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# L10_10 - ESI 6314 Deterministic Methods in Operations...

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1/30 ESI 6314 Deterministic Methods in Operations Research Lecture Notes University of Florida Department of Industrial and Systems Engineering / REEF

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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 0 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 0 where c T = [3 , 2 , 4 , 0 , 0 , 0] A = 1 1 2 1 0 0 2 0 3 0 1 0 2 1 3 0 0 1 , b = 4 5 7

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4/30 The Revised Simplex Method: Example Initialization: c T = [3 , 2 , 4 , 0 , 0 , 0] A = 1 1 2 1 0 0 2 0 3 0 1 0 2 1 3 0 0 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 = 0 0 0 , 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 0 3 0 0 3 0 1 , A N = 1 1 0 2 0 1 2 1 0 , c B = 4 0 0 , c N = 3 2 0 , ¯ x B = 5 / 3 2 / 3 2

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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 0 0 ⇐⇒ y = 0 4 / 3 0
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 0 2 0 1 2 1 0 = [1 / 3 , 2 , - 4 / 3] and the entering variable is x 2 . To perform the ratio test, we only need ¯ x B and the second column d of the matrix B - 1 A N , which equals to d = B - 1 × (column 2 of A N ) . Denoting the second column of A N by a , we now need to solve the system Bd = a for d : 2 1 0 3 0 0 3 0 1 d 1 d 2 d 3 = 1 0 1 ⇐⇒ d = 0 1 1

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8/30 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 } We compare the ratios of components of ¯ x B and d : d = 0 1 1 ¯ x B = 5 / 3 2 / 3 2 ratios : - 2 / 3 2 x 4 wins .
9/30 z = c T B B - 1 b + ( c T N - c T B B - 1 A N ) x N x B = B -

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L10_10 - ESI 6314 Deterministic Methods in Operations...

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