Lecture 4

# The second x x line is the assumption that x

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Unformatted text preview: ions, oftentimes constraints are linear. In that case GCQ is automatically satisﬁed, so you don’t need to check it (exercise). It is known that the GCQ is the weakest possible condition [1]. 5 Suﬃcient condition The Karush-Kuhn-Tucker theorem provides necessary conditions for optimalSecond Order ity: if the constraint qualiﬁcation holds, then a local solution must satisfy Sufficient Condition the Karush-Kuhn-Tucker conditions (ﬁrst-order conditions and complementary Not needed slackness conditions). Note that the KKT conditions are equivalent to ∇x L(¯, λ, µ) = 0, x (4) where L(x, λ, µ) is the Lagrangian. (4) is the ﬁrst-order necessary condition of the unconstrained minimization problem min L(x, λ, µ). (5) x ∈ RN Below I give a suﬃcient condition for optimality. Proposition 6. Suppose that x is a solution to the unconstrained minimization ¯ problem (5) for some λ ∈ RI and µ ∈ RJ . If gi (¯) ≤ 0 and λi gi (¯) = 0 for all x x + i and hj (¯) = 0 for all j , then x is a solution to the constrained minimization x ¯ problem (1). Proof. Take any x such that gi (x) ≤ 0 for all i and hj (x) = 0 for all j . Then J I µj hj (¯) x λi gi (¯) + x f (¯) = f (¯) + x x j =1 i=1 = L(¯, λ, µ) ≤ L(x, λ, µ) x J I µj hj (x) ≤ f (x). λi gi (x) + = f (x) + i=1 j =1 The ﬁrst line is due to λi gi (¯) = 0 for all i and hj (¯) = 0 for all j . The second x x line is the assumption that x minimizes L(·, λ, µ). The third line is due to λi ≥ 0 ¯ and gi (x) ≤ 0 for all i and hj (x) = 0 for all j . 6 Constrained maximization Finally, we brieﬂy discuss maximization. Although maximization is equivalent to minimization by ﬂipping the sign of the objective function, doing so every time is awkward. So consider the maximization problem maximize f (x) subject to gi (x) ≥ 0 hj (x) = 0 (i = 1, . . . , I ) (j = 1, . . . , J ). (6) 2014W Econ 172B Operations Research (B) Alexis Akira Toda (6) is equivalent to the minimization problem minimize − f (x) subject to − gi (x) ≤ 0 − hj (x) = 0 (i = 1, . . . , I ) (j = 1, . . . , J ). (7) Assuming that x is a local solution and the constraint qualiﬁcation holds, then ¯ the KKT conditions are I − ∇f (¯) − x J λi ∇gi (¯) − x i=1 µj ∇hj (¯) = 0, x (8a) j =1 (∀i) λi (−gi (¯)) = 0. x (8b) But (8) is equivalent to (3). For this reason, it is customary to formulate a maximization problem as in (6) so that the inequality constraints are always “greater than or equal to zero”. As an example, consider a consumer with utility function u(x) = α log x1 + (1 − α) log x2 , where 0...
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## This document was uploaded on 02/18/2014 for the course ECON 172b at UCSD.

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