Linear Algebra, Theory And Applications, 2012a

148 LINEAR PROGRAMMING
Now to maximize, you try to get rid of the negative entries in the bottom left row. The
most negative entry is the -1 in the fifth column. The pivot is the 1 in the third row of this
column. The new tableau is
⎛
1
0 1 1
3
0 0 − 2 ⎞
1
5
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 ⎟
⎝
5
0 0 1
3
0 1 − 1 3
− 7 31
3
0 0 ⎠
3
0 0 1 − 4 3
0 0 − 4 3
− 1 31
3
0 1
3
Consider the fourth column. The pivot is the top 1/3. The new tableau is
⎛
⎞
0 3 3 1 0 0 −2 1 0 0 5
1 −1 −1 0 0 0 1 −1 0 0 1
0 6 7 0 1 0 −5 2 0 0 11
⎜ 0 1 1 0 0 0 −1 0 1 0 2
⎟
⎝ 0 −5 −4 0 0 1 3 −4 0 0 2 ⎠
0 4 5 0 0 0 −4 1 0 1 17
There is still a negative in the bottom, the -4. The pivot in that column is the 3. The
algorithm yields
⎛
0 − 1 1
2
3 3
1 0
3
0 − 5 ⎞
19
3
0 0
3
2 1
1
3 3
0 0 − 1 1
1
3
0
3
0 0 3
0 − 7 1
5
3 3
0 1
3
0 − 14 43
3
0 0 3
⎜ 0 − 2 3
− 1 1
3
0 0
3
0 − 4 8
3
1 0 3 ⎟
⎝ 0 − 5 3
− 4 1
3
0 0
3
1 − 4 2
3
0 0 ⎠
3
0 − 8 3
− 1 4
3
0 0
3
0 − 13 59
3
0 1
3
Note how z keeps getting larger. Consider the column having the −13/3 in it. The pivot is
the single positive entry, 1/3. The next tableau is
⎛
⎞
5 3 2 1 0 −1 0 0 0 0 8
3 2 1 0 0 −1 0 1 0 0 1
14 7 5 0 1 −3 0 0 0 0 19
⎜ 4 2 1 0 0 −1 0 0 1 0 4
.
⎟
⎝ 4 1 0 0 0 −1 1 0 0 0 2 ⎠
13 6 4 0 0 −3 0 0 0 1 24
There is a column consisting of all negative entries. There is therefore, no maximum. Note
also how there is no way to pick the pivot in that column.
Example 6.3.4 Minimize 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.
There exists an answer because the region defined by the constraints is closed and
bounded. Adding in slack variables you get the following augmented matrix corresponding
to the constraints. ⎛
⎞
1 1 1 1 0 0 0 10
⎜ 1 1 1 0 −1 0 0 2
⎟
⎝ 1 1 3 0 0 1 0 8 ⎠
1 2 1 0 0 0 1 7