Linear Algebra, Theory And Applications, 2012a

6.3. THE SIMPLEX ALGORITHM 149
Of course there is a problem with the obvious solution obtained by setting to zero all
variables corresponding to a nonsimple column because of the simple column which has the
−1 in it. Therefore, I will use the simplex algorithm to make this column non simple. The
third column has the 1 in the second row as the pivot so I will use this column. This yields
⎛
⎜
⎝
0 0 0 1 1 0 0 8
1 1 1 0 −1 0 0 2
−2 −2 0 0 3 1 0 2
0 1 0 0 1 0 1 5
⎞
⎟
⎠ (6.17)
and the obvious solution is feasible. Now it is time to assemble the simplex tableau. First
add in the bottom row and second to last column corresponding to the equation for z. This
yields
⎛
⎞
0 0 0 1 1 0 0 0 8
1 1 1 0 −1 0 0 0 2
⎜ −2 −2 0 0 3 1 0 0 2
⎟
⎝ 0 1 0 0 1 0 1 0 5 ⎠
−1 3 −1 0 0 0 0 1 0
Next you need to zero out the entries in the bottom row which are below one of the simple
columns in (6.17). This yields the simplex tableau
⎛
⎞
0 0 0 1 1 0 0 0 8
1 1 1 0 −1 0 0 0 2
⎜ −2 −2 0 0 3 1 0 0 2
⎟
⎝ 0 1 0 0 1 0 1 0 5 ⎠ .
0 4 0 0 −1 0 0 1 2
The desire is to minimize this so you need to get rid of the positive entries in the left bottom
row. There is only one such entry, the 4. In that column the pivot is the 1 in the second
row of this column. Thus the next tableau is
⎛
⎜
⎝
0 0 0 1 1 0 0 0 8
1 1 1 0 −1 0 0 0 2
0 0 2 0 1 1 0 0 6
−1 0 −1 0 2 0 1 0 3
−4 0 −4 0 3 0 0 1 −6
There is still a positive number there, the 3. The pivot in this column is the 2. Apply the
algorithm again. This yields
⎛
1
1
2
0
2
1 0 0 − 1 ⎞
13
2
0
2
1
1
1
7
2
1
2
0 0 0
2
0 2
1
5
⎜ 2
0
2
0 0 1 − 1 9
2
0 2 ⎟
⎝ − 1 2
0 − 1 1
3
2
0 1 0
2
0 ⎠ .
2
− 5 2
0 − 5 2
0 0 0 − 3 2
1 − 21 2
Now all the entries in the left bottom row are nonpositive so the process has stopped. The
minimum is −21/2. It occurs when x 1 =0,x 2 =7/2,x 3 =0.
Now consider the same problem but change the word, minimize to the word, maximize.
Example 6.3.5 Maximize z = x 1 − 3x 2 + x 3 subject to the constraints x 1 + x 2 + x 3 ≤
10,x 1 + x 2 + x 3 ≥ 2, x 1 + x 2 +3x 3 ≤ 8 and x 1 +2x 2 + x 3 ≤ 7 with all variables nonnegative.
⎞
⎟
⎠