Linear Programming Lecture Notes - Penn State Personal Web Server

here we are interested only in directions satisfying e T d = 1. This is a normalizing constraint
that will chose only vectors whose components sum to 1.
Theorem 4.39. A direction d ∈ D is an extreme direction of P if and only if d is an
extreme point of D when D is taken as a polyhedral set.
Proof. (⇒)Suppose that d is an extreme point of D (as a polyhedral set) and not an
extreme direction of P . Then there exist two directions d1 and d2 of P and two constants λ1
and λ2 with λ1, λ2 ≥ 0 so that d = λ1d1 + λ2d2. Without loss of generality, we may assume
that d1 and d2 are vectors satisying e T di = 1 (i = 1, 2). If not, then we can scale them so
their components sum to 1 and adjust λ1 and λ2 accordingly. But this implies that:
1 = e T d = λ1e T d1 + λ2e T d2 = λ1 + λ2
Further, the fact that d1 and d2 are directions of P implies they must be in D. Thus we have
found a convex combination of element of D that equals d, contradicting our assumption
that d was an extreme point.
(⇐)Conversely, suppose that d is an extreme direction of P whose components sum to
1 and not an extreme point of D. (Again, we could scale d if needed.) Then d cannot be
recovered from a positive combination of directions d1 and d2. But if d is not an extreme
point of D, then we know λ1d1 +λ2d2 = d for d1, d2 ∈ D and λ1 +λ2 = 1 and λ1, λ2 ∈ (0, 1).
This is clearly contradictory. Every strict convex combination is a positive combination and
therefore our assumption that d was an extreme direction was false.
Example 4.40. Let’s consider Example 4.26 again. The polyhedral set in this example
was defined by the A matrix:
1 −1
A =
−2 −1
and the b vector:
1
b =
−6
If we assume that P = {x ∈ R n : Ax ≤ b, x ≥ 0}, then the set of extreme directions of P
is the same as the set of extreme points of the set
D = {d ∈ R n : Ad ≤ 0, d ≥ 0, e T d = 1}
Then we have the set of directions d = [d1, d2] T so that:
d1 − d2 ≤ 0
−2d1 − d2 ≤ 0
d1 + d2 = 1
d1 ≥ 0
d2 ≥ 0
The feasible region (which is really only the line d1 + d2 = 1) is shown in red in Figure
4.10. The critical part of this figure is the red line. It is the true set D. As a line, it has
two extreme points: (0, 1) and (1/2, 1/2). Note that (0, 1) as an extreme point is one of the
direction [0, 1] T we illustrated in Example 4.26.
63