Linear Programming Lecture Notes - Penn State Personal Web Server

We can choose either s1 or s2 as a leaving variable. For the sake of argument, suppose
we choose s2, the variable that’s most negative as the leaving variable. Then our entering
variable is chosen by comparing:
z1 − c1
|a21|
z2 − c2
|a22|
1
=
| − 1|
1
=
| − 2|
Clearly, 1/2 < 1 and therefore, x2 is our entering variable.
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z x1 x2 s1 s2 RHS
z ⎢ 1 1/2 0 0 1/2 −1 ⎥
⎢
⎥
s1 ⎣ 0 -3/2 0 1 −1/2 −3 ⎦
x2 0 1/2 1 0 −1/2 1
At this point, we see we have maintained dual feasibility, but we still do not have primal
feasibility. We can therefore choose a new leaving variable (s1) corresponding to the negative
element in the RHS. The minimum ratio test shows that this time x1 will enter and the final
simplex tableau will be:
z
x1
x2
⎡
⎢
⎣
z x1 x2 s1 s2 RHS
1 0 0 1/3 1/3 −2
0 1 0 −2/3 1/3 2
0 0 1 1/3 −2/3 0
It’s clear this is the optimal solution to the problem since we’ve achieved primal and dual
feasibility and complementary slackness. It’s also worth noting that this optimal solution is
degenerate, since there is a zero in the right hand side.
Exercise 70. Prove that the minimum ratio test given in the dual simplex algorithm
will maintain dual feasibility from one iteration of the simplex tableau to the next. [Hint:
Prove that the reduced costs remain greater than or equal to zero, just as we proved that b
remains positive for the standard simplex algorithm.]
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