In order to apply the dual simplex method, convert minZ to maxZ:
maxZ=−2x1−2x2−4x3
subject to
2x1+3x2+5x3≤2
3x1+x2+7x3≤3
x1+4x2+6x3≤5
x1,x2,x3≥0
The problem is converted to canonical form by adding slack, surplus and artificial variables:
- as the constraint-1 is of type '≤ ' we should add slack variable S1
- as the constraint-2 is of type '≤ ' we should add slack variable S2
- as the constraint-1 is of type '≤ ' we should add slack variable S3
After introducing slack variables:
maxZ=−2x1−2x2−4x3+0S1+0S2+0S3
2x1+3x2+5x3+S1=2
3x1+x2+7x3+S2=3
x1+4x2+6x3+S3=5
x1,x2,x3,S1,S2,S3≥0
Since all Zj−Cj≥0 and all XBi≥0 thus the current solution is the optimal solution:
x1=x2=x2=0
Comments