Optimization

283Integer Programming

and the following output is obtained.

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basic

3 4 5

nonbasic_set =

1 2

Initial_Table =

1 0 0 2 5 16

0 1 0 2 -3 7

0 0 1 0 1 1

Cost =

0 0 0 -4 -5 0

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basic_set =

3 4 2

nonbasic_set =

1 5

Table =

1 0 0 2 -5 11

0 1 0 2 3 10

0 0 1 0 1 1

Cost =

0 0 0 -4 5 5

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basic_set =

3 1 2

nonbasic_set =

4 5

Table =

1 0 0 -1 -8 1

0 1 0 1/2 3/2 5

0 0 1 0 1 1

Cost =

0 0 0 2 11 25

— — —SOLUTION— — —

basic_set =

3 1 2

xb =

1

5

1

zz =

-25

The optimal value of the objective function is −25 and occurs at x

1

= 5 and

x

2

= 1 (Figure 10.6). Because this subprob ...