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. Thus1 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.