hw6soln_fa09

# hw6soln_fa09 - UC Berkeley Department of Electrical...

This preview shows pages 1–3. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

{[ snackBarMessage ]}

### Page1 / 7

hw6soln_fa09 - UC Berkeley Department of Electrical...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online