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THE UNIVERSITY OF NEW SOUTH WALES, SYDNEY SCHOOL OF MATHEMATICS AND STATISTICS MATH3161/MATH5165 — OPTIMIZATION – Term 1, 2019 Problem Sheet 4 Solutions – Unconstrained Problems 1. Find all the stationary points of the following functions and identify, if possible, the local minimizers, local maximizers and saddle points. (a) f ( x ) = x 2 1 + x 2 2 + x 2 3 + x 1 x 2 + x 1 x 3 + x 2 x 3 - 7 x 1 - 8 x 2 - 9 x 3 + 101 Answer The gradient and Hessian of f are f ( x ) = 2 x 1 + x 2 + x 3 - 7 x 1 + 2 x 2 + x 3 - 8 x 1 + x 2 + 2 x 3 - 9 , 2 f ( x ) = 2 1 1 1 2 1 1 1 2 The linear system f ( x ) = 0 corresponds to the augmented matrix 2 1 1 7 1 2 1 8 1 1 2 9 R 1 R 3 1 1 2 9 1 2 1 8 2 1 1 7 R 2 R 2 - R 1 R 3 R 3 - 2 R 1 1 1 2 9 0 1 - 1 - 1 0 - 1 - 3 - 11 R 3 R 3 + R 1 1 1 2 9 0 1 - 1 - 1 0 0 - 4 - 12 upon using row operations to reduce it to row-echelon form. Back-substitution now gives - 4 x 3 = - 12 = x 3 = 3 x 2 - x 3 = - 1 = x 2 = 2 x 1 + x 2 + 2 x 3 = 9 = x 1 = 1 As the Hessian 2 f ( x * ) is positive definite (eigenvalues 1 , 1 , 4), the stationary point x * = [ 1 2 3 ] T is a strict local minimizer of f . Moreover as the Hessian does not depend on the variables x it is positive definite for all x R 3 , so f is strictly convex on R 3 . Hence x * is the unique global minimizer of f over R 3 . Note that, for the 3 by 3 matrix G = 2 f ( x ), det ( G - λI ) = 2 - λ 1 1 1 2 - λ 1 1 1 2 - λ = - λ 3 +6 λ 2 - 9 λ +4 = - ( λ - 1) 2 ( λ - 4) = 0 gives the eigenvalues. (b) f ( x ) = x 1 ( x 2 - 1) + x 3 ( x 2 3 - 3) 1
Answer The gradient and Hessian of f are f ( x ) = x 2 - 1 x 1 3 x 2 3 - 3 , 2 f ( x ) = 0 1 0 1 0 0 0 0 6 x 3 The stationary points are the solutions to f ( x ) = 0 , giving x * = 0 1 1 , x = 0 1 - 1 .

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