Linear Programming Lecture Notes - Penn State Personal Web Server

Since bi ≥ 0, we will require xai ≥ 0. If xai = 0, then this is simply the original constraint.
Thus if we can find values for the ordinary decision variables x so that xai = 0, then constraint
i is satisfied. If we can identify values for x so that all the artificial variables are zero and
m variables of x are non-zero, then the modified constraints described by Equation 6.2 are
satisfied and we have identified an initial basic feasible solution.
Obviously, we would like to penalize non-zero artificial variables. This can be done by
writing a new linear programming problem:
⎧
⎪⎨
(6.3) P1
⎪⎩
min e T xa
s.t. Ax + Imxa = b
x, xa ≥ 0
Remark 6.1. We can see that the artificial variables are similar to slack variables, but
they should have zero value because they have no true meaning in the original problem P .
They are introduced artificially to help identify an initial basic feasible solution to Problem
P .
Lemma 6.2. The optimal objective function value in Problem P1 is bounded below by 0.
Furthermore, if the optimal solution to problem P1 has xa = 0, then the values of x form a
feasible solution to Problem P .
Proof. Clearly, setting xa = 0 will produce an objective function value of zero. Since
e > 0, we cannot obtain a smaller objective function value. If at optimality we have xa = 0,
then we know that m of the variables in x are in the basis and the remaining variables (in
both x and xa) are not in the basis and hence at zero. Thus we have found a basic feasible
solution to Problem P .
Example 6.3. Consider the following problem:
min x1 + 2x2
s.t. x1 + 2x2 ≥ 12
(6.4)
2x1 + 3x2 ≥ 20
x1, x2 ≥ 0
We can convert the problem to standard form by adding two surplus variables:
min x1 + 2x2
s.t. x1 + 2x2 − s1 = 12
(6.5)
2x1 + 3x2 − s2 = 20
x1, x2, s1, s2 ≥ 0
It’s not clear what a good basic feasible solution would be for this. Clearly, we cannot set
x1 = x2 = 0 because we would have s1 = −12 and s2 = −20, which is not feasible. We can
introduce two artificial variables (xa1 and xa2) and create a new problem P1.
min xa1 + xa2
s.t. x1 + 2x2 − s1 + xa1 = 12
(6.6)
2x1 + 3x2 − s2 + xa2 = 20
x1, x2, s1, s2, xa1, xa2 ≥ 0
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