Linear Programming Lecture Notes - Penn State Personal Web Server

Matlab Solution
Figure 3.1. The feasible region for the diet problem is unbounded and there are
alternative optimal solutions, since we are seeking a minimum, we travel in the
opposite direction of the gradient, so toward the origin to reduce the objective
function value. Notice that the level curves hit one side of the boundary of the
feasible region.
This is not an accident, it’s because we started with a negative identity matrix inside the
augmented matrix. The point x1 = 4, x2 = 4 is a point of intersection, shown in Figure 3.1.
It also happens to be one of the alternative optimal solutions of this problem. Notice in
Figure 3.1 that the level curves of the objective function are parallel to one of the sides of
the boundary of the feasible region. If we continued in this way, we could actually construct
all the points of intersection that make up the boundary of the feasible region. We’ll can do
one more, suppose xB = [x1 s1] T . Then we would use Gauss-Jordan elimination to obtain:
1 2 0 −1 12
0 1 1 −2 4
Notice there are now columns of the identity matrix in the columns corresponding to s1 and
x1. That’s how we know we’re solving for s1 and x2. We have x1 = 12 and s1 = 4. By
definition x1 = s2 = 0. This corresponds to the point x1 = 12, x2 = 0 shown in Figure 3.1.
Let’s use Matlab to solve this problem. Our original problem is:
⎧
⎪⎨
⎪⎩
min x1 + 1.5x2
s.t. 2x1 + 3x2 ≥ 20
x1 + 2x2 ≥ 12
x1, x2 ≥ 0
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