# L11_gap - Lecture 11 Strong Duality Results Lecture 11...

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Unformatted text preview: Lecture 11: Strong Duality Results February 26, 2007 Lecture 11 Outline • Slater Condition and its Variations • Convex Objective with Linear Inequality Constraints • Quadratic Objective over Quadratic Constraints • Recession Cone Conditions • Representation Issue • Multiple Dual Choices Convex Optimization 1 Lecture 11 General Convex Problem Primal Problem minimize f ( x ) subject to g j ( x ) ≤ , j = 1 ,...,m a T i x = b i , i = 1 ,...,r x ∈ X Assumptions throughout the lecture • The problem is convex • The objective f and all g j are convex • The set X ⊆ R n is nonempty and convex • The problem is feasible C = { x ∈ R n | g ( x ) , Ax = b, x ∈ X } 6 = ∅ where g = ( g 1 ,...,g m ) T and A is a matrix with rows a T i , i = 1 ,...,r • The problem is meaningful , i.e., dom f ∩ C 6 = ∅ Convex Optimization 2 Lecture 11 Slater Condition for Inequality Constrained Problem Convex Primal Problem minimize f ( x ) subject to g j ( x ) ≤ , j = 1 ,...,m x ∈ X Slater Condition There exists a vector ¯ x ∈ R n such that g j (¯ x ) < for all j and ¯ x ∈ X ∩ dom f Observation: The Slater condition is a condition on the constraint set only when dom f = R n Theorem Let the Slater condition hold and assume that f * is finite. Then: • There is no duality gap, i.e., q * = f * • The set of dual optimal solutions is nonempty and bounded Convex Optimization 3 Lecture 11 Proof Consider the set V ⊂ R m × R given by V = { ( u,w ) | g ( x ) u, f ( x ) ≤ w, x ∈ X } The Slater condition is illustrated in the figures below. The set V is convex by the convexity of f , g j s , and X . Convex Optimization 4 Lecture 11 Proof continues The vector (0 ,f * ) is not in the interior of the set V . Suppose it is, i.e., (0 ,f * ) ∈ intV . Then, there exists an > such that (0 ,f *- ) ∈ V contradicting the optimality of f * . Thus, either (0 ,f * ) ∈ bd V or (0 ,f * ) 6∈ V . By the Supporting Hyperplane Theorem, there exists a hyperplane passing through (0 ,f * ) and supporting the set V : there exists ( μ,μ ) ∈ R m × R with ( μ,μ ) 6 = 0 such that μ T u + μ w ≥ μ f * for all ( u,w ) ∈ V (1) This relation implies that μ and μ ≥ . Suppose that μ = 0 . Then, μ 6 = 0 and relation (1) reduces to inf ( u,v ) ∈ V μ T u = 0 On the other hand, by the definition of the set V , since μ and μ 6 = 0 , we have inf ( u,v ) ∈ V μ T u ≤ inf x ∈ X μ T g ( x ) ≤ μ T g (¯ x ) <- a contradiction Convex Optimization 5 Lecture 11...
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## This note was uploaded on 08/22/2008 for the course GE 498 AN taught by Professor Angelianedich during the Spring '07 term at University of Illinois at Urbana–Champaign.

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L11_gap - Lecture 11 Strong Duality Results Lecture 11...

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