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Linear Algebra, Theory And Applications, 2012a

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140 LINEAR PROGRAMMING<br />

Example 6.2.3 Consider z = x 1 −x 2 subject to the constraints, x 1 +2x 2 ≤ 10,x 1 +2x 2 ≥ 2,<br />

and 2x 1 + x 2 ≤ 6,x i ≥ 0. Find a simplex tableau.<br />

Adding in slack variables, an augmented matrix which is descriptive of the constraints<br />

is<br />

⎛<br />

⎝ 1 1 2 2 1 0 0 −1 0 0 10<br />

6<br />

⎞<br />

⎠<br />

2 1 0 0 1 6<br />

The obvious solution is not feasible because of that -1 in the fourth column. When you let<br />

x 1 ,x 2 =0, you end up having x 4 = −6 which is negative. Consider the second column and<br />

select the 2 as a pivot to zero out that which is above and below the 2.<br />

⎛<br />

0 0 1 1 0 4<br />

⎞<br />

⎝ 1<br />

2<br />

1 0 − 1 2<br />

0 3 ⎠<br />

3<br />

1<br />

2<br />

0 0<br />

2<br />

1 3<br />

This one is good. When you let x 1 = x 4 =0, you find that x 2 =3,x 3 =4,x 5 =3. The<br />

obvious solution is now feasible. You can now assemble the simplex tableau. The first step<br />

is to include a column and row for z. This yields<br />

⎛<br />

⎜<br />

⎝<br />

0 0 1 1 0 0 4<br />

1<br />

2<br />

1 0 − 1 2<br />

0 0 3<br />

3<br />

1<br />

2<br />

0 0<br />

2<br />

1 0 3<br />

−1 0 1 0 0 1 0<br />

Now you need to get zeros in the right places so the simple columns will be preserved as<br />

simple columns in this larger matrix. This means you need to zero out the 1 in the third<br />

column on the bottom. A simplex tableau is now<br />

⎛<br />

⎜<br />

⎝<br />

0 0 1 1 0 0 4<br />

1<br />

2<br />

1 0 − 1 2<br />

0 0 3<br />

3<br />

1<br />

2<br />

0 0<br />

2<br />

1 0 3<br />

−1 0 0 −1 0 1 −4<br />

Note it is not the same one obtained earlier. There is no reason a simplex tableau should<br />

be unique. In fact, it follows from the above general description that you have one for each<br />

basic feasible point of the region determined by the constraints.<br />

6.3 The Simplex Algorithm<br />

6.3.1 Maximums<br />

The simplex algorithm takes you from one basic feasible solution to another while maximizing<br />

or minimizing the function you are trying to maximize or minimize. <strong>Algebra</strong>ically,<br />

it takes you from one simplex tableau to another in which the lower right corner either<br />

increases in the case of maximization or decreases in the case of minimization.<br />

I will continue writing the simplex tableau in such a way that the simple columns having<br />

only one entry nonzero are on the left. As explained above, this amounts to permuting the<br />

variables. I will do this because it is possible to describe what is going on without onerous<br />

notation. However, in the examples, I won’t worry so much about it. Thus, from a basic<br />

feasible solution, a simplex tableau of the following form has been obtained in which the<br />

columns for the basic variables, x B are listed first and b ≥ 0.<br />

( )<br />

I F 0 b<br />

0 c 1 z 0 (6.10)<br />

⎞<br />

⎟<br />

⎠<br />

⎞<br />

⎟<br />

⎠ .

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