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Conjugate Duality

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Conjugate Duality LECTURE 5: CONJUGATE DUALITY 1. Primal problem and its conjugate dual 2. Duality theory and optimality conditions 3. Relation to other type of dual problems 4. Linear conic programming problems Motivation of conjugate duality Min f(x) (over R) = Max g(y) (over R) and h(y) = - g(y) • f(x) + h(y) = f(x) – g(y) can be viewed as “duality gap” • Would like to have (i) weak duality 0 (ii) strong duality 0 = f(x*) + h(y*) Where is the duality information • Recall the fundamental result of Fenchel’s conjugate inequality • Need a structure such that in general and at optimal solutions 0 = <x*, y*> = f(x*) + h(y*) Concept of dual cone • Let X be a cone in • Define its dual cone • Properties: (i) Y is a cone. (ii) Y is a convex set. (iii) Y is a closed set. Observations Conjugate (Geometric) duality Dual side information • Conjugate dual function • Dual cone Properties: 1. Y is a cone in 2. Y is closed and convex 3. both are closed and convex. Conjugate (Geometric) dual problem Observations Conjugate duality theory Conjugate duality theory Proof Conjugate duality theory Example – Standard form LP Conjugate dual problem Dual LP Example – Karmarkar form LP Example – Karmarkar form LP Example – Karmarkar form LP • Conjugate dual problem becomes which is an unconstrained convex programming problem. Illustration Example – Posynomial programming • Nonconvex programming problem • Transformation Posynomial programming • Primal problem: Conjugate dual problem: Dual Posynomial Programming h(y) , supx2En [< x; y > −f (x)] < +1 Pn Pn xj Let g(x) = j=1 xj yj − log j=1 cj e ∗ xj @g cj e = 0 ) yj = x∗ @xj Pn j j=1 cj e Pn ) yj > 0 and j=1 yj = 1: ) Ω is closed. n Pn ) Ω = fy 2 E jyj > 0 and j=1 yj = 1g: ∗ ∗ xj Pn xj log(yj ) = log(cj e ) − log j=1 cj e ∗ ∗ Pn xj = log(cj ) + xj − log j=1 cj e ∗ yj n x ) log( ) + log P c e j = x∗ cj j=1 j j ∗ h(y) = supx2 n [< x; y > −f (x)] = g(x ) E ∗ Pn ∗ Pn xj = j=1 xj yj − log j=1 cj e ∗ ∗ n yj n x n x = P y [log( ) + log P c e j ] − log P c e j j=1 j cj j=1 j j=1 j ∗ ∗ n yj n n x n x = P y log( ) + P y log P c e j − log P c e j j=1 j cj j=1 j j=1 j j=1 j ∗ ∗ n yj n n x n x = P y log( ) + (P y )log P c e j − log P c e j j=1 j cj j=1 j j=1 j j=1 j = Pn y log( yj ) j=1 j cj Conjugate dual problem Degree of difficulties • When degree of difficulty = 0, we have a system of linear equations: • When degree of difficulty = k, we have Duality gap • Definition: • Observation: Extremality conditions • Definition: • Corollary: Proof of Corollary Necessary and sufficient conditions • Corollary • Observation When will the duality gap vanish? Nonlinear complementarity problem Lagrangian Function and Saddle Point T T I Let f (x): S X and h(y):Ω Y be a known conjugate pair with X being closed and convex and f 2 C0(S). I The lagrangian function is defined as L(x; y) , f (x)− < x; y > : I A point pair (x¯; y¯) 2 S × Y is called a saddle point of L(x; y) if L(x; y¯) > L(x¯; y¯) > L(x¯; y); 8x 2 S; y 2 Y ; or inf L(x; y¯) = L(x¯; y¯) = sup L(x¯; y): x2S y2Y Saddle Point and Optimality Theorem: (saddle point ) Optimality) If (x¯; y¯) 2 S × Y is a saddle point of L(x; y), then x¯ 2 S ∗ and y¯ 2 T ∗. Proof: By definition, inf L(x; y¯) = L(x¯; y¯) = sup L(x¯; y) x2S y2Y 1 L(x¯; y¯) = infx2S L(x; y¯) = infx2S[f (x)− < x; y¯ >] = − supx2S[< x; y¯ > −f (x)] Hence y¯ 2 Ω and L(x¯; y¯) = −h(y), (now, y¯ 2 Y T Ω = T ) 2 L(x¯; y¯) = supy2Y L(x¯; y) = supy2Y [f (x¯)− < x¯; y >] = f (x¯) + supy2Y [− < x¯; y >] = f (x¯) − infy2Y [< x¯; y >] Since Y is a cone, x¯ 2 dual(Y ) = X and infy2Y [< x¯; y >] = 0. Hence L(x¯; y¯) = f (x¯). (Now, x¯ 2 S T X = S ) 3 Putting 1 and 2 together, we have f (x¯) = L(x¯; y¯) = −h(y¯): Hence f (x¯) + h(y¯) = 0, and x¯ 2 S ∗; y¯ 2 T ∗: Saddle Point and Optimality Theorem (Convex Optimality with No Gap ) Saddle Point) Let f : S T X and h :Ω T Y be a closed convex conjugate dual pair with no duality gap. Then (x¯; y¯) is a saddle point of L(x; y) if and only if x¯ 2 S ∗ and y¯ 2 T ∗: Proof: We need only to prove that “If x¯ 2 S ∗ and y¯ 2 T ∗, then (x¯; y¯) is a saddle point of L(x; y).” When x¯ 2 S ∗ and y¯ 2 T ∗, we have < x¯; y¯ >= 0. Since there is no duality gap, L(x¯; y¯) = f (x¯)− < x¯; y¯ >= −h(y¯). Hence, −h(y¯) = L(x¯; y¯) = f (x¯) | {z } |{z} infx2S L(x;y¯) supy2Y L(x¯;y) Observations 1. If x¯ 2 S ∗ is known, then y¯ 2 T ∗ can be found by solving max L(x¯; y) = f (x¯)− < x¯; y > s. t. y 2 Y When Y is linearly structures, then it is a linearly programming problem. 2. If y¯ 2 T ∗ is known, then x¯ 2 S ∗ can be found by solving min L(x; y¯) = f (x)− < x; y¯ > s. t. x 2 S Linear conic programming problems • Linear Conic Programming (LCoP) A general conic optimization problem is as follows: (P) minimize cx subject to Ax = b x ∈ where is a closed and convex cone and " " is a linear operator like "inner product." Dual linear conic dual problem Duality theorems for Linear CP Duality theorems for linear CP • Theorem 3 (Conic Duality Theorem): (i) If problems (CP) and (CD) are both feasible, then they have optimal solutions. (ii) If one of the two problems has an interior feasible solution with a finite optimal objective value, then the other one is feasible and has the same optimal objective value. (iii) If one of the two problems is unbounded, then the other has no feasible solution. (iv) If (CP) and (CD) both have interior feasible solutions, then they have optimal solutions with zero duality gap. Examples of conic programs Example of conic programs Example of conic programs Semi-definite Programming (SDP) Duality theorems for SDP Quadratic programming problem Conjugate dual QP Conjugate dual QP Conjugate dual QP .
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