Simplex Method
Consider the following linear Linear Programming Problem and its canonical form.

Minimize z = -30 x1 - 6 x2 + 5 x3 - 18 x4
   subject to:
      x1  + 2 x3 + x4 < = 20
- 2 x1 +   x2  - x4 < = 15
   6 x1 + 2 x2 - 3 x3 < = 54
                       x1, x2, x3, x4 > = 0 
   Solve:
    x1 + 2 x3   + x4 + x5 = 20
- 2 x1 + x2 - x4 + x6 = 15
  6 x1 + 2 x2 - 3 x3 + x7 = 54
-30x1 - 6 x2 + 5x3 -18x4 = z, 
                            x1, x2, x3, x4, x5, x6, x7 > = 0,
 where z is minimum. 

The vector X0 = (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 x1, x2 , . . . , x6 are all nonnegative and zero is not the minimum value, then any optimal solution must make some of the variables x1 through x4 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.

A b
c z
x1 x2 x3 x4
x5 x6 x7
b
x5
x6
x7
1 0 2 1
-2 1 0 -1
6* 2 -3 0
1 0 0
0 1 0
0 0 1
20
15
54
c0
-30 -6 5 -18
0 0 0
0
x5
x6
x1
0 -1/3 5/2 1*
0 5/3 -1 -1
1 1/3 -1/2 0
1 0 -1/6
0 1 1/3
0 0 1/6
11
33
9
c1
0 4 -10 -18
0 0 5
270
x4
x6
x1
0 -1/3 5/2 1
0 4/3 3/2 0
1 1/3* -1/2 0
1 0 -1/6
1 1 1/6
0 0 1/6
11
44
9
c2
0 -2 35 0
18 0 2
468
x4*
x6*
x2*
1 0 2 1
-4 0 7/2 0
3 1 -3/2 0
1 0 0
1 1 -1/2
0 0 1/2
27
20
8
c*
6 0 32 0
18 0 3
522
Tableau 1. 
c01, c03 and c04 are negative. We choose c01= - 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 bi/ai1 with positive ai1
min { b1/a11 = 20/1 , b3 /a31 = 54/6 } = 9 
Thus a31 is the pivot entry. Now we use some row operations to change the non-basic variable x1 into a 
basic variable 
e3 = ( 0 , 0 , 1 , 0 )t. 
Tableau 2. 
The component c14 = - 18 is the smallest component and only a13 > 0; so a13 is the pivot entry. Therefore the non-basic variable x4 must be changed into 
e1 = ( 1 , 0 , 0 , 0 )t. 
Tableau 3. 
The only negative component of c2 is c24 = - 2. 
min { b2/a22 = 44/(4/3) , b3 /a32 = 9/(1/3) } = 27 
Thus a32 is the pivot entry. The non-basic variable x2 
must be changed into 
e3 = ( 0 , 0 , 1 , 0 )t. 
Tableau 4. 
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 
X* = ( 0 , 27 , 0 , 20 , 0 , 8 , 0 )t. 

A linear programming problem with feasible solutions but unbounded below.

 The following problem is unbounded below, since c4 = -18 < 0 and none of the a14, a24 and a34 is positive.

A b
c z
x1 x2 x3 x4
x5 x6 x7
b
x5
x6
x7
1 0 2 -1
-2 1 0 -1
6 2 -3 0
1 0 0
0 1 0
0 0 1
20
15
54
c
-30 -6 5 -18
0 0 0
0

Note. Although c1 = -30 < c4 = - 18 and in general we select the smallest component of the vector c to obtain our pivot entry but in this case by choosing c4 instead of c1 we immediately reach to the unboundedness conclusion.