Solutions Manual: Ch19-Learning Curves-s
Review Questions
19.1 Describe the learning curve phenomenon.
Answer: The learning curve phenomenon is the reduction in cycle time that occurs in a
repetitive work activity as the number of cycles increase.
19.2 Define the term learning rate.
Answer: The learning rate is defined as the proportion by which the dependent variable,
usually task time, is multiplied every time the number of task cycles or work units doubles.
19.3 What is the rate of improvement and how is it related to the learning rate?
Answer: The rate of improvement is the proportion by which the dependent variable is
reduced every time the number of task cycles or work units doubles. The rate of
improvement IR is related to the learning rate LR as follows: IR = 1 - LR
19.4 Which value of learning rate means faster learning, 75% or 90%?
Answer: A learning rate of 75% means faster learning that 90%.
19.5 Why is the basic equation in learning curve theory, y = kxm, called a log-linear model?
Answer: The equation y = kxm plots as a straight line on log-log paper.
19.6 What is the basic difference between the Crawford equation and the Wright equation in
learning curve theory?
Answer: The Crawford equation uses unit time as the dependent variable in the log-linear
equation, whereas the Wright equation uses cumulative average time as the dependent
variable.
19.7 Why does the cumulative average time always have a higher value than the unit time in the
log-linear equation?
Answer: Because the cumulative average time includes unit time values for lower values
of N, and these unit time values are always larger, so they increase the average.
19.8 What are the two basic parameters that are required to use either the Crawford or the
Wright learning curve equations?
Answer: The two parameters are the value of the dependent variable for the first unit T1
and the value of the slope m, which is related to the learning rate LR.
19.9 How is the worker able to reduce the cycle time as the number of work units increases?
Answer: The reasons given in the text are the following: (1) The worker becomes familiar
with the task. (2) The worker makes fewer mistakes as the task is repeated. (3) The
worker’s hand and body motions become more efficient, and a rhythm and pattern are
developed in the work cycle. (4) Minor adjustments are made in the workplace layout to
reduce the distances moved. (5) There are fewer delays that interrupt the operation.
19.10 How is the larger organization able to reduce the cycle time as the number of work units
increases?
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Solutions Manual: Ch19-Learning Curves-s
Answer: The reasons given in the text are the following: (1) Methods improvements by
the industrial engineering department or other group whose objective is cost reduction, (2)
fine-tuning of machinery and tooling to eliminate delays and determine optimum operating
conditions, (3) development of special tooling that facilitates performing the task, (4)
technological improvements in processes, tooling, and machinery, (5) product design
improvements that make the product easier to make, (6) improved quality of incoming
materials so that variations and exceptions are reduced, (7) management learning, such as
better planning and scheduling, (8) improved logistical support for the operation; for
example, minimizing delays due to lack of raw materials, and (9) better motivation of
workers.
19.11 What are the two ways to obtain the value of the learning rate for a given application?
Answer: The two ways are (1) use industry averages and (2) analyze data from the actual
application or very similar applications.
19.12 What is the simplest way to estimate the learning rate based on actual data?
Answer: By observing the doubling effect in the data, that is the ratio of T2N divided by TN.
19.13 Why is it reasonable to believe that the learning rate for a worker-machine system will be
higher than for pure manual work?
Answer: Because the work cycle in a worker-machine system consists of a worker-
controlled portion and a machine-controlled portion, and the worker’s manual portion will
be affected by the learning curve phenomenon, but the machine’s portion will not because
it operates on a constant cycle time. Therefore, the overall learning rate in the task will be
greater than for a task that is 100% worker-controlled.
19.14 What is a composite learning curve?
Answer: A composite learning curve refers to the situation in which a learning rate is to be
determined for a large entity of work, such as an entire product, and this learning rate is a
function of the learning rates of the various types of work that contribute to its completion.
19.15 What is remission when used in the context of an interruption in the learning curve?
Answer: Remission in the learning curve is defined as the increase in cycle time that
occurs following an interruption in production.
19.16 How does product complexity affect the learning curve?
Answer: When product complexity is greater, this provides more opportunities for
learning, which results in higher values of first unit time or cost and steeper learning slopes
(lower values of learning rate).
19.17 What is the likely effect of greater pre-production planning on the first unit time T1 and the
slope m in the log-linear learning curve?
Answer: The effect of greater pre-production planning will be that the first unit time T1
will be lower and the slope m will be flatter.
19.18 What is the plateau model in learning curve terminology?
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Solutions Manual: Ch19-Learning Curves-s
Answer: The plateau model refers to the situation in which learning proceeds for a while,
as reflected in a continuing decrease in the unit time as the number of units increase, but
then learning finally stops, resulting in a horizontal line as the number of units continues to
increase.
19.19 What does the term standard reference quantity mean in the context of learning curves?
Answer: The standard reference quantity (SRQ) is the discrete quantity of production
(e.g., 10, 100, 1000 units) for which the standard time for the task applies. It is the
production quantity for which the standard is valid. If the actual quantity of production
varies from the SRQ, then an adjustment in the standard should be made to reflect the
difference.
Problems
Learning Curve Theory
19.1 In a mechanical assembly operation, the first work unit required 7.83 min to complete and
the learning rate for mechanical assembly is 84% in the Crawford model. Using this learning
curve model, determine the unit times to produce (a) the second unit, (b) the 10th unit, and (c)
the 100th unit. (d) What are the total cumulative time and the average cumulative time for the
100 units?
Solution: (a) m = ln(0.84)/ln 2 = -0.25154
T2 = 7.83(2)-0.25154 = 7.84(0.84) = 6.58 min
(b) T10 = 7.83(10)-0.25154 = 7.84(0.5604) = 4.39 min
(c) T100 = 7.83(100)-0.25154 = 7.84(0.3140) = 2.46 min
(d) E(TT100) = 7.83{(1001-0.25154 – 0.51-0.25154)/(1 – 0.25154)}
E(TT100) = 7.83{(31.517 – 0.595)/0.74846} = 7.83(41.314) = 323.49 min = 5.39 hr
T 100 = 323.49/100 = 3.235 min
19.2 In a certain manual operation, the first work unit required 7.83 min to complete and the
learning rate is known to be 84% in the Wright model. Using this learning curve model,
determine the average cumulative times to produce (a) the second unit, (b) the 10th unit, and
(c) the 100th unit. (d) Find the total cumulative time for the 100 units and the unit time for the
100th unit.
Solution: (a) m = ln(0.84)/ln 2 = -0.25154
T 2 = 7.83(2)-0.25154 = 7.84(0.84) = 6.58 min
(b) T 10 = 7.83(10)-0.25154 = 7.84(0.5604) = 4.39 min
(c) T 100 = 7.83(100)-0.25154 = 7.84(0.3140) = 2.46 min
(d) TT100 = 7.83(100)1-0.25154 = 7.83(31.40) = 245.86 min = 4.098 hr
TT99 = 7.83(99)1-0.25154 = 7.83(31.164) = 244.02 min = 4.067 hr
T100 = 245.86 – 244.02 = 1.845 min
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Solutions Manual: Ch19-Learning Curves-s
19.3 The unit time for the first unit of production is 72 hr and the total cumulative time for 50
units is 2,347 hr. Determine the learning rate for (a) the Crawford learning curve model and
(b) the Wright learning curve model.
Solution: (a) Crawford model: E(TT50) = 2,347 = 72{(50.5m+1 – 0.5m+1)/(m+1)}
2,347/72 = 32.5972 = (50.5m+1 – 0.5m+1)/(m+1)
Trial and error solution required.
Try m = -0.30
32.5972 = (50.50.7 – 0.50.7)/(0.7) = 21.36
Try m = -0.20
32.5972 = (50.50.8 – 0.50.8)/(0.8) = 28.09
Try m = -0.15
32.5972 = (50.50.85 – 0.50.85)/(0.85) = 32.34
Try m = -0.14
32.5972 = (50.50.86 – 0.50.86)/(0.86) = 33.27
Interpolating, (32.597 – 32.34)/(33.27 – 32.34) = 0.276
Try m = -0.15 – 0.276(0.01) = -0.147
32.5972 = (50.50.853 – 0.50.853)/(0.853) = 32.61 (Close enough)
LR = 2-0.147 = 0.903 = 90.3%
(b) Wright model: 2,347 = 72(50)m+1
32.5972 = (50)m+1
(m+1) ln 50 = ln 32.597
3.912 m + 3.912 = 3.484
3.912 m = -0.428 m = -0.1094
-0.1094
LR = 2 = 0.927 = 92.7%
19.4 A metal fabrication shop would like to know what the learning rate is for a certain section of
the plant that makes welded assemblies. In one case study, a total of four assemblies were
completed. Records were kept for units 1 and 4, and the times were 60 hours and 48.6 hours,
respectively. Unfortunately, the times for units 2 and 3 were lost. Use the Crawford learning
curve model to determine the learning rate indicated by the data?
Solution: Two solutions:
(1) Note the doubling effect:
48.6 = 60(LR)(LR) = 60(LR)2
LR2 = 48.6/60 = 0.81
LR = 0.81 = 0.90 = 90%
(2) Use the log-linear equation:
T4 = T1(N)m
48.6 = 60(4)m
0.81 = 4m
m ln 4 = ln 0.81
1.3863 m = -0.21072 m = -0.152
-0.152
LR = 2 = 0.90 = 90%
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Solutions Manual: Ch19-Learning Curves-s
19.5 A worker produces 13 parts during the first day on a new job, and the first part takes 46 min
(unit time). On the second day, the worker produces 20 parts, and the 20th part on the second
day takes 18 min. Using the Crawford learning curve model, determine (a) the learning rate
and (b) the total cumulative time to produce all 33 parts.
Solution: (a) Given T1 = 46 min and T13+20 = T33 = 18 min
T33 = T1(33)m
18 = 46(33)m
0.3913 = 33m
m ln 33 = ln 0.3913
3.4965 m = -0.9383 m = -0.2683
LR = 2-0.2683 = 0.830 = 83%
(b) E(TT33) = 46{(33.51-0.2683 – 0.51-0.2683)/(1-0.2683)}
E(TT33) = 46{(13.056- 0.602)/0.7317} = 783 min = 13.05 hr
19.6 A welder produces 7 welded assemblies during the first day on a new job, and the seventh
assembly takes 45 minutes (unit time). The worker produces 10 welded assemblies on the
second day, and the 10th assembly on the second day takes 30 minutes. Given this
information, (a) what is the percent learning rate and (b) what is the total cumulative time to
produce all 17 welded assemblies? Use the Crawford learning curve model.
Solution: (a) Given T7 = 45 min and T7+10 = T17 = 30 min
T7 = T1(7)m = 45
T17 = T1(17)m = 30
T1 = 45/7m = 30/17m
45/30 = 7m/17m = (7/17)m
1.5 = (0.41177)m
ln 1.5 = m ln 0.41177
0.40547 = -0.8873 m m = -0.45696
LR = 2-0.45696 = 0.7285 = 72.9%
(b) T1 = 45/7-0.45696 = 109.5 hr
E(TT17) = 109.5{(17.51-0.45696 – 0.51-0.45696)/(1-0.45696)}
E(TT17) = 109.5{(4.7317- 0.6863)/0.54304} = 815.7 min = 13.595 hr
19.7 Four units of a welded steel product have been completed and are being shipped to the
customer by 18-wheeler tractor-trailer. The same worker welded all units. The total time to
complete all four units was 100 hours. If the learning rate applicable to the fabrication of
products of this type is known to be 84%, what were the unit times for each of the four units?
Use the Crawford learning curve model.
Solution: TT4 = 100 = T1 + T2 + T3 + T4 = T1 + T1(2)m + T1(3)m + T1(4)m
100 = T1(1 + 2m + 3m + 4m)
m = ln(LR)/ln 2 = ln(0.84)/0.69315 = -0.17435/0.69315 = -0.25154
100 = T1(1 + 0.84 + 0.7586 + 0.7056) = T1(3.3042)
T1 = 100/3.3042 = 30.265 hr
T2 = 30.265(0.84) = 25.42 hr
T3 = 30.265(0.7586) = 22.96 hr
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Solutions Manual: Ch19-Learning Curves-s
T4 = 30.265(0.7056) = 21.36
19.8 Solve the previous problem but use the Wright learning curve model instead of the Crawford
model.
Solution: TT4 = 100
TTN = N T N or T N = TTN/N
T 4 = TT4/4 = 100/4 = 25 hr
m = ln(LR)/ln 2 = ln(0.84)/0.69315 = -0.17435/0.69315 = -0.25154
TTN = T1Nm+1
100 = T1(4)1-0.25154 = 2.8224 T1
T1 = 100/2.8224 = 35.43 hr
T2 = TT2 – T1 = T1(2)1-0.25154 – T1 = 35.43(20.74846 – 1) = 35.43(.680) = 24.09 hr
T3 = TT3 – TT2 = T1(31-0.25154 – 21-0.25154) = 35.43(2.2756 – 1.680) = 21.10 hr
T4 = TT4 – TT3 = T1(41-0.25154 – 31-0.25154) = 35.43(2.8224 – 2.2756) = 19.37 hr
19.9 A new commercial aircraft is being produced by Penn Airplane Co. Records indicate that the
first unit of the new plane required 100,000 hrs (unit time) to complete, and a cumulative
total of 534,591 hrs had been spent by the time the eighth aircraft was completed. Assuming
a constant learning rate was in effect, determine: (a) the percent learning rate, and (b) the
most likely unit time required to build the 20th airplane. Use the Crawford learning curve
model.
Solution: Given T1 = 100,000 hr and TT8 = 534,591 hr
(a) Use the estimating equation for TT8 for the Crawford model:
E(TT8) = 534,591 = 100,000{(8.5m+1 – 0.5m+1)/(m+1)}
5.34591 = (8.5m+1 – 0.5m+1)/(m+1)
Trial and error solution required:
Try m = -0.3, m+1 = 0.70
5.34591 = (8.50.7 – 0.50.7)/0.7 = (4.4729 – 0.6156)/0.7 = 5.5105
Try m = -0.35, m+1 = 0.65
5.34591 = (8.50.65 – 0.50.65)/0.65 = (4.0190 – 0.6373)/0.65 = 5.203
Interpolate: (5.3459 – 5.203)/(5.5105 – 5.203) = 0.465
Try m = -0.3 – 0.465(0.05) = -0.323, m+1 = 0.677
5.34591 = (8.50.677 – 0.50.677)/0.677 = (4.2581 – 0.6255)/0.677 = 5.366
Try m = -0.325, m+1 = 0.675
5.34591 = (8.50.675 – 0.50.675)/0.675 = (4.2399 – 0.6263)/0.675 = 5.3534
Try m = -0.326, m+1 = 0.674
5.34591 = (8.50.674 – 0.50.674)/0.674 = (4.2309 – 0.6268)/0.674 = 5.3473 (Close enough)
LR = 2-0.326 = 0.798 = 79.8%
(b) T20 = T1(20)m = 100,000(20)-0.326 = 100,000(0.3766) = 37,660 hr
19.10 Solve the previous problem but use the Wright learning curve model instead of the Crawford
model.
Solution: Given T1 = 100,000 hr and TT8 = 534,591 hr
(a) TTN = T1Nm+1
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Solutions Manual: Ch19-Learning Curves-s
534,591 = 100,000(8)m+1
5.34591 = 8m+1
(m+1) ln 8 = ln 5.34591
2.07944(m+1) = 1.67633
2.07944 m + 2.07944 = 1.67633
2.07944 m = 1.67633 - 2.07944 = -0.40311 m = -0.19385
LR = 2-0.19385 = 0.874= 87.4%
(b) T20 = TT20 – TT19
TT20 = T1(20)1-0.19385 = 1,118,988 hr
TT19 = T1(19) 1-0.19385 = 1,073,661 hr
T20 = 1,118,988 - 1,073,661 = 45,327 hr
19.11 Fifty units of a special pump product are scheduled to be made for a Middle Eastern country
to move water across the desert. A team of four workers will make the pumps, and they have
already produced three units. Time records for the first unit were not kept; however, the
second and third units took 15.0 hr and 13.4 hr, respectively. Using the Crawford learning
curve model, determine: (a) the percent learning rate; (b) the most likely time it took to do
the first unit; and (c) if the learning rate continues, how long it will take to complete the
entire quantity of 50 units.
Solution: Given T2 = 15.0 hr and T3 = 13.4 hr
TN = T1(N)m
15.0 = T1(2)m
13.4 = T1(3)m
T1 = 15.0/2m and T1 = 13.4/3m
15.0/2m = 13.4/3m
3m/2m = (1.5)m = 13.4/15 = 0.89333
m ln 1.5 = ln 0.89333
0.40546 m = -0.112796 m = -0.2782
LR = 2-0.2782 = 0.8246 = 82.46%
(b) T1 = 15.0/2-0.2782 = 15/0.8246 = 18.2 hr
(c) E(TT50) = 18.2{(50.5)1-0.2782 –(0.5)1-0.2782)/(1-0.2782)}
E(TT50) = 18.2(16.96 – 0.606)/0.7218 = 412.36 hr
19.12 An assembled product is built by ten workers who coordinate their tasks. A total of 100 units
will be made. Time records indicate that the ten workers took a total of 95 hrs (unit time) to
complete the first unit of the product. Times to complete the second and third units of work
are not available; however, the fourth work unit took 71 hrs to complete. Determine: (a) the
percent learning rate and (b) the most likely times required to complete the second and third
units. (c) If the learning rate continues, how long will it take to complete the 100th unit? (d)
What is the expected total time that will be expended by the ten workers on all 100 units?
Solution: Given T1 = 95 hr and T4 = 71 hr
TN = T1(N)m
71 = 95(4)m
71/95 = 0.74737 = (4)m
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Solutions Manual: Ch19-Learning Curves-s
ln 0.74737 = m ln 4
1.3863 m = -0.2912 m = -0.210
LR = 2-0.210 = 0.8645 = 86.45%
(b) T2 = 95(2)-0.210 = 95(0.8645) = 82.13 hr
T3 = 95(3)-0.210 = 95(0.794) = 75.42 hr
(c) T100 = 95(100)-0.210 = 95(0.3802) = 36.11 hr
(d) E(TT100) = 95{(100.51-0.21 – 0.51-0.21)/(1-0.21)}
E(TT100) = 95(38.150 – 0.578)/0.790 = 95(47.574) = 4519.5 hr
19.13 A new product will be produced in the factory by six workers, each contributing to its total
work content. A total of 500 units will be made. Time records indicate that the six workers
took a total of 8.25 hrs (unit time) to complete the first unit of the product. The time to
complete the second unit of work is not available; however, the third work unit took 6.50 hrs
to complete. Using the Crawford learning curve model, determine: (a) the percent learning
rate; (b) the most likely time it took to do the second unit; and (c) if the learning rate
continues, how long will it 5take to complete the last unit (500th unit)? (d) What is the
expected total cumulative time to complete all 500 units?
Solution: Given T1 = 8.25 hr and T3 = 6.50 hr
TN = T1(N)m
6.5 = 8.25(3)m
6.5/8.25 = 0.78788 = (3)m
ln 0.78788 = m ln 3
1.0986 m = -0.23841 m = -0.217
-0.217
LR = 2 = 0.8604 = 86.04%
(b) T2 = 8.25(2)-0.217 = 8.25(0.8604) = 7.10 hr
(c) T500 = 8.25(500)-0.210 = 8.25(0.2596) = 2.14 hr
(d) E(TT500) = 8.25{(500.51-0.217 – 0.51-0.217)/(1-0.217)}
E(TT500) = 8.25(129.907 – 0.5812)/0.783 = 8.25(165.167) = 1,362.6 hr
Interruptions in Learning
19.14 An assembled product is to be made in batches. The first unit of production takes 92.7 min
to complete and the learning rate for this type of part is known to be 85% in the Crawford
learning curve model. The first batch was for 100 units. The second order is expected in six
months for another 100 units. Determine (a) the cycle time for the first unit in the second
batch, (b) the setback, and (c) the total time to produce the second batch compared to the
total time to produce that batch if no interruption had occurred.
Solution: (a) m = ln(0.85)/ln 2 = -0.2345
First batch T100 = 92.7(100)-0.2345 = 92.7(0.33967) = 31.49 min
Remission: T101 = (92.7 – 31.49)(6 – 1)/11 = 61.21(5/11) = 27.82 min
T101 = 31.49 + 27.82 = 59.31 min
(b) Setback: 59.31 = 92.7(N)-0.2345
59.31/92.7 = 0.6398 = N-0.2345
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Solutions Manual: Ch19-Learning Curves-s
-0.2345 ln N = ln 0.6398 = -0.4466
ln N = (-0.4466)/(-0.2345) = 1.9047 N = 6.7 rounded to 7
Setback = 101-7 = 94 units
(c) E(TT101,200) = 92.7{((200.5 – 94)1-0.2345 – (100.5 – 94)1-0.2345)/(1-0.2345)}
E(TT101,200) = 92.7{(35.389 – 4.1910)/0.7655} = 92.7(40.753) = 3777.8 min = 62.96 hr
With no interruption,
E(TT101,200) = 92.7{(200.51-0.2345 – 100.51-0.2345)/(1-0.2345)}
E(TT101,200) = 92.7{(57.856 – 34.098)/0.7655} = 92.7(31.035) = 2876.9 min = 47.95 hr
Increase in production time caused by the interruption = 62.96 – 47.95 = 14.92 hr
Learning Curve Applications
19.15 A firm specializes in customizing vans and trucks for its clients. For a certain order for 5
customized vans, it took 90 hours to remake the first unit, but time records were not kept for
the other four vehicles. The firm knows that a learning rate of 82% is typical for its jobs. (a)
If the firm charges $40.00/hr, how much should it charge the client for this order? (b) If labor
cost = $20/hr and average materials cost per van = $800, how much profit did the firm
realize on the five units? (c) On the other hand, if the firm prices the first job at cost (labor
cost and materials) plus 15%, and makes its profit on the remaining models on the basis of
the learning curve effect, what is the firm’s total profit for all five vans? Use the Crawford
learning curve model.
Solution: (a) m = ln(0.82)/ln 2 = -0.2863
T2 = 90(0.82) = 73.8 hr
T3 = 90(3)-0.2863 = 65.71 hr
T4 = 90(0.82)(0.82) = 60.52 hr
T5 = 90(5)-0.2863 = 56.77 hr
TT5 = T1 + T2 + T3 + T4 + T5 = 346.8 hr
Charge = $40(346.8) = $13,872
(b) Profit = 13,872 – (346.8($20) + 5($800)) = $2,936
(c) Revenue = 5(90($20) + 800)(1 + 15%) = $14,950
Profit = 14,950 – (346.8($20) + 5($800)) = $4,014
19.16 An automobile-customizing firm makes alterations of new cars to satisfy individual
requirements of its clients. Jobs include customizing of cars for police and fire departments,
limousine fleets, etc. For a certain order for 4 customized cars, it took 100 hours to remake
the first unit, but time cards were lost for the remaining three vehicles. However, the firm
knows that a learning rate of 80% is typical for its jobs. If the firm charges $30.00/hr, how
much should it charge the client for this order? Use the Crawford learning curve model.
Solution: m = ln(0.80)/ln 2 = -0.32193
T2 = 100(0.80) = 80 hr
T3 = 100(3)-0.32193 = 70.21 hr
T4 = 100(0.80)(0.80) = 64 hr
TT4 = T1 + T2 + T3 + T4 = 314.21 hr
Charge = $30(314.21) = $9,426.30
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Solutions Manual: Ch19-Learning Curves-s
19.17 A team of workers produced a total of 12 special processing machines for a large
pharmaceutical company. Unfortunately, time records were documented for only the 12th
unit, and the labor hours expended on this unit totaled 137 hours. Although the firm did not
record the labor hours for the previous 11 units, it knows from past experience that a learning
rate of 86% is typical for worker teams producing machines of this general type. Cost of
component parts per machine = $3000, and the labor cost = $18/hr. In setting the price for
the 12 machines, the policy is to charge $35 per labor hour and to mark up the component
parts cost by 20%. (a) How much should the pharmaceutical company be charged for the 12
machines? (b) How much profit was realized on the 12 machines? Use the Crawford learning
curve model.
Solution: (a) m = ln(0.86)/ln 2 = -0.2176
T12 = 137 = T1(12)-0.2176 = T1(0.5823)
T1 = 137/0.5823 = 235.3 hr
E(TT12) = 235.3{(12.51-0.2176 – 0.5-0.2176)/(1 – 0.2176)}
E(TT12) = 235.3(7.2149 – 0.5814)/0.7824 = 1,994.6 hr
Charge = $35(1994.6) + 12($3,000)(1.20) = 69,811 + 43,200 = $113,011
(b) Profit = 113,011 – ($18(1,994.6) + 12($3,000)) = $41,108
19.18 The learning curve phenomenon is one of the reasons why an assembly line with n stations is
capable of out-producing n single workstations, where each single station does the entire
work content of the job. Consider the case of a product whose theoretical work content time
for the first unit is 20 min. The effect of an 85 percent learning rate is to be compared for two
cases, using the Crawford learning curve model: (i) 10 single workstations, each doing the
entire assembly task, and (ii) one perfectly balanced 10-station assembly line, where each
station does 2.0 minutes of the total work content. For the 1000th unit produced, determine
the rate of production of: (a) the 10 single workstations, and (b) the 10-station assembly line.
Use the Crawford learning curve model.
Solution: (a) 10 single stations, T1 = 20 min
m = ln(0.85)/ln 2 = -0.2345
With 10 single stations, each station has produced 100 units by the time a total of 1,000 units
are produced.
T100 = T1Nm = 20(100)-0.2345 = 6.794 min
The rate of production at each station Rp = 60/6.794 = 8.83 units per hour
With 10 stations, Rp = 10(8.83) = 88.3 units per hour
(b) A 10-station assembly line. Assume perfect balance so that at each station, T1 = 20/10 =
2.0 min. By the time a total of 1,000 units have been produced, each worker has performed
his/her respective task on 1,000 units.
T1000 = 2.0(1000)-0.2345 = 0.396 min
Production rate of the line Rp = 60/0.396 = 151.5 units/hr
19.19 An important customer is considering the ABC Company as a contractor for an assembly
job to produce 500 units of a certain subassembly that is used on one of its products. The
customer will let the contract to ABC for an average cumulative time of 10 hours per unit
at an allowed labor rate that is attractive to ABC. The customer has provided two sets of
component parts for the subassembly for analysis of the job. After a minimum of
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Solutions Manual: Ch19-Learning Curves-s
preplanning, ABC sets up a trial assembly workstation and its assembly team is able to
complete the first unit in 27.4 hr and the second unit in 21.8 hr. Using the Wright learning
curve model, evaluate whether the terms of the contract are favorable to ABC.
Solution: Using the Wright model, T 2 = (T1 + T2)/2 = (27.4 + 21.8)/2 = 24.6 hr
LR = 24.6/27.4 = 0.898 = 89.8%
m = ln(0.898)/ln 2 = -0.1555
T 500 = 27.4(500)-0.1555 = 27.4(0.3804) = 10.42 hr
TT500 = 500(10.42) = 5,212 hr
This compares with the customer’s contract offer of 500(10) = 5,000 hr. Based on these
learning curve computations using the Wright model, the company would have to swallow
the difference of 212 hr, which makes the terms of the contract unfavorable to ABC.
19.20 Solve the previous problem but use the Crawford learning curve model instead of the
Wright model.
Solution: Using the Crawford model, LR = 21.8/27.4 = 0.7956 = 79.56%
m = ln(0.7956)/ln 2 = -0.3298, m+1 = 0.6702
E(TT500) = 27.4(500.50.6702 – 0.50.6702)/0.6702 = 27.4(64.419 – 0.6284)/0.6702 = 2,608 hr
This compares with the customer’s contract offer of 500(10) = 5,000 hr. Based on these
learning curve computations using the Crawford model, the ABC Company would enjoy a
very large profit on the terms of the contract. The company is being paid for 5,000 hr for a
job that only requires 2,608 hr.
Comment: Comparing these values with the calculated values using the Wright model in the
previous problem, we see a two-to-one discrepancy in the results. It also shows the risk in
basing these kinds of decisions on only two data points.
19.21 The ABC Company uses a screening test for job applicants in its assembly department.
Since the typical assembly work in the department is batch assembly in quantities between
10 and 100 units per batch, the screening test requires the applicant to achieve a unit time
of 15 min on the tenth unit for a certain assembly task that has been designed specifically
for the test. However, although this is the test that all concerned parties in the company
have agreed to, it has been decided that requiring the worker to perform ten repetitions of
the assembly task would be too time-consuming, not only for the applicant but for ABC
personnel as well. An alternative test has been proposed in which the applicant would
perform only two cycles of the test assembly task, and the results would be analyzed to
determine if the applicant is likely to be capable of finishing the tenth cycle in 15 min.
Determine which of the following applicants passes the screening test: (a) this applicant
finished the first and second units in 40.0 min and 30.4 min, respectively, or (b) this
applicant finished the first and second units in 29.0 min and 23.5 min, respectively? Use
the Crawford learning curve model in your analysis.
Solution: (a) Given T1 = 40 min and T2 = 30.4 min
LR = 30.4/40 = 0.76
m = ln(0.76)/ln 2 = -0.396
T10 = 40(10)-0.396 = 40(0.40186) = 16.07 min
This candidate does not pass the test of T10 = 15 min.
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Solutions Manual: Ch19-Learning Curves-s
(b) Given T1 = 29 min and T2 = 23.5 min
LR = 23.5/29 = 0.81
m = ln(0.81)/ln 2 = -0.303
T10 = 29(10)-0.303 = 29(0.4973) = 14.4 min
This candidate passes the test. Although the applicant’s learning rate seems to be less than
the first candidate’s, this may be due to previous experience in similar assembly operations.
This may explain why the first unit time value (T1 = 29 min) was significantly less than the
other candidate’s first unit time (T1 = 40 min).
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