We have X x z x greaterorequalslant 0 z greaterorequalslant R m n The

# We have x x z x greaterorequalslant 0 z

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We have X = { ( x, z ): x greaterorequalslant 0, z greaterorequalslant 0 } R m + n . The Lagrangian is given by L (( x, z ) , λ )= c T x - λ T ( Ax - z - b )=( c T - λ T A ) x + λ T z + λ T b and has a finite minimum over X if and only if λ Y = { µ R m : c T - µ T A greaterorequalslant 0, µ greaterorequalslant 0 } . 10 2 · Convex and Linear Optimization For λ Y , the minimum of L (( x, z ) , λ ) is attained when both ( c T - λ T A ) x = 0 and λ T z = 0 , and thus g ( λ )= inf ( x,z ) X L (( x, z ) , λ )= λ T b. We obtain the dual max { b T λ : A T λ lessorequalslant c, λ greaterorequalslant 0 } . (2.3) The dual of (2.2) can be determined analogously as max { b T λ : A T λ lessorequalslant c } . 2.4 Complementary Slackness An important relationship between primal and dual solutions is provided by conditions known as complementary slackness . Complementary slackness requires that slack does not occur simultaneously in a variable, of the primal or dual, and the corresponding constraint, of the dual or primal. Here, a variable is said to have slack if its value is non-zero, and an inequality constraint is said to have slack if it does not hold with equality. It is not hard to see that complementary slackness is a necessary condition for optimality. Indeed, if complementary slackness was violated by some variable and the corresponding contraint, reducing the value of the variable would reduce the value of the Lagrangian, contradicting optimality of the current solution. The following result formalizes this intuition. Theorem 2.4 . Let x and λ be feasible solutions for the primal (2.1) and the dual (2.3) , respectively. Then x and λ are optimal if and only if they satisfy complementary slackness, i.e., if ( c T - λ T A ) x = 0 and λ T ( Ax - b )= 0. Proof. If x and λ are optimal, then c T x = λ T b = inf x X ( c T x - λ T ( Ax - b ) ) lessorequalslant c T x - λ T ( Ax - b ) lessorequalslant c T x. Since the first and last term are the same, the two inequalities must hold with equality. Therefore, λ T b = c T x - λ T ( Ax - b ) = ( c T - λ T A ) x + λ T b , and thus ( c T - λ T A ) x = 0 . Furthermore, c T x - λ T ( Ax - b )= c T x , and thus λ T ( Ax - b )= 0 . If on the other hand ( c T - λ T A ) x = 0 and λ T ( Ax - b )= 0 , then c T x = c T x - λ T ( Ax - b )=( c T - λ T A ) x + λ T b = λ T b, and by weak duality x and λ must be optimal. 2.5 · Shadow Prices 11 2.5 Shadow Prices A more intuitive understanding of Lagrange multipliers can be obtained by again viewing (1.1) as a family of problems parameterized by b R m . As before, let φ ( b ) = inf { f ( x ) : h ( x ) = b, x R n } . It turns out that at the optimum, the Lagrange multipliers equal the partial derivatives of φ .  #### You've reached the end of your free preview.

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