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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.
 Fall '13

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