The problem is:
Maximize 4 X1 + 1 X2 + 5 X3 + 3 X4 subject to 1 X1 - 1 X2 - 1 X3 + 3 X4 <= 1 5 X1 + 1 X2 + 3 X3 + 8 X4 <= 55 -1 X1 + 2 X2 + 3 X3 - 5 X4 <= 3 X1, X2, X3 >= 0 The initial dictionary: X5 = 1 - 1 X1 + 1 X2 + 1 X3 - 3 X4 X6 = 55 - 5 X1 - 1 X2 - 3 X3 - 8 X4 X7 = 3 + 1 X1 - 2 X2 - 3 X3 + 5 X4 ----------------------------------------------- z = 0 + 4 X1 + 1 X2 + 5 X3 + 3 X4
Minimize Y1 + 55 Y2 + 3 Y3 subject to 1 Y1 + 5 Y2 - 1 Y3 >= 4 -1 Y1 + 1 Y2 + 2 Y3 >= 1 -1 Y1 + 3 Y2 + 3 Y3 >= 5 3 Y1 + 8 Y2 - 5 Y3 >= 3 Y1, Y2, Y3 >= 0 In standard form: Maximize -1 Y1 -55 Y2 -3 Y3 subject to -1 Y1 - 5 Y2 + 1 Y3 <= -4 1 Y1 - 1 Y2 - 2 Y3 <= -1 1 Y1 - 3 Y2 - 3 Y3 <= -5 -3 Y1 - 8 Y2 + 5 Y3 <= -3 Y1, Y2, Y3 >= 0Looking at the two optimal solutions:
From the primal: X4 = 5 - 1 X1 - 1 X3 - 2 X5 - 1 X7 X6 = 1 + 5 X1 + 9 X3 + 21 X5 + 11 X7 X2 = 14 - 2 X1 - 4 X3 - 5 X5 - 3 X7 ----------------------------------------------- z = 29 - 1 X1 - 2 X3 - 11 X5 - 6 X7 The optimal solution: 28.999992 X1 = 0 X2 = 14 X3 = 0 X4 = 5 X5 = 0 X6 = 1 X7 = 0 From the dual: Y3 = 6 - 11 Y2 + 3 Y5 + 1 Y7 Y1 = 11 - 21 Y2 + 5 Y5 + 2 Y7 Y4 = 1 - 5 Y2 + 2 Y5 + 1 Y7 Y6 = 2 - 9 Y2 + 4 Y5 + 1 Y7 ----------------------------------------------- z = -29 - 1 Y2 - 14 Y5 - 5 Y7 The optimal solution: -28.999989 Y1 = 11 Y2 = 0 Y3 = 6 Y4 = 1 Y5 = 0 Y6 = 2 Y7 = 0Note: The optimal solution to the dual from the computer program is -29. This means the solution to the original problem is 29 because of the conversion from a minimization to a maximization problem.