The problem. Needed for contract. 10% Reduction. Resources available

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1 As part of May 2010 s P2 paper, a question was asked (Q6) which required candidates to prepare a graph showing the optimum production plan for the organisation in question. In his post examination guide (PEG), available on the CIMA website, the examiner identified several deficiencies in the responses attempted by candidates. This article revisits that question and, in a detailed analysis of the answer, aims to inform future candidates of a better way of producing a high scoring answer. The problem Firstly, we need to remind ourselves of the initial problem via this extract from the full question (please refer to the May 2010 paper for the whole question): RT produces two products from different quantities of the same resources using a just-in-time (JIT) production system. The selling price and resource requirements of each of the products are shown below: Product R T Unit selling price ($) Resources per unit: Direct labour ($8 per hour) 3 hours 5 hours Material A ($3 per kg) 5 kgs 4 kgs Material B ($7 per litre) 2 litres 1 litre Machine hours ($10 per hour) 3 hours 4 hours Market research shows that the maximum demand for products R and T during June 2010 is 500 units and 800 units respectively. This does not include an order that RT has agreed with a commercial customer for the supply of 250 units of R and 350 units of T. At a recent meeting of the purchasing and production managers to discuss the production plans of RT for June, the following resource restrictions for June were identified: Direct labour hours Material A Material B Machine hours 7,500 hours 8,500 kgs 3,000 litres 7,500 hours It then becomes apparent that the predicted resource restrictions were rather optimistic and that their availability could be as much as 10% lower than their original predictions. When we look at the production which RT wishes to undertake, including the fulfilment of the contract, and compare it to the resources at its disposal, we can see that there is more than one scarce resource. Resource Initial quantity 10% Reduction Needed for contract Resources available Resources needed for maximum demand Direct labour 7, ,500 4,250 5,500 Material A 8, ,650 5,000 5,700 Material B 3, ,850 1,800 Machine hours 7, ,150 4,600 4,700 For illustration of this table, let s look at the direct labour hours. Originally it was thought that 7,500 labour hours would be available but now the forecast has been reduced by 10%. The contract requires the following hours: R 250 units x 3 hours = 750 hours T 350 units x 5 hours = 1,750 hours 2,500 hours

2 This leaves 7, ,500 = 4,250 hours available. Maximum demand requires (500 x 3) + (800 x 5) = 5,500 hours, hence the direct labour is a scarce resource. Similar calculations show that material A and machine hours are also scarce. Having more than one scarce resource means that we cannot solve the problem using contribution per unit of scarce resource and then ranking the products as we could in parts (a) and (b) of this question (this was also covered in my earlier Financial Management articles of November / December 2009 and March 2010). The examiner has told you this by telling you to use graphical linear programming to show the optimum production plan to answer part (c). Preparing the solution When preparing solutions to graphical linear programming problems, I recommend a five step approach. Step 1 Define the variables We must define the variables (letters) which we are going to use. For example: Let R be the number of units of product R to be produced in June Let T be the number of units of product T to be produced in June Step 2 Determine the objective function Next we need to determine the objective of the exercise and express it as an equation. RT wishes to maximise total contribution (TC) where TC = 47R + 61T You may wish to confirm for yourself that the variable cost of a unit of R is $83 and hence its contribution per unit is $(130 83) = $47. You can do the same for T. Step 3 Establish the constraints This is where we express the resource constraints in the form of inequalities. It sounds difficult, but if you get the first one they should be logical to follow. Looking at direct labour: Each unit of R takes 3 hours, each unit of T takes 5 hours and we know that only 4,250 hours available. Therefore the total usage of the labour hours must be less than (<) or equal (=) to the total labour hours available. We can show this as follows: 3R + 5T <= 4,250 (labour hours) The same logic can then be applied to the other three resource constraints: 5R + 4T <= 5,000 (Material A) 2R + 1T <= 1,850 (Material B) 3R + 4T <= 4,600 (Machine hours)

3 That s the resource constraints dealt with, but let s not forget that there are two other restrictions on the values that R and T can take. Since we are using a just-in-time production system, we will not produce any more units than we are selling. Therefore, there is an upper limit of 500 units on the production of R, and 800 units for T. i.e. R <= 500 (production) T <= 800 (production) More precisely, we should also define the lower limit to the values for R and T, which in this scenario is zero: 0 <= R <= <= T <= 800 Taking all of this together, we have now defined our problem which is that we wish to maximise contribution TC = 47R + 61T subject to the following (numbered) constraints 3R + 5T <= 4, R + 4T <= 5, R + 1T <= 1, R + 4T <= 4, <= R <= <= T <= Step 4 Graph the constraints (by converting them into straight line equations) This seems to be the most difficult area for students and was where many candidates let themselves down in the examination. Regarding May 2010 s candidates answers, the examiner said in the PEG, many candidates earned only half the available marks due to a number of issues including poorly constructed graphs, presenting only an inaccurate sketch graph and not putting forward a final answer...many graphs were poorly constructed; the lines and the axes were not labelled and the graphs were very untidy. Clearly, this needs some attention. In order to draw the graph, you need to accurately plot each of the inequalities expressed above as straight line equations. So, for example, constraint 1 becomes: 3R + 5T = 4,250 To plot this line, you need only 2 points. I recommend plotting the two extreme points in this manner: Putting R = 0, then the equation simplifies to 5T = 4,250 And T = 850. Similarly, putting instead T = 0, the equation simplifies to 3R = 4,250 And R = 1,

4 This not only gives us two points on the graph, and hence the first constraint line, but it gives us the first indication of the scale needed for the graph. Too many students in my experience produce graphs that are either too large in that the constraint lines cannot be shown on the page or alternatively too small such that they are squashed postage stamp sized in the corner of the page. The figures for constraint 1 suggest that a scale of around 1 centimetre for 100 units may suffice (performing similar calculations on other constraints may confirm this). Graph 1 showing labelled scaled axes (as per examiners answer)

5 Plotting the figures for constraint 1 gives us the following two points: Graph 2 showing two points of the first constraint

6 And hence the two points can be joined to form the first constraint line, which should then be labelled. Graph 3 showing the first constraint

7 Moving onto constraint 2 for material A which becomes: 5R + 4T = 5,000 Find the 2 points: Putting R = 0, then the equation simplifies to 4T = 5,000 And T = 1,250. Similarly, putting instead T = 0, the equation simplifies to 5R = 5,000 And R = 1,000. Now we have the second constraint line on the diagram (and confirmation that our scale seems appropriate tip, do all of these workings first and know the maximum figure for both R and T) Graph 4 showing the second constraint

8 If we repeat this exercise for the other two resource equations and the two (simpler) demand equations, then we get the fuller picture regarding the constraints. Graph 5 showing all 6 constraints and feasible area OABCD These lines represent the maximum possible use of each of these constraints. They form a boundary beyond which production cannot take place. In fact, production can only take place in the area formed by points OABCD. This is called the Feasible Region and should be labelled on your graph. Step 5 Determine the optimum solution Before stepping into the intricacies of finding the optimum point, a little common sense can simplify matters. Point B must be a better outcome than point A. At A, units of R are 0, whilst units of T are 800. At point B, we have the same 800 units of T, but clearly some units of R, and therefore the total contribution at point B must be bigger than it is at point A. Similarly, point C is bound to be a better outcome than point D. At D, units of T are 0, whilst units of R are 500. At point D, we have the same 500 units of R, but over 500 units of T. This logic points to the optimum solution being either at B or C or even at both and at all points in between. This is where a good, accurate graph is really beneficial, as you can simply read off the coordinates of points B and C. Point B has T = 800 and R = (provable mathematically by placing T = 800 into the labour constraint: 3R + (5 x 800) = 4,250 3R = 250 R = 83.33)

9 Point C has R = 500 and T = 550. (provable mathematically by placing R = 500 into the labour constraint: (3 x 500) + 5T = 4,250 5T = 2,750 T = 550) The total contribution we can earn at each point will tell us which is better, B or C, and we do this by using the objective function established in Step 2. At B, Total contribution = (47 x 83.33) + (61 x 800) = 52, At C, Total contribution = (47 x 500) + (61 x 550) = 57,050. Therefore, the optimum point is C with production of 500 units of R and 550 units of T yielding a total contribution of $57,050. The textbook way of arriving at the optimum point is to use what is known as the Iso-contribution function referred to in the examiners answer. With this, we plot the objective function Total contribution = 47R + 61T, which reflects the relative unit contributions of the products namely $47 for a unit of R, $61 for a unit of T. In order to do this, we can take any value for total contribution. Students often find it baffling that we can seemingly invent a value for total contribution, but the point is that the line will move anyway as we seek to find the best possible point of production. My tip here is to select a value for total contribution which is a multiple of the individual contributions i.e. 47 x 61 = $2,867 so that it makes it easy to plot your line. This seems a little low in the context of the question, so I shall scale it up by 10 to $28,670. Therefore, we want to plot the line for: 28,670 = 47R + 61T.

10 Once again, Putting R = 0, then the equation simplifies to 28,670 = 61T 470 = T Similarly, putting instead T = 0, the equation simplifies to 28,670 = 47R 610 = R Adding this to our graph gives us Graph 6 showing all 6 constraints plus dotted Iso-contribution line as per examiners answer

11 If you place your ruler along this line and slide it outwards, but keeping it parallel to the initial line (this represents what would happen when output is increased: the line would move outwards), you should be able to see the furthest possible point which can be reached while keeping part of it within the feasible region is point C. This is therefore the optimum point as we proved by inspection earlier. Summary Graph 7 showing all 6 constraints lines plus the new Iso-contribution line, parallel to the original one, but going through C In summary, once the problem has been formulated as in steps 1 to 3 above, you need to: Draw on a graph two axes to represent the two decision variables, R and T in this case Plot all of the constraints of the model as straight lines by evaluating where the limiting equations intersect the axes. Identify the area on the graph (known as the feasible region) that satisfies all of the constraints. Find the optimum point. If the problem is one of maximisation, as in this case, then the values of the decision variables, R and T, computed above either by inspection or graphically, that yields the largest figure for the objective function (Total contribution), constitute the combination which makes best use of the scarce resources. Ian Janes is CIMA Course Leader at Newport Business School, where his colleague Rosemary Eaton assisted in the technical production of the graphs.

12 Question for students (answer to be given in the April 2011 edition of Velocity) The final part of the question in the May 2010 examination, part (d), asked candidates to consider how the graph in your solution to (c) above can be used to help to determine the optimum production plan for June 2010 if the actual resource availability lies somewhere between the managers optimistic and pessimistic predictions If we isolate the effect of having more direct labour hours available, chosen since this is the binding resource constraint, then we can see from graph 7 that an expansion of hours will mean an outward movement of that line and hence point C will also move. This will bring about increased production of units of T, with units of R remaining the same. This will mean more total contribution. With this in mind: (a) Calculate the shadow price of direct labour. (b) Assuming that it is just more direct labour hours which become available, and there s no further availability of the other three scarce resources, state the upper limit of units and hours over which the shadow price in (a) remains the same, and state (and explain) the shadow price of labour beyond that limit