Linear Programming Lecture Notes - Penn State Personal Web Server

It is clear that P is bounded. In fact, if P is bounded, then P = P . Furthermore P is
a polyhedral set contained in P and therefore the extreme points of P are also the extreme
points of P . Define
Ep = {x1, . . . , xk, . . . , xk+u}
as the extreme points of P . By Theorem 4.41 we know that 0 ≤ u < ∞. If x ∈ Ep,
then x can be written as a convex combination of the elements of Ep. Therefore, assume
that x ∈ Ep. Now, suppose that the system of equations Gy = g represents the binding
hyperplanes (constraints) of P that are active at x. Clearly rank(G) < n (otherwise, x
would be an extreme point of Ep).
Let d = 0 be a solution to the problem Gd = 0 and compute γ1 = max{γ : x+γd ∈ X}.
Since X is bounded and x is not in Ep, then 0 < γ1 < ∞. Let y = x + γ1d. Just as in
the proof of Lemma 4.41, at y, we now have (at least one) additional linearly independent
binding hyperplane of P . If there are now n binding hyperplanes, then y is an extreme point
of P . Otherwise, we may repeat this process until we identify such an extreme point. Let
y1 be this extreme point. Clearly Gy1 = g. Now define:
γ2 = max{γ : x + γ(x − y1) ∈ P }
The value value x − y1 is the direction from y1 to x and γ2 may be thought of as the size of
a step that one can from x away from y1 along the line from y1 to x. Let
(4.32) y2 = x + γ2(x − y1)
Again γ2 < ∞ since P is bounded and further γ2 > 0 since:
(4.33) G (x + γ2(x − y1) = g
for all γ ≥ 0 (as Gx = Gy1). As we would expect, Gy2 = g and there is at least one
additional hyperplane binding at y2 (as we saw in the proof of Lemma 4.41). Trivially, x is
a convex combination of y1 and y2. Specifically, let
x = δy1 + (1 − δ)y2
with δ ∈ (0, 1) and δ = γ2/(1 + γ2). This follows from Equation 4.32, by solving for x.
Now if y2 ∈ Ep, then we have written x as a convex combination of extreme points of P .
Otherwise, we can repeat the process we used to find y2 in terms of y3 and y4, at which an
additional hyperplane (constraint) is binding. We may repeat this process until we identify
elements of EP . Reversing this process, we can ultimately write x as a convex combination
of these extreme points. Thus we have shown that we can express x as a convex combination
of the extreme points of EP .
Based on our deduction, we may write:
(4.34)
k+u
x =
i=1
k+u
1 =
i=1
δixi
δi
δi ≥ 0 i = 1, . . . , k + u
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