Question: Back in Problem 2 the initial Max simplex tableau, for the objective function is z = 60x1 + 30x2 + 20a3 was z 11

Back in Problem 2 the initial Max simplex tableau, for the objective

Back in Problem 2 the initial Max simplex tableau, for the objective function is z = 60x1 + 30x2 + 20a3 was z 11 12 T3 81 82 83 84 RHS 8 6 1 1 0 0 0 48 4. 2 1.5 1 0 0 20 1.5 0.5 0 1 8. 0 0 1 0 0 0 1 5 1 -60 -30 -20 0 0 0 0 o and had optimal tableau equal to I3 81 82 83 S4 RHS -2 1 2 -8 24 -2 1 -4 8 1 1.25 -0.5 1.5 0 0 1 0 0 0 0 1 1 0 5 0 0 10 10 0 280 I2 is a NBV in this optimal feasible solution (meaning not profitable to produce). find the tableau we would have if we change the objective function to z = 60r, + 35x2 + 20x1 (just raising price of x2) The B-matrix method tells us that if we use the B-1 for the above optimal tableau, then the constraint rows do not change So, we just replace row0 in the above optimal tableau by the new objective function and then fix rowo this tableau should be optimal and should have an alternate bfs Using a simplex step, Find the alternate bfs: (*1, T2, 23) equals ( LO 2.

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Answer Solution Table with X 601 3512 20x3 will be 60 35 20 2 24 20 2 4 8 1 25 05 15 2 15 ... View full answer

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