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Unformatted text preview: UC Berkeley Department of Electrical Engineering and Computer Science EECS 227A Nonlinear and Convex Optimization Solutions 6 Fall 2009 Solution 6.1 (a) p ∗ = 1 (b) The Lagrangian is L ( x,y,λ ) = e − x + λx 2 /y . The dual function is g ( λ ) = inf x,y> ( e − x + λx 2 /y ) = braceleftbigg if λ ≥ −∞ otherwise so we can write the dual problem as maximize subject to λ ≥ with optimal value d ∗ = 0. The optimal duality gap is p ∗ − d ∗ = 1 (c) Slater’s condition is not satisfied. (d) p ∗ ( u ) = 1 if u = 0 ,p ∗ ( u ) = 0 if u > 0 and p ∗ ( u ) = ∞ if u < Solution 6.2 Suppose x is feasible. Since f i are convex and f i ( x ) ≤ 0, we have ≥ f i ( x ) ≥ f i ( x ∗ ) + ∇ f i ( x ∗ ) T ( x − x ∗ ) ,i = 1 ,...,m Using λ ∗ i ≥ 0, we conclude that ≥ m summationdisplay i =1 λ ∗ i ( f i ( x ∗ ) + ∇ f i ( x ∗ ) T ( x − x ∗ ) = m summationdisplay i =1 λ ∗ i ( f i ( x ∗ ) + m summationdisplay i =1 ∇ f i ( x ∗ ) T ( x − x ∗ ) = −∇ f ( x ∗ ) T ( x − x ∗ ) In the last line, we use the complementary slackness condition λ ∗ i f i ( x ∗ ) = 0, and the last KKT condition. This show that ∇ f ( x ∗ ) T ( x − x ∗ ) ≥ 0, i.e. ∇ f ( x ∗ ) defines a supporting hyperplane to feasible set at x ∗ Solution 6.3 (a) Follows from tr( Wxx T ) = x T Wx and ( xx T ) ii = x 2 i (b) It gives a lower bound because we minimize the same objective function over a larger set. If X is rank one, it is optimal. 1 (c) We write the problem as a minimization problem minimize 1 T ν subject to W + diag ( ν ) followsequal Introducing a Lagrange multiplier X ∈ S n for the matrix inequality, we obtain the Lagrangian L ( ν,X ) = 1 T ν − tr( X ( W + diag ( ν ))) = 1 T ν − tr( XW ) − ∑ n i =1 ν i X ii = − tr( XW ) + ∑ n i =1 ν i (1 − X ii ) This is bounded below as a function of ν only if X ii = 1 for all i , so we obtain the dual problem maximize − tr( WX ) subject to X followsequal X ii = 1 ,i = 1 ,...,n Changing the sign again, and switching from maximization to minimization, yields the problem in part (a) Solution 6.4 (a) We introduce the new variables, and write the problem as minimize c T x subject to bardbl y i bardbl 2 ≤ t i ,i = 1 ,...,m y i = A i x + b i ,i = 1 ,...,m t i = c T i x + d i ,i = 1 ,...,m The Lagrangian is L ( x,y,t,λ,ν,μ ) = c T x + m summationdisplay...
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 Spring '11
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 Electrical Engineering, Optimization, Yi, Dual problem, dual newton method, primal newton method

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