Decision Maths Linear Programming

Linear ProgrammingDecision 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.

ProblemA 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.

ProblemAll of this can be summarised in a table.

ProblemThe 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.

Problem - Step 1Step 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.

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 A`s must require 2x hours. Using a similar idea for shed B you can get total time for y shed B`s will be 3y.

From this table it is now easy to write down the algebraic equations.

Problem Step 2Now 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 30The 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

Problem Step 2Non-Negativity constraintsIn 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 305x + 5y 60x 0, y 0

Problem Step 3Step 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 = PWhere P stands for profit.

Final - ProblemThe original problem can now be summarised in algebraic form.Maximise the profit function.P = 60x + 84ySubject to the constraints

2x + 3y 305x + 5y 60x 0, y 0

Question 1Allwood 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.

Answer 1X = number of type A tablesY = number of type B tables

Maximise Z = 15x + 17ySubject to 20x + 12y 48010x + 15y 330x 0, y 0

Question 2To 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 and . Food provides 30 units of vitamin A, 20 units of vitamin B and 10 units of vitamin C per 100g. Food provides 10 units of vitamin A, 25 units of vitamin B and 40 units of vitamin C per 100g. Food costs 40p per 100g and food costs 30p per 100g. How many grams of food and food should she purchase daily if the food bill is to be kept to a minimum? Formulate this as a linear programming problem.

Answer 2x = (hundred) grams of y = (hundred) grams of

Minimise C = 40x + 30ySubject to 30x + 10y 2520x + 25y 3010x + 40y 15x 0, y 0

Question 3A 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.

Answer 3X = number of type A madeY = number of type B madeZ = number of type C made

Maximise P = 15x + 20y + 25z

Subject to9x + 6y + 4z 36002x + 9y + 12z 360018x + 4y + 6z 36006x + 9y + 8z 3600x 0, y 0, z 0