Step-1 :
The problem is converted to canonical form by adding slack, surplus and artificial variables as appropiate.
2x1+3x2+s1=62x1+x2+s2=4
The problem is converted to canonical form by adding slack, surplus and artificial variables as appropiate.
Z′−2x1−2x2+x3=0
2x1+3x2+s1=62x1+x2+s2=4
Now represent the new system of constraint equations in the matrix form
⎣⎡100−222−131100010001⎦⎤⎣⎡Z′x1x2x3s1s2⎦⎤=⎣⎡064⎦⎤
or [−10−cA][zx]=[0b]≥0
wheree=β0,a3=β1,a4=β2
Step 2:The basis matrix B1 of order(2+1)=3can be expressed asβ1=[β0β1β2]=⎣⎡100010001⎦⎤
Iteration=1 : Repeat steps 3 to 5 to get new solution
Step-3: To select the vector corresponding to a non-basic variable to enter into the basis, we compute
ck−zk=Max{(cj−zj)>0;j=1,2}
=Max{−(1st row of β1−1)(Columns aj not in basis)}
=max{−[100]⎣⎡222−131100⎦⎤}
=Max{−[−2−11]}=Max{[21−1]}=2(correspnds to c1−z1)
Thus, vector x1 is selected to enter into the basis, for k=1
Step-4: To select a basic variable to leave the basis, we compute yk for k=1, as follows
y1=β1−1a1=⎣⎡100010001⎦⎤⎣⎡−222⎦⎤=⎣⎡−222⎦⎤and XB=⎣⎡064⎦⎤
Now, calculate the minimum ratio to select the basic variable to leave the basis
yrkxβr=min{yikxβi,yik>0}
=min{26,24}=min{3,2}
=2(corresponds toy21xβ2)
Thus, vector S2 is selected to leave the basis, for r=2
The table with new entries in column Y1 and the minimum ratio
The table solution is now updated by replacing variable S2 with the variable X1 into the basis.
For this we apply the following row operations in the same way as in the simplex method
The improved solution is
Iteration=2 : Repeat steps 3 to 5 to get new solution
Step-3: To select the vector corresponding to a non-basic variable to enter into the basis, we compute
ck−zk=Max{(cj−zj)>0;j=1,2}
=Max{−(1st row of β1−1)(Columns aj not in basis)}
=max{−[101]⎣⎡001−131100⎦⎤}
=Max{−[−101]}=Max{[−10−1]}
Since all Zj−Cj≥0
Hence,optimal solution is arrived with value of variables as:x1=2,x2=0,x12=0
MaxZ=4
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