2. Consider the following primal LP: max z = - 4x1 - x2 s.t; 4x1 + 3x2 2 6 X1 + 2x2 < 3 3x1 + x2 = 3 x1,x220 After subtr

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2 Consider The Following Primal Lp Max Z 4x1 X2 S T 4x1 3x2 2 6 X1 2x2 3 3x1 X2 3 X1 X220 After Subtr 1 (94.75 KiB) Viewed 38 times

2. Consider the following primal LP: max z = - 4x1 - x2 s.t; 4x1 + 3x2 2 6 X1 + 2x2 < 3 3x1 + x2 = 3 x1,x220 After subtracting an excess variable e from the first constraint, adding a slack variable są to the second constraint, and adding artificial variables a and az to the first and third constraints, the optimal tableau for this primal LP is as shown below. Z ei 1 0 0 0 X1 0 0 1 X2 0 1 0 0 0 0 0 S2 1/5 3/5 -1/5 a1 M 0 0 -1 a2 M-7/5 -1/5 2/5 1 Rhs -18/5 6/5 3/5 0 0 a. Find the dual to this LP and its optimal solution. b. In the primal LP, find the range for b3 (RHS value of the third constraint, currently b3=3) for which the current basis remains optimal. What is the shadow price for this constraint? What would the new z-value be if b3 were 8? c. If we added a new variable xx3 and changed the primal LP to max z = -4x1 X2 X3 s.t; 4x1 + 3x2 + x3 2 6 X1 + 2x2 + x3 <3 3x1 + x2 + x3 = 3 X1, X2, X3 20 would the current optimal solution remain optimal? (HINT: Use the relation between primal optimality and dual feasibility.)