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Unformatted text preview: 10725 Optimization, Spring 2008: Homework 4 Solutions Due: Wednesday, April 2, beginning of the class 1 Lagrangian relaxation of Boolean LP [Han, 12 points] 1. [6 pts] The Lagrangian is L ( x,μ,v ) = c T x + μ T ( Ax b ) ν T x + x T diag( ν ) x (1) = x T diag( ν ) x + ( c + A T μ v ) T x b T μ (2) Minimizing over x gives the dual function g ( μ,v ) = b T μ (1 / 4) ∑ n i =1 ( c i + a T i ν i ) 2 /ν i ν ≥∞ o.w. (3) where a i is the i th column of A , and we adopt the convention that a 2 / 0 = ∞ if a 6 = 0, and a 2 / 0 = 0 if a = 0. The resulting dual problem is maximize b T μ (1 / 4) n X i =1 ( c i + a T i ν i ) 2 /ν i (4) s.t.ν ≥ ,μ ≥ . (5) In order to simplify this dual, we optimize analytically over ν , by noting that sup ν i ≥ ( c i + a T i ν i ) 2 ν i = 4( c i + a T i μ ) c i + a T i μ ≤ c i + a T i μ ≥ (6) = min { , 4( c i + a T i μ ) } . (7) This allows us to eliminate ν from the dual problem, and simplify it as maximize b T μ + n X i =1 min { ,c i + a T i μ } (8) s.t.μ ≥ . (9) 2. [6 pts] We follow the hint. The Lagrangian and dual function of the LP relaxation are L ( x,u,v,w ) = c T x + u T ( Ax b ) v T x + w T ( x 1) (10) = ( c + A T u v + w ) T x b T u 1 T w (11) g ( u,v,w ) = b T u 1 T w A T u v + w + c = 0∞ o.w. (12) The dual problem is maximize b T μ 1 T w (13) s.t.A T u v + w + c = 0 (14) u ≥ ,v ≥ ,w ≥ (15) which is equivalent to the Lagrange relaxation problem derived above. We conclude that the two relaxations give the same value. 1 2 A convex problem in which strong duality fails [Han, 8 points] 1. [3 pts] It’s straightforward to verity that this is a convex optimization problem. Since x 2 /y ≤ 0 is not affine and there is not a point in relint( D ), such that x 2 /y < 0, the Slater’s condition does not hold. 2. [5 pts] It’s trivially to see that the primal optimal p * = 1, while the Lagrangian dual problem can be formulated as maximize g ( λ ) s.t. λ ≥ (16) where g ( λ ) = inf x ∈ R ,y> { e x + λx 2 /y } . The dual optimal is d * = 0. Therefore, the duality gap is 1. 3 Multiobjective optimization [Han, 15 points] 1. [3 pts] Assume that x P is not a Pareto optimal solution. Then there is a vector ¯ x ∈ D such that either g 1 (¯ x ) ≤ g 1 ( x P ) , g 2 (¯ x ) < g 2 ( x P ) (17) or g 1 (¯ x ) < g 1 ( x P ) , g 2 (¯ x ) ≤ g 2 ( x P ) (18) In either case, by using the fact that μ * 1 > 0 and μ * 2 > 0, we have μ * 1 g 1 (¯ x ) + μ * 2 g 2 (¯ x ) < μ * 1 g 1 ( x P ) + μ * 2 g 2 ( x P ) , (19) yielding a contradiction. Therefore x P is a Pareto optimal solution. 2. [7 pts] Let A = { ( z 1 ,z 2 ) ∃ x ∈ D , s.t. g 1 ( x ) ≤ z 1 , g 2 ( x ) ≤ z 2 } . (20) We first show that A is convex. Indeed, let ( a 1 ,a 2 ), ( b 1 ,b 2 ) be elements of A , and let ( c 1 ,c 2 ) = θ ( a 1 ,a 2 ) + (1 θ )( b 1 ,b 2 ), for any θ ∈ [0 , 1]. then for some x a ∈ D , x b ∈ D , we have g 1 ( x a ) ≤ a 1 , g 2 ( x b ) ≤...
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This note was uploaded on 05/25/2008 for the course MACHINE LE 10708 taught by Professor Carlosgustin during the Spring '07 term at Carnegie Mellon.
 Spring '07
 CarlosGustin

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