v2010.10.26 - Convex Optimization

4.6. CARDINALITY AND RANK CONSTRAINT EXAMPLES 367max cut statement (818) is the same as, for A = [a ij ]∈ S n1maximizex 4 〈11T − xx T , A〉subject to δ(xx T ) = 1Because of Boolean assumption δ(xx T ) = 1so problem (820) is the same as(820)〈11 T − xx T , A〉 = 〈xx T , δ(A1) − A〉 (821)1maximizex 4 〈xxT , δ(A1) − A〉subject to δ(xx T ) = 1This max cut problem is combinatorial (nonconvex).(822)Because an estimate of upper bound to max cut is needed to ascertainconvergence when vector x has large dimension, we digress to derive thedual problem: Directly from (822), its Lagrangian is [61,5.1.5] (1418)L(x, ν) = 1 4 〈xxT , δ(A1) − A〉 + 〈ν , δ(xx T ) − 1〉= 1 4 〈xxT , δ(A1) − A〉 + 〈δ(ν), xx T 〉 − 〈ν , 1〉= 1 4 〈xxT , δ(A1 + 4ν) − A〉 − 〈ν , 1〉(823)where quadratic x T (δ(A1+ 4ν)−A)x has supremum 0 if δ(A1+ 4ν)−A isnegative semidefinite, and has supremum ∞ otherwise. The finite supremumof dual functiong(ν) = supL(x, ν) =x{ −〈ν , 1〉 , A − δ(A1 + 4ν) ≽ 0∞otherwiseis chosen to be the objective of minimization to dual (convex) problemminimize −ν T 1νsubject to A − δ(A1 + 4ν) ≽ 0(824)(825)whose optimal value provides a least upper bound to max cut, but is nottight ( 1 4 〈xxT , δ(A1)−A〉< g(ν) , duality gap is nonzero). [157] In fact, wefind that the bound’s variance with problem instance is too large to beuseful for this problem; thus ending our digression. 4.404.40 Taking the dual of dual problem (825) would provide (826) but without the rankconstraint. [150] Dual of a dual of even a convex primal problem is not necessarily thesame primal problem; although, optimal solution of one can be obtained from the other.