Linear Programming Lecture Notes - Penn State Personal Web Server

This leads to the following linear programming problem:
⎧
min x1 + 1.5x2
⎪⎨ s.t. 2x1 + 3x2 ≥ 20
(3.60)
x1 + 2x2 ≥ 12
⎪⎩
x1, x2 ≥ 0
Before turning to Matlab, let’s investigate this problem in standard form. To transform
the problem to standard form, we introduce surplus variables s1 and s2 and our problem
becomes:
⎧
min x1 + 1.5x2
⎪⎨ s.t. 2x1 + 3x2 − s1 = 20
(3.61)
x1 + 2x2 − s2 = 12
⎪⎩
x1, x2, s1, s2 ≥ 0
This leads to a set of two linear equations with four variables:
2x1 + 3x2 − s1 = 20x1 + 2x2 − s2 = 12
We can look at the various results from using Expression 3.51 and Definition 3.47. Let:
2 3 −1 0 20
(3.62) A =
b =
1 2 0 −1 12
Then our vector of decision variables is:
⎡ ⎤
⎢
(3.63) x = ⎢
⎣
x1
x2
s1
s2
⎥
⎦
We can use Gauss-Jordan elimination on the augmented matrix:
2 3 −1 0 20
1 2 0 −1 12
Suppose we wish to have xB = [s1 s2] T . Then we would transform the matrix as:
−2 −3 1 0 −20
−1 −2 0 1 −12
This would result in x1 = 0, x2 = 0 (because xN = [x1 x2] T and s1 = −20 and s2 = −12.
Unfortunately, this is not a feasible solution to our linear programming problem because
we require s1, s2 ≥ 0. Alternatively we could look at the case when xB = [x1 x2] T and
xN = [s1 s2] T . Then we would perform Gauss-Jordan elimination on the augmented matrix
to obtain: 1 0 −2 3 4
0 1 1 −2 4
That is, x1 = 4, x2 = 4 and of course s1 = s2 = 0. Notice something interesting:
−1 −2 3
1 −2
= −
2 3
1 2
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