final09sols

final09sols - L. Vandenberghe EE103 12/10/09 Final Exam...

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L. Vandenberghe 12/10/09 EE103 Final Exam Solutions Problem 1 (10 points). Let P be the set of vectors x = ( x 1 , x 2 , x 3 , x 4 ) in R 4 that satisfy the conditions x 2 + x 3 + x 4 = 1 , 2 x 1 + x 3 = 1 , x 1 0 , x 2 0 , x 3 0 , x 4 0 . 1. Write P as a the solution set of linear inequalities Ax b . 2. Which of the following three vectors are extreme points of P ? x = (1 / 2 , 0 , 0 , 1) , x = (0 , 0 , 1 , 0) , x = (1 / 4 , 1 / 2 , 1 / 2 , 0) . Solution. 1. A = 0 1 1 1 0 1 1 1 2 0 1 0 2 0 1 0 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 , b = 1 1 1 1 0 0 0 0 . 2. The matrix of active constraints at x = (1 / 2 , 0 , 0 , 1) is 0 1 1 1 0 1 1 1 2 0 1 0 2 0 1 0 0 1 0 0 0 0 1 0 . This matrix has a zero nullspace. Therefore x is an extreme point. The matrix of active constraints at x = (0 , 0 , 1 , 0) is 0 1 1 1 0 1 1 1 2 0 1 0 2 0 1 0 1 0 0 0 0 1 0 0 0 0 0 1 . This matrix has a zero nullspace. Therefore x is an extreme point.
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The matrix of active constraints at x = (1 / 4 , 1 / 2 , 1 / 2 , 0) is 0 1 1 1 0 1 1 1 2 0 1 0 2 0 1 0 0 0 0 1 . This matrix has a nonzero nullspace. For example, the vector (1 / 2 , 1 , 1 , 0) is in the nullspace. Therefore x is not an extreme point. Problem 2 (10 points). The derivative of a function f at a point ˆ x can be approximated as f x ) f x + h ) f x h ) 2 h for small positive h . The righthand side is known as a fnite-diFerence approximation
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This note was uploaded on 02/03/2011 for the course EE 103 taught by Professor Vandenberghe,lieven during the Spring '08 term at UCLA.

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final09sols - L. Vandenberghe EE103 12/10/09 Final Exam...

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