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Unformatted text preview: CS 435 : Linear Optimization Fall 2008 Lecture 2: Linear Algebra: Solving Ax = b via Gaussian Elimination Lecturer: Sundar Vishwanathan Computer Science & Engineering Indian Institute of Technology, Bombay 1 Formulation A linear optimization problem can be formulated as max c T x Ax b (1) where, A is an m ¢ n matrix, c a n ¢ 1 vector, b a m ¢ 1 vector and x a n ¢ 1 vector. We are given as input c , A , and b . As output, among all x that satis es Ax b , we wish to nd one which maximises c T x . Our rst goal is to understand the set f x : Ax b g . We begin with a simpler set f x : Ax = b g . Before we begin to understand what this set looks like, we recall procedures we have learnt for nding at least one solution or recognising instances which do not have solutions. 2 Solving Ax = b Ax = b is short hand notation for the following set of equalities given below. A 11 x 1 + A 12 x 2 + A 13 x 3 + ¡ ¡ ¡ + A 1 n x n = b 1 A 21 x 1 + A 22 x 2 + A 23 x 3 + ¡ ¡ ¡ + A 2 n x n = b 2 . . . A m 1 x 1 + A m 2 x 2 + A m 3 x 3 + ¡ ¡ ¡ + A mn x n = b m We can solve such a system of equations using Gaussian Elimination. Here is an example. Example 1 2 x + 7 y = 13 (I) x + 3 y = 4 (II) Replacing II by - 1 2 ¡ I + II gives 7 2 y + 3 y = 13 2 + 4 y = 5 (2) Now one can solve for x using the rst equation. 3 Gaussian Elimination Gaussian Elimination starts with a set of equations. There are two phases. Both phases are iterative. We discuss the rst phase which is the crux. In each step of the rst phase these set of equations are 1 2 replaced with an equivalent set of equations. By equivalent, we mean that the solution sets of the two sets are the same. We also make sure that the new set is simpler to analyse....
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