v2008.02.02 - Convex Optimization

v2008.02.02 - Convex Optimization v2008.02.02 - Convex Optimization

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178 CHAPTER 2. CONVEX GEOMETRYFrom optimality condition (309),because∇f(x ⋆ ) T (Z(ÂZ)† b − x ⋆ )≥ 0 ∀b ≽ 0 (396)−∇f(x ⋆ ) T Z(ÂZ)† (b − b ⋆ )≤ 0 ∀b ≽ 0 (397)x ⋆ ∆ = Z(ÂZ)† b ⋆ ∈ K (398)From membership relation (392) and Example 2.13.10.1.1〈−(Z T Â T ) † Z T ∇f(x ⋆ ), b − b ⋆ 〉 ≤ 0 for all b ∈ R m−l+⇔(399)−(Z T Â T ) † Z T ∇f(x ⋆ ) ∈ −R m−l+ ∩ b ⋆⊥Then the equivalent necessary and sufficient conditions for optimality of theconic program (395) with pointed polyhedral feasible set K are: (confer (315))(Z T Â T ) † Z T ∇f(x ⋆ ) ≽R m−l+0, b ⋆ ≽R m−l+0, ∇f(x ⋆ ) T Z(ÂZ)† b ⋆ = 0 (400)When K = R n + , in particular, then C =0, A=Z =I ∈ S n ; id est,minimize f(x)xsubject to x ≽ 0 (401)The necessary and sufficient conditions become (confer [46,4.2.3])R n +∇f(x ⋆ ) ≽ 0, x ⋆ ≽ 0, ∇f(x ⋆ ) T x ⋆ = 0 (402)R n +R n +2.13.10.1.3 Example. Linear complementarity. [202] [235]Given matrix A ∈ R n×n and vector q ∈ R n , the complementarity problem isa feasibility problem:find w , zsubject to w ≽ 0z ≽ 0w T z = 0w = q + Az(403)

2.13. DUAL CONE & GENERALIZED INEQUALITY 179Volumes have been written about this problem, most notably by Cottle [59].The problem is not convex if both vectors w and z are variable. But if one ofthem is fixed, then the problem becomes convex with a very simple geometricinterpretation: Define the affine subsetA ∆ = {y ∈ R n | Ay=w − q} (404)For w T z to vanish, there must be a complementary relationship between thenonzero entries of vectors w and z ; id est, w i z i =0 ∀i. Given w ≽0, thenz belongs to the convex set of feasible solutions:z ∈ −K ⊥ R n + (w ∈ Rn +) ∩ A = R n + ∩ w ⊥ ∩ A (405)where KR ⊥ (w) is the normal cone to n Rn + + at w (394). If this intersection isnonempty, then the problem is solvable.2.13.11 Proper nonsimplicial K , dual, X fat full-rankAssume we are given a set of N conically independent generators 2.61 (2.10)of an arbitrary polyhedral proper cone K in R n arranged columnar inX ∈ R n×N such that N > n (fat) and rankX = n . Having found formula(362) to determine the dual of a simplicial cone, the easiest way to find avertex-description of the proper dual cone K ∗ is to first decompose K intosimplicial parts K i so that K = ⋃ K i . 2.62 Each component simplicial conein K corresponds to some subset of n linearly independent columns from X .The key idea, here, is how the extreme directions of the simplicial parts mustremain extreme directions of K . Finding the dual of K amounts to findingthe dual of each simplicial part:2.61 We can always remove conically dependent columns from X to construct K or todetermine K ∗ . (F.2)2.62 That proposition presupposes, of course, that we know how to perform simplicialdecomposition efficiently; also called “triangulation”. [228] [123,3.1] [124,3.1] Existenceof multiple simplicial parts means expansion of x∈ K like (353) can no longer be uniquebecause N the number of extreme directions in K exceeds n the dimension of the space.

178 CHAPTER 2. CONVEX GEOMETRYFrom optimality condition (309),because∇f(x ⋆ ) T (Z(ÂZ)† b − x ⋆ )≥ 0 ∀b ≽ 0 (396)−∇f(x ⋆ ) T Z(ÂZ)† (b − b ⋆ )≤ 0 ∀b ≽ 0 (397)x ⋆ ∆ = Z(ÂZ)† b ⋆ ∈ K (398)From membership relation (392) and Example 2.13.10.1.1〈−(Z T Â T ) † Z T ∇f(x ⋆ ), b − b ⋆ 〉 ≤ 0 for all b ∈ R m−l+⇔(399)−(Z T Â T ) † Z T ∇f(x ⋆ ) ∈ −R m−l+ ∩ b ⋆⊥Then the equivalent necessary and sufficient conditions for optimality of theconic program (395) with pointed polyhedral feasible set K are: (confer (315))(Z T Â T ) † Z T ∇f(x ⋆ ) ≽R m−l+0, b ⋆ ≽R m−l+0, ∇f(x ⋆ ) T Z(ÂZ)† b ⋆ = 0 (400)When K = R n + , in particular, then C =0, A=Z =I ∈ S n ; id est,minimize f(x)xsubject to x ≽ 0 (401)The necessary and sufficient conditions become (confer [46,4.2.3])R n +∇f(x ⋆ ) ≽ 0, x ⋆ ≽ 0, ∇f(x ⋆ ) T x ⋆ = 0 (402)R n +R n +2.13.10.1.3 Example. Linear complementarity. [202] [235]Given matrix A ∈ R n×n and vector q ∈ R n , the complementarity problem isa feasibility problem:find w , zsubject to w ≽ 0z ≽ 0w T z = 0w = q + Az(403)

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