Final Assessment
ACCA
Paper F5
Performance Management December 2008
Final Assessment – Answers
To gain maximum benefit, do not refer to these answers until you have completed the final assessment questions and submitted them for marking.
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ACCA F5 Performance Management
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Final Assessment
ANSWER 1
(a) (i) OAR = activity of level Budgeted
overheads Budgeted = £8m
£880,000 = 11% of list price
Selling and distribution charge for Order A = 11% of £1,200 = £132 Selling and distribution charge for Order B = 11% of £900 = £99
(ii) Cost driver rates
Invoice processing
Cost per invoice = invoices 8,000£280,000 25%× = £8.75 per invoice
Cost per invoice line = lines invoice 28,000
£280,000 75%× = £7.50 per invoice line
Packing £32 for large packages and £25 for small packages. Delivery
Loading costs = journeys 1,000
£40,000 = £40 per journey
There are 12 small packages to a lorry and so the loading costs are £40/12 = £3.33 per small package. There are 6 large packages to a lorry and so the loading costs are £6.67 per large package.
Mileage costs = miles 350,000£40,000 £180,00 − = £0.40 per mile
Other overheads
Cost per order = orders 8,000
£200,000 = £25 per order
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ACCA F5 Performance Management
A B £ £
Invoice processing 8.75 8.75 £7.50 × 2 15.00 £7.50 × 8 60.00 Packing 25.00 32.00 Delivery 3.33 6.67 £0.40 × 8 3.20 £0.40 × 40 16.00 Other overheads 25.00 _____ 25.00 ______ Charge for selling and distribution 80.28 _____ 148.42 ______
(b) REPORT
To Management From Management Accountant Date May 20X5 Subject Proposed ABC system (i) Strengths and weaknesses of proposed system
The present system is very simple but makes no attempt to link the selling and distribution costs to the factors which cause those costs. The present system simply charges all orders a blanket rate of 11% on list price. The proposed ABC system is still very simple but makes some effort to determine the cost drivers, i.e. those factors which are most closely related to the way in which the costs of an activity are incurred. For instance, it has been found for the invoice processing costs that the costs are affected by the number of invoices issued, but also by how complicated the invoices are, i.e. how many different lines there are on the invoice. Charging out the invoice processing costs on the basis of the two cost drivers above will result in more accurate costs and will give more information about the cost structure and the cost drivers in order to improve cost control. Once the cost structure is known, efforts can be made to reduce the volume of activity of the cost driver (e.g. the 28,000 invoice lines) and/or the cost of the cost driver (e.g. the £7.50 per invoice line). It is argued by ABC supporters that the better costs calculated under ABC can then be used as the basis for fixing selling prices and that these selling prices relate to the true cost of the order and thus will prevent loss-making orders. F plc can also compare the true costs of the different elements of the system against the costs of outsourcing. The more accurate costs determined under an ABC system can be used to justify selling prices or selling price increases to customers. Whilst it is undoubtedly true that ABC gives more accurate costs, it is also true that it will be more expensive to implement and the benefit may not exceed the cost.
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Final Assessment
The proposed system, here, is still very simple and it is possible that a more detailed analysis would provide further useful information. ABC attempts to find the cost driver for each type of cost and thus avoid the arbitrariness of absorption costing but, here, some costs are still charged on an arbitrary basis, e.g. the other overheads.
(ii) Recommendations
In the light of the ABC information it should be determined whether any orders/customers are loss making and, if so, whether those types of order or customer should be dropped or if the costs can be reduced or the selling prices increased.
The cost per invoice line is very expensive. It should be investigated as to whether this figure can be reduced, perhaps by the purchase of a special software package.
The costs for long distance deliveries are also very high. F plc could consider only accepting orders below a certain maximum distance, or charging a selling price which takes into account the delivery costs of long distance journeys, or outsourcing the deliveries.
The present system for setting selling prices is cost based; whilst this is a simple method of setting selling prices, it does not normally lead to optimum selling prices. The proposed ABC system could be used as the basis for charging selling prices to individual customers. Whilst the proposed system is very simple, it is probably still too complex to be used for charging prices. For instance, if a new customer wants to place a new order, we would need to know how far away they are before we can calculate the cost of the order and thus the price to quote.
MARKING GUIDE
Marks
(a) (i) OAR 1 Order A costs 1 Order B costs 1 _
Total 3 (ii) Cost driver rates Invoice processing 2 Packing 1 Delivery 3 Other 1 Order A costs (0.5 per cost element) 3 Order B costs (0.5 per cost element) 3 _
Total 13 (b) 1 marks for each point made, to a maximum of 5 for section (i) and 4 for section (ii)
9 __
Total 25 __
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ACCA F5 Performance Management
ANSWER 2 (a) Step 1: determine limiting factor
Resources needed Resources available
Cleaning materials 0.2 (W) × 8,000 + 0.3 (W) × 10,500 4,750 5,000 Direct labour 0.2 (W) × 8,000 + 3
1 (W) × 10,500 5,100 6,000
Machine hours 6
1 (W) × 8,000 + 0.5 (W) × 10,500 6,583 5,000
∴ Machine hours are the limiting factor.
Workings
Laundry Dry cleaning
Cleaning materials $10/litre × Number of litresBal 2.00 3.00 Direct labour $6/hour × Number of hoursBal 1.20 2.00 Variable machine cost $3/hour × Number of hoursBal 0.50 1.50
We can work out the balancing figures.
Laundry Dry cleaning $ $
Cleaning materials $10/litre × 0.2 litres 2.00 $10/litre × 0.3 litres 3.00 Direct labour $6/hour × 0.2 hours 1.20 $6/hour × 3
1 hour 2.00
Variable machine costs $3/hour × 6
1 hour 0.50 $3/hour × 0.5 hour 1.50
Step 2: Key factor analysis
Laundry Dry cleaning $ $ Selling price 7.00 12.00 Cleaning materials 2.00 3.00 Direct labour 1.20 2.00 Variable machine cost 0.50 ____ 1.50 ____ Contribution per unit 3.30 5.50 ÷ Number of machine hours per unit ÷ 6
1 ÷ 0.5
Contribution per machine hour 19.80 11.00 Priority 1st 2nd
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Final Assessment
Units Machine hours Contribution £
Laundry 8,000 × 61 hour 1,333.33 26,400
Dry cleaning 7,333.33Bal 2 × 0.5 hour 3,666.67_________ Bal 40,333 ______
5,000 _________ 66,733 ______
Tutorial note: Notice that the minimum order does not come into it. The optimum plan as shown above fulfils the minimum requirement.
Unusually we were not actually asked for contribution or profit but will need it later on.
(b) (i)
Laundry Dry cleaning $ $ Selling price 5.60 13.20 Cleaning materials 2.00 3.00 Direct labour 1.20 2.00 Variable machine cost 0.50 ____ 1.50 ____ Contribution per unit 1.90 6.70
Let L = Number of laundry services Let D = Number of dry cleaning services Objective function: Maximise contribution = 1.9L + 6.7D Subject to: 1 0.2L + 0.3D ≤ 5,000 (Cleaning materials) 2 0.2L + 3
1 D ≤ 6,000 (Direct labour)
3 61 L + 0.5D ≤ 5,000 (Machine hours)
4 L ≥ 1,200 (Minimum requirement for laundry) 5 D ≥ 2,000 (Minimum requirement for dry cleaning) 6 L ≤ 14,000 (Maximum demand for laundry) 7 D ≤ 9,975 (Maximum demand for dry cleaning) 8 L, D ≥ 0 (Non-negativity constraint)
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ACCA F5 Performance Management
(ii) Tutorial note:
In order to plot the graph we must turn the constraints into equations. Then we can plot the lines which the equations represent on the graph. Constraints 4 – 7 are the easiest and simply represent horizontal and vertical lines on the graph. L = 1,200 D = 2,000 L = 14,000 D = 9,975
9,975
2,000
1,200 14,000
D
L
The first constraint is a little bit more difficult. The topic is LINEAR programming and we are therefore talking about straight lines. We need two points to define a straight line:
0.2L + 0.3D = 5,000
The easiest thing to do usually is to make one of the variables equal to zero and then find the value of the second variable. Then make the second variable equal to zero and find the value of the first.
If L = 0, then 0.2L = 0 and the equation becomes:
0.3D = 5,000
∴ D = 0.3
5,000 = 16,667
If D = 0, then 0.3D = 0 and the equation becomes:
0.2L = 5,000
∴ L = 0.2
5,000 = 25,000
The straight line therefore connects 16,667 on the D axis with 25,000 on the L axis.
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Final Assessment
Similarly, for the second and third constraints:
0.2L + 31 D = 6,000
61 L + 0.5D = 5,000
Linear programming solution for W plc
18,000
16
b
a
c
d
,000
14,000
12,000
10,000
8,000
6,000
4,000
2,000
0
L = 1,200 L = 14
0 5,000 10,000 15,000 20,000 25,000 30,000
,000
D = 9,975
Direct labour 0.2L + 0.3D = 5,000
Number of dry cleaning
services
Number of laundry services
Machine hours 1/6L + 0.5D = 5,000
Direct labour 0.2L + 1/3D = 6,000
ISO – contribution line
D = 2,000
Feasible region
16,667
25,000
D
L
D L
30,000 0 10,000
0
D L
30,000 0 18,000
0
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ACCA F5 Performance Management
Tutorial note:
The axes could have been shown the other way around, i.e. with laundry services on the vertical axis and dry cleaning services on the horizontal axis.
Examination tip: The optimum solution can be found by one of two main methods – the iso-contribution line or ‘testing each corner’. Choose whichever method you prefer. Testing each corner The optimum solution has to be a ‘corner’, i.e. an intersection of two or more lines. The four corners on the graph have been labelled a, b, c and d. (These are the four corners of the feasible region.) We can eliminate node (a) as a possible optimum solution because it is not as good as node (b). Node (b) has the same number of laundry services (1,200) as (a) but it has more dry cleaning services and therefore more contribution and is better. Similarly, node (d) is no good. Node (c) is better. Node (c) has the same number of laundry services (14,000) as (d) but has more dry cleaning services and therefore more contribution. The choice is therefore between node (b) and (c). We need to find the values for L and D at those two points and calculate the contributions. We need to solve simultaneous equations. Luckily they are fairly simple. Node (b) Node (b) lies on the intersection of two lines:
61 L + 0.5D = 5,000
L = 1,200
We can substitute this value for L into the first equation and find D:
61 × 1,200 + 0.5 D = 5,000
200 + 0.5D = 5,000 0.5D = 5,000 − 200 0.5D = 4,800
∴ D = 0.5
4,800 = 9,600
Contribution = 1,200 × 1.9 + 9,600 × 6.7 = 66,600
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Final Assessment
Node (c) Node (c) lies on two lines:
61 L + 0.5D = 5,000
L = 14,000
Substituting the value of L into the first equation:
61 × 14,000 + 0.5D = 5,000
2,333.33 + 0.5D = 5,000 0.5D = 2,666.67
∴ D = 0.5
2,666.67 = 5,333.33 units
Contribution = 14,000 × 1.9 + 5,333.33 × 6.7 = 62,333 The optimum plan according to the linear programming solution is to undertake 1,200 laundry services and 9,600 dry cleaning services. This gives a contribution of £66,600. This contribution is not as good as the £66,733 contribution that we got with the original plan. Therefore we should stick with the original selling prices. Tutorial note: The iso-contribution line The optimum point could also be found by means of an iso-contribution line. An iso-contribution line is a line connecting points representing combinations of L and D which give the same contribution. If the iso-contribution line is then moved out from the origin parallel to its original position, then this will represent a better line with more L and/or more D and therefore more contribution. The idea then is to move the line out as far as possible until it is just about to leave the feasible region. The last point it touches before it leaves the feasible region is the optimum solution. The first step is to set up the iso-contribution line. L has a contribution of 1.9 D has a contribution of 6.7 1.9 × 6.7 = 12.73 To get a contribution of 12.73 we would need 6.7 Ls (12.73/1.9) or 1.9Ds (12.73/6.7). If we draw a line connecting 6.7 on the L axis with 1.9 on the D axis then this would be an iso-contribution line with every point giving a contribution $12.73.
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ACCA F5 Performance Management
After this first step the line is often not that useful, i.e. is difficult to plot on the graph. We simply multiply both numbers by 2 or 5 or 10 or 100 or 1,000 or whatever until it is easy to plot the line. We will multiply by 1,000 and therefore the line will connect 6,700 on the L axis with 1,900 on the D axis. Once the line has been drawn it can be moved out parallel from its original position until it leaves the feasible region. The last point it touches is the optimum solution which is node (b).
MARKING GUIDE
Marks
(a) Determining limiting factor (½ mark per balancing figure on cost card and 1 mark per resource required figure)
6
Key factor analysis (1 mark for ranking, 2 for application) 3 _
9 (b) (i) 1 mark per resource constraint, ½ mark each for
other constraints, 1 mark for objective 6
(ii) Drawing graph 7 Identifying optimal point 2 Answering question 1 _
16 __ Total 25 __
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Final Assessment
ANSWER 3 Working
Original standard Revised standard Actual
Fruit Price £0.16 £0.19 £0.18 Quantity 400 kg 400 kg 428 kg Syrup Price £0.10 £0.12 £0.12 Quantity 700 kg 700 kg 742 kg Pectin Price £0.332 £0.332 £0.328 Quantity 99 kg 99 kg 125 kg Citric Acid Price £2.00 £2.00 £0.95 Quantity 1 kg 1 kg 1 kg
(a) The planning variances relate only to fruit and syrup prices.
Fruit Syrup £ £
RSQ × RSP 76 84 RSQ × SP 64 __ 70 __ 12 __ A 14 __ A
Total planning variance = £12 + £14 = £26 A (b)
Fruit Syrup Pectin Citric Acid £ £ £ £
AQ × AP 77.04 89.04 41.00 0.95 AQ × SP 81.32 89.04 41.50 2.00 SQ × SP 76.00 84.00 32.87 2.00 Price variance 4.28 F 0 0.50 F 1.05 F Usage variance 5.32 A 5.04 A 8.63 A 0
Total price variance = £5.83 F Total usage variance = £18.99 A Total ingredients variance = £13.16 A
(c) The main problem with conventional variance analysis for budgetary control reporting
purposes is that its emphasis on comparison between actual and planned performances results in a disregard for changes in these planned results. Because standards become out of date and unrealistic, the traditional accounting model does not serve as an opportunity cost system.
The operational variance measures management's operating efficiency by comparing actual results with revised standard. This variance reflects opportunity costs. That is, the gain or loss as a result of actual performance differing from a realistic standard. Hence operational variances tend to be controllable. The planning variance compares the original budget with the revised budget. This variance reflects planning error.
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ACCA F5 Performance Management
However, it may not be practical to find all possible perfect substitutes for the resources actually used. It may also be difficult to obtain accurate revised standard costs for the resources actually used and their substitutes. In general, there is a resistance to change. Most companies still use the traditional approach.
(d)
Workings Fruit Syrup Pectin Citric Acid Total
AQ × AM 428 742 125 1 1,296 AQ × SM 432 756 106.92 1.08 1,296 SQ × SM 400 700 99 1 1,164/0.97 = 1,200 SP (£) 0.19 0.12 0.332 2 Variances Mix variance (£) 0.76 F 1.68 F 6.00 A 0.16 F 3.40 A Yield variance (£) 6.08 A 6.72 A 2.68 A 0.16 F 15.59 A
Check: Mix + Yield = £3.40 A + £15.59 A = £18.99 A = Usage variance (e) The mix has been changed so that more of the expensive ingredient pectin has been
used, and less of the cheaper ingredients, fruit and syrup, have been used. This has caused the overall adverse mix variance.
The yield variance means that more input was required for the given output. This means there was an abnormal loss, causing the adverse variance.
MARKING GUIDE
Marks
(a) Choosing only fruit and syrup prices 1 Identification of the original and revised standard prices 1 Fruit variance 2 Syrup variance 2 _
6 (b) Each ingredient, 1½ marks for the variances × 4 6 (c) 1 – 2 marks for each valid point, to a maximum of 4 (d) Mix variance 3 Yield variance 3 _
6 (e) 1 – 2 marks for each reason, to a maximum of 3 __ Total 25 __
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Final Assessment
ANSWER 4 (a) Budgetary control statement
Flexed budget Actual Variance £ £ £
Direct materials (W1) 34,000 33,500 500 F Direct labour (W2) 45,900 44,000 1,900 F Production overhead(W3) 50,800 46,250 4,550 F Selling overhead (W4) 16,600 16,150 450 F Administration overhead 28,000 _______ 27,800 _______ 200 _____ F
175,300 _______ 167,700 _______ 7,600 _____ F
Workings
(W1) (30000/60) × 68 = 34,000 (W2) (40,500/60) × 68 = 45,900 (W3) Variable cost = (58,000 – 46,000)/20 = 600 per percentage 68% is an extra 8% above 60%, making overhead cost − £46,000 + 8 × 600 = £50,800 (W4) Variable cost = (19,000 – 15,000)/20 = 200 per percentage Selling overhead = 15,000 + 8 × 200 = 16,600
(b) A fixed budget is a budget which shows the costs and revenues for a single level of
activity. A flexible budget shows costs and revenues for more than one level of activity so that costs etc. can be predicted for other activity levels. Fixed budgets are useful for controlling costs where the objective is to limit (cap) the level of expenditure – for example in the public sector ore in charities. Flexible budgets are used to control costs where the objectives are the efficient procurement and utilisation of resources.
(c) The principle budget factor is the factor which limits the activities of the organisation
during the budget period.
It is important that the principal budget factor is identified at the outset of the budget preparation process. If this is not done any budgets which are prepared will be impossible to achieve. If budgets are impossible to achieve, any subsequent performance evaluation will be meaningless.
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ACCA F5 Performance Management
(d) The role of the budget committee is to communicate with budget holders and coordinate the budget setting process.
The committee will produce a budget manual which provides detailed guidance for managers in preparing their budgets. The committee will communicate with the Board of Directors, advising them of the progress being made.
MARKING GUIDE
Marks
(a) Flexed budget Materials 1 Labour 1 Production overhead 2 Selling overhead 2 Admin overhead 1 _
7 Variances, 1 mark each variance 5 _
12 (b) 1 – 2 marks each point, to a maximum of 6 (c) 1 – 2 marks for each valid point, to a maximum 4 (d) 1 – 2 marks for each valid point, to a maximum of 3 ___ Total 25 ___
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