learning curve analysis
DESCRIPTION
Learning Curve Analysis. Supplement G. 0.30 – 0.25 – 0.20 – 0.15 – 0.10 – 0.05 – 0 –. Learning curve. Process time per unit (hr). |||||| 50100150200250300. Cumulative units produced. Learning Curves. 0.30 – 0.25 – 0.20 – 0.15 – 0.10 – 0.05 – 0 –. - PowerPoint PPT PresentationTRANSCRIPT
© 2007 Pearson Education
Learning CurveAnalysis
Supplement GSupplement G
© 2007 Pearson Education
Learning Curves
0.30 0.30 –
0.25 0.25 –
0.20 0.20 –
0.15 0.15 –
0.10 0.10 –
0.05 0.05 –
0 0 –| | | | | |
5050 100100 150150 200200 250250 300300
Learning curveLearning curve
Cumulative units producedCumulative units produced
Pro
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s t
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(h
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tim
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hr)
© 2007 Pearson Education
Learning Curves
0.30 0.30 –
0.25 –0.25 –
0.20 0.20 –
0.15 0.15 –
0.10 0.10 –
0.05 0.05 –
0 0 –
| | | | | |5050 100100 150150 200200 250250 300300
Learning curveLearning curve
Cumulative units producedCumulative units produced
Pro
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s t
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(h
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Learning period
Showing the learning period
© 2007 Pearson Education
Learning Curves
0.30 0.30 –
0.25 0.25 –
0.20 0.20 –
0.15 0.15 –
0.10 0.10 –
0.05 0.05 –
0 0 –| | | | | |
5050 100100 150150 200200 250250 300300
Learning curveLearning curve
Cumulative units producedCumulative units produced
Pro
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s t
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nit
(h
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tim
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Learning Learning periodperiod
Standard time
Showing the learning period and the time when standards are calculated
© 2007 Pearson Education
Developing Learning Curves
In developing learning curves we make the following assumptions:
The direct labor required to produce the n + 1st unit will always be less than the direct time of labor required for the nth unit.
Direct labor requirements will decrease at a declining rate as cumulative production increases.
The reduction in time will follow an exponential curve.
kn = k1nb
where k1 = direct labor hours for the 1st unit
n = cumulative number of units produced b = log r / log 2
r = learning rate
© 2007 Pearson Education
80% Conversion Factors for the Cumulative Average Number of
Direct Labor Hours per Unit
© 2007 Pearson Education
90% Conversion Factors for the Cumulative Average Number of
Direct Labor Hours per Unit
© 2007 Pearson Education
Example G.1 Developing the 80% Learning Curve
Manufacturer of diesel locomotives:
Labor hours required for first unit = 50,000
Learning rate = 80%
50 50 –
40 40 –
30 30 –
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10 10 –
0 0 –| | | | | | |
4040 8080 120120 160160 200200 240240 280280Cumulative units producedCumulative units produced
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© 2007 Pearson Education
Example G.1 Estimating Direct Labor Requirements
50 50 –
40 40 –
30 30 –
20 20 –
10 10 –
0 0 –| | | | | | |
4040 8080 120120 160160 200200 240240 280280Cumulative units producedCumulative units produced
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Labor hours required for first unit = 50,000
Learning rate = 80%
© 2007 Pearson Education
Example G.1 using the formula
Labor hours required for 40th unit
k40 = 50,000(40)(log 0.8)/(log 2)
Labor hours required for first unit = 50,000
Learning rate = 80%
50 50 –
40 40 –
30 30 –
20 20 –
10 10 –
0 0 –| | | | | | |
4040 8080 120120 160160 200200 240240 280280Cumulative units producedCumulative units produced
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© 2007 Pearson Education
50 50 –
40 40 –
30 30 –
20 20 –
10 10 –
0 0 –| | | | | | |
4040 8080 120120 160160 200200 240240 280280Cumulative units producedCumulative units produced
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k40 = 50,000(40)-0.322
Labor hours required for first unit = 50,000
Learning rate = 80%
Example G.1 using the formula
© 2007 Pearson Education
50 50 –
40 40 –
30 30 –
20 20 –
10 10 –
0 0 –| | | | | | |
4040 8080 120120 160160 200200 240240 280280Cumulative units producedCumulative units produced
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k40 = 50,000(0.30488)
Labor hours required for first unit = 50,000
Learning rate = 80%
Example G.1 using the formula
© 2007 Pearson Education
50 50 –
40 40 –
30 30 –
20 20 –
10 10 –
0 0 –| | | | | | |
4040 8080 120120 160160 200200 240240 280280Cumulative units producedCumulative units produced
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k40 = 15,244 hours
Labor hours required for first unit = 50,000
Learning rate = 80%
Example G.1 using the formula
© 2007 Pearson Education
50 50 –
40 40 –
30 30 –
20 20 –
10 10 –
0 0 –| | | | | | |
4040 8080 120120 160160 200200 240240 280280Cumulative units producedCumulative units produced
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k40 = 15,244 hours
Labor hours required for first unit = 50,000
Learning rate = 80%
Example G.1 using the formula
© 2007 Pearson Education
50 50 –
40 40 –
30 30 –
20 20 –
10 10 –
0 0 –| | | | | | |
4040 8080 120120 160160 200200 240240 280280Cumulative units producedCumulative units produced
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k40 = 15,244 hours
Labor hours required for first unit = 50,000
Learning rate = 80%
Example G.1 using the formula
© 2007 Pearson Education
50 50 –
40 40 –
30 30 –
20 20 –
10 10 –
0 0 –| | | | | | |
4040 8080 120120 160160 200200 240240 280280Cumulative units producedCumulative units produced
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Cumulative average labor hours =
Labor hours required for first unit = 50,000
Learning rate = 80%n
1 1.000002 0.900003 0.83403. .. .. .
38 0.4363439 0.4330440 0.4298464 0.37382
128 0.30269
80% Learning Rate(n = cumulative production)
Example G.1 using Conversion Factors
© 2007 Pearson Education
50 50 –
40 40 –
30 30 –
20 20 –
10 10 –
0 0 –| | | | | | |
4040 8080 120120 160160 200200 240240 280280Cumulative units producedCumulative units produced
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Labor hours required for first unit = 50,000
Learning rate = 80%
Cumulative average labor hours = 50,000(0.42984)
n
1 1.000002 0.900003 0.83403. .. .. .
38 0.4363439 0.4330440 0.4298464 0.37382
128 0.30269
80% Learning Rate(n = cumulative production)
Example G.1 using Conversion Factors
© 2007 Pearson Education
50 50 –
40 40 –
30 30 –
20 20 –
10 10 –
0 0 –| | | | | | |
4040 8080 120120 160160 200200 240240 280280Cumulative units producedCumulative units produced
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Labor hours required for first unit = 50,000
Learning rate = 80%
Cumulative average labor hours = 21,492 hours
n
1 1.000002 0.900003 0.83403. .. .. .
38 0.4363439 0.4330440 0.4298464 0.37382
128 0.30269
80% Learning Rate(n = cumulative production)
Example G.1 using Conversion Factors
© 2007 Pearson Education
Labor hours required for first unit = 50,000
Learning rate = 80%
50 50 –
40 40 –
30 30 –
20 20 –
10 10 –
0 0 –| | | | | | |
4040 8080 120120 160160 200200 240240 280280Cumulative units producedCumulative units produced
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Example G.1 using Unit-doublings
© 2007 Pearson Education
50 50 –
40 40 –
30 30 –
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10 10 –
0 0 –| | | | | | |
4040 8080 120120 160160 200200 240240 280280Cumulative units producedCumulative units produced
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Labor hours required for first unit = 50,000
Learning rate = 80%
Example G.1 using Unit-doublings
© 2007 Pearson Education
50 50 –
40 40 –
30 30 –
20 20 –
10 10 –
0 0 –| | | | | | |
4040 8080 120120 160160 200200 240240 280280Cumulative units producedCumulative units produced
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Labor hours required for first unit = 50,000
Learning rate = 80%
Second unit = 50,000(80%)
Example G.1 using Unit-doublings
© 2007 Pearson Education
50 50 –
40 40 –
30 30 –
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10 10 –
0 0 –| | | | | | |
4040 8080 120120 160160 200200 240240 280280Cumulative units producedCumulative units produced
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Labor hours required for first unit = 50,000
Learning rate = 80%
Second unit = 40,000 hours
Example G.1 using Unit-doublings
© 2007 Pearson Education
50 50 –
40 40 –
30 30 –
20 20 –
10 10 –
0 0 –| | | | | | |
4040 8080 120120 160160 200200 240240 280280Cumulative units producedCumulative units produced
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Labor hours required for first unit = 50,000
Learning rate = 80%
Fourth unit = 40,000(80%)
Example G.1 using Unit-doublings
© 2007 Pearson Education
50 50 –
40 40 –
30 30 –
20 20 –
10 10 –
0 0 –| | | | | | |
4040 8080 120120 160160 200200 240240 280280Cumulative units producedCumulative units produced
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Labor hours required for first unit = 50,000
Learning rate = 80%
Fourth unit = 32,000 hours
Example G.1 using Unit-doublings
© 2007 Pearson Education
50 50 –
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0 0 –| | | | | | |
4040 8080 120120 160160 200200 240240 280280Cumulative units producedCumulative units produced
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The 80% Learning Curve forExample G.1
© 2007 Pearson Education
50 50 –
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10 10 –
0 0 –| | | | | | |
4040 8080 120120 160160 200200 240240 280280Cumulative units producedCumulative units produced
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The 80% Learning Curve forExample G.1
© 2007 Pearson Education
Application G.1 Estimating Direct Labor Requirements
The 1st unit of a new product is expected to take 1,000 hours. The learning rate is 80%, how much time should the 50th unit take?
© 2007 Pearson Education
Units perUnits per CumulativeCumulativeMonthMonth MonthMonth UnitsUnits
11 22 2222 33 5533 55 101044 88 181855 1212 3030
Example G.2Estimating Labor Requirements
© 2007 Pearson Education
Example G.2Estimating Labor Requirements
Units perUnits per CumulativeCumulativeMonthMonth MonthMonth UnitsUnits
11 22 2222 33 5533 55 101044 88 181855 1212 3030
Cumulative CumulativeAverage Time Total Hours
Month per Unit for All Units
n
1 1.000002 0.950003 0.915404 0.889055 0.86784. .. .. .
30 0.6909064 0.62043
128 0.56069
90% Learning Rate(n = cumulative production)
© 2007 Pearson Education
Units perUnits per CumulativeCumulativeMonthMonth MonthMonth UnitsUnits
11 22 2222 33 5533 55 101044 88 181855 1212 3030
Cumulative CumulativeAverage Time Total Hours
Month per Unit for All Units
1 30,000(0.95000) = 28,500
n
1 1.000002 0.950003 0.915404 0.889055 0.86784. .. .. .
30 0.6909064 0.62043
128 0.56069
90% Learning Rate(n = cumulative production)
Example G.2Estimating Labor Requirements
© 2007 Pearson Education
Units perUnits per CumulativeCumulativeMonthMonth MonthMonth UnitsUnits
11 22 2222 33 5533 55 101044 88 181855 1212 3030
Cumulative CumulativeAverage Time Total Hours
Month per Unit for All Units
1 30,000(0.95000) = 28,500 28,500(2) = 57,0002 30,000(0.86784) = 26,035
n
1 1.000002 0.950003 0.915404 0.889055 0.86784. .. .. .
30 0.6909064 0.62043
128 0.56069
90% Learning Rate(n = cumulative production)
Example G.2Estimating Labor Requirements
© 2007 Pearson Education
Units perUnits per CumulativeCumulativeMonthMonth MonthMonth UnitsUnits
11 22 2222 33 5533 55 101044 88 181855 1212 3030
Cumulative CumulativeAverage Time Total Hours
Month per Unit for All Units
1 30,000(0.95000) = 28,500 28,500(2) = 57,0002 30,000(0.86784) = 26,035 26,035(5) = 130,1753 30,000(0.79945) = 23,983 23,983(10) = 239,8304 30,000(0.74080) = 22,224 22,224(18) = 400,0325 30,000(0.69090) = 20,727 20,727(30) = 621,810
n
1 1.000002 0.950003 0.915404 0.889055 0.86784. .. .. .
30 0.6909064 0.62043
128 0.56069
90% Learning Rate(n = cumulative production)
Example G.2Estimating Labor Requirements
© 2007 Pearson Education
Units perUnits per CumulativeCumulativeMonthMonth MonthMonth UnitsUnits
11 22 2222 33 5533 55 101044 88 181855 1212 3030
Cumulative CumulativeAverage Time Total Hours
Month per Unit for All Units
1 30,000(0.95000) = 28,5002 30,000(0.86784) = 26,0353 30,000(0.79945) = 23,9834 30,000(0.74080) = 22,2245 30,000(0.69090) = 20,727
Example G.2Estimating Labor Requirements
© 2007 Pearson Education
Units perUnits per CumulativeCumulativeMonthMonth MonthMonth UnitsUnits
11 22 2222 33 5533 55 101044 88 181855 1212 3030
Cumulative CumulativeAverage Time Total Hours
Month per Unit for All Units
1 30,000(0.95000) = 28,500 28,500(2) = 57,0002 30,000(0.86784) = 26,0353 30,000(0.79945) = 23,9834 30,000(0.74080) = 22,2245 30,000(0.69090) = 20,727
Example G.2Estimating Labor Requirements
© 2007 Pearson Education
Units perUnits per CumulativeCumulativeMonthMonth MonthMonth UnitsUnits
11 22 2222 33 5533 55 101044 88 181855 1212 3030
Cumulative CumulativeAverage Time Total Hours
Month per Unit for All Units
1 30,000(0.95000) = 28,500 28,500(2) = 57,0002 30,000(0.86784) = 26,035 26,035(5) = 130,1753 30,000(0.79945) = 23,9834 30,000(0.74080) = 22,2245 30,000(0.69090) = 20,727
Example G.2Estimating Labor Requirements
© 2007 Pearson Education
Units perUnits per CumulativeCumulativeMonthMonth MonthMonth UnitsUnits
11 22 2222 33 5533 55 101044 88 181855 1212 3030
Cumulative CumulativeAverage Time Total Hours
Month per Unit for All Units
1 30,000(0.95000) = 28,500 28,500(2) = 57,0002 30,000(0.86784) = 26,035 26,035(5) = 130,1753 30,000(0.79945) = 23,983 23,983(10) = 239,8304 30,000(0.74080) = 22,224 22,224(18) = 400,0325 30,000(0.69090) = 20,727 20,727(30) = 621,810
Example G.2Estimating Labor Requirements
© 2007 Pearson Education
Units perUnits per CumulativeCumulativeMonthMonth MonthMonth UnitsUnits
1 2 22 3 53 5 104 8 185 12 30
Cumulative CumulativeAverage Time Total Hours
Month per Unit for All Units
1 30,000(0.95000) = 28,500 28,500(2) = 57,0002 30,000(0.86784) = 26,035 26,035(5) = 130,1753 30,000(0.79945) = 23,983 23,983(10) = 239,8304 30,000(0.74080) = 22,224 22,224(18) = 400,0325 30,000(0.69090) = 20,727 20,727(30) = 621,810
Month 1:
Example G.2Estimating Labor Requirements
© 2007 Pearson Education
Units perUnits per CumulativeCumulativeMonthMonth MonthMonth UnitsUnits
1 2 22 3 53 5 104 8 185 12 30
Cumulative CumulativeAverage Time Total Hours
Month per Unit for All Units
1 30,000(0.95000) = 28,500 28,500(2) = 57,0002 30,000(0.86784) = 26,035 26,035(5) = 130,1753 30,000(0.79945) = 23,983 23,983(10) = 239,8304 30,000(0.74080) = 22,224 22,224(18) = 400,0325 30,000(0.69090) = 20,727 20,727(30) = 621,810
Month 1: 57,000 – 0 = 57,000 hours
Example G.2Estimating Labor Requirements
© 2007 Pearson Education
Units perUnits per CumulativeCumulativeMonthMonth MonthMonth UnitsUnits
1 2 22 3 53 5 104 8 185 12 30
Cumulative CumulativeAverage Time Total Hours
Month per Unit for All Units
1 30,000(0.95000) = 28,500 28,500(2) = 57,0002 30,000(0.86784) = 26,035 26,035(5) = 130,1753 30,000(0.79945) = 23,983 23,983(10) = 239,8304 30,000(0.74080) = 22,224 22,224(18) = 400,0325 30,000(0.69090) = 20,727 20,727(30) = 621,810
Month 1: 57,000 – 0 = 57,000 hoursMonth 2: 130,175 – 57,000 = 73,175 hours
Example G.2Estimating Labor Requirements
© 2007 Pearson Education
Units perUnits per CumulativeCumulativeMonthMonth MonthMonth UnitsUnits
1 2 22 3 53 5 104 8 185 12 30
Cumulative CumulativeAverage Time Total Hours
Month per Unit for All Units
1 30,000(0.95000) = 28,500 28,500(2) = 57,0002 30,000(0.86784) = 26,035 26,035(5) = 130,1753 30,000(0.79945) = 23,983 23,983(10) = 239,8304 30,000(0.74080) = 22,224 22,224(18) = 400,0325 30,000(0.69090) = 20,727 20,727(30) = 621,810
Month 1: 57,000 – 0 = 57,000 hoursMonth 2: 130,175 – 57,000 = 73,175 hoursMonth 3: 239,830 – 130,175 = 109,655 hoursMonth 4: 400,032 – 239,830 = 160,202 hoursMonth 5: 621,810 – 400,032 = 221,778 hours
Example G.2Estimating Labor Requirements
© 2007 Pearson Education
Units perUnits per CumulativeCumulativeMonthMonth MonthMonth UnitsUnits
1 2 22 3 53 5 104 8 185 12 30
Cumulative CumulativeAverage Time Total Hours
Month per Unit for All Units
1 30,000(0.95000) = 28,500 28,500(2) = 57,0002 30,000(0.86784) = 26,035 26,035(5) = 130,1753 30,000(0.79945) = 23,983 23,983(10) = 239,8304 30,000(0.74080) = 22,224 22,224(18) = 400,0325 30,000(0.69090) = 20,727 20,727(30) = 621,810
Month 1: 57,000 – 0 = 57,000 /150 = 380 employeesMonth 2: 130,175 – 57,000 = 73,175 /150 = 488 employeesMonth 3: 239,830 – 130,175 = 109,655 /150 = 731 employeesMonth 4: 400,032 – 239,830 = 160,202 /150 = 1068 employeesMonth 5: 621,810 – 400,032 = 221,778 /150 = 1479 employees
Example G.2Estimating Labor Requirements
© 2007 Pearson Education
Application G.2 Estimating Cumulative Labor Hours
An example of using the learning model to test budget constraints:
A company has a contract to make a product for the
first time. The total budget for the 38-unit job is 15,000
hours. The first unit took 1000 hours, and the rate of
learning is expected to be 80 percent.
Do you think the 38-unit job can be completed within
the 15,000-hour budget?
How many additional hours would you need for a
second job of an 26 additional units?
© 2007 Pearson Education
Application G.2First 38-unit Job
© 2007 Pearson Education
Application G.2Second additional 26-unit Job