Given the following problem min Z = 7x_1 - 3x_3 s.t. 5x_1 - x_3 = -16 4x_1 + 9x_

Question

Given the following problem min Z = 7x_1 - 3x_3 s.t. 5x_1 - x_3 = -16 4x_1 + 9x_2 lessthanorequalto 15 2x_1 + 3x_2 + x_3 lessthanorequalto 21 x_1 greaterthanorequalto 0, x_2 greaterthanorequalto 0, x_3 greaterthanorequalto 8 Give the initial simplex tableau, presented as a single matrix that includes the right-hand side (Use the big-M method if appropriate; in that case keep M, don't replace it by an actual large number). Specify which variables serve as the initial basic variables (at the start of the Simplex method). And specify which variable will be the first entering basic variable. Without formulating the dual problem, by inspection of the primal problem, state whether the dual is unbounded, the dual is infeasible, the dual and primal have an equal objective value in the optimum, Furthermore, state the property on which you base this choice and explain why that property holds in general. Make only one substitution in the given formulation. Then give the dual problem formulation directly from that. So do not rewrite to standard form; also do not use the results from parts a, b, or c.

Explanation / Answer

Solution : Part a)

The leaving variable is P6 and entering variable is P3.

Pivot row (Row 1):
16 / 1 = 16
-5 / 1 = -5
0 / 1 = 0
1 / 1 = 1
0 / 1 = 0
0 / 1 = 0
1 / 1 = 1

Row 2:
15 - (0 * 16) = 15
4 - (0 * -5) = 4
9 - (0 * 0) = 9
0 - (0 * 1) = 0
1 - (0 * 0) = 1
0 - (0 * 0) = 0
0 - (0 * 1) = 0

Row 3:
21 - (1 * 16) = 5
2 - (1 * -5) = 7
3 - (1 * 0) = 3
1 - (1 * 1) = 0
0 - (1 * 0) = 0
1 - (1 * 0) = 1
0 - (1 * 1) = -1

Row Z:
-16 - (-1 * 16) = 0
5 - (-1 * -5) = 0
0 - (-1 * 0) = 0
-1 - (-1 * 1) = 0
0 - (-1 * 0) = 0
0 - (-1 * 0) = 0
0 - (-1 * 1) = 1

There is any possible solution for the problem, so we can continue to Phase II to calculate it.

Remove the columns corresponding to artificial variables.

Modify the row of the objective function for the original problem.

Calculate the Z line:
-(0) + (3 * 16) + (0 * 15) + (0 * 5) = 48
-(-7) + (3 * -5) + (0 * 4) + (0 * 7) = -8
-(0) + (3 * 0) + (0 * 9) + (0 * 3) = 0
-(3) + (3 * 1) + (0 * 0) + (0 * 0) = 0
-(0) + (3 * 0) + (0 * 1) + (0 * 0) = 0
-(0) + (3 * 0) + (0 * 0) + (0 * 1) = 0

The leaving variable is P5 and entering variable is P1.

Pivot row (Row 3):
5 / 7 = 0.71428571428571
7 / 7 = 1
3 / 7 = 0.42857142857143
0 / 7 = 0
0 / 7 = 0
1 / 7 = 0.14285714285714

Row 1:
16 - (-5 * 0.71428571428571) = 19.571428571429
-5 - (-5 * 1) = 0
0 - (-5 * 0.42857142857143) = 2.1428571428571
1 - (-5 * 0) = 1
0 - (-5 * 0) = 0
0 - (-5 * 0.14285714285714) = 0.71428571428571

Row 2:
15 - (4 * 0.71428571428571) = 12.142857142857
4 - (4 * 1) = 0
9 - (4 * 0.42857142857143) = 7.2857142857143
0 - (4 * 0) = 0
1 - (4 * 0) = 1
0 - (4 * 0.14285714285714) = -0.57142857142857

Row Z:
48 - (-8 * 0.71428571428571) = 53.714285714286
-8 - (-8 * 1) = 0
0 - (-8 * 0.42857142857143) = 3.4285714285714
0 - (-8 * 0) = 0
0 - (-8 * 0) = 0
0 - (-8 * 0.14285714285714) = 1.1428571428571

The optimal solution value is Z = -53.714285714286
X1 = 0.71428571428571
X2 = 0
X3 = 19.571428571429

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