Programming To Pass
Ian Janes looks at an area often found problematic by candidates, graphical linear programming,
by revisiting the May 2010 P2 paper and advising on the approach to take in such questions. This is the full
article, an abridged version of which was published on p.44 of March 2011 Financial Management magazine.
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
Unit selling price ($)
Resources per unit:
Direct labour ($8 per hour)
Material A ($3 per kg)
Material B ($7 per litre)
Machine hours ($10 per hour)
R
130
T
160
3 hours
5 kgs
2 litres
3 hours
5 hours
4 kgs
1 litre
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
Direct labour
Material A
Material B
Machine hours
Initial
quantity
7,500
8,500
3,000
7,500
10%
Reduction
750
850
300
750
Needed for
contract
2,500
2,650
850
2,150
Resources
available
4,250
5,000
1,850
4,600
Resources needed for
maximum demand
5,500
5,700
1,800
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
1,750 hours
T 350 units x 5 hours =
2,500 hours
Programming To Pass
Ian Janes looks at an area often found problematic by candidates, graphical linear programming,
by revisiting the May 2010 P2 paper and advising on the approach to take in such questions. This is the full
article, an abridged version of which was published on p.44 of March 2011 Financial Management magazine.
This leaves 7,500 – 750 – 2,500 = 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)
Programming To Pass
Ian Janes looks at an area often found problematic by candidates, graphical linear programming,
by revisiting the May 2010 P2 paper and advising on the approach to take in such questions. This is the full
article, an abridged version of which was published on p.44 of March 2011 Financial Management magazine.
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 <= 500
0 <= 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 <= 4,250
5R + 4T <= 5,000
2R + 1T <= 1,850
3R + 4T <= 4,600
0 <= R <= 500
0 <= T <= 800.
1
2
3
4
5
6
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 = 1,416.67.
Programming To Pass
Ian Janes looks at an area often found problematic by candidates, graphical linear programming,
by revisiting the May 2010 P2 paper and advising on the approach to take in such questions. This is the full
article, an abridged version of which was published on p.44 of March 2011 Financial Management magazine.
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)
Programming To Pass
Ian Janes looks at an area often found problematic by candidates, graphical linear programming,
by revisiting the May 2010 P2 paper and advising on the approach to take in such questions. This is the full
article, an abridged version of which was published on p.44 of March 2011 Financial Management magazine.
Plotting the figures for constraint 1 gives us the following two points:
Graph 2 showing two points of the first constraint
Programming To Pass
Ian Janes looks at an area often found problematic by candidates, graphical linear programming,
by revisiting the May 2010 P2 paper and advising on the approach to take in such questions. This is the full
article, an abridged version of which was published on p.44 of March 2011 Financial Management magazine.
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
Programming To Pass
Ian Janes looks at an area often found problematic by candidates, graphical linear programming,
by revisiting the May 2010 P2 paper and advising on the approach to take in such questions. This is the full
article, an abridged version of which was published on p.44 of March 2011 Financial Management magazine.
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
Programming To Pass
Ian Janes looks at an area often found problematic by candidates, graphical linear programming,
by revisiting the May 2010 P2 paper and advising on the approach to take in such questions. This is the full
article, an abridged version of which was published on p.44 of March 2011 Financial Management magazine.
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 = 83.33.
(provable mathematically by placing T = 800 into the labour constraint:
3R + (5 x 800) = 4,250
3R = 250
R = 83.33)
Programming To Pass
Ian Janes looks at an area often found problematic by candidates, graphical linear programming,
by revisiting the May 2010 P2 paper and advising on the approach to take in such questions. This is the full
article, an abridged version of which was published on p.44 of March 2011 Financial Management magazine.
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) = 52,716.67
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.
Programming To Pass
Ian Janes looks at an area often found problematic by candidates, graphical linear programming,
by revisiting the May 2010 P2 paper and advising on the approach to take in such questions. This is the full
article, an abridged version of which was published on p.44 of March 2011 Financial Management magazine.
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
Programming To Pass
Ian Janes looks at an area often found problematic by candidates, graphical linear programming,
by revisiting the May 2010 P2 paper and advising on the approach to take in such questions. This is the full
article, an abridged version of which was published on p.44 of March 2011 Financial Management magazine.
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.
Graph 7 showing all 6 constraints lines plus the new Iso-contribution line,
parallel to the original one, but going through C
Summary
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.
Programming To Pass
Ian Janes looks at an area often found problematic by candidates, graphical linear programming,
by revisiting the May 2010 P2 paper and advising on the approach to take in such questions. This is the full
article, an abridged version of which was published on p.44 of March 2011 Financial Management magazine.
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