24.03.2013 Views

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

SHOW MORE
SHOW LESS

Create successful ePaper yourself

Turn your PDF publications into a flip-book with our unique Google optimized e-Paper software.

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

Hooray! Your file is uploaded and ready to be published.

Saved successfully!

Ooh no, something went wrong!