Linear Programming Lecture Notes - Penn State Personal Web Server

2x1 + x2 =6
Extreme direction
x1 − x2 =1
∇z(x1,x2) =(2, −1)
Figure 5.2. Unbounded Linear Program: The existence of a negative column aj
in the simplex tableau for entering variable xj indicates an unbounded problem and
feasible region. The recession direction is shown in the figure.
Proof. The fact that d is a direction is easily verified by the fact there is an extreme
point x = [xB xN] T and for all λ ≥ 0 we have:
(5.36) x + λd ∈ X
Thus it follows from the proof of Theorem 4.24 that Ad ≤ 0. The fact that d ≥ 0 and d = 0
follows from our assumptions. Now, we know that we can write A = [B|N]. Further, we
know that aj = B −1 A·j. Let us consider Ad:
−aj
(5.37) Ad = [B|N]
ek
= −BB −1 A·j + Nek
Remember, ek is the standard basis vector that has have 1 precisely in the position corresponding
to column A·j in matrix N, so A·j = Nej. Thus we have:
(5.38) −BB −1 A·j + Nek = −A·j + A·j = 0
Thus, Ad = 0. We can scale d so that e T d = 1. We know that n − m − 1 elements of
d are zero (because of ek) and we know that Ad = 0. Thus d can be made to represent
the intersection of n-hyperplanes in R n . Thus, d is an extreme point of the polyhedron
D = {d ∈ R n : Ad ≤ 0, d ≥ 0, e T d = 1}. It follows from Theorem 4.39, we know that d is
an extreme direction of X.
Exercise 51. Consider the problem
⎧
min z(x1, x2) = 2x1 − x2
⎪⎨ s.t. x1 − x2 + s1 = 1
⎪⎩
2x1 + x2 − s2 = 6
x1, x2, s1, s2 ≥ 0
83