Answer to Question #272108 in Operations Research for Prathibha Rose

Question

Use big 𝑀 method to solve Minimize 𝑍 = 6𝑥1 + 3𝑥2 + 4𝑥3 𝑆𝑢𝑏𝑗𝑒𝑐𝑡 𝑡𝑜

𝑥1 ≥ 30; 𝑥2 ≥ 50; 𝑥3 ≥ 20; 𝑥1 + 𝑥2 + 𝑥3 = 120

Answer 1

Minimize $𝑍 = 6𝑥_1 + 3𝑥_2 + 4𝑥_3$

𝑆𝑢𝑏𝑗𝑒𝑐𝑡 𝑡𝑜

$𝑥1 ≥ 30$

$𝑥2 ≥ 50$

$𝑥3 ≥ 20$

$𝑥_1 + 𝑥_2 + 𝑥_3 = 120$

After introducing surplus,artificial variables

Max Z=6x₁+3x₂+4x₃+0S₁+0S₂+0S₃-MA₁-MA₂-MA₃-MA₄

subject to x₁-S₁+A₁=30

x₂-S₂+A₂=50

x₃-S₃+A₃=20

x₁+x₂+x₃+A₄=120

and x₁,x₂,x₃,S₁,S₂,S₃,A₁,A₂,A₃,A₄≥0

Negative minimum Z_j-C_j is -2M-6 and its column index is 1. So, the entering variable is x₁.

Minimum ratio is 30 and its row index is 1. So, the leaving basis variable is A₁.

∴ The pivot element is 1.

Entering =x₁, Departing =A₁, Key Element =1

Negative minimum Z_j-C_j is -2M-4 and its column index is 3. So, the entering variable is x₃.

Minimum ratio is 20 and its row index is 3. So, the leaving basis variable is A₃.

∴ The pivot element is 1.

Entering =x₃, Departing =A₃, Key Element =1

Negative minimum Z_j-C_j is -2M-3 and its column index is 2. So, the entering variable is x₂.

Minimum ratio is 50 and its row index is 2. So, the leaving basis variable is A₂.

∴ The pivot element is 1.

Entering =x₂, Departing =A₂, Key Element =1

Negative minimum Z_j-C_j is -M-6 and its column index is 4. So, the entering variable is S₁.

Minimum ratio is 20 and its row index is 4. So, the leaving basis variable is A₄.

∴ The pivot element is 1.

Entering =S₁, Departing =A₄, Key Element =1

Since all Z_j-C_j≥0

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

x₁=50,x₂=50,x₃=20

Max Z=530