Predicting the Number of Repetitions Needed to Achieve a Specified Unit Time

Use learning curve theory to predict the number of repetitions (units) that will be needed for a trainee to achieve a unit time of 6 minutes if the trainee took 10 minutes to do the first unit and a learning curve of 90 percent is operative.

a. Use the learning table.

b. Use the log formula.

Question

Predicting the Number of Repetitions Needed to Achieve a Specified Unit Time

Use learning curve theory to predict the number of repetitions (units) that will be needed for a trainee to achieve a unit time of 6 minutes if the trainee took 10 minutes to do the first unit and a learning curve of 90 percent is operative.

a. Use the learning table.

b. Use the log formula.

Accepted Answer

a. The table approach can be used for the learning percentages that are listed across the top of the table, such as the 90 percent curve in this example. The table approach is based on Formula 7S–2:

T_n = T₁  × Unit time factor (7S–2)

T_n = T₁ × Unit table factor

Setting T_n equal to the specified time of 6 minutes and solving for the unit table factor yields

6 min = 10 min × Unit table factor

Solving,

Unit table factor = 6 min ÷ 10 min = .600

From Table 7S.1, under 90% in the Unit Time column, we find .599 at 29 units.

TABLE 7S.1 Learning curve coefficients

	70%		75%		80%		85%		90%
Unit Number	Unit Time	Total Time	Unit Time	Total Time	Unit Time	Total Time	Unit Time	Total Time	Unit Time	Total Time
1	1.000	1.000	1.000	1.000	1.000	1.000	1.000	1.000	1.000	1.000
2	.700	1.700	.750	1.750	.800	1.800	.850	1.850	.900	1.900
3	.568	2.268	.634	2.384	.702	2.502	.773	2.623	.846	2.746
4	.490	2.758	.562	2.946	.640	3.142	.723	3.345	.810	3.556
5	.437	3.195	.513	3.459	.596	3.738	.686	4.031	.783	4.339
6	.398	3.593	.475	3.934	.561	4.299	.657	4.688	.762	5.101
7	.367	3.960	.446	4.380	.534	4.834	.634	5.322	.744	5.845
8	.343	4.303	.422	4.802	.512	5.346	.614	5.936	.729	6.574
9	.323	4.626	.402	5.204	.493	5.839	.597	6.533	.716	7.290
10	.306	4.932	.385	5.589	.477	6.315	.583	7.116	.705	7.994
11	.291	5.223	.370	5.958	.462	6.777	.570	7.686	.695	8.689
12	.278	5.501	.357	6.315	.449	7.227	.558	8.244	.685	9.374
13	.267	5.769	.345	6.660	.438	7.665	.548	8.792	.677	10.052
14	.257	6.026	.334	6.994	.428	8.092	.539	9.331	.670	10.721
15	.248	6.274	.325	7.319	.418	8.511	.530	9.861	.663	11.384
16	.240	6.514	.316	7.635	.410	8.920	.522	10.383	.656	12.040
17	.233	6.747	.309	7.944	.402	9.322	.515	10.898	.650	12.690
18	.226	6.973	.301	8.245	.394	9.716	.508	11.405	.644	13.334
19	.220	7.192	.295	8.540	.388	10.104	.501	11.907	.639	13.974
20	.214	7.407	.288	8.828	.381	10.485	.495	12.402	.634	14.608
21	.209	7.615	.283	9.111	.375	10.860	.490	12.892	.630	15.237
22	.204	7.819	.277	9.388	.370	11.230	.484	13.376	.625	15.862
23	.199	8.018	.272	9.660	.364	11.594	.479	13.856	.621	16.483
24	.195	8.213	.267	9.928	.359	11.954	.475	14.331	.617	17.100
25	.191	8.404	.263	10.191	.355	12.309	.470	14.801	.613	17.713
26	.187	8.591	.259	10.449	.350	12.659	.466	15.267	.609	18.323
27	.183	8.774	.255	10.704	.346	13.005	.462	15.728	.606	18.929
28	.180	8.954	.251	10.955	.342	13.347	.458	16.186	.603	19.531
29	.177	9.131	.247	11.202	.338	13.685	.454	16.640	.599	20.131
30	.174	9.305	.244	11.446	.335	14.020	.450	17.091	.596	20.727

Hence, approximately 29 units will be required to achieve the specified time.

b. Using the log formula,

(1) Compute the ratio of specified time to first unit time: 6 min ÷ 10 min = .600.

(2) Compute the ratio of ln learning percentage to ln 2: ln .90 ÷ ln 2 = –0.1053605 ÷ 0.6931472 = –0.1520.

(3) Find n such that n^–.1520 = .600: [latex]^{-.1520}\sqrt{.600}[/latex] = 28.809. Round to 29. Hence, 29 units (repetitions) will be needed to achieve a time of 6 minutes.

The learning percentage can be estimated from data on repetition times. The procedure for doing this is illustrated in Solved Problem 2.

Question 7S.4: Predicting the Number of Repetitions Needed to Achieve a Spe...

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