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A learning curve is addressed in percentage terms, depending upon the relationship between the cumulative average times when the cumulative quantities are.
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FINANCIAL 44 MANAGEMENT April 2005
The principle underlying learning curves is generally well understood: if we perform tasks of a repetitive nature, the time we take to complete subsequent tasks reduces until it can reduce no more. This is relevant to management accounting in the two key areas of cost estimation and standard costing. Before we look at these we need to understand the maths. Imagine that we have collected the following information for the production of eight units of a product: it takes 1,000 hours to produce the first unit; 600 hours to produce the second unit; 960 hours to produce the third and fourth units; and 1,536 hours to produce the remaining four units. There is clearly a learning curve effect here, as the production time per unit is reducing from the initial 1,000 hours. Learning curves are initially concerned with the relationship between cumulative quantities and cumulative average times (total cumulative time divided by cumulative quantity). The relationship in this case is shown in table 1. Notice that, as the cumulative quantity doubles, the cumulative average time reduces by 20 per cent. In other words, subsequent cumulative average times can be obtained by multiplying the previous cumulative average time by 80 per cent. This is an example of an 80 per cent learning curve. A learning curve is addressed in percentage terms, depending upon the relationship between the cumulative average times when the cumulative quantities are doubling. For example, if the cumulative average time were 1,000 hours at the production of the first unit, 700 hours at the production of the second, 490 hours at the fourth, 343 hours at the eighth and so on, this would be a 70 per cent learning curve.
The learning curve formula is needed when dealing with situations that do not fit into this doubling-up pattern. A learning curve is geometric with the general form Y = aX b. Y = cumulative average time per unit or batch. a = time taken to produce initial quantity. X = the cumulative units of production or, if in batches, the cumulative number of batches. b = the learning index or coefficient, which is calculated as: log learning curve percentage ÷ log 2. So b for an 80 per cent curve would be log 0.8 ÷ log 2 = – 0.322.
Example one: unit accumulation The first unit took 100 hours to produce. It is expected that an 80 per cent learning curve will apply. You are required to estimate the following times:
Cumulative quantity Cumulative production time Cumulative average production time per unit 1 unit 1,000 hours 1,000 hours 2 units 1,600 hours 800 hours 4 units 2,560 hours 640 hours 8 units 4,096 hours 512 hours
FINANCIAL April 2005 MANAGEMENT 45
a The cumulative average time per unit to produce three units. b The total time it will take to produce three units. c The incremental time for the fourth and fifth units, in total.
The solutions are as follows: a Y = 100(3 – 0.322^ ) = 70.2 hours per unit. b We need to multiply Y by the cumulative number of units (X) to derive the total time: 70.2 x 3 = 210.6 hours. c The incremental time to produce the fourth and fifth units equates to the total time to produce five units minus the total time to produce three units. Cumulative average time to produce five units: Y = 100(5 – 0.322^ ) = 59.56 hours per unit. Total time for five units: 59.56 x 5 = 297.8 hours. Incremental time: 297.8 – 210.6 = 87.2 hours.
Example two: batch accumulation The first batch of a new product has just been made. The batch size was 20 units and the total time taken was 200 hours – ie, an average of 10 hours per unit. A 90 per cent learning curve is expected to apply. You are required to estimate the following: a The cumulative average time for the first two batches. b The total time to produce 40 units. c The incremental time for 41 to 60 units – ie, a third batch of 20 units.
The solutions are as follows: a This is a batch situation, so the Y value will be the cumulative average time per batch for two batches of 20 units. a = 20 x 10 = 200 hours (ie, the time for the first batch). b = log 0.9 ÷ log 2 = – 0.152. X = 40 ÷ 20 = 2 (ie, two cumulative batches). Y = 200(2 – 0.152^ ) = 180 hours per batch. We could have avoided using the formula here and simply multiplied 200 by 0.9, because it was a doubling-up situation. b A total time is needed so, as in example one, all we need to do is multiply Y by X: 2 batches x 180 hours per batch = 360 hours.
c The incremental time for the third batch equals the total time for 60 units minus the total time for 40 units. We already have the total time for 40, so we need the total time for 60 (ie, X = 3): Y = 200(3 – 0.152^ ) = 169.24 hours per batch. Total time for 60 units = 169.24 x 3 = 507.72 hours. Incremental time = 507.72 – 360 = 147.72 hours.
Now that we’ve looked at the mechanics, let’s consider a couple of examples of how the learning curve can be applied.
Example three: cost estimation BB plc uses a marginal costing system. You have been asked to provide calculations of total variable costs for a contract for one of its products, based on the following alternative situations: 1 A contract for one order of 600 units. 2 Contracts for a sequence of individual orders of 200, 100, 100 and 200 units. Four separate costings are required. It’s expected that the average unit variable cost data for an initial batch of 200 units will be as follows: Direct material: 15m^2 at £8 per m 2. Direct labour: department A: 8 hours at £8 per hour; and department B: 100 hours at £10 per hour. Variable overhead: 25 per cent of labour. Labour times in department A are expected to follow an 80 per cent learning curve. Department B labour times are expected to follow a 70 per cent learning curve.
The cost estimates are as follows: 1 For one order for 600 units, we first need to define the appropriate values of a. We are dealing with a batch situation, so we need to define a as the time for the first batch of 200 units: Department A: 200 x 8 = 1,600 hours. Department B: 200 x 100 = 20,000 hours. Next, we calculate the values of b: Department A: log 0.8 ÷ log 2 = – 0.322. Department B: log 0.7 ÷ log 2 = – 0.515.