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Unformatted text preview: 1/30 ESI 6314 Deterministic Methods in Operations Research Lecture Notes University of Florida Department of Industrial and Systems Engineering / REEF 2/30 The Revised Simplex Method The revised simplex method carries out the steps of simplex method in the matrix form. z = c T B B 1 b + ( c T N c T B B 1 A N ) x N x B = B 1 b B 1 A N x N Instead of computing B 1 b directly, we can solve the system By = b to find y = B 1 b . Instead of computing c T B B 1 A N , we can solve the system c T B = v T B (which is equivalent to B T v = c B ) to find v T = c T B B 1 , and then compute v T A N to obtain c T B B 1 A N . 3/30 The Revised Simplex Method: Example Consider the LP max 3 x 1 + 2 x 2 + 4 x 3 s . t . x 1 + x 2 + 2 x 3 4 2 x 1 + 3 x 3 5 2 x 1 + x 2 + 3 x 3 7 x 1 , x 2 , x 3 After introducing the slack variables x 4 , x 5 , x 6 , we can write this problem in the form max z = c T x s . t . Ax = b x where c T = [3 , 2 , 4 , , , 0] A = 1 1 2 1 2 3 1 2 1 3 1 , b = 4 5 7 4/30 The Revised Simplex Method: Example Initialization: c T = [3 , 2 , 4 , , , 0] A = 1 1 2 1 2 3 1 2 1 3 1 , b = 4 5 7 BV = { x 4 , x 5 , x 6 } , NV = { x 1 , x 2 , x 3 } , x B = x 4 x 5 x 6 , x N = x 1 x 2 x 3 , c B = , c N = 3 2 4 , x B = 4 5 7 Iteration 1: Entering variable: x 3 Ratio test: we need to check the ratios of components of x B and the third column of A : 2 3 3 4 5 7 ratios : 2 5 / 3 7 / 3 x 5 wins . 5/30 z = c T B B 1 b + ( c T N c T B B 1 A N ) x N x B = B 1 b B 1 A N x N BV = { x 3 , x 4 , x 6 } , NV = { x 1 , x 2 , x 5 } , New x B = [ x 3 , x 4 , x 6 ] T : x 3 = 5 / 3 (min ratio value) x 4 = 4 2(5 / 3) = 2 / 3 x 6 = 7 3(5 / 3) = 2 . B = 2 1 3 3 1 , A N = 1 1 2 1 2 1 , c B = 4 , c N = 3 2 , x B = 5 / 3 2 / 3 2 6/30 z = c T B B 1 b + ( c T N c T B B 1 A N ) x N x B = B 1 b B 1 A N x N BV = { x 3 , x 4 , x 6 } NV = { x 1 , x 2 , x 5 } Iteration 2: To find the entering variable, we need to compute c T N = c T N c T B B 1 A N Instead of computing c T B B 1 directly, we can denote by y T = c T B B 1 and then find y by solving the system y T B = c T B , or, equivalently B T y = c B . We have 2 3 3 1 0 0 0 0 1 y 1 y 2 y 3 = 4 y = 4 / 3 7/30 z = c T B B 1 b + ( c T N c T B B 1 A N ) x N x B = B 1 b B 1 A N x N BV = { x 3 , x 4 , x 6 } NV = { x 1 , x 2 , x 5 } c T N = c T N c T B B 1 A N = [3 , 2 , 0] [0 , 4 / 3 , 0] 1 1 2 1 2 1 = [1 / 3 , 2 , 4 / 3] and the entering variable is x 2 ....
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This note was uploaded on 12/14/2010 for the course ESI 6314 taught by Professor Vladimirlboginski during the Fall '09 term at University of Florida.
 Fall '09
 VLADIMIRLBOGINSKI
 Operations Research, Systems Engineering

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