Slide 1
Simplex Method
Finite Math
Section 5.1
Slide 2
Standard Maximization
Problems
A linear programming problem is a standard
maximization problem if the following
conditions are met:
The objective function is linear and is to be
maximized.
The variables are all nonnegative
(i.e., x ≥ 0, y ≥ 0, z ≥ 0, …).
The structural constraints are all of the
form ax + by + … ≤ c, where c ≥ 0.
Slide 3
Examples
The following constraints are in the form
appropriate for a standard maximization
problem:
2x – 3y ≤ 9
–5x + 2y ≤ 11
x + 5y + 2z ≤ 8
The following constraints are not in the
form appropriate for a standard maximization
problem:
x + 4y ≥ 3
2x + y ≤ - 4
–2x + 3z ≥ y + 11
Slide 4
Slack Variables
The first step in the simplex method is to
convert each structural constraint into an
equality by adding a slack variable to
the left side and replacing the inequality
symbol with an equal sign.
Slide 5
Slack Variables
Each constraint requires a different slack variable.
A slack variable “takes up the slack” of the inequality
and ensures equality.
For any point in the feasible region of a standard
maximization problem, the value of each slack
variable is nonnegative.
When a point of intersection of two boundary lines
lies outside the feasible region, at least one of the
slack variables is negative.
If the optimum value of the objective function has
been reached and a slack variable still has a positive
value, there is a surplus equal to that positive value.
Slide 6
Values of Slack Variables
Note for a slack variable s:
s = 0 for all points on a border line of
the feasible region.
s > 0 for all points inside the feasible
region
s < 0 for all points outside the feasible
region
Slide 7
Example
Given the following standard maximization
problem:
Maximize f = 2x + 3y subject to
2x + y < 4
x+y <3
x>0
y>0
Insert slack variables where appropriate in
the constraints and form the system of slack
equations.
Slide 8
Example Continued
For each point on the accompanying
graph of the feasible region, find the
values of x, y, s1, s2, and f.
Slide 9
Basic and Nonbasic Variables
Example #1
Identify the basic and nonbasic variables
in the following matrix.
4 0  2 1 5


1
1
3
0
6


Slide 10
Basic and Nonbasic Variables
identity or unit column: contains a 1 in
the pivot position and zeros above and below
the 1
basic variable: a variable with an identity
or unit column. The value of a basic variable
is found by locating the row with the number
1 & reading the constant in the last column of
that row.
non-basic variable: a variable with a nonidentity column. The value of a non-basic
variable is zero.
Slide 11
Basic and Nonbasic Variables
Example #2
Identify the basic and nonbasic variables
in the following matrix.
 1 3 0 2 0  2 6


6 2
0  1 0 1 1
0 5 1 4 0
1 7
Slide 12
The Geometry of Pivoting
Example #1
Write the solution from the following
matrix and decide if the solution is a
corner point of the feasible region.
 3 1 0 2  2


9
 1 0 1 4
Slide 13
The Geometry of Pivoting
If a pivot operation is performed on any
nonzero element in an augmented matrix of
slack variables, and if the nonbasic variables are
set equal to zero in the resulting matrix, then
the resulting solution is a point where two
boundary lines of the feasible region intersect.
If all of the variables in the resulting matrix are
non-negative, the solution is a corner point of
the feasible region.
Slide 14
The Geometry of Pivoting
Example #2
Write the solution from each the following
matrix and decide if the solution is a
corner point of the feasible region.
 1 0 0 1 0 3 2


5  1 1 0 0  1 12 
0 5 0 0 1 1 3
Slide 15
Smallest-Quotient Rule
Example #1
Use the smallest-quotient rule to identify
the first pivot for the following matrix.
 2 0 1 1 0 40 


 1 1 0 3 0 60 
 1 0 0 5 1 38 
Slide 16
Smallest-Quotient Rule
Assuming
a) we have started with the augmented matrix for a
system of slack equations obtained from any
standard maximization problem, and
b) we have obtained a feasible solution by setting the
nonbasic variables equal to zero, and
c) the current augmented matrix has a solution that
reflects a corner point of the feasible region,
then the following rule takes us from one corner point
to another, each time getting closer to the optimal
solution.
Slide 17
Smallest-Quotient Rule
Select the pivot column by finding the most negative
indicator.
Divide each positive number in that column into the
corresponding number in the constant column.
Select as the pivot row the row corresponding to the
smallest nonnegative quotient, zero included.
Pivoting on the element where the pivot column and
the pivot row intersect will always result in a solution
that occurs at a corner point of the feasible region.
Such a solution is called a basic feasible solution
of the linear programming problem.
Note: If, after pivoting, you have a negative number in
the constant column of your matrix, you have chosen
the wrong pivot element.
Slide 18
Smallest-Quotient Rule
Example #2
Use the smallest-quotient rule to identify
the first pivot for the following matrix.
0 0
4 0 0 10 
1


0
3 1
0 0 0 8


0
4 0
1 3 0 2


 3  2 1  3 0 1 24 