Chapter 07S – Learning Curves
7S–11
Education.
Time for Unit 5:
Tn = T1 x Unit time factor
We do not know the Unit time factor so we must use the following approach:
13. Given: Kara’s learning curve percentage is supposed to equal 82%. Her times for the first four
units were 30.5, 28.4, 27.2, & 27.0 minutes.
Step 1: Estimate the times for Units 2, 3, 4, & 5.
Time for Unit 2:
Tn = T1 x Unit time factor
We do not know the Unit time factor so we must use the following approach:
(round to 2 decimals)
Time for Unit 4:
Tn = T1 x Unit time factor
We do not know the Unit time factor so we must use the following approach:
Tn = T1 x nb where b = ln (Learning %) / ln (2)
Chapter 07S – Learning Curves
Step 2: Compare the actual times to the expected times.
Unit
Actual time
Expected time
Actual/Expected
1
30.5
–
2
28.4
25.01
+3.39
3
27.2
22.27
+4.93
4
27.0
20.51
+6.49
Based on the differences, it does not appear that the actual learning curve percentage is 82%.
Although we would not expect the differences to be zero, we would expect the differences to be
close to zero, and a mix of positive and negative differences. All of the actual times exceed the
expected times.
Alternate Solution:
14. Given: The time for the 5th unit of a 10-unit job = 5 hours. The 6th unit has been worked on for
2.00 hours, but is not yet finished. Learning curve percentage = 75%.
Additional amount of time needed to finish the 10-unit job = Remaining Time for Unit 6 + Time
for Unit 7 + Time for Unit 8 + Time for Unit 9 + Time for Unit 10)
Step 1: Determine the time for Unit 1.
Unit time factor = .513 (Table 7S.1)
Step 2: Determine the times for Units 6-10.
Unit 6: Unit time factor = .475 (Table 7S.1)
Tn = T1 x Unit time factor
T6 = 9.75 x .475
T6 = 4.63 hours (round to 2 decimals)
Remaining time = 4.63 – 2.00 = 2.63 hours
Chapter 07S – Learning Curves
7S–13
Education.
Unit 7: Unit time factor = .446 (Table 7S.1)
Tn = T1 x Unit time factor
T7 = 9.75 x .446
T7 = 4.35 hours (round to 2 decimals)
Unit 8: Unit time factor = .422 (Table 7S.1)
Tn = T1 x Unit time factor
T9 = 9.75 x .402
T9 = 3.92 hours (round to 2 decimals)
Unit 10: Unit time factor = .385 (Table 7S.1)
15. Given: We are given the times below for 3 trainees for completing two repetitions.
Trainee
T1 hours
T2 hours
Art
11
9.9
Sherry
10.5
8.4
Dave
12
10.2
Determine how much time will be required for each to reach a time of 7 hours per unit:
Step 1: Estimate the learning curve percentage for each trainee.
Chapter 07S – Learning Curves
7S–14
Education.
Step 2: For each of the learning curve percentages above, determine what Unit Factor x the time
for Unit 1 will equal 7 hours.
Art: Learning rate = .90.
Tn = T1 x Unit time factor
7 = 11 x Unit time factor
Unit time factor = 7 / 11 = .636
Look up this value in Table 7S.1: Closest value ≤ .636 = .634, which corresponds to n = 20
repetitions.
Sherry: Learning rate = .80.
Tn = T1 x Unit time factor
Tn = T1 x Unit time factor
7 = 12 x Unit time factor
Unit time factor = 7 / 12 = .583
Look up this value in Table 7S.1: Closest value ≤ .583 = .583, which corresponds to n = 10
repetitions.
16. Given: Standard = 18 minutes per repetition. Irene took 30 minutes for first repetition and 25
minutes for the second.
Number of repetitions needed for her to achieve standard of 18 minutes:
Step 1: Estimate the learning curve percentage for Irene.
T2 / T1 = 25 / 30 = .8333 (round to 4 decimals)
This value does not fall in Table 7S.1; therefore, we must use the formula.
Step 2: Solve for the unit number needed to achieve the standard time.
Tn = T1 x nb where b = ln (Learning %) / ln (2)
Chapter 07S – Learning Curves
7S–15
Education.
17. Given: We are given the times for two repetitions for each of three workers below. Standard time
= 25 minutes.
Worker
T1 minutes
T2 minutes
Beverly
36
31
Max
40
36
Antonio
37
30
Step 1: Estimate the learning curve percentage for each trainee.
Step 2: Solve for the unit number needed to achieve the standard time.
Beverly:
Tn = T1 x nb where b = ln (Learning %) / ln (2)
Max:
Tn = T1 x nb where b = ln (Learning %) / ln (2)
Antonio:
Tn = T1 x nb where b = ln (Learning %) / ln (2)
Chapter 07S – Learning Curves
7S–16
Education.
18. Given: Desired time = 19 minutes. A researcher’s times for performing searches are shown
below:
Request
no.
1
2
3
4
5
6
7
8
Time
(min.)
55.0
41.0
35.2
31.0
28.7
26.1
24.8
23.5
Step 1: Estimate learning curve percentage based on the doubling concept.
Take the average of the values above (round to 4 decimals :
(.7455 + .7561 + .7415 + .7581) / 4 = .7503
Step 2: Solve for the unit number needed to achieve the desired time.
Tn = T1 x nb where b = ln (Learning %) / ln (2)
Chapter 07S – Learning Curves
7S–17
Education.
19. Given: Learning curve percentage = 85%. Time for Unit 1 is unknown. However, Units 2-4
took a total of 28.14 hours to complete.
Time for Unit 5 = Time for Unit 1 x Unit Time Factor.
Step 1: Determine total time for first 4 units.
Step 3: Set the two equations equal and solve for T1.
T1 x 3.345 = T1 + 28.14
3.345T1 – T1 = 28.14
Chapter 07S – Learning Curves
Education.
Case: Product Recall
The case has numerous possible dimensions that students can explore. When I have used this in class, I
found much interest among students, and usually I have to cut off the discussion because it runs long.
1. Questions that need to be addressed to analyze this case:
a. The recall
1) What is the expected number of recalled cars?
2) Are there geographical concentration considerations?
7) Is the learning rate the same for all models?
2. Observations regarding the information provided in the case:
a. There seems to be a significant disparity between the engineers’ learning rate (90%) and the
mechanics’ rate (75% = 7.2 / 9.6).
b. Paying a fixed rate for repairs has pros and cons. On the plus side, the manufacturer and the
Chapter 07S – Learning Curves
d. If more than one mechanic does the work, each would have to repair 24 cars.
3. Discussion and comments on major points:
a. Dealers would do best to limit one or a few mechanics to do repairs.
b. Collect more data from company and other dealerships before concluding average time of 4