The Simplex Method
The graphical method of solving linear programming problems is
useful only for problems involving two decision variables and
relatively few problem constraints.
What happens when we need more decision variables and more problem constraints?
We use an algebraic method called the simplex method, which
was developed by George B. DANTZIG (1914-2005) in 1947 while
on assignment with the U.S. Department of the air force.
Standard Maximization Problems in
Standard Form
A linear programming problem is said to be a standard maximization problem in
standard form if its mathematical model is of the following form:
Maximize the objective function
Zmax P c1x1 c2x2 ... cnxn
Subject to problem constraints of the form
a1x1 a2x2 ... anxn b ,b 0
With non-negative constraints
x1, x2,..., xn 0
Slack Variables
“A mathematical representation of surplus
resources.” In real life problems, it’s unlikely
that all resources will be used completely, so
there usually are unused resources.
Slack variables represent the unused resources
between the left-hand side and right-hand
side of each inequality.
Basic and Nonbasic Variables
Basic variables are selected arbitrarily with the restriction that
there be as many basic variables as there are equations. The
remaining variables are non-basic variables.
x1 2x2 s1 32
3x1 4x2 s2 84
This system has two equations, we can select any two of the four
variables as basic variables. The remaining two variables are
then non-basic variables. A solution found by setting the two
non-basic variables equal to 0 and solving for the two basic
variables is a basic solution. If a basic solution has no negative
values, it is a basic feasible solution.
SIMPLEX METHOD
Step-1
Step 4
Write the Step 2 Are there Step-5
standard Are there any positive Select the
maximization any Step-3 elements in Y pivot
problem in negative Y Select the pivot element
indicators the column and
standard form,
in the pivot above the perform
introduce slack column
bottom dashed the pivot
variables to form
row? line? operation
the initial system,
and write the
initial tableau. NO NO
STOP
STOP The linear programming
The optimal solution has been problem has no optimal
found. solution
Simplex algorithm for standard maximization problems
To solve a linear programming problem in standard form, use the following steps.
1- Convert each inequality in the set of constraints to an equation by adding slack
variables.
2- Create the initial simplex tableau.
3- Select the pivot column. ( The column with the “most negative value” element
in the last row.)
4- Select the pivot row. (The row with the smallest non-negative result when the
last element in the row is divided by the corresponding in the pivot column.)
5-Use elementary row operations calculate new values for the pivot row so that
the pivot is 1 (Divide every number in the row by the pivot number.)
6- Use elementary row operations to make all numbers in the pivot column equal
to 0 except for the pivot number. If all entries in the bottom row are zero or
positive, this the final tableau. If not, go back to step 3.
7- If you obtain a final tableau, then the linear programming problem has a
maximum solution, which is given by the entry in the lower-right corner of
the tableau.
Pivot
Pivot Column: The column of the tableau
representing the variable to be entered into
the solution mix.
Pivot Row: The row of the tableau representing
the variable to be replaced in the solution mix.
Pivot Number: The element in both the pivot
column and the pivot row.
Simplex Tableau
Most real-world problems are too complex to
solve graphically. They have too many corners
to evaluate, and the algebraic solutions are
lengthy. A simplex tableau is a way to
systematically evaluate variable mixes in order
to find the best one.
Initial Simplex Tableau
All variables Solution
Basic variables coefficients
0
EXAMPLE
The Cannon Hill furniture Company produces tables
and chairs. Each table takes four hours of labor
from the carpentry department and two hours of
labor from the finishing department. Each chair
requires three hours of carpentry and one hour
of finishing. During the current week, 240 hours
of carpentry time are available and 100 hours of
finishing time. Each table produced gives a profit
of $70 and each chair a profit of $50. How many
chairs and tables should be made?
STEP 1
All information about example
Resource Table s ( x1 ) Chairs (x2 ) Constraints
Carpentry (hr) 4 3 240
Finishing (hr) 2 1 100
Unit Profit $70 $50
Objective Function P 70x1 50x2
Carpentry Constraint 4x1 3x2 240
Finishing Constraint 2x1 1x2 100
Non-negativity conditions x1, x2 0
The first step of the simplex method requires that each inequality
be converted into an equation. ”less than or equal to”
inequalities are converted to equations by including slack
variables.
Suppose s1 carpentry hours and s2 finishing hours remain unused
in a week. The constraints become;
4x1 3x2 s1 240 4x1 3x2 s1 0s2 240
or
2x1 x2 s2 100 2x1 x2 0s1 s2 100
As unused hours result in no profit, the slack variables can be
included in the objective function with zero coefficients:
P 70x1 50x2 0s1 0s2
P 70x1 50x2 0s1 0s2 0
The problem can now be considered as solving a system of 3 linear
equations involving the 5 variables x1, x2, s1, s2, P in such a way
that P has the maximum value;
4x1 3x2 s1 0s2 240
2x1 x2 0s1 s2 100
P 70x1 50x2 0s1 0s2 0
Now, the system of linear equations can be written in matrix form
or as a 3x6 augmented matrix. The initial tableau is;
STEP 2
Right
Basic
x1 x2 S1 S2 P Hand
Variables
Side
S1 4 3 1 0 0 240
S2 2 1 0 1 0 100
P -70 -50 0 0 1 0
The tableau represents the initial solution;
x1 0, x2 0, s1 240, s2 100, P 0
The slack variables S1 and S2 form the initial solution mix. The initial
solution assumes that all avaliable hours are unused. i.e. The slack variables
take the largest possible values.
Variables in the solution mix are called basic variables. Each basic
variables has a column consisting of all 0’s except for a single 1.
all variables not in the solution mix take the value 0.
The simplex process, a basic variable in the solution mix is
replaced by another variable previously not in the solution
mix. The value of the replaced variable is set to 0.
STEP 3
Select the pivot column (determine which variable to enter into the
solution mix). Choose the column with the “most negative”
element in the objective function row.
Right
Basic
x1 x2 S1 S2 P hand
Variables
side
S1 4 3 1 0 0 240
S2 2 1 0 1 0 100
P -70 -50 0 0 1 0
Pivot column
x1 should enter into the solution mix because each unit of x1 (a table)
contributes a profit of $70 compared with only $50 for each unit of x1 (a
chair)
STEP 5
Select the pivot row (determine which variable to replace in the solution mix).
Divide the last element in each row by the corresponding element in the
pivot column. The pivot row is the row with the smallest non-negative
result.
Enter
Right
Basic
x1 x2 S1 S2 P hand
Variables
side
S1 4 3 1 0 0 240 240/4 60
Exit S2 2 1 0 1 0 100 100/2 50
P -70 -50 0 0 1 0
Pivot row
Pivot column
Pivot number
Should be replaced by x1 in the solution mix. 60 tables can be made with 240
unused carpentry hours but only 50 tables can be made with 100 finishing
hours. Therefore we decide to make 50 tables.
Now calculate new values for the pivot row. Divide every number in the row
by the pivot number.
Right
Basic
x1 x2 S1 S2 P hand
Variables
side
S1 4 3 1 0 0 240
R2
x1 1 1/2 0 1/2 0 50 2
P -70 -50 0 0 1 0
Use row operations to make all numbers in the pivot column equal to 0 except
for the pivot number which remains as 1.
Right
Basic
x1 x2 S1 S2 P hand
Variables
side
S1 0 1 1 -2 0 40 4.R2 R1
x1 1 1/2 0 1/2 0 50
P 0 -15 0 35 1 3500 70.R2 R3
If 50 tables are made, then the unused carpentry hours are reduced by 200
hours (4 h/table multiplied by 50 tables); the value changes from 240 hours to 40
hours. Making 50 tables results in the profit being increased by $3500; the value
changes from $0 to $3500.
In this case, x1 50, x2 0, s1 40, s2 0, P 3500
Now repeat the steps until there are no negative numbers in the last row.
Select the new pivot column. x2 should enter into the solution mix.
Select the new pivot row. S1 should be replaced by x2 in the solution mix.
Enter
Right
Basic
x1 x2 S1 S2 P hand
Variables
side
Exit S1 0 1 1 -2 0 40 40/1 40
x1 1 1/2 0 1/2 0 50 50/0,5 100
P 0 -15 0 35 1 3500
New pivot row
New pivot
column
Calculate new values for the pivot row. As the pivot number is already 1,
there is no need to calculate new values for the pivot row.
Use row operations to make all numbers in the pivot column equal to
except for the pivot number.
Right
Basic
x1 x2 S1 S2 P hand
Variables
side
x2 0 1 1 -2 0 40
1
x1 1 0 -1/2 3/2 0 30 .R1 R2
2
P 0 0 15 5 1 4100 15.R1 R3
If 40 chairs are made, then the number of tables are reduced by
20 tables (1/2 table/chair multiplied by 40 chairs); the value
changes from 50 tables to 30 tables. The replacement of 20
tables by 40 chairs results in the profit being increased by
$600; the value changes from $3500 to $4100.
As the last row contains no negative numbers, this solution gives
the maximum value of P.
Result
This simplex tableau represents the optimal
solution to the LP problem and is interpreted
as:
x1 30, x2 40, s1 0, s2 0
and profit or P=$4100
The optimal solution (maximum profit to be
made) is to company 30 tables and 40 chairs
for a profit of $4100.
Example-2
A farmer owns a 100 acre farm and plans to plant at
most three crops. The seed for crops A,B, and C costs
$40, $20, and $30 per acre, respectively. A maximum
of $3200 can be spent on seed. Crops A,B, and C
require 1,2, and 1 workdays per acre, respectively,
and there are maximum of 160 workdays available. If
the farmer can make a profit of $100 per acre on
crop A, $300 per acre on crop B, and $200 per acre
on crop C, how many acres of each crop should be
planted to maximize profit?
