The problem is:
Maximize 2 X1 - 1 X2 + 3 X3 subject to 2 X1 + 4 X2 - 1 X3 <= 3 1 X1 - 1 X2 + 0 X3 <= 5 2 X1 + 1 X2 + 1 X3 <= 2 X1, X2, X3 >= 0
Question 1: What is the dual?
Question 2: Restate the dual in our standard form.
Question 3: Solve the primal and dual problems using the Simplex method.
Question 4: Explain how to get the primal solution from the dual and the dual solution from the primal.
Solution 1: What is the dual?Minimize 3 Y1 + 5 Y2 + 2 Y3 subject to 2 Y1 + 1 Y2 + 2 Y3 >= 2 4 Y1 - 1 Y2 + 1 Y3 >= -1 -1 Y1 + 0 Y2 + 1 Y3 >= 3 Y1, Y2, Y3 >= 0
Solution 2: Restate the dual in our standard formMaximize -3 Y1 - 5 Y2 - 2 Y3 subject to -2 Y1 - 1 Y2 - 2 Y3 <= -2 -4 Y1 + 1 Y2 - 1 Y3 <= 1 1 Y1 - 0 Y2 - 1 Y3 <= -3 Y1, Y2, Y3 >= 0
Solution 3: Solve the primal problem.X4 = 3 - 2 X1 - 4 X2 + 1 X3 X5 = 5 - 1 X1 + 1 X2 + 0 X3 X6 = 2 - 2 X1 - 1 X2 - 1 X3 ------------------------------------------------- z = -0 + 2 X1 - 1 X2 + 3 X3 X3 enters. X6 leaves. z = 0 After 1 pivot: X4 = 5 - 4 X1 - 5 X2 - 1 X6 X5 = 5 - 1 X1 + 1 X2 + 0 X6 X3 = 2 - 2 X1 - 1 X2 - 1 X6 ------------------------------------------------- z = 6 - 4 X1 - 4 X2 - 3 X6 The optimal solution: 6 X1 = 0 X2 = 0 X3 = 2 X4 = 5 X5 = 5 X6 = 0 Solution 3: Solve the dual problem. Phase 1: Before pivoting to make feasible. Y4 = -2 + 2 Y1 + 1 Y2 + 2 Y3 + 1 Y0 Y5 = 1 + 4 Y1 - 1 Y2 + 1 Y3 + 1 Y0 Y6 = -3 - 1 Y1 + 0 Y2 + 1 Y3 + 1 Y0 ------------------------------------------------- z = -0 + 0 Y1 + 0 Y2 + 0 Y3 - 1 Y0 Y0 enters. Y6 leaves. z = 0 The initial feasible dictionary for phase 1: Y4 = 1 + 3 Y1 + 1 Y2 + 1 Y3 + 1 Y6 Y5 = 4 + 5 Y1 - 1 Y2 + 0 Y3 + 1 Y6 Y0 = 3 + 1 Y1 + 0 Y2 - 1 Y3 + 1 Y6 ------------------------------------------------- z = -3 - 1 Y1 + 0 Y2 + 1 Y3 - 1 Y6 Y3 enters. Y0 leaves. z = -3 After 1 pivot: Y4 = 4 + 4 Y1 + 1 Y2 + 2 Y6 - 1 Y0 Y5 = 4 + 5 Y1 - 1 Y2 + 1 Y6 + 0 Y0 Y3 = 3 + 1 Y1 + 0 Y2 + 1 Y6 - 1 Y0 ------------------------------------------------- z = 0 + 0 Y1 + 0 Y2 + 0 Y6 - 1 Y0 The optimal solution: 0 Y1 = 0 Y2 = 0 Y3 = 3 Y4 = 4 Y5 = 4 Y6 = 0 The initial dictionary: Y4 = 4 + 4 Y1 + 1 Y2 + 2 Y6 Y5 = 4 + 5 Y1 - 1 Y2 + 1 Y6 Y3 = 3 + 1 Y1 + 0 Y2 + 1 Y6 ------------------------------------------------- z = -6 - 5 Y1 - 5 Y2 - 2 Y6 The optimal solution: -6 Y1 = 0 Y2 = 0 Y3 = 3 Y4 = 4 Y5 = 4 Y6 = 0
Solution 4: Explain how to get the primal solution from the dual and the dual solution from the primal.The final dictionary for the primal problem: X4 = 5 - 4 X1 - 5 X2 - 1 X6 X5 = 5 - 1 X1 + 1 X2 + 0 X6 X3 = 2 - 2 X1 - 1 X2 - 1 X6 ------------------------------------------------- z = 6 - 4 X1 - 4 X2 - 3 X6 The optimal solution: 6 X1 = 0 X2 = 0 X3 = 2 X4 = 5 X5 = 5 X6 = 0 Answer: look at the z row of final dictionary Y1 = -1 * (coeff. of first slack, X4)= 0 Y2 = -1 * (coeff. of second slack, X5)= 0 Y3 = -1 * (coeff. of third slack, X6)= 3
Y4 = 4 + 4 Y1 + 1 Y2 + 2 Y6 Y5 = 4 + 5 Y1 - 1 Y2 + 1 Y6 Y3 = 3 + 1 Y1 + 0 Y2 + 1 Y6 ------------------------------------------------- z = -6 - 5 Y1 - 5 Y2 - 2 Y6 The optimal solution: -6 Y1 = 0 Y2 = 0 Y3 = 3 Y4 = 4 Y5 = 4 Y6 = 0 Answer: look at the z row of final dictionary X1 = -1 * (coeff. of first slack, Y4)= 0 X2 = -1 * (coeff. of second slack, Y5)= 0 X3 = -1 * (coeff. of third slack, Y6)= 2The primal solution is : (0, 0, 2) The dual solution is: (0, 0, 3)
We know that both of these are optimal solutions because: