v2009.01.01 - Convex Optimization

2.13. DUAL CONE & GENERALIZED INEQUALITY 185
(Z T Â T ) † Z T ∇f(x ⋆ ) ≽
R m−l
+
0, b ⋆ ≽
R m−l
+
0, ∇f(x ⋆ ) T Z(ÂZ)† b ⋆ = 0 (412)
When K = R n + , in particular, then C =0, A=Z =I ∈ S n ; id est,
minimize f(x)
x
subject to x ≽ 0 (413)
The necessary and sufficient conditions become (confer [53,4.2.3])
R n +
∇f(x ⋆ ) ≽ 0, x ⋆ ≽ 0, ∇f(x ⋆ ) T x ⋆ = 0 (414)
R n +
R n +
2.13.10.1.3 Example. Linear complementarity. [235] [269]
Given matrix A ∈ R n×n and vector q ∈ R n , the complementarity problem is
a feasibility problem:
find w , z
subject to w ≽ 0
z ≽ 0
w T z = 0
w = q + Az
(415)
Volumes have been written about this problem, most notably by Cottle [76].
The problem is not convex if both vectors w and z are variable. But if one of
them is fixed, then the problem becomes convex with a very simple geometric
interpretation: Define the affine subset
A ∆ = {y ∈ R n | Ay=w − q} (416)
For w T z to vanish, there must be a complementary relationship between the
nonzero entries of vectors w and z ; id est, w i z i =0 ∀i. Given w ≽0, then
z belongs to the convex set of feasible solutions:
z ∈ −K ⊥ R n + (w ∈ Rn +) ∩ A = R n + ∩ w ⊥ ∩ A (417)
where KR ⊥ (w) is the normal cone to n Rn + + at w (406). If this intersection is
nonempty, then the problem is solvable.