Question

Let’s assume the following LP model. Provide the optimal values of the decision variables (X1 and X2) and the optimal value of the objective function. Show your work as much as you can or send a picture of your work if you want to get partial points when your final answers are not correct.

objective function: Min 2 X1+X2
operational constraints: X1_>20; X2>_10
non-negativity constraints:X1,X2_>0

Answer 1

Using Graphical Method,

MINIMIZE: Z = 2 X1 + 1 X2
1 X1 + 0 X2 ≥ 20
0 X1 + 1 X2 ≥ 10
X1, X2 ≥ 0

The problem is unbounded but since it is a minimizing problem can find a solution.

Graph

Thus, X1 = 20 and X2 = 10, So Objective Function is = 2 X1 + 1 X2 = 2*20 + 1*10 = 50,

Using Two Phase simplex method:

The problem is converted to canonical form by adding slack, surplus and artificial variables as appropiate (show/hide details)

• As the constraint 1 is of type '≥' we should add the surplus variable X3 and the artificial variable X5.
• As the constraint 2 is of type '≥' we should add the surplus variable X4 and the artificial variable X6

We'll build the first tableau of Phase I from Two Phase Simplex method.

The leaving variable is P5 and entering variable is P1.

Intermediate operations

Pivot row (Row 1):
20 / 1 = 20
1 / 1 = 1
0 / 1 = 0
-1 / 1 = -1
0 / 1 = 0
1 / 1 = 1
0 / 1 = 0

Row 2:
10 - (0 * 20) = 10
0 - (0 * 1) = 0
1 - (0 * 0) = 1
0 - (0 * -1) = 0
-1 - (0 * 0) = -1
0 - (0 * 1) = 0
1 - (0 * 0) = 1

Row Z:
-30 - (-1 * 20) = -10
-1 - (-1 * 1) = 0
-1 - (-1 * 0) = -1
1 - (-1 * -1) = 0
1 - (-1 * 0) = 1
0 - (-1 * 1) = 1
0 - (-1 * 0) = 0

The leaving variable is P6 and entering variable is P2.

Intermediate operations:

Pivot row (Row 2):
10 / 1 = 10
0 / 1 = 0
1 / 1 = 1
0 / 1 = 0
-1 / 1 = -1
0 / 1 = 0
1 / 1 = 1

Row 1:
20 - (0 * 10) = 20
1 - (0 * 0) = 1
0 - (0 * 1) = 0
-1 - (0 * 0) = -1
0 - (0 * -1) = 0
1 - (0 * 0) = 1
0 - (0 * 1) = 0

Row Z:
-10 - (-1 * 10) = 0
0 - (-1 * 0) = 0
-1 - (-1 * 1) = 0
0 - (-1 * 0) = 0
1 - (-1 * -1) = 0
1 - (-1 * 0) = 1
0 - (-1 * 1) = 1

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

Intermediate operations :

Remove the columns corresponding to artificial variables.

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

Calculate the Z line:
-(0) + (-2 * 20) + (-1 * 10) = -50
-(-2) + (-2 * 1) + (-1 * 0) = 0
-(-1) + (-2 * 0) + (-1 * 1) = 0
-(0) + (-2 * -1) + (-1 * 0) = 2
-(0) + (-2 * 0) + (-1 * -1) = 1

The optimal solution value is Z = 50
X1 = 20
X2 = 10