Question #272108

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



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

1
Expert's answer
2021-11-29T16:56:36-0500

Minimize 𝑍=6𝑥1+3𝑥2+4𝑥3𝑍 = 6𝑥_1 + 3𝑥_2 + 4𝑥_3

𝑆𝑢𝑏𝑗𝑒𝑐𝑡 𝑡𝑜

𝑥130𝑥1 ≥ 30

𝑥250𝑥2 ≥ 50

𝑥320𝑥3 ≥ 20

𝑥1+𝑥2+𝑥3=120𝑥_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 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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