ACCA F5 Performance Management - 2010 - Study text - Emile Woolf Publishing
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(c) Labour is in short supply and restricted to 2,125 hours
Casing A
B
C D
Contribution/unit $360
$520
$300 $540
Labour hours/unit 7.5
7.5
6.25 12.5
Contribution per hour $48.00
$69.33
$48.00 $43.20
Ranking for manufacture 2nd
1st
2nd 4th
Profit-maximising production schedule
Labour hours
A
B C
D
units
units units
units
Special order 812.5
30
20 30
20
Remaining hours 1,312.5
175
Total hours 2,125.0
Total production/sales
30
195 30
20
(d) Make or buy decision
A
B
C D
$
$
$ $
Contribution if made 360
520
300 540
Contribution if bought in 285
475
250 490
Extra contribution if made 75
45
50 50
Labour hours 7.5
7.5
6.25 12.5
Extra contribution per hour $10
$6
$8 $4
Ranking/priority for making 1st
2nd
3rd 4th
Profit-maximising production schedule
Casing Hours
A
B
C D
Special order 1,625
30
20
30 20
Remaining hours 2,625
175
Total hours 4,250
Made internally
205
20
30 20
Purchased externally
95
180
270 180
Total sales
300
200
300 200
11 Proglin
Let the number of units of Mark 1 be x
Let the number of units of Mark 2 be y.
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The objective function is to maximise total contribution: 10x + 15y.
Subject to the following constraints:
Directmaterials 2x+4y≤24,000
Directlabour 3x+2y≤18,000
Salesdemand,Mark1 x≤5,000
Non‐negativity x,y≥ 0
These constraints are shown in the graph below. The graph also shows an iso-
contribution line 10x + 15y = 60,000.
The feasible solutions are shown by the area 0ABCD in the graph.
Using the slope of the iso-contribution line, it can be seen that contribution is
maximised at point B on the graph.
At point B, we have the following simultaneous equations:
(1)
2x + 4y = 24,000
(2) 3x + 2y = 18,000
Multiply (2) by 2
(3) 6x + 4y = 36,000
Subtract (1) from (3)
4x = 12,000
Therefore x = 3,000
Substitute in equation (1)
2 (3,000) + 4y = 24,000
4y = 18,000
y = 4,500
The objective in this problem is to maximise 10x + 15y.
5,000 6,000 12,000
$
9,000
6,000
4,000
0
x = 5,000
3x + 2y = 18,000
A
B
C
D
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The total contribution where x = 3,000 and y = 4,500 is as follows:
$
3,000 units of Mark 1 (× $10)
30,000
4,500 units of Mark 2 (× $15)
67,500
Total contribution 97,500
(b) The graph for a linear programming solution is virtually identical to the graph
shown above, and the solution is still at point B. However, at point B:
(1) 2x + 4y = 24,001
(2) 3x + 2y = 18,000
Multiply (2) by 2
(3) 6x + 4y = 36,000
Subtract (1) from (3)
4x = 11,999
Therefore x = 2,999.75
Substitute in equation (2)
3 (2,999.75) + 2y = 18,000
2y = 9,000.75
y = 4,500.375
The objective in this problem is to maximise 10x + 15y.
The total contribution where x = 2,999.75 and y = 4,500.375 is as follows:
$
2,999.75 units of Mark 1 (× $10)
29,997.500
4,500.375 units of Mark 2 (× $15)
67,505.625
Total costs 97,503.125
The effect of having $1 more of direct materials would be an increase of $3.125
in total contribution.
The shadow price of direct materials is therefore $3.125 per $1 of direct materials.
12 Price
(a) If the profit margin is one-sixth of the selling price:
$
Selling price 100.00
Profit as a % of sales price 16.67
Full cost as % of sales price 83.33
Profit as % of full cost = (16.67/83.33) × 100% = 20%.
$
Full cost 100
Profit margin (20%) 20
Selling price 120
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(b)
$
Direct materials:
Material M1 30
Material M2 4
Direct labour 30
Variable cost 64
Selling price 120
Mark-up 56
The required mark-up on variable cost would be 56/64 = 0.875 or 87.5%.
(c) Alternative pricing strategies are price skimming and penetration pricing.
Premium pricing would also be an acceptable solution.
Price discrimination, with different prices charged for the same product in
different geographical markets, would also be acceptable.
13 Marginal
When p = 250, q = 400,000.
The quantity sold will fall by 20,000 for every $10 increase in price.
Therefore the price when q = 0 will be 250 + [(400,000/20,000)
× $10] = 450
The price function can be stated as:
P
q
= 450 – (10/20,000)q = 450 – 0.0005q.
Total revenue = pq = (450 – 0.0005q)q = 450q – 0.0005q
2
Marginal revenue = 450 – (2
× 0.0005 × q) = 450 – 0.001q
Marginal cost = variable cost per unit = 210.
Profit is maximised where MR = MC:
450 – 0.001q = 210
0.001q = 240
q = 240,000 units.
At this volume of sales, p = 450 – 0.0005 (240,000) = 330.
Profit is maximised at a price of $330 per unit, and annual sales will be 240,000
units.
$000
Sales (240,000 × $330)
79,200
Variable costs (240,000 × $210)
50,400
Contribution 28,800
Fixed costs 20,000
Profit 8,800
The annual profit will be $8,800,000.
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14 MC = MR
(a) The demand falls by 1,000 units for every $20 increase in price.
Demand will be 0 when the sales price is $50 + $20
× (1,000/1,000) = $70.
The quantity demanded rises by 20/1,000 = 0.02 for every unit of demand.
The
demand curve can therefore be expressed as:
P = 70 – 0.02Q
(b)
Total revenue = PQ = (70 – 0.02Q)Q
= 70Q – 0.02Q
2
(c) Marginal revenue = 70 – 0.04Q
Marginal cost = 8
Profit is maximised where MC = MR
8 = 70 – 0.04Q
Q = 62/0.04 = 1,550
P = 70 – 0.02Q = 70 – 0.02 (1,550) = 39.
The profit is maximised at a price of $39, when demand will be 1,550 units.
Contribution per unit will be $31 ($39 – $8).
Total contribution will be $48,050 ($31
× 1,550).
15 Snapco
(a) (i) Full cost pricing plus, on the current basis
$000
Direct materials 12
Direct labour 8
Direct cost 20
Production overheads (100% of direct cost) 20
Full production cost 40
Administrative and marketing overheads (25%) 10
Full cost of sale 50
Profit (20% of full cost of sale) 10
Selling price for order (5,000 units) 60
Sales price per unit = $12.
(ii) Price to maintain budgeted profit of $50,000
If sales of SP8 falls are 25% below budget, the expected profit will be as
follows.
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$000
Direct costs
75%
× 100,000
75.0
Variable overheads
75%
× 50,000
37.5
Total variable costs 112.5
Fixed production overhead 50.0
Other fixed overheads 50.0
Total costs 212.5
Sales
75%
× 300,000
225.0
Profit 12.5
In order to make a profit of $50,000 for the year, the simplified units of
SP8 must make $37,500 contribution to profit. The price for the order
should be as follows:
$000
Materials 12.0
Labour 8.0
Direct cost 20.0
Variable overhead (see note) 10.0
Variable cost 30.0
Contribution 37.5
Selling price 67.5
Note: Variable overheads are assumed to vary with direct materials cost.
In the original budget, variable overheads are $50,000 and direct
material costs are $60,000. Variable overheads are therefore 50/60
×
material costs. For the simplified units, variable overheads will be
$12,000
× 50/60 = $10,000.
The selling price per unit would need to be $67,500/5,000 units = $13.50.
(b) Advice
The price of SP8 is currently $15 ($300,000/20,000 units). The price for the
simplified units of SP8 must be lower than this; otherwise the customer will
not buy them.
A price of $13.50 is needed to achieve the budgeted profit, but the customer
may be unwilling to pay this amount. A price of $12 will give a profit of 20%
on full cost, using the budgeted absorption rates for overhead, but there will
be some under-absorbed overheads.
Any price in excess of the minimum price of $30,000 ($6 per unit) will make
profit higher than it will be if the simplified units are not sold to the customer.
However, the company must think of the longer term. Will the customer want
to buy more units of the simplified product next year? If so, the company will
want to charge a price at which it will make satisfactory profits.
The recommendation should therefore be to negotiate with the customer. If
the agreed price is lower than $13.50, Snapco might want to warn the
customer that more units of the simplified SP8 might not be available in the
future, except at a higher price.
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16 Bridon
(a) Budgeted variable production costs
Cost in
current year
Inflation
Cost next year
Current
material
New
material
$
%
Direct material 30
20
36.0 31.25
Direct labour 10
5
10.5 10.50
Variable o’hd 10
5
10.5 10.50
50
57.0 52.25
Cost of rejected units (5%)
2.75
55.00
Using the cheaper substitute material will reduce the variable unit cost by $2,
but fixed costs will increase by $30,000. Since annual sales at the current price
will be 20,000 units, it will be more profitable to use the cheaper material.
The unit variable cost will be $55.
Annual fixed costs will be:
$
Costs in the current year ($10
× 20,000 units)
200,000
Add inflation (10%) 20,000
Inspection costs 30,000
Total fixed costs for the year 250,000
(b) Optimal selling price
Method 1
It is assumed that the only sales prices to be considered are those in the sales
manager’s estimates of sales demand.
The variable cost will be $55 per unit.
Price
Contribution/
unit
Sales
Total contribution
$
$
000 units
$000
80
25
25
625
84
29
23
667
88
33
21
693
90
35
20
700
92
37
19
703
96
41
17
697
100
45
15
675
Contribution will be maximised at a price of $92, and sales will be 19,000
units.
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$
Total contribution 703,000
Fixed costs 250,000
Profit 453,000
Method 2
It is assumed that the sales manager has identified a straight-line demand
curve, and any sales price on this curve may be selected.
Sales demand falls by 1,000 units for every $2 increase in the sales price.
Sales demand will be 0 when the sales price is $130 [$100 + (15
× $2)]
Demand curve:
P = 130 – 2Q, where Q is sales demand in 000s units
Total revenue TR = (130 – 2Q)Q = 130Q – 2Q
2
Marginal revenue MR = 130 – 4Q
Marginal cost MC = variable cost per unit = 55.
Profit is maximised when MR = MC
130 – 4Q = 55
Q = 18.75
Price = 130 – 2(18.75) = $92.50.
Profit will be maximised at a price of $92.50, and the contribution per unit will
be $37.50.
$
Total contribution (18,750 × $37.50)
703,125
Fixed costs 250,000
Profit 453,125
17 Materials and relevant costs
Material X: This material is in regular use. Its relevant cost is therefore its current
replacement cost, because any existing inventory will be replaced if it is used on the
job.
Materials Y and Z: The relevant cost of the additional quantities that will have to be
purchased is their current replacement cost.
Material Y: units already held in inventory. The relevant cost of these units is their
opportunity cost, which is the cash that could be obtained by disposing of them.
Material Z: units already held in inventory. The relevant cost of these units is the
higher value of their disposal value ($20 per unit) and the contribution that they
would earn if they are used as a substitute material on a different job ($25 per unit)
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Relevant costs
$ $
Material X: 800 units × $23
18,400
Material Y:
Opportunity cost of units in inventory = disposal value
(400 units
× $12) 4,800
Purchase cost of additional units (200 units
×
$19)
3,800
8,600
Material Z:
Opportunity cost of units in inventory = (300 units
×
$25)
7,500
Purchase cost of additional units (200 units
×
$40)
8,000
15,500
Total relevant cost of materials 42,500
18 Printing leaflets
(a) Tutorial note: The volume of sales required to achieve a target profit is an
application of CVP analysis.
$
$
Target profit
5,400
General fixed costs
12,000
Specific fixed costs:
Leaflet Type A
7,200
Leaflet Type B
12,000
Leaflet Type C
28,500
Total contribution required
65,100
Contribution from:
50,000 Leaflets Type A: (50
× (300 – 120))
9,000
50,000 Leaflets Type B: (50
× (660 – 210))
22,500
31,500
Contribution required from Leaflets Type C
33,600
The contribution from Leaflets Type C is $(1,350 – 390) = $960 per 1,000 leaflets.
The sales quantity of Leaflets Type C required to achieve a target profit of $5,400
each month is therefore $33,600/$960 per 1,000 = 35,000 leaflets.
(b)
Relevant costs
$
Materials
To be purchased: 10,000
×
$500/1,000
5,000
Currently held in inventory 6,000
(Relevant cost = higher of [$1,500 and (20,000
×
$300/1,000)]
Variable costs of labour/overheads 2,400
(30,000
×
$80/1,000)
Total relevant costs 13,400
Contract price 25,000
Incremental profit 11,600
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19 JB
The contract should be accepted if the revenue from the contract will exceed the
relevant costs of the contract.
Workings
Material M1
. This material is in continuous/regular use. The relevant cost of the
1,000 kilograms is their replacement cost.
Relevant cost = 400 components × 3 kilos × $5.50 per kilo = $6,600.
Material P2. The material held in inventory has a relevant cost that is the higher of
its scrap value ($1.50) and the costs saved by putting it to an alternative use, which
is $2 ($3.60 – $1.60).
There are more units held in stock than are needed for the contract. The excess
quantity should be ignored.
Relevant cost of material in stock = 400 components × 2 kilos × $2 per kilo = $1,600.
Part 678. Relevant cost = 400 components × $50 = $20,000.
Skilled labour
. The relevant cost of skilled labour is the extra cash that would have to
be spent to hire additional labour.
Relevant cost = 400 components × 5 hours per component × $4 per hour = $8,000.
Semi-skilled labour. Relevant cost = 400 components × 5 hours per component × $3
per hour = $6,000.
Variable overheads. It is assumed that the overhead absorption rate for variable
overheads is the rate at which cash expenditure is incurred on variable overheads.
Relevant cost = 400 components × 4 machine hours per component × $7 per
machine hour = $11,200.
Relevant cost statement
$
Material M1 6,600
Material P2 1,600
Part 678 20,000
Skilled labour 8,000
Semi-skilled labour 6,000
Variable overheads 11,200
Incremental fixed costs 3,200
Total relevant costs 56,600
Contract sales value (400 × $145)
58,000
Incremental profit 1,400