5. Linear Programming

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5. Linear Programming
Objectives

1. Problem formulation
2. Solving an LP problem graphically
3. Bounded regions and corner points

Refs BZ 5.2.
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Example 1
See handout for description.
Our first task is to formulate a mathematical
model of this problem.
In this example, we are asked to determine the
number of ads to place in two different
newspapers.
So let
x be the number of ads placed in Yarra News
y be the number of ads placed in Sun City Paper
The variables x and y are called
decision variables.
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Our objective is to maximize the number of people
who see our ad.
The exposure depends on the number of people
who read each newspaper.
For each ad we place in Yarra News, we can
guarantee that 50,000 people see it.
As the number of ads placed in Yarra News is
represented by x we see that
50,000x
people see our ad.
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For each ad we place in Sun City Paper we can
guarantee that 20,000 people see it.
As the number of ads placed in Sun City Paper is
represented by y we can see that
20,000y
people see this ad.
So our exposure statement can be written as
P50,000x20,000y
This is called the
objective function.
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Obviously, making x and y as large as possible
will increase the value of P. But there are
bounds (or constraints) on the values of x and y.
The first constraint concerns the total amount
of money which the company can spend on
advertising.
We are told that there is an upper limit of
9,000.
So we need to formulate an equation which
represents the cost of advertising.
1. Each ad in Yarra News will cost 300.
Since x represents the number of ads we place in
Yarra News, we can represent this part of our
cost by
300x
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2. Similarly, each ad in Sun City Paper will
cost us 100.
Since y represents the number of ads we place in
Sun City Paper, we can represent this component
of cost by
100y
So in total, the cost of advertising is
300x 100y.
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To satisfy the conditions of our problem, we
must ensure that
300x 100y 9,000.
The second constraint concerns the number of
ads we place each month. This can be described as
x y 30
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A final obvious (but necessary) constraint is
that the number of ads we place in each paper
cannot be negative.
So
x 0,
y 0.
These are called
non-negativity constraints.
We are now ready to give a mathematical
description of our problem.
maximize
P50,000x 20,000y
Subject to
300x 100y 9,000
x y 30
x 0
y 0
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To solve such a problem we must firstly determine
which pairs of points (x, y) satisfy all of the
above constraints.
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This region is called the feasible region and any
point in it is called a feasible point.
A feasible region is either bounded (which means
it can be enclosed within a circle) or unbounded
(it cannot be enclosed within a circle).
Each point (x,y) in the feasible region will give
us a value for P. Our task is to determine which
point, or points, gives the maximum value of P on
the feasible region.
remember this is our objective function
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1. P 800,000 50,000x 20,000y
800,000
2. P 1,200,000 50,000x 20,000y 1,200,000
3. P 1,600,000 50,000x 20,000y 1,600,000
4. P 2,000,000 50,000x 20,000y 2,000,000
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Notice that we can write the equation
P 50,000x 20,000y
So we see that changing the value of P does not
change the slope of the lines - they are
parallel.
Also note that as P increases the y-intercept
increases.

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Notice that the line 50,000x 20,000y
2,000,000 does not intersect the feasible region.
Since increasing P will move the line further
from the feasible region, we can see that our
solution must have
Plt 2,000,000.
The line 50,000x 20,000y 1,600,000 does
intersect The feasible region - any point in the
feasible region which is on this line will give
P 1,600,000.
Since there are feasible points to the right of
this line we conclude that we can find an (x, y)
which will give us a value of
P gt 1,600,000.
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So now we have
1,600,000 lt P lt 2,000,000
but how do we find P - the maximum value of P ?
Place a ruler on the line
50,000x 20,000y P
(any P will do) and slide it along, parallel to
this line (in the direction of increasing P)
until you are just touching the edge of the
feasible region.
Draw the line.
What has happened?
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(x, y) (0, 90)
This line determines the maximum value of P and
intersects The feasible region at one point only.
To determine this value of P we simply evaluate
the objective function at the point of
intersection (x, y).
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Point of intersection.
Easy in this case because it is on the axis.
At this point
P 50,000x 20,000y
50,000(0) 20,000(90)
1,800,000.
So the optimal solution is
(x, y) (0,90) and
max P 1,800,000.
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