Linear Programming Lecture Notes - Penn State Personal Web Server

Again, we chose to enter variable x4 as it has the most positive reduced cost. Variables x1
and x2 tie in the minimum ratio test. So we consider a new minimum ratio test on the first
column of the tableau:
1 0
(7.7) min ,
1/4 1/2
From this test, we see that x2 is the leaving variable and we pivot on element 1/2 as indicated
in the tableau. Note, we only need to execute the minimum ratio test on variables x1 and x2
since those were the tied variables in the standard minimum ratio test. That is, I0 = {1, 2}
and we construct I1 from these indexes alone. In this case I1 = {2}. Pivoting yields the new
tableau:
Tableau II:
⎡
⎤
z x1 x2 x3 x4 x5 x6 x7 RHS
z ⎢ 1 0 −3/2 0 0 −2 5/4 −21/2 0 ⎥
x1 ⎢ 0 1 −1/2 0 0 −2 −3/4 15/2 0 ⎥
x4 ⎣ 0 0 2 0 1 −24 −1 6 0 ⎦
x3 0 0 0 1 0 0 1 0 1
There is no question this time of the entering or leaving variable, clearly x6 must enter and
x3 must leave and we obtain 1 :
Tableau III:
⎡
z
x1
x4
x6
⎢
⎣
z x1 x2 x3 x4 x5 x6 x7 RHS
1 0 −3/2 −5/4 0 −2 0 −21/2 −5/4
0 1 −1/2 3/4 0 −2 0 15/2 3/4
0 0 2 1 1 −24 0 6 1
0 0 0 1 0 0 1 0 1
Since this is a minimization problem and the reduced costs of the non-basic variables are
now all negative, we have arrived at an optimal solution. The lexicographic minimum ratio
test successfully prevented cycling.
2.2. Convergence of the Simplex Algorithm Under Lexicographic Minimum
Ratio Test.
Lemma 7.7. Consider the problem:
⎧
⎪⎨ max c
P
⎪⎩
T x
s.t. Ax = b
x ≥ 0
Suppose the following hold:
(1) Im is embedded in the matrix A and is used as the starting basis,
(2) a consistent entering variable rule is applied (e.g., largest reduced cost first), and
(3) the lexicographic minimum ratio test is applied as the leaving variable rule.
Then each row of the sequence of augmented matrices [b|B −1 ] is lexicographically positive.
Here B is the basis matrix and b = B −1 b.
1 Thanks to Ethan Wright for finding a small typo in this example, that is now fixed.
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⎤
⎥
⎦