# Paper P2 Performance Management: Many Students Tend to Find Graphical Linear Programming Difficult, So It's Well Worth Revisiting May 2010's Paper to Determine How Best to Tackle Questions That Require It

## Article excerpt

Question 6 of May 2010's P2 paper required candidates to "prepare a graph showing the optimum production plan" for a particular manufacturer. In his post-exam guide, the examiner identified several deficiencies in candidates' answers. I will revisit that question and, through detailed analysis, aim to help candidates score better marks on future questions of this type.

First, here's a reminder of the initial problem in the following extract from question 6 (for more details refer to the May 2010 paper, which is available via your "My CIMA" account). A company called RT makes two products - R and T - using varying quantities of two raw materials known as a A and B. It uses a just-in-time (JIT) production system in both cases. The selling price and resource requirements for the two products are shown in the table at the top of the next page.

[ILLUSTRATION OMITTED]

Market research has indicated that the maximum demand for R and T in June 2010 will be 500 and 800 units respectively, not including a separate order that has been agreed with a commercial customer for 250 units of R and 350 units of T.

```Product                         R          T

Unit selling price              \$130       \$160

Resources used per unit:

Direct labour (@ \$8 per hour)   3hours     5hours

Material A (@ \$3 per kg)        5kg        4kg

Material B (@ \$7 per litre)     2litres    1litre

Machine hours (@ \$10 per hour)  3          4
```

At a recent meeting the purchasing and production managers identified the following resource restrictions for June:

* Direct labour: 7,500 hours.

* Material A: 8,500kg.

* Material B: 3,000 litres.

* Machine hours: 7,500.

But it has since emerged that the managers' forecasts were too optimistic: the availability of these resources could be as much as ten per cent lower than they had first predicted. When you look at the company's production targets, including the fulfilment of the contract, and compare them with the resources at its disposal, you can see that there is a shortage of more than one of those resources. Taking direct labour hours, for example: fulfilling the contract requires 750 hours of labour to make 250 units of R and 1,750 hours to make 350 units of T, equating to a total of 2,500 hours. This leaves 7,500 - 750 - 2,500 = 4,250 hours available. The maximum demand requires (500 x 3) + (800 x 5) = 5,500 direct labour hours. Since this is greater than the number of hours available, direct labour is a scarce resource.

Similar calculations for material A and machine hours show that they are also scarce resources: 5,700kg of A are needed to satisfy maximum demand, but 5,000kg are available; 4,700 machine hours are needed to satisfy maximum demand, but 4,600 hours are available. (Material B is not scarce, since 1,800 litres are needed to satisfy maximum demand and 1,850 litres are available.)

Having more than one scarce resource means that you cannot solve the problem by simply using "contribution per unit of scarce resource" and then ranking the products, as you could in parts (a) and (b) of question 6. (My earlier FM articles of March 2010 and November/December 2009 also covered that aspect.) The examiner has indicated this by requiring you to use "graphical linear programming to show the optimum production plan" in your answer to part (c).

I recommend that candidates take the following five-step approach to solving graphical linear programming problems.

Step one: define the variables.

First you need to define the variables (letters) that you're going to use. For example, let R be the number of units of product R to be made in June and T be the number of units of product T.

Step two: work out the objective function.

Next you need to determine the objective of the exercise and express it as an equation as follows: RT wishes to maximise the total contribution (TC), where TC = 47R + 61T. …