Topic 19 - Intro to Lagrangian Duality

# Minimize subject to in this case we would take re

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Unformatted text preview: plot the relationship between z and y for any feasible . 211 We have mapped all required that to the . In the plane. To be feasible to the overall problem, we also space above, indicate which points are feasible to P. Among all feasible points, P seeks the point with the minimal above, indicate the optimal point to P. value. In the space optimal value of Now let us visualize what the Lagrangian dual problem is doing. Recall that it is given by D: maximize subject to where , { } { } { }. Consider the objective Under our new mapping, we have contours for this problem. For points with objective values equal to , we have or . What do these contours look like? Therefore, we can visualize the Lagrangian dual as follows. In the inner problem, the slope is fixed at a value of – (which is , since ). With this fixed slope, we will push the line down until it supports G. The inner objective value is the intercept of this line. The outer loop searches for good values of . It wants to find the value of that maximizes the intercept value, since that is the objective value of the inner problem. Overall process: Find a supporting hyperplane (with negative slope) that has the maximum intercept. 212 This general concept is illustrated in Figure 6.1 of the BSS textbook. z A X (g , f ) x G [ g(x), f (x)] B ( y, z ) z uy (u) v(z) Slope u Slope u y Figure 6.1 (BSS textbook, page 260) We can also illustrate this concept for a discrete problem. (Lagrangian duality can be very effective for integer programs.) Example 5: Consider the following problem. Minimize subject to binary Recall that we want to construct the set { Let us take to represent the binary conditions. That is, need to map each point in to the -space. }. { (0,0) (0,1) (1,0) (1,1) Therefore, G is given by a series of points, which are shown below. 213 }. We z = f(x) 1 -2 -1 1 y = g(x) -1 -2 -3 Therefore, we have mapped all problem, we also required that are feasible to P. to the . In the plane. To be feasible to the overall space above, indicate which points Among all feasible points, P seeks that point with the minimal space above, indicate the optimal point to P. value. In the optimal value of Now let us visualize what the Lagrangian dual problem is doing. Recall that it is looking for a hyperplane that supports all the points. Among those, it wants the one that has the highest intercept value, which equals the optimal value of . On the graph above, sketch this supporting hyperplane, and indicate the optimal solution for the Lagrangian dual. maximum value of the intercept = optimal value of Lesson Learned: In integer programming problems, there can be a duality gap, meaning that the optimal objective value of the primal problem need not equal the optimal value of its Lagrangian dual. 214 Note that we could have formulated and solved the Lagrangian dual directly, without transforming to the space. D: maximize subject to , { where...
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## This document was uploaded on 11/28/2013.

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