L28 - CO350 Linear Programming Chapter 9 The Revised...

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CO350 Linear Programming Chapter 9: The Revised Simplex Method 11th July 2005
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Chapter 9: Revised Simplex Method 1 Example of infeasibility Solve the LP using revised two-phase method with smallest- subscript rules. ( P ) max ( z =) 2 x 1 - x 2 s.t. 3 x 1 + 10 x 2 + 7 x 3 + 4 x 4 = 2 2 x 1 + 5 x 2 + 3 x 3 + x 4 = 3 x 1 , x 2 , x 3 , x 4 0 The auxiliary problem is ( P 0 ) max g T x s.t. Dx = f x 0 where D = 2 4 3 10 7 4 1 0 2 5 3 1 0 1 3 5 , f = 2 4 2 3 3 5 and g = [0 , 0 , 0 , 0 - 1 , - 1] T
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Chapter 9: Revised Simplex Method 2 Iteration 1: B = { 5 , 6 } , x * B = 2 4 2 3 3 5 , A B = 2 4 1 0 0 1 3 5 . Solve A T B y = c B = 2 4 - 1 - 1 3 5 to get y = 2 4 - 1 - 1 3 5 . Compute ¯ c 1 = c 1 - A T 1 y = 0 - [ 3 2 ] y = 5 > 0 . x 1 enters. Solve A B d = A 1 to get d = 2 4 3 2 3 5 . t = min { 2 3 , 3 2 } = 2 3 . x 5 leaves. Update x * 1 = t = 2 3 , x * 6 = 3 - (2)( 2 3 ) = 5 3 . Drop artificial x 5 . Iteration 2: B = { 1 , 6 } , x * B = 2 4 2 / 3 5 / 3 3 5 , A B = 2 4 3 0 2 1 3 5 . Solve
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This note was uploaded on 06/11/2011 for the course C 350 taught by Professor Wolkowicz during the Fall '97 term at Waterloo.

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L28 - CO350 Linear Programming Chapter 9 The Revised...

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