Linear Programming Lecture Notes - Penn State Personal Web Server

Based on this information, we can construct the tableau for this problem as:
⎡
⎤
z x1 x2 s1 s2 RHS
z ⎢
4 −1 10
(5.34) ⎢ 1 0 0 ⎥
3 3 3 ⎥
⎣
1 −1 7
0 1 0 ⎦
x1
x2 0 0 1
3
−2
3
3
−1
3
3
4
3
We see that s2 should enter the basis because cBB−1A·4 − c4 < 0. But the column
corresponding to s2 in the tabluau is all negative. Therefore there is no minimum ratio test.
We can let s2 become as large as we like and we will keep increasing the objective function
without violating feasibility.
What we have shown is that the ray with vertex
⎡ ⎤
7/3
⎢
x0 = ⎢4/3
⎥
⎣ 0 ⎦
0
and direction:
⎡ ⎤
1/3
⎢
d = ⎢1/3
⎥
⎣ 0 ⎦
1
is entirely contained inside the polyhedral set defined by Ax = b. This can be see from the
fact that:
xB = B −1 b − B −1 NxN
When applied in this case, we have:
We know that
xB = B −1 b − B −1 A·4s2
−B −1 A·4 =
1/3
1/3
We will be increasing s2 (which acts like λ in the definition of ray) and leaving s1 equal to
0. It’s now easy to see that the ray we described is contained entirely in the feasible region.
This is illustrated in the original constraints in Figure 5.2.
Based on our previous example, we have the following theorem that extends Theorem
5.7:
Theorem 5.12. In a maximization problem, if aji ≤ 0 for all i = 1, . . . , m, and zj − cj <
0, then the linear programming problem is unbounded furthermore, let aj be the j th column
of B −1 A·j and let ek be a standard basis column vector in R m×(n−m) where k corresponds to
the position of j in the matrix N. Then the direction:
−aj
(5.35) d =
ek
is an extreme direction of the feasible region X = {x ∈ R n : Ax = b, x ≥ 0}.
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