Unformatted text preview: ) / 3 10 M/ 3 + 41/31 12/3 10/3 1 2/3 1/3 4/3 This is an optimal tableau, but a 1 is still positive, so the original problem is not feasible. #4, 4.12: Twophase simplex method. Initial Phase I tableau is w x 1 x 2 e 1 a 1 a 2 RHS 1 5 31 10 2 11 1 6 3 2 1 4 Now enter x 1 in the basis in row 2. w x 1 x 2 e 1 a 1 a 2 RHS 11/315/3 10/31/31 12/3 10/3 1 2/3 1/3 4/3 This is an optimal tableau to phase I, but there is still an arti±cial variable a 1 which is positive, so the original problem is not feasible. Note, in this case there is no phase II....
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 Spring '08
 Arkin,E
 Optimization, X1, Simplex algorithm

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