Chapter 3: Using Graphs

1. Chapter 3: Using Graphs

2. Objectives • Create graphs of times series data • Illustrate break-even analysis • Show a feasible area • Solve two variable linear programming problems

3. Time Series • Probably the most common graph • Very simple to construct • By hand • By computer • Very simple to understand • Works for annual, quarterly, monthly weekly, daily or even hourly data

4. Annual Data • Graph has years on X-axis • Data values on the y-axis • Annual data smooths out short-term effects • Often used to consider long-term trends in the data • If they exist Data used by kind permission of the National Gallery, New Media Department: http://www.nationalgallery.org.uk/default.htm

5. Quarterly Data • Quarterly data will show seasonal patterns • Many of these are obvious, eg. coat sales • Knowledge of patterns helps in planning for the business

6. Short time periods

7. Break-even Analysis • Tries to answers the fundamental question • “How many do we have to make/serve to cover our costs?” • Any business which cannot do this will, in the long run, fail • Not to mention the cash flow problems in the short term. • For a single product company, the calculation is simple • Much more difficult of a large, multi-product company since you then need to address the accounting question of allocation of overheads.

8. Graphing Break-even Revenues £ Firstly we need to identify costs Then revenues In the simplest case, both of these will be linear functions Costs Output Break-even Break-even is where the two functions cross

9. Calculating Break-even It will usually be easier, and more convenient, to calculate the break-even figure If we sell the product at £5, then the Revenue function is 5X If the fixed cost is 120 and unit cost is 2, then the Cost function is 120 + 2X To do this we use X for the output and set up cost and revenue functions Revenue = Cost 5X = 120 + 2X 3X = 120 X = 40 Now put them equal to each other to find the X-value This is the break-even production figure.

10. Calculating Break-even (2) An alternative, and quicker, way to calculate break-even is to use the accounting concept ofcontribution First step is to find the difference between the price per unit and the cost per unit eg. If P = 40 and C = 25 then the contribution (from each unit sold) is 15 Then divide the Fixed Cost by the contribution eg. If Fixed Cost is 3000 then the break-even is 3000/15 = 200

11. Break-even with Non-linear functions • If the cost function is non-linear, then we can still graph the cost and revenue functions • Break-even will still be where R = C • For a quadratic cost function, there may be two break-even points

12. Break-even with Non-linear functions (2) • An alternative is to define a Profit Function as • Profit = Total Revenue – Total Cost • Then graph this function • Break-even is where it crosses the X-axis • (if it does)

13. Linear Programming Linear programming is a technique which seeks the optimum allocation of scarce resources between competing products or activities It has been used in a wide range of situations in business, government and industry. Examples include: optimum product mix media selection share portfolio selection

14. Feasible Area We are trying to create a graph which showsallfeasible mixes of the products, media types or shares. We will limit our analysis to onlytwoitems, but you should note that the techniques will work in much more complex situations The first step is always toformulatethe problem i.e. to write out equations

15. An Example A small company (Singletons & Co.) make two products. They are asking for your advice on what mix of products to make, and have been able to provide the following information: Ambers require1hour of labour time Zeonites require2hours of labour time Total labour hours per week is40 Ambers require6litres of moulding fluid Zeonites require5litres of moulding fluid Maximum moulding fluid per week is150litres Profit contribution from Ambers is£2 Profit contribution from Zeonites is£3

16. Formulating the Problem What are we trying to achieve? Probably maximum profit Where does this profit come from? The two products we produce How much profit do we make? Profit = £2A + £3Z

17. Limitations If there were a plentiful supply of everything we needed, then there would be no problem! This is never the case! Labour We only have 40 hours per week of labour available We know that Ambers take 1 hour each And Zeonites take 2 hours each Total hours used will be: A + 2Z But this total must be less than or equal to 40 So:A + 2Z <= 40

18. What does this look like? Z 30 A + 2Z = 40 20 10 A 20 40 Feasible area

19. More Limitations Moulding Fluid: We only have 150 litres per week of moulding fluid available We know that Ambers take 6 litres each And Zeonites take 5 litres each But this must be less than or equal to 150 litres Total Fluid used will be: 6A + 5Z So:6A + 5Z <= 150

20. The Graph Z 30 6A + 5Z = 150 20 10 A 20 40 Feasible area

21. An Assumption We assume that it is only possible to get answers which are either zero or are positive This means that: A >= 0 andZ >= 0 In terms of a graph, this means that we work in the first quadrant i.e. The one where both variables are positive.

22. Output Combinations We need to find combinations of outputs which are feasible underallconstraints i.e. those which use no more than thelabouravailable and no more than themoulding fluidavailable Since we have graphs of each constraint, we can bring these together to find The feasible area for the whole problem

23. Feasible Solutions Z 30 A + 2Z = 40 Feasible Area 20 6A + 5Z = 150 10 A 20 40

24. How many do we produce? On the graph, the feasible area has several “corners” One where we produceonlyZeonites (0,20) One where we produceonlyAmbers (25, 0) And one where we produce acombinationof the two where the two constraints cross: 6A + 5Z =150 A + 2Z = 40 6A + 12Z = 240 Multiply by 6 Subtract the first equation from this one 7Z = 90 Z = 12.857 A = 14.2857 Substituting gives:

25. Profit Levels We can evaluate the profit contribution at each “corner” of the feasible area. Profit Contribution = 2A + 3Z 60 50 67.14 Highest

26. LP - Minimisation The previous example tried to maximise profit contribution but the technique can also be used for finding minimum cost solutions

27. LP – Minimisation (2) EXAMPLE: A company has 2 machines, A & B which can each produce either the MINI or MAXI version of their product A can produce 5 MINI or 1 MAXI per session B can produce 2 MINI or 3 MAXI per session Contracts dictate that the minimum number of MINI’s must be 100 of MAXI’s must be 90 The cost of running machine A is £1000 per session The cost of running machine B is £2000 per session What is the minimum cost number of sessions for each machine?

28. Formulation We can construct equations as follows: Number of MINI’s 5A + 2B >= 100 Number of MAXI’s A + 3B >=90 A, B >= 0 Costs: Minimise 1000A + 2000B Again we can use a graph.

29. Graphical Representation B 50 Feasible Area 30 20 90 A

30. Corners & Solution The corners of the feasible area are at: A = 0, B = 50 A = 90, B = 0 And A = 9.23, B = 26.9 The cost function is: 1000A + 2000B £100,000 £90,000 £63,030 MINIMUM

31. Conclusions • Linear programming provides a method of solution for a wide range of problems • It is not limited to two items and a few constraints, as in our example • Computer basedsolutions are easily available • - for small problems you can use an add-in to Excel • - for large problems there is specialist software • It provides a short to medium term solution , but • in the long run, managers need to address theresource constraintsthemselves if they wish to increase production levels