Nonlinear-Programming-Bertsekas-solutions (1)

# Nonlinear-Programming-Bertsekas-solutions (1) - Nonlinear...

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Nonlinear Programming 2nd Edition Solutions Manual Dimitri P. Bertsekas Massachusetts Institute of Technology Athena Scientifc, Belmont, Massachusetts 1

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NOTE This solutions manual is continuously updated and improved. Portions of the manual, involving primarily theoretical exercises, have been posted on the internet at the book’s www page http://www.athenasc.com/nonlinbook.html Many thanks are due to several people who have contributed solutions, and particularly to Angelia Nedic, Asuman Ozdaglar, and Cynara Wu. Last Updated: May 2005 2
Section 1.1 Solutions Chapter 1 SECTION 1.1 1.1.9 www For any x, y ∈R n , from the second order expansion (see Appendix A, Proposition A.23) we have f ( y ) f ( x )=( y x ) 0 f ( x )+ 1 2 ( y x ) 0 2 f ( z )( y x ) , (1) where z is some point of the line segment joining x and y . Setting x = 0 in (1) and using the given property of f , it can be seen that f is coercive. Therefore, there exists x n such that f ( x ) = inf x ∈R n f ( x ) (see Proposition A.8 in Appendix A). The condition m || y || 2 y 0 2 f ( x ) y, x, y n , is equivalent to strong convexity of f . Strong convexity guarantees that there is a unique global minimum x . By using the given property of f and the expansion (1), we obtain ( y x ) 0 f ( x m 2 || y x || 2 f ( y ) f ( x ) ( y x ) 0 f ( x M 2 || y x || 2 . Taking the minimum over y n in the expression above gives min y ∈R n ³ ( y x ) 0 f ( x m 2 || y x || 2 ´ f ( x ) f ( x ) min y ∈R n µ ( y x ) 0 f ( x M 2 || y x || 2 . Note that for any a> 0 min y ∈R n ³ ( y x ) 0 f ( x a 2 || y x || 2 ´ = 1 2 a ||∇ f ( x ) || 2 , and the minimum is attained for y = x f ( x ) a . Using this relation for a = m and a = M ,we obtain 1 2 m ||∇ f ( x ) || 2 f ( x ) f ( x ) ≤− 1 2 M ||∇ f ( x ) || 2 . The ±rst chain of inequalities follows from here. To show the second relation, use the expansion (1) at the point x = x , and note that f ( x ) = 0, so that f ( y ) f ( x )= 1 2 ( y x ) 0 2 f ( z )( y x ) . The rest follows immediately from here and the given property of the function f . 3

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Section 1.1 1.1.11 www Since x is a nonsingular strict local minimum, we have that 2 f ( x ) > 0. The function f is twice continuously diFerentiable over < n , so that there exists a scalar δ> 0 such that 2 f ( x ) > 0 , x, with || x x || ≤ δ. This means that the function f is strictly convex over the open sphere B ( x ) centered at x with radius δ . Then according to Proposition 1.1.2, x is the only stationary point of f in the sphere B ( x ). If f is not twice continuously diFerentiable, then x need not be an isolated stationary point. The example function f does not have the second derivative at x = 0. Note that f ( x ) > 0 for x 6 = 0, and by de±nition f (0) = 0. Hence, x = 0 is the unique (singular) global minimum.
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## This note was uploaded on 12/19/2009 for the course INDUSTRIAL ie500 taught by Professor Mathematicalprogrammibg during the Spring '09 term at Bilkent University.

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Nonlinear-Programming-Bertsekas-solutions (1) - Nonlinear...

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