Lecture on 3-19 to 3-26
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SA305 Spring 2014
Canonical Form
To construct the simplex method we need to put our linear programs all in a similar form
so that the algorithm is standardized and can use the mechanics of the extreme points.
Definition 1.1. A linear program with n variables is in canonical form if it is of the following
form
max ~c>~x
A~x = ~b
~x ≥ 0
where A = (aij ) is a m × n matrix, m ≤ n, and the rows of A are linearly independent.
Can every linear program be put in canonical form? First let’s look at an example.
Example 1.2. Let
x + 2y ≤ 1
2x + y ≥ 1
x, y ≥ 0
be the constraints of a LP.
For each constraint where there might be slack or surplus you add or subtract a slack
or surplus variable to amke the constraint equality and then append the LP with the nonnegativity of these new variables.
So our new constraints look like
x + 2y + s1
2x + y
x, y, s1 , s2 ≥ 0.
−s2
Now let’s put this in matrix form. Let


x
 y 

~x = 
 s1 
s2
A=
1 2 1 0
2 1 0 −1
and
1
=1
=1
~b =
1
1
.
Then we can view the constraints as
A~x = ~b
~x ≥ 0.
This might seem ridiculous but actually this setting makes it much easier to deal with
basic feasible solutions.
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Basic Variables
How do we find basic feasible solutions? This is one of the main questions in designing our
algorithm because the fundamental theorem says that we should only focus on these extreme
points.
With our linear program in canonical form it makes the job a lot easier. Let’s first look
at this example.
Example 2.1. Have constraints
A~x = ~b
~x ≥ 0
where


x
 y 

~x = 
 s1 
s2
1 2 1 0
A=
2 1 0 −1
and
~b =
1
1
.
Since the number of variables n = 4 and the number of rows of A is m = 2 this means
that for a solution ~x0 to be a basic feasible solution it must have n − m = 4 − 2 of the
non-negativity constraints be active. For example we can choose x = 0 and y = 0. Then
solve for s1 and s2 and we get that s1 = 1 and s2 = −1. So a basic feasible solution is
~x0 = [0, 0, 1, −1]. Then we can list all the other basic feasible solutions in this way:
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basic solution
not active non-neg. constraint active non-negativity constraint
[0, 0, 1, −1]NOT feasible
s1 , s2
x, y
y, s2
x, s1
[0, 1/2, 0, −3/2]NOT feasible
[0, 1, −1, 0]NOT feasible
s1 , y
x, s2
x, s2
y, s1
[1, 0, 0, 1]
[1/2, 0, 1/2, 0]
s1 , x
y, s2
[1/3, 1/3, 0, 0]
x, y
s1 , s2
Now we note that this happens in general.
Theorem 2.2. Suppose that ~x = [x1 , . . . , xn ] is a basic feasible solution to an LP in canonical
form. Then there exists m indices i1 , . . . im (or variables xi1 , . . . , xim ) such that
(1) The columns i1 , . . . , im of A are linearly independent.
(2) The other variables are zero xj = 0 for j 6= i1 , . . . , im .
Because of the conclusion (1) and (2) in the Theorem since there are m non-zero variables
that must satisfy m linearly independent conditions we can find a unique solution that
satisfies these conditions.
Because of this decomposition we create explicit notation for these indices.
Definition 2.3. Let ~x be a basic feasible solution to an LP in canonical form.
• A variable xk is basic for ~x if k is one of the indices in the above theorem (equivalently
that coordinatein ~x is not zero).
• A variable xk is nonbasic for ~x if it is not basic (equivalently that coordinate in vecx
is zero).
• The set of basic variables for ~x is called the basis for ~x.
Using this notation we can rename our previous table in the exmaple.
Example 2.4.
basic variables nonbasic variables
basic solution
[0, 0, 1, −1]NOT feasible
s1 , s2
x, y
[0, 1/2, 0, −3/2]NOT feasible
y, s2
x, s1
[0, 1, −1, 0]NOT feasible
s1 , y
x, s2
x, s2
y, s1
[1, 0, 0, 1]
[1/2, 0, 1/2, 0]
s1 , x
y, s2
[1/3, 1/3, 0, 0]
x, y
s1 , s2
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