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Linear%20Optimisation

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Title: PowerPoint Presentation Author: MAGS Last modified by: Joanna McHardy Created Date: 11/7/2004 3:39:32 AM Document presentation format: On-screen Show (4:3) – PowerPoint PPT presentation

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Title: Linear%20Optimisation


1
Linear Optimisation
2
Example 1 Bursary 1991 Question 2
A video chain has been conducting research for a
new outlet. The outlet can stock old classic
films only, new releases only, or a combination
of these. The maximum number of old classics
available is 4500 the maximum number of new
releases available is 8000. There is shelf space
in the outlet for no more than 9000 videos. The
cost of stocking each old classic is 50, and for
each new release it is 30. The maximum budget
for stocking is 300,000. The profit on each old
classic is 40 and on each new release it is
60. The video chain wants to find how many of
each type of video it should stock to maximise
its profit.
3
Step 1 Define your variables x Old
Classics y New Releases
4
Step 2 Write the inequations for the constraints.
5
Step 3 Draw the graph and establish the feasible
region
6
Step 4 Write the objective function
7
Step 5 List coordinates of vertices in the
feasible region and substitute values into
objective function to find the maximum profit.
Conclusion Buy 1000 Old Classics and 8000 New
Releases for a maximum profit of 520,000
8
Example 2 1995 Bursary
Louise makes and sells two types of herbal
hand-cream rosemary and lavender. Each jar of
rosemary cream takes 4 minutes to make and
requires 60 grams of lanolin. Each jar of
lavender cream takes 12 minutes to make and
requires 100 grams of lanolin. Each day Louise
can spend at most 180 minutes making hand-cream
and can obtain up to 1800 grams of lanolin. Her
friend Camille does the packing and can pack at
most 25 jars a day. The rosemary cream sells at a
profit of 4 a jar and the lavender cream sells
at a profit of 5 a jar. Find the solution to the
linear programming problem that maximises the
total profit.
9
Step 1 Define your variables
Let x represent the number of jars of rosemary
cream, and y represent the number of jars of
lavender cream.
10
Step 2 Write inequations for the constraints
11
Step 3 Graph constraints and identify the
feasible region.
12
Step 4 Write the objective function
13
Step 5 List coordinates of vertices in the
feasible region and substitute values into
objective function to find the maximum profit.
Conclusion make 17.5 and 7.5 resp. of each for a
maximum profit of 107.50
14
Problem What if Louise can make only whole
numbers of jars each day
Solution Analyse all integer values around the
solution to find the maximum. Remember to
eliminate those that do not meet the constraints.
Conclusion Make 18 jars of Rosemary and 7 of
Lavendar for maximum profit of 107
15
Example 3 Bursary 1994
The school principal has arranged for a day out
and needs transport for the journey. Provision
must be made to transport at least 400 students,
36 teachers and 120 boxes of supplies. The local
rental company can provide vans at 50 each and
buses at 160 each. Each van can take 10
students, 1 teacher and 4 boxes each bus can
take 40 students, 3 teachers and a certain number
of boxes.
16
Step 1 Define the variables
-vans (x) and - buses (y)
17
Step 2 Write the inequations for the constraints.
Notes We do not need the non-negatives as
everything is greater than. We do not know how
many boxes a bus can take- information will come
from the graph
18
Graph is given. Step 3 Identify the feasible
region (Region 1) and check the intercepts of
boxes line (y-intercept is 15 and hence bus can
take 120/15 8 boxes.
19
Step 4 Write the objective function (Cost to be
minimised in this case)
20
Step 5 Identify the vertices in the feasible
region and evaluate.
Conclusion Minimise cost by using 24 vans and 4
buses
21
Problem The day before the trip the rental
company informs the principal that while the cost
of a van was correctly given as 50, there was a
mistake in giving the cost of a bus. What range
of costs for bus hire will still give the same
optimum solution found.
Solution For the same solution the gradient of
the objective function must lie between the
gradients of the two adjacent lines.
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