Linear Programming Lecture Notes - Penn State Personal Web Server

If δi = 0 if i > k, then we have succeeded in expressing x as a convex combination of the
extreme points of P . Suppose not. Consider xv with v > k (so xv is an extreme point of P
but not P ). Then it follows that e T xv = M must hold (otherwise, xv is an extreme point of
P ). Since there are n binding hyperplanes at xv, there must be n − 1 hyperplanes defining
P binding at xv. Thus, there is an edge connecting xv and some extreme point of P (i.e.,
there is an extreme point of P that shares n − 1 binding hyperplanes with xv). Denote this
extreme point as xi(v); it is adjacent to xv. Consider the direction xv − xv(i). This must be a
recession direction of P since there is no other hyperplane that binds in this direction before
e T x = M. Define:
and let
θv = e T (xv − xv(i))
d = xv − xv(i)
θv
then e T d = 1 as we have normalized the direction elements and therefore d ∈ D. Again,
let Gy = g be the system of n − 1 binding linear hyperplanes shared by xv and xv(i). Then
trivially:
G(xv − xv(i)) = Gd = 0
and therefore, there are n−1 linearly independent binding hyperplanes in the system Ad ≤ 0,
d ≥ 0. At last we see that with e T d = 1 that d must be an extreme point of D and therefore
an extreme direction of P . Let dj(v) = d be this extreme direction. Thus we have:
xv = xi(v) + θvdj(v)
At last we can see that by substituting this into Expression 4.34 for each such v and arbitrarily
letting i(v) = j(v) = 1 if δv = 0 (in which case it doesn’t matter), we obtain:
k k+u
k+u
(4.35) x = δjxj + δvxi(v) +
i=1
v=k+1
v=k+1
δvθvdj(v)
We have now expressed x as desired. This completes the proof.
Example 4.45. The Cartheodory Characterization Theorem is illustrated for a bounded
and unbounded polyhedral set in Figure 4.11. This example illustrates simply how one could
construct an expression for an arbitrary point x inside a polyhedral set in terms of extreme
points and extreme directions.
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