optimal solution
The second
simplex table was not optimal because there was positive value in the _{} row under the x_{1
}column. It mean that x_{1} will be the entering value and
according to the value _{} x_{4} will be
the leaving value. The pivot row is the x_{4} row. To get 1 to the
pivot position we must divide whole x_{4} row by the pivot element  in
this case by 4/3
New pivot
line 
6 
_{} 
1 
0 
1/4 
3/4 
6 

Table 1: New
Pivot Line
The element
above the pivot is 2/3 in x_{2 }row_{
}and should be 0, so we have to multiply the new pivot row by 2/3 and this
new row subtract from the old x_{2}
row. Note the changes in objective function coefficients in the very left
column – changes together with the changes of solution variables.
Pivot
(A) New pivot line multiplied by 2/3 
2/3 
0 
1/6 
1/2 
4 
(B ) Old x_{2} row 
2/3 
1 
1/3 
0 
8 
(B)(A) 
0 
1 
1/2 
1/2 
4 
Table 2: Getting new x_{1 }Row
The new
simplex tableau is optimal because the _{} row contains no
positive variable see Table3:


6 
7 
0 
0 
b 


_{} 
_{} 
_{} 
_{} 


7 
_{} 
0 
1 
½ 
1/2 
4 

6 
_{} 
1 
0 
1/4 
3/4 
6 


_{} 
0 
0 
2 
1 
64 

Table 3:
Optimal Simplex Tableau
Compare with
the optimal solution got by graphical method  optimum point.