Fit the rate to the pilot, cost the increment rather than the average, and know which of the two learning curve models you are using.
Under an 80 per cent learning curve, every time cumulative output doubles, the cumulative average time per unit falls to 80 per cent of what it was: total hours for x units are a × x1+b, with b = log(0.8) / log(2) = −0.3219. A part whose first unit took 250 hours averages 128 hours over a pilot of 8, but units 9 to 32 need only 1,597 hours between them, 66.56 hours each. Quote the follow-on order at the pilot average and the unit cost reads 9,096 instead of 5,594, 62.6 per cent too high.
Worked in full in Cost Accounting by Julian R. Sterling, with every figure reproduced in a free workbook.See the book on Amazon →
The error is common because the pilot average is the only actual figure the cost office has. It is a real number, correctly measured, and it describes units that will never be made again.
| Input | Value |
|---|---|
| Direct labour hours for the first unit | 250 |
| Learning rate (cumulative average model) | 80% |
| Pilot batch, units 1 to 8 | 8 |
| Follow-on order, units 9 to 32 | 24 |
| Direct labour rate per hour | 42.00 |
| Variable overhead per labour hour | 15.00 |
| Conversion cost per labour hour | 57.00 |
| Material per unit (does not learn) | 1,800.00 |
Material is held flat on purpose. Learning acts on labour and on the overhead that moves with labour hours, not on the bar stock, so applying the curve to the whole unit cost is the first thing to avoid.
The rate is read off the pilot, not chosen from a textbook. The time clock gave these cumulative figures:
| Cumulative units | Cumulative average hours | Total hours | Time of the last unit |
|---|---|---|---|
| 1 | 250.00 | 250.00 | 250.00 |
| 2 | 200.00 | 400.00 | 150.00 |
| 4 | 160.00 | 640.00 | 113.42 |
| 8 | 128.00 | 1,024.00 | 88.64 |
| 16 | 102.40 | 1,638.40 | 70.15 |
| 32 | 81.92 | 2,621.44 | 55.83 |
250 to 200 to 160 to 128: each doubling multiplies the cumulative average by 0.80. That is the definition of an 80 per cent curve in the Wright, or cumulative average, model, which is the version most management accounting syllabuses and most quoting spreadsheets use. Rows 16 and 32 are the forecast.
b = log(learning rate) / log(2) = log(0.80) / log(2) = −0.3219
Cumulative average Y = a × xb, so total hours T(x) = a × x1+b = 250 × x0.6781
T(8) = 250 × 80.6781 = 1,024.00
T(32) = 250 × 320.6781 = 2,621.44
Units 9 to 32 = T(32) − T(8) = 2,621.44 − 1,024.00 = 1,597.44 hours
Per unit = 1,597.44 / 24 = 66.56 hours
Conversion = 66.56 × 57.00 = 3,793.92; plus material 1,800.00 = 5,593.92 per unit
Excel: =250*(32^(1+LOG(0.8,2))-8^(1+LOG(0.8,2)))/24, with the first-unit hours in place of 250
The subtraction in step 3 is the whole technique. The cumulative average for 32 units is 81.92 hours, but that average still contains the expensive pilot. The new order carries only the increment, and the increment is cheaper than any average that includes the early units.
| Basis | Hours per unit | Unit cost | Order hours | Order cost |
|---|---|---|---|---|
| Pilot average (units 1 to 8) | 128.00 | 9,096 | 3,072 | 218,304 |
| Last pilot unit (unit 8) | 88.64 | 6,853 | 2,127 | 164,464 |
| Learning curve (units 9 to 32) | 66.56 | 5,594 | 1,597 | 134,254 |
The pilot average overstates the order by 1,475 hours, or 84,050 of conversion cost, and puts 3,502 too much on every unit. On a cost-plus quote that gap goes straight into the price, and the customer comparing bids sees a supplier who is 62.6 per cent dearer on cost before margin. The second row is the instinct of an experienced planner, "use what the last one took", and it is still 22.5 per cent too high, because every unit after the eighth is faster than the eighth: unit 9 already takes 85.14 hours and unit 32 takes 55.83. The mark-up arithmetic that follows is in how to calculate a cost-plus price for a target margin.
The rate is an estimate from eight units, so test it. A lower percentage means faster learning.
| Learning rate | Index b | Pilot average hours | Order hours | Hours per unit | Unit cost |
|---|---|---|---|---|---|
| 75% | −0.4150 | 105.47 | 1,055 | 43.95 | 4,305 |
| 80% | −0.3219 | 128.00 | 1,597 | 66.56 | 5,594 |
| 85% | −0.2345 | 153.53 | 2,321 | 96.72 | 7,313 |
| 90% | −0.1520 | 182.25 | 3,266 | 136.08 | 9,557 |
| 100% (no learning) | 0.0000 | 250.00 | 6,000 | 250.00 | 16,050 |
Five points of learning rate move the unit cost by 1,289 on the way down and 1,719 on the way up. That is why the rate should be fitted to the actual pilot hours rather than borrowed, and why a quote should say which rate it assumes.
Learning also stops. If the cell reaches its steady state at unit 16, every later unit takes the unit 16 time of 70.15 hours, the order needs 1,737 hours (139 more), and the unit cost rises to 5,925. Machine-paced work plateaus early; hand assembly and fitting keep learning for longer.
The quote follows the increment, not the average. The cumulative average always includes the pilot. A follow-on order should be costed as T(end) − T(start), divided by the units in the order.
There are two learning curve models, and an 80 per cent rate means different things in each.
Plug the same 250 hours and 80 per cent into the unit time model and units 9 to 32 need 2,333 hours, 46.1 per cent more than the Wright answer, for a unit cost of 7,341. Neither model is wrong, but the rate has to be fitted in the model you forecast with. Fitted to the same 1,024 pilot hours, the unit time model implies a first unit of 191.55 hours and a follow-on of 1,788 hours, 74.49 per unit and 6,046 a unit. The gap between 5,594 and 6,046 is model risk, and it belongs in the quote's contingency, not in a hidden buffer on the rate.
Fit the rate to the pilot, subtract cumulative totals to get the hours of the new order, apply the curve to labour and labour-driven overhead only, and state the model. On this part the honest cost of the follow-on is 5,594 a unit, not the 9,096 the pilot average suggests, and the difference decides whether the order is won. The same logic of a standard that no longer describes the work runs through the free workbooks for Cost Accounting; for the related decision of pricing work below a full-cost figure, see should you accept a special order below full cost.
Cumulative average time per unit is Y = a × x^b, where a is the first unit's time, x the cumulative units and b = log(learning rate) / log(2). For an 80 per cent curve b is −0.3219. Total time is a × x^(1+b), so with a first unit of 250 hours, 8 units take 1,024 hours and 32 units take 2,621.44 hours.
In the cumulative average model, subtract consecutive totals: time of unit x = T(x) − T(x−1). With a 250-hour first unit at 80 per cent, unit 8 takes 1,024.00 − T(7), which is 88.64 hours, and unit 32 takes 55.83 hours. The marginal unit is always faster than the cumulative average, which is 128.00 hours at 8 units.
Wright applies the learning rate to the cumulative average time; Crawford applies it to the time of each individual unit. With the same 250 hours and 80 per cent, units 9 to 32 take 1,597 hours under Wright and 2,333 under Crawford, 46.1 per cent more. Fit the rate in the model you forecast with.
This article is one calculation from Cost Accounting. The book takes the same case from first principles to the decision, chapter by chapter, and every figure it prints is a live formula in the free companion workbooks.
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