Linear Programming Lecture Notes - Penn State Personal Web Server

generality, suppose that γ is determined by x − γd. (That is, x − (γ + ɛ)d ∈ P for ɛ > 0).
Let x1 = x − γd. Since r hyperplanes are binding at x (and y1 and y2) it is clear that these
same hyperplanes are binding at x1 and at least one more (because of how we selected γ).
Thus there are at least r + 1 binding hyperplanes at x1. If r + 1 = n, then we have identified
an extreme point. Otherwise, we may repeat the process until we find an extreme point.
To show that the number of extreme points is finite, we note that every extreme point
is the intersection of n linearly independent hyperplanes defining P . There are n + m
hyperplanes defining P and therefore the number of possible extreme points is limited by
. This completes the proof.
n+m
n
Lemma 4.42. Let P be a non-empty polyhedral set. Then the set of directions of P is
empty if and only if P is bounded.
Proof. Clearly if P is bounded then it cannot have a direction. If P were contained in
a ball Br(x0) then we know that for every x ∈ P we have |x − x0| < r. If d is a direction
of P , then we have x + λd for all λ > 0. We simply need to choose λ large enough so that
|x + λd − x0| > r.
If P has no directions, then there is some absolute upper bound on the value of |x| for
all x ∈ P . Let r be this value. Then trivially, Br+1(0) contains P and so P is bounded.
Lemma 4.43. Let P be a non-empty unbounded polyhedral set. Then the number extreme
directions of P is finite and non-zero.
Proof. The result follows immediately from Theorem 4.39 and Lemma 4.41.
Theorem 4.44. Let P be a non-empty, unbounded polyhedral set defined by:
P = {x ∈ R n : Ax ≤ b, x ≥ 0}
(where we assume A is not an empty matrix). Suppose that P has extreme points x1, . . . , xk
and extreme directions d1, . . . , dl. If x ∈ P , then there exists constants λ1, . . . , λk and
µ1, . . . , µl such that:
(4.30)
x =
k
λixi +
i=1
k
λi = 1
i=1
l
j=1
µjdj
λi ≥ 0 i = 1, . . . , k
µj ≥ 0 1, . . . , l
Proof. Let x ∈ P . We will show that we can identify λ1, . . . , λk and µ1, . . . , µl making
Expression 4.30 true. Define the set:
(4.31) P = P ∩ {x ∈ R n : e T x ≤ M}
where M is a large constant so that e T xi < M for i = 1, . . . , k and e T x < M. That is, M
is large enough so that the sum of the components of any extreme point is less than M and
the sum of the components of x is less than M.
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