Use big 𝑀 method to solve Minimize 𝑍 = 6𝑥1 + 3𝑥2 + 4𝑥3 𝑆𝑢𝑏𝑗𝑒𝑐𝑡 𝑡𝑜
𝑥1 ≥ 30; 𝑥2 ≥ 50; 𝑥3 ≥ 20; 𝑥1 + 𝑥2 + 𝑥3 = 120
Minimize
𝑆𝑢𝑏𝑗𝑒𝑐𝑡 𝑡𝑜
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 Zj-Cj 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 Zj-Cj 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 Zj-Cj 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 Zj-Cj 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 Zj-Cj≥0
Hence, optimal solution is arrived with value of variables as :
x1=50,x2=50,x3=20
Max Z=530
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