Linear Algebra, Theory And Applications, 2012a

142 LINEAR PROGRAMMING
for all positive F pi the smallest. This will work because for F ki > 0,
b p
F pi
≤ b k
F ki
⇒ b k ≥ F kib p
F pi
Having gotten a new simplex tableau, you do the same thing to it which was just done
and continue. As long as b > 0, so you don’t encounter the degenerate case, the values
for z associated with setting x F = 0 keep getting strictly larger every time the process is
repeated. You keep going until you find c ≥ 0. Then you stop. You are at a maximum.
Problems can occur in the process in the so called degenerate case when at some stage of
the process some b j =0. In this case you can cycle through different values for x with no
improvement in z. This case will not be discussed here.
Example 6.3.1 Maximize 2x 1 +3x 2 subject to the constraints x 1 + x 2 ≥ 1, 2x 1 + x 2 ≤
6,x 1 +2x 2 ≤ 6, x 1 ,x 2 ≥ 0.
The constraints are of the form
x 1 + x 2 − x 3 = 1
2x 1 + x 2 + x 4 = 6
x 1 +2x 2 + x 5 = 6
where the x 3 ,x 4 ,x 5 are the slack variables. An augmented matrix for these equations is of
the form
⎛
1 1 −1 0 0 1
⎞
⎝ 2 1 0 1 0 6 ⎠
1 2 0 0 1 6
Obviously the obvious solution is not feasible. It results in x 3 < 0. We need to exchange
basic variables. Lets just try something.
⎛
⎝ 1 1 −1 0 0 1 ⎞
0 −1 2 1 0 4 ⎠
0 1 1 0 1 5
Now this one is all right because the obvious solution is feasible. Letting x 2 = x 3 =0,
it follows that the obvious solution is feasible. Now we add in the objective function as
described above.
⎛
⎞
1 1 −1 0 0 0 1
⎜ 0 −1 2 1 0 0 4
⎟
⎝ 0 1 1 0 1 0 5 ⎠
−2 −3 0 0 0 1 0
Then do row operations to leave the simple columns the same. Then
⎛
⎞
1 1 −1 0 0 0 1
⎜ 0 −1 2 1 0 0 4
⎟
⎝ 0 1 1 0 1 0 5 ⎠
0 −1 −2 0 0 1 2
Now there are negative numbers on the bottom row to the left of the 1. Lets pick the first.
(It would be more sensible to pick the second.) The ratios to look at are 5/1, 1/1 sopickfor