Question

Consider the following problem

    Maximize Z=2x₁ + 5x₂

subject to

               4x₁+ 2x₂ ≤ 6

                x₁ + x₂ ≥ 2

                x_{i ≥}0 for i=1,2

1. Inserting slack, excess, or artificial variables, construct the initial simplex tableau.
2. Identify the corresponding initial (artificial) basic feasible solution including the objective function value.
3. Identify the entering basic variable and the leaving basic variable for the next iteration.

Answer 1

Maximize Objective function Z=2x₁ + 5x₂

subject to constraints:

               4x₁+ 2x₂ ≤ 6

                x₁ + x₂ ≥ 2

                x_{i ≥}0 for i=1,2

I have converted this problem to canonical form, here, I have added slack, surplus and artificial variables as needed

slack variable X3 is added as As the constraint 1 includes sign '≤'

surplus variable X4 and artificial variable X5 are added as constraint 2 includes sign '≥'

First tableau of Phase I

leaving variable is P3

entering variable is P1

Intermediate operations

Pivot row (Row 1):
6 / 4 = 1.5
4 / 4 = 1
2 / 4 = 0.5
1 / 4 = 0.25
0 / 4 = 0
0 / 4 = 0

Row 2:
2 - (1 * 1.5) = 0.5
1 - (1 * 1) = 0
1 - (1 * 0.5) = 0.5
0 - (1 * 0.25) = -0.25
-1 - (1 * 0) = -1
1 - (1 * 0) = 1

Row Z:
-2 - (-1 * 1.5) = -0.5
-1 - (-1 * 1) = 0
-1 - (-1 * 0.5) = -0.5
0 - (-1 * 0.25) = 0.25
1 - (-1 * 0) = 1
0 - (-1 * 0) = 0

Leaving variable is P5
Entering variable is P2

Intermediate calculations

Pivot row (Row 2):
0.5 / 0.5 = 1
0 / 0.5 = 0
0.5 / 0.5 = 1
-0.25 / 0.5 = -0.5
-1 / 0.5 = -2
1 / 0.5 = 2

Row 1:
1.5 - (0.5 * 1) = 1
1 - (0.5 * 0) = 1
0.5 - (0.5 * 1) = 0
0.25 - (0.5 * -0.5) = 0.5
0 - (0.5 * -2) = 1
0 - (0.5 * 2) = -1

Row Z:
-0.5 - (-0.5 * 1) = 0
0 - (-0.5 * 0) = 0
-0.5 - (-0.5 * 1) = 0
0.25 - (-0.5 * -0.5) = 0
1 - (-0.5 * -2) = 0
0 - (-0.5 * 2) = 1

Phase II

Intermediate calculations

Remove the columns corresponding to artificial variables.

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

Calculate the Z line:
-(0) + (2 * 1) + (5 * 1) = 7
-(2) + (2 * 1) + (5 * 0) = 0
-(5) + (2 * 0) + (5 * 1) = 0
-(0) + (2 * 0.5) + (5 * -0.5) = -1.5
-(0) + (2 * 1) + (5 * -2) = -8

Leaving variable is P1
Entering variable is P4

Intermediate calculations

Pivot row (Row 1):
1 / 1 = 1
1 / 1 = 1
0 / 1 = 0
0.5 / 1 = 0.5
1 / 1 = 1

Row 2:
1 - (-2 * 1) = 3
0 - (-2 * 1) = 2
1 - (-2 * 0) = 1
-0.5 - (-2 * 0.5) = 0.5
-2 - (-2 * 1) = 0

Row Z:
7 - (-8 * 1) = 15
0 - (-8 * 1) = 8
0 - (-8 * 0) = 0
-1.5 - (-8 * 0.5) = 2.5
-8 - (-8 * 1) = 0

Optimal solution Objective Function Z = 15
X1 = 0
X2 = 3