Answer to Question #190708 in Operations Research for Raa

Question

To use the simplex method to solve the following LLP :

Maximise z = 4x₁ + 3x₂

Subject to :

2x₁ + x₂ <= 1000

x₁ + x₂ <= 800

x₁ <= 400

x₂ <= 700

x₁ , x₂ >= 0

Answer 1

Max $Z=0.4x_1+0.3x_2$

subject to-

$2x_1+x_2\le 1000\\ x_1+x_2\le 800\\ x_1\le400\\ x_2\le 700\\ x_1,x_2\ge 0$

After introducing slack variables -

$Max Z=0.4x_1+0.3x_2+0s_1+0s_2+0s_3+0s_4$

subject to-

$2x_1+x_2+s_1=1000\\ x_1+x_2+s_2=800\\ x_1+s_3=400\\ x_2+s_4=700\\$

and $s_2,s_2,s_3,s_4,x_1,x_2\ge 0$

Negative minimum $Z_j-C_j$ is -0.4 and its column index is 1. So the next value is x_1

Minimum ration is 400 and its row index is 3. so the leaving basis variable is $s_3.$

The pivot element is 1. So entering $x_1$ , departing s_3 and key element =1

Since all $Z_j-C_j\ge 0$

Hence ,Optimal solution is arrived with value of $x_1=200,x_2=600$

Max $Z=260$