Linear Algebra, Theory And Applications, 2012a

6.3. THE SIMPLEX ALGORITHM 143
the pivot the 1 in the second column and first row. This will leave the right column above
the lower right corner nonnegative. Thus the next tableau is
⎛
⎞
1 1 −1 0 0 0 1
⎜ 1 0 1 1 0 0 5
⎟
⎝ −1 0 2 0 1 0 4 ⎠
1 0 −3 0 0 1 3
There is still a negative number there to the left of the 1 in the bottom row. The new ratios
are 4/2, 5/1 so the new pivot is the 2 in the third column. Thus the next tableau is
⎛
⎞
1
1
2
1 0 0
2
0 3
3
⎜ 2
0 0 1 − 1 2
0 3
⎟
⎝ −1 0 2 0 1 0 4 ⎠
− 1 3
2
0 0 0
2
1 9
Still, there is a negative number in the bottom row to the left of the 1 so the process does
not stop yet. The ratios are 3/ (3/2) and 3/ (1/2) and so the new pivot is that 3/2 in the
first column. Thus the new tableau is
⎛
⎞
⎜
⎝
0 1 0 − 1 3 3
0 2
3
2
0 0 1 − 1 2
0 3
2 2
0 0 2
3 3
0 6
1 4
0 0 0
3 3
1 10
Now stop. The maximum value is 10. This is an easy enough problem to do geometrically
and so you can easily verify that this is the right answer. It occurs when x 4 = x 5 =0,x 1 =
2,x 2 =2,x 3 =3.
6.3.2 Minimums
How does it differ if you are finding a minimum? From a basic feasible solution, a simplex
tableau of the following form has been obtained in which the simple columns for the basic
variables, x B are listed first and b ≥ 0.
( )
I F 0 b
0 c 1 z 0 (6.14)
Let x 0 i = b i for i =1, ··· ,m and x 0 i =0fori>m.Then ( x 0 ,z 0) is a solution to the above
system and since b ≥ 0, it follows ( x 0 ,z 0) is a basic feasible solution. So far, there is no
change.
Suppose first that some c i > 0andF ji ≤ 0foreachj. Then let x ′ F consist of changing x i
by making it positive but leaving the other entries of x F equal to 0. Then from the bottom
row,
z = −c i x i + z 0
and you let x ′ B = b − F x′ F ≥ 0. Thus the constraints continue to hold when x i is made
increasingly positive and it follows from the above equation that there is no minimum for
z. You stop when this happens.
Next suppose c ≤ 0. Then in this case, z = z 0 − cx F and from the constraints, x F ≥ 0
and so −cx F ≥ 0andsoz 0 is the minimum value and you stop since this is what you are
looking for.
What do you do in the case where some c i > 0andsomeF ji > 0? In this case, you use
the simplex algorithm as in the case of maximums to obtain a new simplex tableau in which
2
⎟
⎠