Linear Algebra, Theory And Applications, 2012a

6.3. THE SIMPLEX ALGORITHM 147
I need to do the row operations on a nonsimple column which has the pivot in the fourth
row. Such a column is the second to the last. The pivot is the 3. The new matrix is
⎛
⎞
7
1 1 2
0 1 0
3
−1 0
3 3
0
3
1
1
1 0 0
3
0 0
3
− 2 8
3
0 3
⎜ 0 0 1 −2 1 0 −1 0 0 1
⎟
⎝ 0 0 0 − 1 3
0 0 − 1 3
− 1 1
3
1 ⎠ . (6.15)
3
11
2
0 0 0
3
−1 1
3
− 7 28
3
0
3
Now the obvious basic solution is feasible. You let x 4 =0=x 5 = x 7 = x 8 and x 1 =
8/3,x 2 =2/3,x 3 =1, and x 6 =28/3. You don’t need to worry too much about this. It is
the above matrix which is desired. Now you can assemble the simplex tableau and begin
the algorithm. Remember C ≡ 2x 1 +3x 2 +2x 3 +3x 4 . First add the row and column which
deal with C. This yields
⎛
⎜
⎝
⎞
7
1 1
2
0 1 0
3
−1 0
3 3
0 0
3
1
1
1 0 0
3
0 0
3
− 2 8
3
0 0 3
0 0 1 −2 1 0 −1 0 0 0 1
0 0 0 − 1 3
0 0 − 1 3
− 1 1
3
1 0 3 ⎟
11
2
0 0 0
3
−1 1
3
− 7 28
3
0 0 ⎠
3
−2 −3 −2 −3 0 0 0 0 0 1 0
(6.16)
Now you do row operations to keep the simple columns of (6.15) simple in (6.16). Of course
you could permute the columns if you wanted but this is not necessary.
This yields the following for a simplex tableau. Now it is a matter of getting rid of the
positive entries in the bottom row because you are trying to minimize.
⎛
⎞
7
1 1
2
0 1 0
3
−1 0
3 3
0 0
3
1
1
1 0 0
3
0 0
3
− 2 8
3
0 0 3
0 0 1 −2 1 0 −1 0 0 0 1
⎜ 0 0 0 − 1 3
0 0 − 1 3
− 1 1
3
1 0 3 ⎟
⎝
11
2
0 0 0
3
−1 1
3
− 7 28
3
0 0 ⎠
3
2
0 0 0
3
−1 0 − 1 3
− 1 28
3
0 1
3
The most positive of them is the 2/3 and so I will apply the algorithm to this one first. The
pivot is the 7/3. After doing the row operation the next tableau is
⎛
3
0
7
0 1 − 3 ⎞
1 1
2
7
0
7 7
0 0
7
1 − 1 1
2
7
0 0
7
0
7
− 5 18
7
0 0 7
6
1
0
7
1 0
7
0 − 5 2
11
7 7
0 0 7
1
⎜ 0
7
0 0 − 1 7
0 − 2 7
− 2 3
7
1 0 7 ⎟
⎝ 0 − 11 4
1
7
0 0
7
1
7
− 20
58
7
0 0 ⎠
7
0 − 2 7
0 0 − 5 7
0 − 3 7
− 3 64
7
0 1
7
and you see that all the entries are negative and so the minimum is 64/7 and it occurs when
x 1 =18/7,x 2 =0,x 3 =11/7,x 4 =2/7.
There is no maximum for the above problem. However, I will pretend I don’t know this
and attempt to use the simplex algorithm. You set up the simiplex tableau the same way.
Recall it is ⎛
⎞
7
1 1
2
0 1 0
3
−1 0
3 3
0 0
3
1
1
1 0 0
3
0 0
3
− 2 8
3
0 0 3
0 0 1 −2 1 0 −1 0 0 0 1
⎜ 0 0 0 − 1 3
0 0 − 1 3
− 1 1
3
1 0 3 ⎟
⎝
11
2
0 0 0
3
−1 1
3
− 7 28
3
0 0 ⎠
3
2
0 0 0
3
−1 0 − 1 3
− 1 28
3
0 1
3