Question

7. Use dual simplex method to solve the following LPP. (10)

Min 1 2 2 3 3

z = x + x + x

Subject to

x1 − x2 + x3 ≥

x1 + x2 + 2x3 ≤ 8

x1 − x3 ≥ 2

x1

, x2

, x3 ≥ .0

Answer 1

Min $Z = x_1+2x_2+3x_3$

Subject to :

$x_1 − x_2 + x_3 \ge 2$

$x_1 + x_2 + 2x_3 \le 8$

$x1 − x3 \le 2$

$x_1,x_2,x_3 \ge 0$

After introducing slack,surplus, artificial variables

Max $Z = x_1+2x_2+3x_3+0S_1+0S_2+ 0S_3 -MA_1 - MA_2$

Subject to

$x_1 - x_2+ x_3 - S_1+ A_1 = 2$

$x_1+x_2+2x_3+S_2 = 8$

$x_1-x_3-S_3+A_2 = 2$

and all are greater than zero.

Negative number $Z_j - C_j$ is -$2M-1$ and its column index is 1. So, the entering variable is $x_1$ .

Minimum ratio is 2 and its row index is 1. So, the leaving basis variable is $A_1$ .

Therefore, the pivot element is 1.

Entering = $x_1$ , Departing = A_1 , Key element = 1

$R_1(new) = R_1(old)$

$R_2(new) = R_2(old) - R_1(new)$

$R_3(new) = R_3(old)- R_1(new)$

Negative number $Z_j - C_j$ is $-M-3$ and its column index is 2. So, the entering variable is $x_2$ .

Minimum ratio is 0 and its row index is 3. So, the leaving basis variable is $A_2$ .

Therefore, the pivot element is 1.

Entering = $x_2$ , Departing = $A_2$ , Key element = 1

$R_3(new) = R_3(old)$

$R_1(new) = R_1(old) + R_1(new)$

$R_2(new) = R_2(old)- 2R_3(new)$

Negative number $Z_j - C_j$ is -8 and its column index is 3. So, the entering variable is $x_3$ .

Minimum ratio is 1.2 and its row index is 2. So, the leaving basis variable is $S_2$ .

Therefore, the pivot element is 5.

Entering = $x_3$ , Departing = $S_2$ , Key element = 5

$R_2(new) = \dfrac{R_2(old)}{5}$

$R_1(new) = R_1(old) + R_2(new)$

$R_3(new) = R_3(old)+2R_2(new)$

Since all $Z_j-C_j \ge 0.$

Hence, the optimal solution is arrived with value of variables as:

$x_1 = 3.2$

$x_2 = 2.4$

$x_3 = 1.2$

Max $Z = 11.6$