Hw6_Sol

# Hw6_Sol - Math464 HW 6 Due on Thursday Feb 25 1 Linear...

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Unformatted text preview: Math464 - HW 6 Due on Thursday, Feb 25 1 Linear Optimization (Spring 2010) Brief solutions to Homework 6 1. To get a contradiction, assume that there is a point x ∈ S with f ( x ) < f ( x * ). Notice that x has to be non-local to x * , i.e., k x- x * k > , else it will violate the fact that x * is a local optimum. The line segment joining x * and x will lie inside S (as S is convex), but will partly be outside the sphere of radius centered at x * . Consider a point y on this line segment that is inside this sphere, i.e., k y- x * k < . Thus, y is a convex combination of x * and x , and is also local to x * . Since it is local to x * , we must have f ( y ) ≥ f ( x * ) , (1) by the definition of local minimum. With y = λ x * +(1- λ ) x for some 0 < λ < 1, we use the fact that f is a convex function to write f ( y ) = f ( λ x * + (1- λ ) x ) ≤ λf ( x * ) + (1- λ ) f ( x ) . (2) Using the result in Equation (1), we get λf ( x * ) + (1- λ ) f ( x ) ≥ f ( y ) ≥ f ( x * ) ,...
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## This note was uploaded on 02/15/2011 for the course EECS 6.231 taught by Professor Bertsekas during the Spring '10 term at MIT.

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Hw6_Sol - Math464 HW 6 Due on Thursday Feb 25 1 Linear...

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