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Unformatted text preview: 7. GENERAL NONLINEAR PROGRAMMING In this chapter we develop necessary and sufficient conditions for optimality for the general nonlinear programming problem minimize f ( x ) subject to g i ( x ) = 0 , for i = 1 ,...,m, ( NLP ) g i ( x ) , for i = m + 1 ,...,p. 7.1 Equality-Constrained Problems Lagrange Multipliers We begin with the special case of linear-equality constrained problems: ( LEP ) minimize f ( x ) subject to Ax = b. This is the case of ( NLP ) in which g ( x ) = Ax- b and p = m . It provides a simple prototype of the results in this chapter. Theorem 7.1.1. Suppose f : R n R is differentiable at x and assume that A x = b . (a) If x is a local minimizer for ( LEP ) , then f ( x ) range( A T ) . (b) If f is convex and f ( x ) range( A T ) , then x is a global minimizer for ( LEP ) . Observation 7.1.2. The condition f ( x ) range( A T ) can be rewritten as f ( x ) + A T = 0 . In other words, f ( x ) + m X i =1 i a ( i ) = 0 , where A T = [ a (1) ,...,a ( m ) ] . We call i the Lagrange multiplier for the constraint a ( i ) x = b i . Proof of Theorem 7.1.1. Part (a): Assume x is a local minimizer for ( LEP ). To show f ( x ) range( A T ) = null( A ) , we must show that f ( x ) d = 0 whenever Ad = 0. (See Figure 7.1.) Consider d null( A ) and define d ( t ) = f ( x + td ). Because A ( x + td ) = A x + Ad = b + 0 = b , we see that x + td satisfies the constraints of ( LEP ) for all t . Thus t = 0 is a local minimizer for d , and hence 0 = d (0) = f ( x ) d as desired. Part (b): Assume that f is convex, A x = b , and f ( x ) = A T . By convexity, each vector x must satisfy f ( x ) f ( x )+ f ( x ) ( x- x ) = f ( x )+( A T ) ( x- x ) = f ( x )+ A ( x- x ). If x is feasible (so Ax = b = A x ), then this implies that f ( x ) f ( x ) + 0 = f ( x ). Hence x is a global minimizer for ( LEP ). 73 f = 2 . 8 f = 3 . 4 f = 3 . 9 f = 4 . 5 f = 5 . Ax = b x f ( x ) arrownortheast Figure 7.1 Level curves of f with f ( x ) null( A ) Example 7.1.3 (Equality-constrained linear least squares) . Consider the problem of minimizing k Cx- d k subject to Ax = b , which was discussed at the end of Chapter 6. This is equivalent to minimizing f ( x ) = (1 / 2) k Cx- d k 2 subject to Ax = b . We write f ( x ) out as f ( x ) = 1 2 k Cx- d k 2 = 1 2 ( Cx- d ) ( Cx- d ) = 1 2 [ x C T Cx- 2( C T d ) x + d d ] and differentiate to get f ( x ) = C T Cx- C T d and Hf ( x ) = C T C . The necessary and sufficient conditions for optimality are therefore A x = b and C T ( C x- d ) = A T u for some u , which recovers the result found in Section 6.5....
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- Spring '12
- Linear Programming