Decision Maths

1
Decision Maths

• Lesson 7 Linear Programming

2
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.
• First we are going to look at how to turn
  problems into algebraic equations.

3
Problem

• A company makes two types of garden shed, A and
  B.
• Both types require processing in two departments.
• Department 1 is where machines are used to
  produce the wood to the required lengths.
• Department 2 is where the craftsmen work on and
  produce the shed.
• Shed A requires 2 hours of machine time and 5
  hours of craftsman time. When sold it will earn
  60 profit.
• Shed B requires 3 hours of machine time and 5
  hours of craftsman time. When sold it will earn
  84 profit.
• Each day there are 30 hours of machine time and
  60 hours of craftsman time available.
• We want to find out how many of each shed we need
  to make each day in order to maximise our profit.

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Problem

• All of this can be summarised in a table.

5
Problem

• The first step in formulating a linear
  programming problem is to determine which
  quantities you need to know to solve the problem.
• These are called the Decision variables.
• The second is to decide what the Constraints are
  in the problem.
• What is it that will hold up production and
  prevent products being made.
• The third step is to decide what the objective to
  be achieved is.
• The function of the decision variables that is to
  be optimised is called the objective function.

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Problem - Step 1

• Step 1 Decide on the Decision variables.
• In this case it is clear that what we want to
  know is how many of each type of shed to make.
• Let x Number of Shed A made.
• Let y Number of Shed B made.

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Problem Step 2

• Step 2 Decide on the Constraints.
• Consider the work that needs to be done in the
  machine room.
• From the table we can see that each day there are
  30 hours available for work.
• One of shed A requires 2 hours. So x shed As
  must require 2x hours.
• Using a similar idea for shed B you can get total
  time for y shed Bs will be 3y.
• From this table it is now easy to write down the
  algebraic equations.

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Problem Step 2

• Now the total work done in the machine room on
  both sheds cannot exceed 30 hours in one day.
• Therefore we can formulate the algebraic
  equation.
• 2x 3y 30
• The other constraint in this problem is the work
  that needs to be completed in the craft shop.
• The work in the craft room cannot exceed 60
  hours.
• So 5x 5y 60

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Problem Step 2

• Non-Negativity constraints
• In addition to the two constraints that we have
  just looked at, it is obvious that both x and y
  must be positive numbers.
• Now the whole problem can be written like so
• 2x 3y 30
• 5x 5y 60
• x 0, y 0

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Problem Step 3

• Step 3 decide on the objective function.
• The whole problem is about selling sheds.
• We need to know how many of each of the sheds to
  sell to make maximum profit.
• This gives us the function 60x 84y P
• Where P stands for profit.

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Final - Problem

• The original problem can now be summarised in
  algebraic form.
• Maximise the profit function.
• P 60x 84y
• Subject to the constraints
• 2x 3y 30
• 5x 5y 60
• x 0, y 0

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Question 1

• Allwood PLC plans to make two kinds of table.
• For table A the cost of the materials is 20, the
  number of person-hours needed to complete it is
  10 and the profit, when sold, is 15.
• For table B the cost of materials is 12, the
  number of person-hours needed to complete it is
  15 and the profit, when sold, is 17.
• The total money available for materials is 480
  and the labour available is 330 person-hours.
• Formulate this as a linear programming problem.

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Answer 1

• X number of type A tables
• Y number of type B tables
• Maximise Z 15x 17y
• Subject to 20x 12y 480
• 10x 15y 330
• x 0, y 0

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Question 2

• To ensure that her family has a healthy diet Mrs
  Brown decides that the familys daily intake of
  vitamins A, B and C should not fall below 25
  units, 30 units and 15 units respectively.
• To provide these vitamins she relies on two fresh
  foods a and ß.
• Food a provides 30 units of vitamin A, 20 units
  of vitamin B and 10 units of vitamin C per 100g.
• Food ß provides 10 units of vitamin A, 25 units
  of vitamin B and 40 units of vitamin C per 100g.
• Food a costs 40p per 100g and food ß costs 30p
  per 100g.
• How many grams of food a and food ß should she
  purchase daily if the food bill is to be kept to
  a minimum?
• Formulate this as a linear programming problem.

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Answer 2

• x (hundred) grams of a
• y (hundred) grams of ß
• Minimise C 40x 30y
• Subject to 30x 10y 25
• 20x 25y 30
• 10x 40y 15
• x 0, y 0

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Question 3

• A chair supplier makes three types of wooden
  chairs. Each type is manufactured in a four-stage
  process. The company is able to obtain all the
  raw materials it needs. The available production
  capacity during the 60-hour production working
  week is as follows
• It is assumed that there are 60 hours of labour
  available for each process. The profits on each
  of the three types of chair are 15, 20 and 25
  respectively. Formulate this as a linear
  programming problem, given that the profit is to
  be maximised.

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Answer 3

• X number of type A made
• Y number of type B made
• Z number of type C made
• Maximise
• P 15x 20y 25z

• Subject to 9x 6y 4z 3600
• 2x 9y 12z 3600
• 18x 4y 6z 3600
• 6x 9y 8z 3600
• x 0, y 0, z 0