Minimize z = 30 x_{1}  6
x_{2} + 5 x_{3}  18 x_{4}
subject to:

Solve:
where z is minimum. 
The vector X_{0} = (0 , 0 , 0 , 0 , 20 , 15 , 54 )^{t} is a basic solution to the canonical form but not an optimal solution, since some of the components of the vector c in the canonical form are negative; so we need to find new feasible solutions to the system which lower the value of z until it reaches its minimum. Since x_{1}, x_{2} , . . . , x_{6} are all nonnegative and zero is not the minimum value, then any optimal solution must make some of the variables x_{1} through x_{4} basic and must produce a new objective function with nonnegative components. To obtain new basic feasible solutions, we use a row reduction method known as the Simplex Method.

Tableau 1.
c_{01}, c_{03} and c_{04} are negative. We choose c_{01}=  30. ( In general we should select the smallest component). The pivot entry will be an entry of the matrix A which produces the smallest value of b_{i}/a_{i1} with positive a_{i1}. basic variable The component c_{14} =  18 is the smallest component and only a_{13} > 0; so a_{13} is the pivot entry. Therefore the nonbasic variable x_{4} must be changed into The only negative component of c_{2} is c_{24} =  2. must be changed into Since all the components of the vector c* are nonnegative, this tableau will be our final tableau which indicates that the minimum value of z =  ( 522 ) and an optimal solution is 
The following problem is unbounded below, since c_{4} = 18 < 0 and none of the a_{14}, a_{24} and a_{34} is positive.



b  



 
c 

 0 