Revised Simplex Method

[pinki82]

Use the Revised Simplex Method to find the optimal value and one
optimal solution to the problem,
Max ( x_1 - x_3 )
x_1 - 2x_2 - x_3 + x_4 <= 1
-x_1 + 4x_2 - x_3 <= 2
-x_1 - 3x_2 + x_3 <= 3
x_1, x_2, x_3, x_4 >= 0

b) Prove that the optimal value in a) is correct.
c) Why is the solution to the Dual Unique?
d) Why is it not a surprise that the dual solution is Degenerate?
e) Show that although the dual optimal solution is degenerate, there is
only one primal optimal solution.
f) Also find 6 basic Feasible Solutions.





My work: FOr part a, i got optimal value as 4 and optimal solution x_1=4,
x_2 = 3/2 , x_3 = 0 , x_4 = 0

But i don't understnad how to do other ones..
any hint or help on anyone of them please..thanks a lot!

 

[mighty2000]

a) To solve this problem using the Revised Simplex Method, we first need to convert the problem into standard form. This can be done by introducing slack variables and converting all inequalities into equations. The standard form of the problem is:

Max ( x_1 - x_3 )
x_1 - 2x_2 - x_3 + x_4 + x_5 = 1
-x_1 + 4x_2 - x_3 + x_6 = 2
-x_1 - 3x_2 + x_3 + x_7 = 3
x_1, x_2, x_3, x_4, x_5, x_6, x_7 >= 0

Next, we need to create the initial tableau by writing down the coefficients of the variables and the objective function. This will help us in determining the pivot element and performing the necessary operations to reach the optimal solution. The initial tableau for this problem is:

| Basic Variables | x_1 | x_2 | x_3 | x_4 | x_5 | x_6 | x_7 | RHS |
| x_1 | 1 | -2 | -1 | 1 | 0 | 0 | 0 | 1 |
| x_2 | -1 | 4 | -1 | 0 | 1 | 0 | 0 | 2 |
| x_3 | -1 | -3 | 1 | 0 | 0 | 1 | 0 | 3 |
| z | 1 | 0 | -1 | 0 | 0 | 0 | 0 | 0 |

To find the optimal value and optimal solution, we need to perform the Revised Simplex Method. This involves selecting the pivot element, performing row operations to make it the leading entry in its column, and then substituting the values of the basic variables into the objective function to determine the optimal value. The optimal solution is then obtained by setting the remaining non-basic variables to 0.

Following the Revised Simplex Method, the optimal value is 4 and the optimal solution is x_1 = 4, x_2 = 3/2, x_3 = 0, x_4 = 0. This can be verified by substituting these values into the objective function, which gives us 4 as the