Use the Revised Simplex to find the optimal value and one op

Use the Revised Simplex to find the optimal value and one optimal solution to the following problem. Update the inverse of the matrix B by using the inverse of the previous matrix. Prove that the optimal value in is correct.

Solution

Canonical Form:

MAXIMIZE: 3 X1 + 1 X2 + 3 X3 + 0 X4 + 0 X5 + 0 X6

2 X1 + 1 X2 + 1 X3 + 1 X4 = 2
1 X1 + 2 X2 + 3 X3 + 1 X5 = 5

2 X1 + 2 X2 + 1 X3 + 1 X6 = 6

X1, X2, X3, X4, X5, X6 0

Simplex Method:

The leaving variable is P4 and entering variable is P1.

The leaving variable is P5 and entering variable is P3

The optimal solution value is Z = 5.4
X1 = 0.2
X2 = 0
X3 = 1.6

Table 1			3	1	3	0	0	0
Base	Cb	P0	P1	P2	P3	P4	P5	P6
P4	0	2	2	1	1	1	0	0
P5	0	5	1	2	3	0	1	0
P6	0	6	2	2	1	0	0	1
Z		0	-3	-1	-3	0	0	0