MIN Zx=20x1+30x2+50x3+40x4Subject to:4x1+6x2+x3+2x4≥122x1+x2+6x3+5x4≥14x1+2x2+4x3+3x4≥8and x1,x2,x3,x4≥0;
Dual is (Solution steps of Dual by BigM method)MAX Zy=12y1+14y2+8y3Subject to:4y1+2y2+y3≤206y1+y2+2y3≤30y1+6y2+4y3≤502y1+5y2+3y3≤40and y1,y2,y3≥0;
Max:Z=12y1+14y2+8y3subject to:4y1+2y2+y3≤206y1+y2+2y3≤30y1+6y2+4y3≤502y1+5y2+3y3≤40and y1,y2,y3≥0
The problem is converted to canonical form by adding slack, surplus, and artificial variables as appropriate
1. As the constraint-1 is of type '≤' we should add slack variable S1
2. As the constraint-2 is of type '≤' we should add slack variable S2
3. As the constraint-3 is of type '≤' we should add slack variable S3
4. As the constraint-4 is of type '≤' we should add slack variable S4
After introducing slack variables
MaxZ=12y1+14y2+8y3+0S1+0S2+0S3+0S4subject to:4y1+2y2+y3+S1=206y1+y2+2y3+S2=30y1+6y2+4y3+S3=502y1+5y2+3y3+S4=40 and y1,y2,y3,S1,S2,S3,S4≥0
The negative minimum Zj-Cj is -14 and its column index is 2. So, the entering variable is y2 .
The minimum ratio is 8 and its row index is 4. So, the leaving basis variable is S4 .
∴ The pivot element is 5.
Entering =y2 , Departing =S4 , Key Element =5
The negative minimum Zj-Cj is -6.4 and its column index is 1. So, the entering variable is y1.
The minimum ratio is 1.25 and its row index is 1. So, the leaving basis variable is S1.
∴ The pivot element is 3.2.
Entering =y1, Departing =S1, Key Element =3.2
Since all Zj-Cj≥0
Hence, the optimal solution arrives with value of variables as :
y1=1.25,y2=7.5,y3=0
Max Z=120
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