(i) Minimize Z=0.2x1+0.1x2+0.3x3
Subject to constraint
0.5x1+0.2x2+0.7x3=0.4200.3x1+0.2x2+0.5x3≥0.280x1,x2,x3≥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 artificial variable A1
2. As the constraint-2 is of type '≥' we should subtract surplus variable S1 and add artificial variable A2
After introducing surplus, artificial variablesMin Z=0.2x1+0.1x2+0.3x3+0S1+MA1+MA2subject to0.5x1+0.2x2+0.7x3+A1=0.420.3x1+0.2x2+0.5x3−S1+A2=0.28and x1,x2,x3,S1,A1,A2≥0
Positive maximum Zj−Cj is 1.2M-0.3 and its column index is 3. So, the entering variable is x3.
Minimum ratio is 0.56 and its row index is 2. So, the leaving basis variable is A2.
∴ The pivot element is 0.5.
Entering =x3, Departing =A2, Key Element =0.5
Positive maximum Zj-Cj is 1.4M-0.6 and its column index is 4. So, the entering variable is S1.
Minimum ratio is 0.02 and its row index is 1. So, the leaving basis variable is A1.
∴ The pivot element is 1.4.
Entering =S1, Departing =A1, Key Element =1.4
Positive maximum Zj-Cj is 0.0143 and its column index is 1. So, the entering variable is x1.
Minimum ratio is 0.35 and its row index is 1. So, the leaving basis variable is S1.
∴ The pivot element is 0.0571.
Entering =x1, Departing =S1, Key Element =0.0571
Since all Zj-Cj≤0
Hence, optimal solution is arrived with value of variables as :
x1=0.35,x2=0,x3=0.35
Min Z=0.175
(ii) Dual is :
Max Z=0.42y1+0.28y2subject to 0.5y1+0.3y2≤0.20.2y1+0.2y2≤0.10.7y1+0.5y2≤0.3and y2≥0; y1 unrestricted in signSince y1 is unrestricted in sign,introduce the non−negative variables y1′,y1′′so that y1=y1′−y1′′; y1′,y1′′≥0.
The standard form of the LP problem becomes
Max Z=0.42y1′−0.42y1′′+0.28y2subject to0.5y1′−0.5y1′′+0.3y2≤0.20.2y1′−0.2y1′′+0.2y2≤0.10.7y1′−0.7y1′′+0.5y2≤0.3and y1′,y1′′,y2≥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
After introducing slack variables
Max Z=0.42y1′−0.42y1′′+0.28y2+0S1+0S2+0S3subject to0.5y1′−0.5y1′′+0.3y2+S1=0.20.2y1′−0.2y1′′+0.2y2+S2=0.10.7y1′−0.7y1′′+0.5y2+S3=0.3and y1′,y1′′,y2,S1,S2,S3≥0
Negative minimum Zj-Cj is -0.42 and its column index is 1. So, the entering variable is y1′.
Minimum ratio is 0.4 and its row index is 1. So, the leaving basis variable is S1.
∴ The pivot element is 0.5.
Entering =y1′, Departing =S1, Key Element =0.5
Negative minimum Zj-Cj is -0.028 and its column index is 3. So, the entering variable is y2.
Minimum ratio is 0.25 and its row index is 2. So, the leaving basis variable is S2.
∴ The pivot element is 0.08.
Entering = y2, Departing =S2, Key Element =0.08
Since all Zj−Cj≥0
Hence, optimal solution is arrived with value of variables as :
y1′=0.25,y1′′=0,y2=0.25
Max Z=0.175
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