Question

Use big 𝑀 method solve. 𝑀𝑖𝑛𝑖𝑚𝑖𝑧𝑒 𝑍 = 2𝑦1 + 4𝑦2 𝑆𝑢𝑏𝑗𝑒𝑐𝑡 𝑡𝑜 2𝑦1 – 3𝑦2 ≥ 2,

−𝑦1 + 𝑦2 ≥ 3; 𝑦1

, 𝑦2 ≥ 0

Answer 1

𝑀𝑖𝑛𝑖𝑚𝑖𝑧𝑒 $𝑍 = 2𝑦_1 + 4𝑦_2$

𝑆𝑢𝑏𝑗𝑒𝑐𝑡 𝑡𝑜

$2𝑦_1 – 3𝑦_2 ≥ 2$

$−𝑦_1 + 𝑦_2 ≥ 3$

$𝑦_1, 𝑦_2 ≥ 0$

After introducing surplus,artificial variables

Min Z=2x₁+4x₂+0S₁+0S₂+MA₁+MA₂

subject to

2x₁-3x₂-S₁+A₁=2-x₁+x₂-S₂+A₂=3

and x₁,x₂,S₁,S₂,A₁,A₂≥0

Positive maximum Z_j-C_j is M-2 and its column index is 1. So, the entering variable is x₁.

Minimum ratio is 1 and its row index is 1. So, the leaving basis variable is A₁.

∴ The pivot element is 2.

Entering =x₁, Departing =A₁, Key Element =2

Since all Z_j-C_j≤0

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

x₁=1,x₂=0

Min Z=2

But this solution is not feasible

because the final solution violates the 2nd constraint -  x₁ +  x₂ ≥ 3.

and the artificial variable A₂ appears in the basis with positive value 4