244 CHAPTER 11. LINEAR PROGRAMMING
73 . The next tableau is
0 207 0 1 2
757 0 − 2
7 − 17 0 3
70 11
7 0 0 − 17
17 1 − 6
7 − 37 0 2
70 − 1
7 1 0 27 − 2
7 0 57 − 1
7 0 37
1 − 47 0 0 1
7 − 17 0 − 1
737 0 5
70 58
7 0 0 37
187 0 11
727 1 64
7
and all the entries in the left bottom row are nonnegative so the answer is 64/7. This is thesame as obtained before. So what values for x are needed? Here the basic variables arey1,y3,y4,y7. Consider the original augmented matrix, one step before the simplex tableau.
1 5 1 2 1 1 0 0 0 0 22 3 2 1 1 0 1 0 0 0 31 2 2 1 1 0 0 1 0 0 23 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. Thus1 1 2 0 5 1 1 0 0 0 22 2 1 1 3 1 0 0 0 0 31 2 1 0 2 1 0 1 0 0 23 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
1 1 2 02 2 1 11 2 1 03 1 1 0
and so B−T equals
− 17 − 2
757
17
0 0 0 1− 1
757 − 2
7 − 67
37 − 1
7 − 17 − 3
7
Also bT
B =(
5 6 7 0)
and so from Corollary 11.5.3,
x =
− 1
7 − 27
57
17
0 0 0 1− 1
757 − 2
7 − 67
37 − 1
7 − 17 − 3
7
5670
=
187011727
which agrees with the original way of doing the problem.