Linear
Programming
The Modelling Cycle in Decision Maths
Accept
solution
Yes
Real life Problem
No
Compare the solution
with reality – is it
realistic?
Interpret the solution in
terms of the original
problem
Review
Make simplifying
assumptions
Define variables and
decide on the
mathematical techniques
to be used
Solve the mathematical
problem
What is linear programming?
A linear programming problem consists of a collection of
linear inequalities of a number of real variables and a given
linear function of these real variables which is to be
maximised or minimised.
Linear Programming
 Decision making is a process that has to be carried out in many areas
of life.
 After the Second World War a group of American mathematicians
developed some mathematical methods to help with decision making.
 They produced mathematical models that turned the requirements,
constraints and objectives of a project into algebraic equations.
 Linear programming is the process of solving these equations by
searching for an Optimal Solution.
 The optimal solution is the maximum or minimum value of a required
function.
 Linear programming methods are some of the most widely used
methods employed to solve management and economic problems, they
can be applied to a variety of contexts, with enormous savings in
money and resources.
Linear Programming
1. Be able to manipulate inequalities algebraically.
2. Be able to illustrate linear inequalities in two variables
graphically.
3. Be able to formulate simple maximisation of profit and
minimisation of cost problems.
•
Formulation of constrained optimisation problems
4. Be able to use graphs to solve 2-D problems, including
integer valued problems.
•
Solution of constrained optimisation problems
5. Be able to interpret solutions, including spare capacities.
•
Algebraic interpretation of the graphical solution in 2-D.
Key Points
1. Formulating
•
Let x be the number of … , let y be the number of …
•
Text into linear constraints
•
Have you used all the information in the question?
•
Don’t forget x ≥ 0, y ≥ 0
2. Graphing
•
Two variables only
•
Use intercepts on axes, where feasible to do so
•
Label equations of lines; show inequalities by shading
Key Points
3. Solving
•
The objective may be to minimise or maximise
•
Plot an example objective function (choose RHS)
•
Test the vertices of the Feasible Region.
•
The solution could be all the points along one edge.
4. Interpreting
•
Does it have to be an integer solution?
•
Are there alternative feasible solutions?
•
You may need to comment on the solution.
Example:
A small factory produces two types of toys: bicycles and
trucks. In the manufacturing process two machines are
used: the lathe and the assembler. The table shows the
length of time needed for each toy:
The lathe can be operated for 16 hours a day and there
are two assemblers which can each be used for 12 hours a
day. Each bicycle gives a profit of £16 and each truck gives
a profit of £14. Formulate and solve a linear programming
problem so that the factory maximises its profit.
Example:
A small factory produces two types of toys: bicycles and
trucks. In the manufacturing process two machines are
used: the lathe and the assembler. The table shows the
length of time needed for each toy:
The lathe can be operated for 16 hours a day and there
are two assemblers which can each be used for 12 hours a
day. Each bicycle gives a profit of £16 and each truck gives
a profit of £14. Formulate and solve a linear programming
problem so that the factory maximises its profit.
Bicycle
Truck
Lathe
2 hours
1 hour
Assembler
2 hours
3 hours
1. Formulate the problem
Let x be number of bicycles made
Let y be number of trucks made.
Objective Function
Maximise P = 16x + 14y
Subject to constraints
2x + y  16 Lathe
2x + 3y  24 Assembler
x, y  0
There are two
assemblers
Define variables and
decide on the
mathematical techniques
to be used
Finding the Optimum Value
Method 1: Tour of vertices
(0,8) profit = £112
(6,4) profit = £152
(8,0) profit = £128
Optimal solution is to
make 6 bicycles and 4 trucks. Profit £152
Method 2: Profit Line
Draw a line through the origin parallel to
the gradient of the profit function. Move
this line up the y-axis until it is just
leaving the feasible region – the point at
which it leaves the feasible region is the
optimum value.
Profit line
y = 1/14 (P – 16x)
Using ICT
Finding the Optimum Value using AUTOGRAPH
If you have Autograph you can use this to solve graphical
LP problems
2. Using the on-line resources (www.mei.org.uk)
LinPro
Integer Prog
What happens next?
In real life you would go back to the
original situation and assess whether
this solution is reasonable and
feasible before deciding whether to
implement it.
Compare the solution with
reality – is it realistic?
Minimisation problems
Minimise C = 3x + 4y
subject to the constraints:
3x - 4y ≤ 12,
x + 2y ≥ 4
x ≥1, y ≥ 0.
The feasible region for this set of
constraints is shown on the graph.
The value of C at each vertex
C = 3x + 4y
(1, 1.5): 3(1)+4(1.5) = 9
(4, 0): 3(4)+4(0) = 12
solution
x = 1, y = 1.5,
C = 9.
January 2010 question 4
Link to the examination paper
Link to FMSP revision recording
January 2010 question 4
x
June 2010 question 4
Link to the examination paper
Link to FMSP revision recording
June 2010 question 4
x
What happens if there are more than 2 variables?
In geometric terms we are considering a closed, convex,
region, P, (known as a polytope), defined by intersecting a
number of half-spaces in n-dimensional Euclidean space
(these are the constraints).
If the objective is to maximise a
linear function L(x), consider the
family of hyperplanes, H(c),
defined by L(x) = c.
As c increases, these form a
parallel family. We want to find the
largest value of c such that H(c)
intersects P.
Optimal
Solution
Starting
Vertex
What is linear programming?
In this case we can show that the optimum value of c is
attained on the boundary of P using the extreme point
theorem
If P is a convex polygon and L(x) is a linear
function then all of the values of L(x) at the points
of P, both maximum and minimum occur at the
extreme points.
Hence, if an LP has a bounded optimal solution
then there exists an extreme point of the feasible
region that is optimal
Convex
region
An Introduction to Linear Programming and the Theory of Games
by Abraham M Glicksman
Published by Dover publications Isbn 0-486-41710-7
Concave
region
Introducing the simplex method
Methods for finding this
optimum point on P work in
several ways: some attempt to
improve a possible point by
moving through the interior
of P (so-called interior point
methods);
others start and remain on the
boundary searching for an
optimum.
Optimal
Solution
Starting
Vertex
Solution by simplex
Graphical method
1. Formulate
the problem
 Objective
Function
 Subject to
constraints
Solve problem
Simplex method
Let x be no. of bicycles made
Let y be no. of trucks made.
Maximise P = 16x + 14y
Lathe
2x + y  16
Assembler
2x + 3y  24
Maximise P – 16x – 14y = 0
2x + y + s1 =16
2x + 3y + s2 = 24
y
16
14
12
10
8
6
4
2
x
0
0
2
4
6
8
10
12
14
Solution: P = 152, x = 6, y = 4
16
p x y
1 -16 -14
0 2 1
s1
0
1
s2
0
0
0
16
0
2
0
1
24 24/2=12
1
0
0
0
1
0
0
0
1
128
8 8/0.5=16
8 8/2=4
1
0
0
0
1
0
3
-6 8
0.5 0.5
2
-1
0
0
1
ratio test
5
3 152
0.75 -0.25 6
-0.5 0.5 4
16/2=8
Jeff Trim
jefftrim@furthermaths.org.uk
Central Coordinator
& Area Coordinator, SE Region
Further Mathematics Support Programme