Linear Programming Lecture Notes - Penn State Personal Web Server

If there is some i such that c T di > 0, then we can simply choose µi as large as we like,
making the objective as large as we like, the problem will have no finite solution.
Therefore, assume that c T di ≤ 0 for all i = 1, . . . , l (in which case, we may simply choose
µi = 0, for all i). Since the set of extreme points x1, . . . xk is finite, we can simply set λp = 1
if c T xp has the largest value among all possible values of c T xi, i = 1, . . . , k. This is clearly
the solution to the linear programming problem. Since xp is an extreme point, we have
shown that if P has a solution, it must have an extreme point solution.
Corollary 5.2. Problem P has a finite solution if and only if c T di ≤ 0 for all i = 1, . . . l
when d1, . . . , dl are the extreme directions of X.
Proof. This is implicit in the proof of the theorem.
Corollary 5.3. Problem P has alternative optimal solutions if there are at least two
extreme points xp and xq so that c T xp = c T xq and so that xp is the extreme point solution
to the linear programming problem.
Proof. Suppose that xp is the extreme point solution to P identified in the proof of
the theorem. Suppose xq is another extreme point solution with c T xp = c T xq. Then every
convex combination of xp and xq is contained in X (since X is convex). Thus every x with
form λxp + (1 − λ)xq and λ ∈ [0, 1] has objective function value:
λc T xp + (1 − λ)c T xq = λc T xp + (1 − λ)c T xp = c T xp
which is the optimal objective function value, by assumption.
Exercise 48. Let X = {x ∈ R n : Ax ≤ b, x ≥ 0} and suppose that d1, . . . dl are the
extreme directions of X (assuming it has any). Show that the problem:
(5.6)
min c T x
s.t. Ax ≤ b
x ≥ 0
has a finite optimal solution if (and only if) c T dj ≥ 0 for k = 1, . . . , l. [Hint: Modify the
proof above using the Cartheodory characterization theorem.]
2. Algorithmic Characterization of Extreme Points
In the previous sections we showed that if a linear programming problem has a solution,
then it must have an extreme point solution. The challenge now is to identify some simple
way of identifying extreme points. To accomplish this, let us now assume that we write X
as:
(5.7) X = {x ∈ R n : Ax = b, x ≥ 0}
Our work in the previous sections shows that this is possible. Recall we can separate A into
an m × m matrix B and an m × (n − m) matrix N and we have the result:
(5.8) xB = B −1 b − B −1 NxN
We know that B is invertible since we assumed that A had full row rank. If we assume that
xN = 0, then the solution
(5.9) xB = B −1 b
70