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Unformatted text preview: KKT Examples Stanley B. Gershwin Massachusetts Institute of Technology The purpose of this note is to supplement the slides that describe the KarushKuhnTucker conditions. Neither these notes nor the slides are a complete description of these conditions; they are only intended to provide some intuition about how the conditions are sometimes used and what they mean. The KKT conditions are usually not solved directly in the analysis of practical large nonlinear programming problems by software packages. Iterative successive approximation methods are most often used. The results, however they are obtained, must satisfy these conditions. Example No constraints: min J = x 2 1 + x 2 2 + x 2 3 2 4 + x The solution is x 1 = x 2 = x 3 = x 4 = and J = 0. Example 1 One equality constraint: min J = x 2 1 + x 2 2 2 3 2 4 + x + x subject to x 1 + x 2 + x 3 + x 4 = 1 (1) 1 KKT Examples Solution: Adjoin the constraint 2 2 2 2 min J ¯ = x 1 + x 2 + x 3 + x 4 + (1 − x 1 − x 2 − x 3 − x 4 ) subject to x 1 + x 2 + x 3 + x 4 = 1 In this context, is called a Lagrange multiplier . The KKT conditions reduce, in this case, to setting ¯ J/ x to zero: ⎨ ⎨ 2 x 1 − ⎩ ⎩ J ¯ ⎩ 2 x 2 − ⎩ = ⎩ = ⎩ (2) x ⎪ 2 x 3 − ⎪ 2 x 4 − Therefore x 1 = x 2 = x 3 = x 4 = 2 so 1 x 1 + x 2 + x 3 + x 4 = 4 = 1 or = 2 2 so 1 1 x 1 = x 2 = x 3 = x 4 = and J = 4 4 x 1 x 4 1 4 , 1 4 , 1 4 , 1 4 x = 1 x = 4 ) ( J=constant 2 equality constraint Figure 1: Example 1, represented in two dimensions Comments • Why should we adjoin the constraints? The answer to this question has two parts: 2 KKT Examples ¯ λ First, it does no harm. There is no difference between J and J for any set of x i that satisfies the problem since the set of x i that satisfies the problem must satisfy (1). Furthermore, it works. This may not feel like a very satisfying answer, but this is also why we plug x = Ce t into a linear differential equation with constant coeﬃcients. We could have done a zillion other things that would have done no harm, but they would not have done any good either. There is not much point in studying them; we only study what works....
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 Fall '10
 StanleyGershwin
 Optimization, lagrange multipliers, Karush–Kuhn–Tucker conditions, b oundary, equality constraint

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