


b 


2 
6 
0 
1 
2 
2 
0  11 
1 
1 
0 

1 
0 
0 
0 
0 
1 
0 
0 
0 
0 
1 
0 
0 
0 
0 
1 


c_{0} 

 0 

1 
2 
6 
0 
1 
2 
0 
4  
0 
19 
6 

1 
0 
0 
0 
0 
1 
0 
0 
2 
0 
1 
0 
1 
0 
0 
1 


c_{1} 

 20 
x_{1}*  x_{5}* 
x_{3}* 
x_{7}* 

1 
22 
0 
0 
9 
0 
0 
4  1 
0 
23 
0 

11 
0 
6 
0 
4 
1 
2 
0 
2 
0 
1 
0 
11 
0 
6 
1 


c* 

 26 

Tableau 1.
The vector b has two negative component b_{1} =  10
and b_{4} =  2 , we choose b_{1} =  10.
( In general we should select the smallest component).
Next we choose an entry of the matrix A which produces the largest value of
c_{0j}/a_{1j} with negative a_{1j}.
max { c_{0}/a_{11} = 2/(1) , c_{03}/a_{31} = 18/(6) } = 2
Thus a_{11} is the pivot entry. Now we use some row operations to change the nonbasic variable
x_{1} into a basic variable
e_{1} = ( 1 , 0 , 0 , 0 )^{t}.
Tableau 2.
The only negative component of b is b_{3} =  1, and only
a_{33} < 0. Thus
a_{33} is the pivot entry. The nonbasic variable
x_{3} must be changed into
e_{3} = ( 0 , 0 , 1 , 0 )^{t}.
Tableau 3.
Since all the components of the vector b* are nonnegative, this is the final tableau; it indicates that the minimum value of z =  (  26 ) = 26 and the optimal solution is
X* = ( 4 , 0 , 1 , 0 , 4 , 0 , 2 )^{t}.
