Strategic Cost Management · Learning Curve

Learning Curve Numerical Problems with Solutions

Updated 11 October 2026 · Fact-checked

A learning curve numerical uses the rule that cumulative average time per unit falls by a fixed percentage each time output doubles. Find the average time for the required units, multiply by the number of units for total time, subtract earlier totals for incremental time, then multiply by the labour rate for cost.

Understand Learning Curve Numerical Problems

When a worker repeats a task, they get faster. Fewer mistakes, better tool handling and smoother methods cut the time needed. The learning curve puts a number on this effect.

The standard model used in exam questions is the cumulative average time model. An 80% curve means that each time cumulative output doubles, the cumulative average time per unit becomes 80% of the earlier average. So if 1 unit takes 100 hours, the average for 2 units is 80 hours, for 4 units 64 hours, and for 8 units 51.2 hours.

Note what falls. It is the average time of all units made so far, not the time of each new unit. The time for later units is found by subtracting: total time for n units minus total time for the earlier units. That gives the incremental time.

The curve applies to labour-intensive, repetitive work, mostly at the start of production of a new product. After a point, learning stops and time per unit becomes steady. Questions usually tell you when this happens, and you must then switch to a fixed time per unit.

Most exam problems are one chain: average time, total time, incremental time, then labour cost. If you keep the chain in this order, the numerical becomes routine.

Key rules to remember

Learning curve model: Y = aXᵇ; Y = cumulative average time per unit for X units; a = time for the first unit; X = cumulative units.
Index b: b = log r ÷ log 2; r is the learning rate as a decimal (0.8 for 80%). b is negative. For 80%, b ≈ -0.3219. For 90%, b ≈ -0.1520.
Doubling rule: Average time for 2X units = r × average time for X units; Use this when units are 1, 2, 4, 8, 16 and so on. No logs are needed.
Total time: Total time for X units = X × Y; Y is the cumulative average time for X units.
Incremental time: Time for units (m+1) to n = Total time for n units − Total time for m units; For the nth unit alone, take total time for n units less total time for (n−1) units.
Labour cost: Labour cost = Time × Rate per hour; Apply the rate to total or incremental hours, as the question asks.

How to solve Learning Curve Numerical Problems questions

Use this order for any learning curve question. It works for 80% or 90% curves and for any number of units.

1Read the question and note the learning rate, the time for the first unit (or first batch), the labour rate and the units asked for.
2Check whether the model is cumulative average time. If the question gives a different basis, follow the question.
3Find the cumulative average time for the target units. If units are 1, 2, 4, 8, multiply by the rate at each doubling. Otherwise use Y = aXᵇ with the log or power value given.
4Multiply the average time by the number of units to get total time.
5For later units, compute the total time for the earlier units the same way and subtract to get incremental time.
6If learning stops after a certain unit, price the remaining units at the steady time per unit.
7Multiply the hours by the labour rate to get cost. Add other costs only if asked.
8Write the answer with units (hours or ₹) and state any assumption.

Quickest way: Doubling table method

When to use it: Use when the units asked are 2, 4, 8, 16 or batches that double. It avoids logs and saves time.

Write a small table with columns: cumulative units, average time, total time.
Fill the first row with 1 unit and the first-unit time.
For each row, double the units and multiply the previous average by the rate.
Compute total time as units × average in each row.
Take differences between rows for incremental time. Multiply by the rate per hour for cost.

Common mistakes in Learning Curve Numerical Problems

Treating the 80% as the time of the second unit, so unit 2 is taken as 80% of unit 1 and unit 4 as 80% of unit 2 only.
Students read the curve as unit-by-unit instead of as a cumulative average.
Fix: Remember that 80% applies to the cumulative average time when output doubles. Find the average first, then total.
Using the average time as the total time.
The curve gives an average, and students forget to multiply by the number of units.
Fix: Always do total time = units × average time before computing cost.
Subtracting averages to get incremental time.
Students subtract 64 from 80 hours and call it the incremental time.
Fix: Subtract totals, not averages. Incremental time = total for n units − total for earlier units.
Using the wrong sign or log value for b.
The index is negative, and 80% and 90% values are mixed up.
Fix: Write b = log r ÷ log 2 and check that b is negative. Use -0.3219 for 80% and -0.1520 for 90%, or the value given in the question.
Continuing the curve after learning has stopped.
Students miss the statement that steady state is reached after a certain number of units.
Fix: Underline any steady-state condition. Apply the curve up to that point and a constant time afterwards.

Worked examples

Example 1

A firm makes a new product. The first unit takes 100 labour hours. An 80% learning curve (cumulative average time model) applies. Labour costs ₹120 per hour. Calculate (a) the average time per unit for 8 units, (b) the total time and labour cost for 8 units, and (c) the time and cost for units 5 to 8.

Show the solution

Use the doubling rule. Average for 1 unit = 100 hours.
Average for 2 units = 100 × 0.8 = 80 hours.
Average for 4 units = 80 × 0.8 = 64 hours.
Average for 8 units = 64 × 0.8 = 51.2 hours.
Total time for 8 units = 8 × 51.2 = 409.6 hours.
Labour cost for 8 units = 409.6 × ₹120 = ₹49,152.
Total time for 4 units = 4 × 64 = 256 hours.
Incremental time for units 5 to 8 = 409.6 − 256 = 153.6 hours.
Cost of units 5 to 8 = 153.6 × ₹120 = ₹18,432.

Answer: (a) 51.2 hours. (b) 409.6 hours and ₹49,152. (c) 153.6 hours and ₹18,432.

Example 2

The first unit of a product takes 50 hours. A 90% learning curve (cumulative average time model) applies. Labour costs ₹200 per hour. Calculate (a) the time and cost for units 3 and 4 together, and (b) the time for the 4th unit alone. Take 3 raised to the power -0.152 as 0.8462.

Show the solution

Average for 1 unit = 50 hours.
Average for 2 units = 50 × 0.9 = 45 hours. Total for 2 units = 90 hours.
Average for 4 units = 45 × 0.9 = 40.5 hours. Total for 4 units = 4 × 40.5 = 162 hours.
Units 3 and 4 together = 162 − 90 = 72 hours.
Cost = 72 × ₹200 = ₹14,400.
For 3 units: average = 50 × 0.8462 = 42.31 hours.
Total for 3 units = 3 × 42.31 = 126.93 hours.
Time for the 4th unit = 162 − 126.93 = 35.07 hours.

Answer: (a) 72 hours, costing ₹14,400. (b) About 35.07 hours for the 4th unit.

Exam tips

Write the doubling table first. Even if you slip later, the method marks are safe.
Check the question for the model. If it states the log value or the power value, use it exactly as given.
Show total time and incremental time as separate lines. Examiners award marks for each.
Look for a steady-state condition or an idle-time or setup cost that sits outside the curve.
In the written part, end with a clear conclusion, such as a price quote or a decision on a repeat order, not only the hours.

Practice questions from Learning Curve

Learning Curve Numerical Problems: frequently asked questions

How do I calculate total time using an 80% learning curve?

Find the cumulative average time for the required units by applying 80% at each doubling of output, or by using Y = aXᵇ. Then multiply the average time by the number of units. For 8 units with a first unit of 100 hours, total time is 8 × 51.2 = 409.6 hours.

How do I find the incremental time for the nth unit?

Compute the total time for n units and for (n−1) units. Subtract the second from the first. The difference is the time for the nth unit alone.

What is the value of b for 80% and 90% curves?

b = log r ÷ log 2. For 80% it is about -0.3219 and for 90% about -0.1520. Use the value given in the question if it is provided.

Does the learning curve apply forever?

No. Learning is strongest early in production. Many questions state that learning stops after a certain number of units, and later units take a constant time. Apply the curve only up to that point.