20
Then
1 0
-3 1
such that
5 |
-1 |
0 |
1 |
-3 |
0 |
Let b1 = 1, δ2 = 2 and b3 = —1. Then A and A correspond to Example 2 and
Example 4, respectively. See Figure 2 and Figure 4.
Example 8. We are given
P = { x E R2 I ajx ≤ bi, i = 1, 2, 3 }
where ɑɪ = (3, — l)τ, α2 = (—3, 2)τ and α3 = ( —1, — l)τ, b1 = 2, δ2 = —1 and
b3 = 0. See Figure 5 where there are three completely labeled simplices. One of
them is of type II. The other two are of type I. We use the following unimodular
transformation
such that
2 -1
-1 2
—2 -1
Now it is easy to check that the resulting polytope generates no completely labeled
simplex of type II. In fact the Labeling Rule results in three completely labeled
simplices of type I in this case, see Figure 6. Compare Figure 5 with Figure 6.