06.09.2021 Views

Linear Algebra, Theory And Applications, 2012a

Linear Algebra, Theory And Applications, 2012a

Linear Algebra, Theory And Applications, 2012a

SHOW MORE
SHOW LESS

Create successful ePaper yourself

Turn your PDF publications into a flip-book with our unique Google optimized e-Paper software.

138 LINEAR PROGRAMMING<br />

a solution to the system of equations corresponding to the system of equations represented<br />

by (6.6) and row operations leave solution sets unchanged. Note how attractive this is. The<br />

z 0 is the value of z at the point x 0 . The augmented matrix of (6.9) is called the simplex<br />

tableau and it is the beginning point for the simplex algorithm to be described a little<br />

later. It is very convenient to express the simplex( tableau ) in the above form in which the<br />

I<br />

variables are possibly permuted in order to have on the left side. However, as far<br />

0<br />

as the simplex algorithm is concerned it is not necessary to be permuting the variables in<br />

this manner. Starting with (6.9) you could permute the variables and columns to obtain an<br />

augmented matrix in which the variables are in their original order. What is really required<br />

for the simplex tableau?<br />

It is an augmented m +1× m + n + 2 matrix which represents a system of equations<br />

which has the same set of solutions, (x,z) T as the system whose augmented matrix is<br />

( )<br />

A 0 b<br />

−c 1 0<br />

(Possibly the variables for x are taken in another order.) There are m linearly independent<br />

columns in the first m + n columns for which there is only one nonzero entry, a 1 in one of<br />

the first m rows, the “simple columns”, the other first m + n columns being the “nonsimple<br />

columns”. As in the above, the variables corresponding to the simple columns are x B ,<br />

the basic variables and those corresponding to the nonsimple columns are x F , the free<br />

variables. Also, the top m entries of the last column on the right are nonnegative. This is<br />

the description of a simplex tableau.<br />

In a simplex tableau it is easy to spot a basic feasible solution. You can see one quickly<br />

by setting the variables, x F corresponding to the nonsimple columns equal to zero. Then<br />

the other variables, corresponding to the simple columns are each equal to a nonnegative<br />

entry in the far right column. Lets call this an “obvious basic feasible solution”. If a<br />

solution is obtained by setting the variables corresponding to the nonsimple columns equal<br />

to zero and the variables corresponding to the simple columns equal to zero this will be<br />

referred to as an “obvious” solution. Lets also call the first m + n entries in the bottom<br />

row the “bottom left row”. In a simplex tableau, the entry in the bottom right corner gives<br />

the value of the variable being maximized or minimized when the obvious basic feasible<br />

solution is chosen.<br />

The following is a special case of the general theory presented above and shows how such<br />

a special case can be fit into the above framework. The following example is rather typical<br />

of the sorts of problems considered. It involves inequality constraints instead of Ax = b.<br />

This is handled by adding in “slack variables” as explained below.<br />

The idea is to obtain an augmented matrix for the constraints such that obvious solutions<br />

are also feasible. Then there is an algorithm, to be presented later, which takes you from<br />

one obvious feasible solution to another until you obtain the maximum.<br />

Example 6.2.2 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 for a problem of the form x ≥ 0,Ax = b<br />

which is equivalent to the above problem.<br />

You add in slack variables. These are positive variables, one for each of the first three constraints,<br />

which change the first three inequalities into equations. Thus the first three inequalities<br />

become x 1 +2x 2 +x 3 =10,x 1 +2x 2 −x 4 =2, and 2x 1 +x 2 +x 5 =6,x 1 ,x 2 ,x 3 ,x 4 ,x 5 ≥ 0.<br />

Now it is necessary to find a basic feasible solution. You mainly need to find a positive so-

Hooray! Your file is uploaded and ready to be published.

Saved successfully!

Ooh no, something went wrong!