# HW 1 - ENGR62/MS&amp;amp;amp;E111 Spring 2008 Introduction...

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Unformatted text preview: ENGR62/MS&amp;E111 Spring 2008 Introduction to Optimization April 11, 2008 Prof. Ben Van Roy Homework Assignment 1: Solutions Network Routing See the Excel solution worksheet on the course web page. Linear Algebra Review (Chapter 2) Problem 1 Expanding out the first four expressions, we have: a T x = 0 x 2 =- 1 2 x 1 (1) a T x = 1 x 2 =- 1 2 x 1 + 1 2 (2) b T x = 0 x 2 =- 2 x 1 (3) b T x = 1 x 2 =- 2 x 1 + 1 (4) Each of these corresponds to a line in the plane. Expanding out expressions 5 and 6, we have: [ a | b ] x = [0 , 1] T 1 2 2 1 x 1 x 2 = 1 (5) [ a | b ] x = [0 , 2] T 1 2 2 1 x 1 x 2 = 2 (6) Each of these is a system with a unique solution. Using pencil and paper we get that x = 2 3- 1 3 and x = 4 3- 2 3 are their respective solutions. We thus obtain the following graph:- 6 x 1 x 2 1: a T x =0 2: a T x =1 3: b T x =0 4: b T x =1 5 6 where the shaded region represents all x R 2 such that a T x 0. 1 Problem 2 Part 1 Finding a b such that Ax = b has no solution is equivalent to finding a vector, b R 3 , such that b is not in C ( A ). One easy choice is: b = 1 (7) The only way to get the first two b values in some linear combination of the columns of A is to set x 1 = 0 ,x 2 = 0. But, this leads to 0 = 1 in the third row equation. So, Ax = b cannot have a solution in this case. Part 2 Finding a non-zero b such that Ax = b has a solution is equivalent to finding a vector, b R 3 , such that b is in C ( A ). One easy choice is: b = 1 2 (8) which leads to a solution of x = 1 ....
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## HW 1 - ENGR62/MS&amp;amp;amp;E111 Spring 2008 Introduction...

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