WUSTL ESE 415 - tutorial5 (4 pages)

Previewing page 1 of 4 page document View the full content.
View Full Document

tutorial5



Previewing page 1 of actual document.

View the full content.
View Full Document
View Full Document

tutorial5

10 views


Pages:
4
School:
Washington University in St. Louis
Course:
Ese 415 - Optimization

Unformatted text preview:

ESE 415 U S Kamilov Tutorial session 2 23 2018 TA Hesam Mazidi Problem 1 Show that the Vandermonde matrix 4 x A x3 x2 x3 x2 x x2 x 1 1 is positive semidefinite PSD but not positive definite PD Solution 1 We note that A is PSD if and only if for any y y1 y2 y3 T we have y T Ay 0 Further A is PD if and only if for any y y1 y2 y3 T 6 0 0 0 T we have y T Ay 0 Notice that for any y y1 y2 y3 T y T Ay y1 x2 y2 x y3 2 2 Thus it immediately follows that A is PSD Now assume y y1 y2 y3 T 6 0 0 0 T In this case we can always find a vector y1 y2 y3 T such that y1 x2 y2 x y3 0 In particular 2 x x2 satisfies this condition concluding that A is not PD as desired Problem2 In each of the following problems fully justify your answer using optimality conditions i Show that f x y x2 4 2 y 2 has two global minima and one stationary point which is neither a local maximum nor a local minimum ii Find all local minima and all local maxima of f x y sin x sin y sin x y within the set 0 x 2 0 y 2 Page 1 of 4 ESE 415 U S Kamilov Tutorial session 2 23 2018 TA Hesam Mazidi Solution 2 i Note that f is a differentiable function with domain R2 This means that any stationary point x y T local maximum global maximum etc must satisfy f x y T 0 That is 2 x 2 4 2x 4 x 3 4x f 0 3 2y 2y The stationary points are easily obtained form eq 3 as x y T 0 0 T 2 0 T or 2 0 T Since f is twice differentiable we can invoke sufficient second order optimality conditions In particular denote H x y T as the Hessian of f obtained at x y T Then if H x y T is positive definite it follows that x y T is a local maximum if H x y T is negative definite x y T is a local minimum H x y T 4 3x 2 4 0 0 2 4 We therefore have the following characterizations 2 0 T is a local minimum since H 2 0 T is positive definite 2 0 T is a local minimum since H 2 0 T is positive definite Note that H 0 0 T is indefinite that is it is neither positive semidefinite nor negative semidefinite By invoking the necessary second order optimality conditions



View Full Document

Access the best Study Guides, Lecture Notes and Practice Exams

Loading Unlocking...
Login

Join to view tutorial5 and access 3M+ class-specific study document.

or
We will never post anything without your permission.
Don't have an account?
Sign Up

Join to view tutorial5 and access 3M+ class-specific study document.

or

By creating an account you agree to our Privacy Policy and Terms Of Use

Already a member?