Linear Algebra, Theory And Applications, 2012a

6.5. DUALITY 155
and there are still negative numbers. Pick the column which has the −13/4. The pivot is
the 3/8 in the top. This yields
⎛
⎞
1 8
1 2
3 3
0 1
3 3
0 − 1 2
3
0 0
3
1 1 0 0 0 0 1 −1 0 0 1
1
⎜ 3
− 1 1
3
1 0
3
− 1 2
2
3
0
3
0 0 3 ⎟
⎝ 7
3
− 4 1
3
0 0
3
− 1 3
0 − 1 5
3
1 0 ⎠
3
− 2 3
26
1
3
0 0
3
8
5
26
3
0
3
0 1
3
which has only one negative entry on the bottom left. The pivot for this first column is the
7
3
. The next tableau is
⎛
20
2 5
0
7
0 1
7 7
0 − 2 7
− 1 ⎞
3
7
0
7
11
0
7
0 0 − 1 1
7 7
1 − 6 7
− 3 2
7
0 7
⎜ 0 − 1 2
7
1 0
7
− 2 5
7
0
7
− 1 3
7
0 7 ⎟
⎝ 1 − 4 1
7
0 0
7
− 1 7
0 − 1 3 5
7 7
0 ⎠
7
58
3 18 11 2 64
0
7
0 0
7 7
0
7 7
1
7
and all the entries in the left bottom row are nonnegative so the answer is 64/7. This is
the same as obtained before. So what values for x are needed? Here the basic variables are
y 1 ,y 3 ,y 4 ,y 7 . Consider the original augmented matrix, one step before the simplex tableau.
⎛
⎜
⎝
1 5 1 2 1 1 0 0 0 0 2
2 3 2 1 1 0 1 0 0 0 3
1 2 2 1 1 0 0 1 0 0 2
3 1 1 1 1 0 0 0 1 0 3
−5 −8 −6 −7 −4 0 0 0 0 1 0
Permute the columns to put the columns associated with these basic variables first. Thus
⎛
⎞
1 1 2 0 5 1 1 0 0 0 2
2 2 1 1 3 1 0 0 0 0 3
⎜ 1 2 1 0 2 1 0 1 0 0 2
⎟
⎝ 3 1 1 0 1 1 0 0 1 0 3 ⎠
−5 −6 −7 0 −8 −4 0 0 0 1 0
⎞
⎟
⎠ .
The matrix B is
and so B −T equals
⎛
⎜
⎝
1 1 2 0
2 2 1 1
1 2 1 0
3 1 1 0
⎞
⎟
⎠
⎛
− 1 7
− 2 5 1
7 7 7
⎜ 0 0 0 1
⎝ − 1 5
7 7
− 2 7
− 6 7
3
7
− 1 7
− 1 7
− 3 7
Also b T B = ( 5 6 7 0 ) and so from Corollary 6.5.3,
⎛
− 1 7
− 2 5 1
7 7 7
x = ⎜ 0 0 0 1
⎝ − 1 5
7 7
− 2 7
− 6 7
3
7
− 1 7
− 1 7
− 3 7
⎞ ⎛
⎟ ⎜
⎠ ⎝
5
6
7
0
⎞
⎟
⎠
⎞ ⎛
⎟
⎠ = ⎜
⎝
18
7
0
11
7
2
7
⎞
⎟
⎠