Differential Equations Solutions 46

Differential Equations Solutions 46 - Then the optimality...

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56 Chapter 10. Solutions: Numerical Methods for Constrained Optimization (a) K is the set of vectors that form nonnegative inner products with every vector that has nonnegative entries. Therefore, K = K = { x : x 0 } , the positive orthant in R n . (b) The dual problem is max w w T b subject to A T w + s = c and s 0 . (c) Since s 0 , the constraint A T w + s = c means that each component of A T w is less than or equal to each component of c , since we need to add s on in order to get equality. Therefore, A T w c . (d) First order optimality conditions: The constraints are x 0 , Ax b = 0 . The derivative matrix for the constraints becomes · I A ¸ . Using a bit of foresight, let’s call the Lagrange multipliers for the inequality con- straints s and those for the equality constraints
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Unformatted text preview: λ . Then the optimality conditions are Is + A T λ = c , s ≥ , x ≥ , Ax = b , s T x = 0 . The central path: Ax = b , A ∗ w + s = c , μ X − 1 e + s = , where s , x ≥ 0 (since the feasible cone is the positive orthant). Both sets of conditions have Ax = b . Setting λ = w shows the equivalence of s + A T λ = c and A ∗ w + s = c , since A ∗ = A T . Multiplying μ X − 1 e + s = by X , we obtain μ e + Xs = , which is equivalent to s T x = 0 when μ = 0, since x and s are nonnegative. Therefore the conditions are equivalent....
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This note was uploaded on 01/21/2012 for the course MAP 3302 taught by Professor Dr.robin during the Fall '11 term at University of Florida.

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