9.3 THE SIMPLEX METHOD: MAXIMIZATION

498 CHAPTER 9 LINEAR PROGRAMMING
x1 x2 s1 s2 s3 b
Basic
Variables
1 1 1 0 0 11 x2 2 0 1 1 0 16 s2 7 0 5 0 1 35 s3 10
0 6 0 0 66
Note that x2 has replaced s1 in the basis column and the improved solution
x 1 , x 2 , s 1 , s 2 , s 3 0, 11, 0, 16, 35
has a z-value of
z 4x 1 6x 2 40 611 66.
In Example 1 the improved solution is not yet optimal since the bottom row still has a
negative entry. Thus, we can apply another iteration of the simplex method to further improve
our solution as follows. We choose x1 as the entering variable. Moreover, the smallest
nonnegative ratio of 111, 162 8, and 357 5 is 5, so s3 is the departing
variable. Gauss-Jordan elimination produces the following.
1
2
7
10
1
0
0
0
1
1
5
6
Thus, the new simplex tableau is as follows.
Basic
x 1 x 2 s 1 s 2 s 3 b Variables
2
0 1 0 16 x2 0 0 1 6 s2 1 0 0
1
5 x1 5
2
7
7
3
7
7
7
10
7
0 0 0 116
8
7
0
1
0
0
1
7
0
0
1
0
11
16
35
66
1
2
1
10
In this tableau, there is still a negative entry in the bottom row. Thus, we choose s1 as the
entering variable and s2 as the departing variable, as shown in the following tableau.
0
0
1
0
1
0
0
0
1
0
0
0
2
7
3
7
5
7
8
7
1
1
5
7
6
0
1
0
0
0
1
0
0
1
7
2
7
1
7
10
7
0
0
1
7
0
16
6
5
116
11
16
5
66