Linear Algebra, Theory And Applications, 2012a

6.2. THE SIMPLEX TABLEAU 139
lution to the equations,
x 1 +2x 2 + x 3 =10
x 1 +2x 2 − x 4 =2
2x 1 + x 2 + x 5 =6
the solution set for the above system is given by
x 2 = 2 3 x 4 − 2 3 + 1 3 x 5,x 1 = − 1 3 x 4 + 10
3 − 2 3 x 5,x 3 = −x 4 +8.
An easy way to get a basic feasible solution is to let x 4 = 8 and x 5 =1. Then a feasible
solution is
(x 1 ,x 2 ,x 3 ,x 4 ,x 5 )=(0, 5, 0, 8, 1) .
( )
A 0 b
It follows z 0 = −5 and the matrix (6.2),
with the variables kept track of
−c 1 0
on the bottom is ⎛
⎞
1 2 1 0 0 0 10
1 2 0 −1 0 0 2
⎜ 2 1 0 0 1 0 6
⎟
⎝ −1 1 0 0 0 1 0 ⎠
x 1 x 2 x 3 x 4 x 5 0 0
and the first thing to do is to permute the columns so that the list of variables on the bottom
will have x 1 and x 3 at the end.
⎛
⎞
2 0 0 1 1 0 10
2 −1 0 1 0 0 2
⎜ 1 0 1 2 0 0 6
⎟
⎝ 1 0 0 −1 0 1 0 ⎠
x 2 x 4 x 5 x 1 x 3 0 0
Next, as described above, take the row reduced echelon form of the top three lines of the
above matrix. This yields
⎛
⎝ 1 0 0 ⎞
1 1
2 2
0 5
0 1 0 0 1 0 8 ⎠ .
3
0 0 1
2
− 1 2
0 1
Now do row operations to
to finally obtain
⎛
⎜
⎝
⎛
⎜
⎝
⎞
1 1
1 0 0
2 2
0 5
0 1 0 0 1 0 8
⎟
3
0 0 1
2
− 1 2
0 1 ⎠
1 0 0 −1 0 1 0
1 1
1 0 0
2 2
0 5
0 1 0 0 1 0 8
3
0 0 1
2
− 1 2
0 1
0 0 0 − 3 2
− 1 2
1 −5
and this is a simplex tableau. The variables are x 2 ,x 4 ,x 5 ,x 1 ,x 3 ,z.
It isn’t as hard as it may appear from the above. Lets not permute the variables and
simply find an acceptable simplex tableau as described above.
⎞
⎟
⎠