Question #190708

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


1
Expert's answer
2021-05-10T18:21:29-0400

Max Z=0.4x1+0.3x2Z=0.4x_1+0.3x_2


subject to-

2x1+x21000x1+x2800x1400x2700x1,x202x_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 -


MaxZ=0.4x1+0.3x2+0s1+0s2+0s3+0s4Max Z=0.4x_1+0.3x_2+0s_1+0s_2+0s_3+0s_4


subject to-


2x1+x2+s1=1000x1+x2+s2=800x1+s3=400x2+s4=7002x_1+x_2+s_1=1000\\ x_1+x_2+s_2=800\\ x_1+s_3=400\\ x_2+s_4=700\\

and s2,s2,s3,s4,x1,x20s_2,s_2,s_3,s_4,x_1,x_2\ge 0




Negative minimum ZjCjZ_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 s3.s_3.

The pivot element is 1. So entering x1x_1 , departing s_3 and key element =1



Since all ZjCj0Z_j-C_j\ge 0

Hence ,Optimal solution is arrived with value of x1=200,x2=600x_1=200,x_2=600


Max Z=260Z=260


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