Question #201595

Solve using dual simplex method


Minimize z 2x₁ + 2x₂ + 4x3


2x+3x2 + 5x3 2 2


Subject to 3x1 + x₂ + 7x3 <3


x1 + 4x₂ + 6x3 ≤ 5


1
Expert's answer
2021-07-27T10:26:33-0400


In order to apply the dual simplex method, convert minZ to maxZ:

maxZ=2x12x24x3maxZ=-2x_1-2x_2-4x_3

subject to

2x1+3x2+5x322x_1+3x_2+5x_3\le2

3x1+x2+7x333x_1+x_2+7x_3\le3

x1+4x2+6x35x_1+4x_2+6x_3\le5

x1,x2,x30x_1,x_2,x_3\ge0


The problem is converted to canonical form by adding slack, surplus and artificial variables:

  •  as the constraint-1 is of type '\le ' we should add slack variable S1
  •  as the constraint-2 is of type '\le ' we should add slack variable S2
  •  as the constraint-1 is of type '\le ' we should add slack variable S3


After introducing slack variables:

maxZ=2x12x24x3+0S1+0S2+0S3maxZ=-2x_1-2x_2-4x_3+0S_1+0S_2+0S_3

2x1+3x2+5x3+S1=22x_1+3x_2+5x_3+S_1=2

3x1+x2+7x3+S2=33x_1+x_2+7x_3+S_2=3

x1+4x2+6x3+S3=5x_1+4x_2+6x_3+S_3=5

x1,x2,x3,S1,S2,S30x_1,x_2,x_3,S_1,S_2,S_3\ge0



Since all ZjCj0Z_j-C_j\ge0 and all XBi0X_{Bi}\ge0 thus the current solution is the optimal solution:

x1=x2=x2=0x_1=x_2=x_2=0



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