1. 7. Linear Programming (Simplex Method)
Objectives:
Slack variables
Basic solutions - geometry
Examples
Refs: B&Z

2. Last week we saw how to solve a Linear Programming
problem geometrically. This method, however, has
limitations.
If we increase the number of constraints we may have
hundreds of corner points.
If we have more than two decision variables we will
not even be able to sketch the graph.
Fortunately, a procedure known as the Simplex Method
will handle all these cases efficiently.

3. Slack Variables In week 1 we learnt techniques to solve systems of
linear equations. However we don’t have good techniques
for handling systems of inequalities.
Our task is to transform a system we can’t handle into one
that we can.
5x1+ 2x2=1,200
200
600
400
Consider the example:
In this example we have 2
decision variables (x1, x2)
and 2 functional constraints.
max
P = 4x1+ 5x2
subj to
2x1+ 3x2 ≤ 600
5x1+ 2x2 ≤ 1,200
x1 ≥ 0
x2 ≥ 0
2x1+ 3x2=600
feasible region

4. We introduce slack variables which turn our original system
2x x2
≤ 600
5x x2
≤ 1,200
into
2x x2
5x x2
+ s1
+ s2
= 600
= 1,200
The slack variables “pick up the slack” on the LHS of
the inequalities. You should use a different slack variable in
each constraint.
Notice that we are adding a positive value on the LHS
to obtain the equality so we should say
s1≥ 0, s2≥ 0

5. So now we have an L.P. problem which is equivalent to
the original (in the sense that they have the same optimal
solution) but with no inequalities.
without slack variables
max
P = 4x1+ 5x2
subj to
2x1+ 3x2 ≤ 600
5x1+ 2x2 ≤ 1,200
x1 ≥ 0
x2 ≥ 0
with slack variables
max
P = 4x1+ 5x2
subj to
2x1+ 3x2 + s1 = 600
5x1+ 2x s2 =1, 200
x1 ≥ 0,
x2 ≥ 0, s1 ≥ 0, s2 ≥ 0

6. Let’s check the values of s1 and s2 at the corner points.
(x1, x2, s1, s2 ) = (0, 0, 600, 1200)
P (0, 200) = 1,000
s1 = 0
s2 = 800
(x1, x2, s1, s2 ) = (0, 200, 0, 800)
P (240, 0) = 960
s1 = 120
s2 = 0
(x1, x2, s1, s2 ) = (240, 0, 120, 0)
P (2400/11, 600/11) = (9, ,000)/11
= 12,600/11
= 1, /11
s1 = 0
s2 = 0
(x1, x2, s1, s2 ) = (2400/11, 600/11, 0, 0)

7. Observation:
In this example there are two zeroes occurring at each
corner point. In general, if we have n variables and
m constraints we will have n-m zeroes at each corner point.
In the previous example we had 4 variables (2 decision
variables and 2 slack variables) and 2 constraints:
n=4, m=2 so n-m=2 zeroes.
These solutions are called basic solutions. In fact if we
have an L.P. problem with n variables and m equations,
and set any n-m variables equal to 0, we have a basic
solution.

8. Some of these points will be non-feasible.
For example: (i)
x2 = s1 = 0
x1 = 300, s2 = -300
x1 = s2 = 0
x2 = 600, s1 = -1200
(ii)
The non-feasibility is indicated by the negativity of
some variables.
The basic solutions consist of all corner points of the
feasible region and some non-feasible points.
The corner points are those with non-negative
co-ordinates. They are called basic feasible solutions.

9. Example 1:
Find all basic feasible solutions of the following system:
Max P = 5x1 + 6x2
Subj to
4x1 + 2x2 ≤ 200
x1 + 3x2 ≤ 150
x1≥ 0
x2≥ 0
First add slack variables so that our new constraints are:
4x1 + 2x2 + s = 200
x1 + 3x s2 = 150
x1≥ 0
x2≥ 0
s1≥ 0
s2≥ 0

10. In this example we have 2 equations and 4 variables.
150
100
4x1+2x2=200
x1+3x2=150
feasible region 50
(30, 40)
In this example we have 2 equations and 4 variables.
We find basic solutions by setting 2 variables at a time
equal to zero.

11. x x s s2
1. feasible
2. Not feasible
3. feasible
4. feasible
5. Not feasible
6. feasible
Basic feasible solutions are:
(0,0,200,150), (0,50,100,0),
(50,0,0,100), (30,40,0,0)

12. To solve the L.P. problem we need to evaluate the
objective function at each of the basic feasible solutions.
However, in practice this becomes impractical. Say
for example we had an L.P. problem with 3 decision
variables and 3 constraints (hence 3 slack variables).
Putting 6-3=3 variables equal to zero at a time gives
basic solutions. For 4 decision variables and 5
constraints, we have
basic solutions………and so on. Luckily there
is a better way!

13. You should now be
able to complete
Q’s 1(a), (b) and (c)
Example Sheet 3
from the Orange Book.