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Decision Maths! Usually there is a particular aim in making one

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					D1/5 LINEAR PROGRAMMING                                      5.1. Formulation and solution
Specification
Be able to manipulate inequalities algebraically (L1)
Be able to illustrate linear inequalities in two variables graphically (L2)
Be able to formulate simple maximisation of profit and minimisation of cost problems (L3)
Be able to use graphs to solve 2D problems: show alternative feasible points and their associated costs/profits
(L4 part)
Learning objectives
Properly understand the terms ‘formulation’, ‘constraints’, ‘objective function’ and ‘feasible region’
Be able to manipulate inequalities algebraically
Be able to illustrate linear inequalities in two variables graphically
Be able to formulate simple maximisation of profit and minimisation of cost problems
Know that a redundant constraint is one which does not border the feasible region
Ch.5 p.138-145                                               Ex.5A Q1,3,4,6
                                                             Multiple Choice Test 1

Decision Maths! Usually there is a particular aim in making one decision rather than another.
This aim might be to maximise profit or minimise cost. Linear programming produces a
mathematical model of a situation in which requirements, constraints and objectives are
expressed as algebraic equations. The process dates from World War 2.

Example 1
A small firm builds two types of garden shed.
Type A requires 2 hours of machine time and 5 hours of craftsman time.
Type B requires 3 hours of machine time and 5 hours of craftsman time.
Each day there are 30 hours of machine time available and 60 hours of craftsman time.
The profit on each type A shed is £60 and on each type B shed is £84.

Formulating the LP problem (Modelling)
1. Identify the variables to be used (decision variables). What must we decide?
    Let x be the number of Type A sheds produced per day
    Let y be the number of Type B sheds produced per day

2.    Constraints
      Machine time: 2x + 3y ≤ 30
      Craftsman time: 5x + 5y ≤ 60
           This can be simplified: x + y ≤ 12
      “Assumed”: x ≥ 0, y ≥ 0 (we cannot produce a negative number of sheds)

3.    Objective function
      We wish to produce sheds so as to maximise the total profit.
      Expression for total profit: P = 60x + 84y

This leads us to a LP expression:
     Maximise P = 60x + 84y
     subject to 2x + 3y ≤ 30
                x + y ≤ 12
                x ≥ 0, y ≥ 0

Solving the LP problem
The constraints can be represented graphically. How do we draw the region corresponding to
2x + 3y ≤ 30? The (very sensible) convention is that we shade out the regions not required.
Autograph:
     15   y




     10




      5


                   FEASIBLE REGION

                                                                                            x

                                    5                           10                          15




Shading out unacceptable points keeps the feasible region clear. The feasible region contains
every possible way the sheds could be produced subject to the constraints.

We wish to maximise the objective function subject to the constraints. Obviously the bigger x
and y are, the bigger P will be, so we look at points on the border of the feasible region away
from O. There are three candidates:

           x       y          P = 60x + 84y
 1         0      10               840
 2         6       6               864                    Solution: x = 6, y = 6, P = 864
 3        12      0                720

So:
•   The best solution will be found at a vertex of the feasible region
•   Find all the vertices and evaluate the objective function at each
•   You could use simultaneous equations to locate the vertices exactly, although “an
    accurate graph is acceptable”.

Example 2
A firm has to move 1200 people in its lorries and vans.
Each lorry can carry 200 parcels and each van 50.
There are 12 drivers, 7 lorries and 15 vans available.
It costs £75 to use a lorry and £25 to use a van. What is the cheapest way to do the job?

1.   Decision variables
     Let x be the number of lorries used
     Let y be the number of vans used

2.   Constraints
     No. of parcels: 200x + 50y ≥ 1200 ⇒ 4x + y ≥ 24
     No. of drivers: x + y ≤ 12
     No. of lorries: x ≤ 7
     No. of vans: y ≤ 15
     Also x ≥ 0, y ≥ 0

3.   Objective function
     Minimise cost C = 75x + 25y

Graph:
     25   y




     20




     15




     10




      5



                                                                                         x

                                     5                        10                         15



NB. y ≤ 15 is a redundant constraint because it does not border the feasible region.

Tour of vertices:

          x         y            C = 75x + 25y
 1        4         8                 £500
 2        6         0                 £450                  Solution: x = 6, y = 0, C = 450
 3        7         0         Must cost more than 2
 4        7         5        Must cost more than 2, 3

Resources
•   PowerPoint “LP example” – maximisation problem
•   LinPro – interactive spreadsheet
•   There is a Flash animation on the MEI DL website.
D1/5   LINEAR PROGRAMMING

Example 1
A small firm builds two types of garden shed.
Type A requires 2 hours of machine time and 5 hours of craftsman time.
Type B requires 3 hours of machine time and 5 hours of craftsman time.
Each day there are 30 hours of machine time available and 60 hours of craftsman time.
The profit on each type A shed is £60 and on each type B shed is £84.
How many of each type should we make?

Example 2
A firm has to move 1200 people in its lorries and vans.
Each lorry can carry 200 parcels and each van 50.
There are 12 drivers, 7 lorries and 15 vans available.
It costs £75 to use a lorry and £25 to use a van. What is the cheapest way to do the job?

				
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Description: Decision Maths! Usually there is a particular aim in making one