and construct two problems by adding the constraints x j ≤ b x * j c and x j ≥ d x * j e in each problem. 1 Example Apply the branch and bound method to find a solution for min x 1-2 x 2 s . t .-4 x 1 + 6 x 2 ≤ 9 x 1 + x 2 ≤ 4 x 1 , x 2 ≥ x 1 , x 2 integer Apply the branch and bound method to find a solution for max 12 x 1 + 8 x 2 + 7 x 3 + 6 x 4 s . t . 8 x 1 + 6 x 2 + 5 x 3 + 4 x 4 ≤ 15 x i ∈ { , 1 } , i = 1 , . . . , 4 2...
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