Gruber P. Convex and Discrete Geometry

v1 = v0 + λx0
The Simplex Algorithm Terminates
v0
We prove the following result.
Fig. 20.1. Simplex algorithm
20 Linear Optimization 341
Theorem 20.2. The simplex algorithm either shows that the supremum in (1) is infinite
or leads to an optimum solution in finitely many steps.
Proof. Assume that, on the contrary, the simplex algorithm applied to (1) does not
terminate. Then it does not show that the supremum in (1) is infinite and produces
sequences of matrices, vectors and reals,
Ak, bk,vk, uk, xk,λk, k = 0, 1,...
We note that no step can lead to an unbounded edge, and that P has only finitely
many vertices. Moving from one vertex to the next one increases the objective function
or leaves it the same if the vertices coincide, we therefore have the following:
from a certain index on, each step leads to case (iib) and always to the same vertex.
Since for the d × d sub-matrices of A there are only finitely many choices, there are
indices k < l, such that
Ak = Al,vk = vk+1 =···=vl,λk = λk+1 =···=λl = 0.
Let n be the largest index of a row which is removed from one of Ak,...,Al in some
step, say from A p. Then A p+1 no longer has this row. Since A p contains this row,
it must have been inserted into one of A p+1,...,Al = Ak, Ak+1,...,Ap−1. Let
Aq+1 be the first matrix in this sequence of matrices which again contains the nth
row. By our choice of n,
(5) None of the rows ai of A p with i > n is removed in any of the following
steps.
By (2) for A p and (4) for Aq instead of A0, wehaveu p Axq = cxq > 0. Thus