Linear Programming Lecture Notes - Penn State Personal Web Server
Linear Programming Lecture Notes - Penn State Personal Web Server
Linear Programming Lecture Notes - Penn State Personal Web Server
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5. Identifying Unboundedness<br />
We have already identified a theorem for detecting unboundedness. Recall Theorem 5.7:<br />
In a maximization problem, if aji < 0 for all i = 1, . . . , m, and zj − cj < 0, then the linear<br />
programming problem is unbounded.<br />
This condition occurs when a variable xj should enter the basis because ∂z/∂xj > 0<br />
and there is no blocking basis variable. That is, we can arbitrarily increase the value of xj<br />
without causing any variable to become negative. We give an example:<br />
Example 5.11. Consider the <strong>Linear</strong> programming problem from Example 2.9:<br />
⎧<br />
max z(x1, x2) = 2x1 − x2<br />
⎪⎨ s.t. x1 − x2 ≤ 1<br />
⎪⎩<br />
2x1 + x2 ≥ 6<br />
x1, x2 ≥ 0<br />
We can convert this problem into standard form by adding a slack variable s1 and a surplus<br />
variable s2:<br />
⎧<br />
⎪⎨<br />
⎪⎩<br />
max z(x1, x2) = 2x1 − x2<br />
s.t. x1 − x2 + s1 = 1<br />
2x1 + x2 − s2 = 6<br />
x1, x2, s1, s2 ≥ 0<br />
This yields the matrices:<br />
⎡ ⎤ ⎡ ⎤<br />
2 x1<br />
⎢<br />
c = ⎢−1⎥<br />
⎢x2⎥<br />
⎥<br />
⎣ 0 ⎦ x = ⎢ ⎥<br />
⎣s1⎦<br />
0<br />
s2<br />
A =<br />
<br />
1 −1 1<br />
<br />
0<br />
2 1 0 −1<br />
We have both slack and surplus variables, so the case when x1 = x2 = 0 is not a valid initial<br />
solution. We can chose a valid solution based on our knowledge of the problem. Assume<br />
that s1 = s2 = 0 and so we have:<br />
B =<br />
<br />
1 −1<br />
2 1<br />
In this case we have:<br />
<br />
This yields:<br />
<br />
x1<br />
xB =<br />
x2<br />
B −1 b =<br />
<br />
7/3<br />
4/3<br />
N =<br />
<br />
s1<br />
xN =<br />
s2<br />
<br />
1 0<br />
0 −1<br />
<br />
B −1 N =<br />
cB =<br />
We also have the cost information:<br />
cBB −1 b = 10<br />
3 cBB −1 N = 4<br />
3<br />
<br />
2<br />
−1<br />
cN =<br />
<br />
1/3<br />
<br />
−1/3<br />
−2/3 −1/3<br />
<br />
0<br />
0<br />
b =<br />
<br />
1<br />
6<br />
<br />
1 − 3 cBB −1 N − cN = 4<br />
3<br />
81<br />
<br />
1 − 3