Linear Programming Lecture Notes - Penn State Personal Web Server

using this information, we can compute:
c T BB −1 A·5 − c5 = 7
c T BB −1 A·2 − c2 = −6
Based on this information, we can only choose x2 to enter the basis to ensure that the
value of the objective function increases. We can perform the minimum ration test to figure
out which basic variable will leave the basis. We know that B −1 A·2 is just the second column
of B−1N which is:
B −1 ⎡
A·2 = ⎣ 1
⎤
2⎦
0
Performing the minimum ratio test, we see have:
15 125
min ,
1 2
In this case, we see that index 1 (15/1) is the minimum ratio. Therefore, variable x2 will
enter the basis and variable s1 will leave the basis. The new basic and non-basic variables
will be: The new basic and non-basic variables will be:
⎡
xB = ⎣ x2
⎤
⎡
s3
s2⎦
xN = cB = ⎣
s1
6
⎤
0⎦
0
cN =
0
7
x1
and the matrices become:
⎡ ⎤ ⎡
1 0 3 0
B = ⎣2 1 1⎦
N = ⎣0 ⎤
1
0⎦
0 0 1 1 0
The derived matrices are then:
B −1 ⎡
b = ⎣ 15
⎤
95⎦
B
35
−1 ⎡
−3
N = ⎣ 5
⎤
1
−2⎦
1 0
The cost information becomes:
c T BB −1 b = 335 c T BB −1 N = −11 6 c T BB −1 N − cN = −11 6
Based on this information, we can only choose s3 to (re-enter) the basis to ensure that
the value of the objective function increases. We can perform the minimum ration test to
figure out which basic variable will leave the basis. We know that B −1 A·5 is just the fifth
column of B−1N which is:
B −1 ⎡
A·5 = ⎣ −3
⎤
5 ⎦
1
Performing the minimum ratio test, we see have:
95 35
min ,
5 1
76