Linear Programming Lecture Notes - Penn State Personal Web Server

was called a basic solution (See Definition 3.47.) Clearly any basic solution satisfies the
constraints Ax = b but it may not satisfy the constraints x ≥ 0.
Definition 5.4 (Basic Feasible Solution). If xB = B −1 b and xN = 0 is a basic solution
to Ax = b and xB ≥ 0, then the solution (xB, xN) is called basic feasible solution.
Theorem 5.5. Every basic feasible solution is an extreme point of X. Likewise, every
extreme point is characterized by a basic feasible solution of Ax = b, x ≥ 0.
Proof. Since Ax = BxB + NxN = b this represents the intersection of m linearly
independent hyperplanes (since the rank of A is m). The fact that xN = 0 and xN contains
n − m variables, then we have n − m binding, linearly independent hyperplanes in xN ≥
0. Thus the point (xB, xN) is the intersection of m + (n − m) = n linearly independent
hyperplanes. By Theorem 4.31 we know that (xB, xN) must be an extreme point of X.
Conversely, let x be an extreme point of X. Clearly x is feasible and by Theorem 4.31
it must represent the intersection of n hyperplanes. The fact that x is feasible implies that
Ax = b. This accounts for m of the intersecting linearly independent hyperplanes. The
remaining n − m hyperplanes must come from x ≥ 0. That is, n − m variables are zero. Let
xN = 0 be the variables for which x ≥ 0 are binding. Denote the remaining variables xB.
We can see that A = [B|N] and that Ax = BxB + NxN = b. Clearly, xB is the unique
solution to BxB = b and thus (xB, xN) is a basic feasible solution.
3. The Simplex Algorithm–Algebraic Form
In this section, we will develop the simplex algorithm algebraically. The idea behind the
simplex algorithm is as follows:
(1) Convert the linear program to standard form.
(2) Obtain an initial basic feasible solution (if possible).
(3) Determine whether the basic feasible solution is optimal. If yes, stop.
(4) If the current basic feasible solution is not optimal, then determine which non-basic
variable (zero valued variable) should become basic (become non-zero) and which
basic variable (non-zero valued variable) should become non-basic (go to zero) to
make the objective function value better.
(5) Determine whether the problem is unbounded. If yes, stop.
(6) If the problem doesn’t seem to be unbounded at this stage, find a new basic feasible
solution from the old basic feasible solution. Go back to Step 3.
Suppose we have a basic feasible solution x = (xB, xN). We can divide the cost vector
c into its basic and non-basic parts, so we have c = [cB|cN] T . Then the objective function
becomes:
(5.10) c T x = c T BxB + c T NxN
We can substitute Equation 5.8 into Equation 5.10 to obtain:
−1 −1
B b − B NxN + cNxN = c T BB −1 b + c T N − c T BB −1 N xN
(5.11) c T x = c T B
Let J be the set of indices of non-basic variables. Then we can write Equation 5.11 as:
(5.12) z(x1, . . . , xn) = c T BB −1 b +
j∈J
cj − c T BB −1
A·j xj
71