lesson 2: learning and experience curves
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LESSON 2: LEARNING AND EXPERIENCE CURVES. Outline Rate of Learning Learning Curve Estimating Parameter Values. Rate of Learning. - PowerPoint PPT PresentationTRANSCRIPT
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LESSON 2: LEARNING AND EXPERIENCE CURVES
Outline
• Rate of Learning • Learning Curve • Estimating Parameter Values
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Rate of Learning
• As workers gain more experience with the requirements of a particular process, or as the process is improved with time, the number of hours required to produce an additional unit declines.
• The learning curve models this relationship.• Rate of learning, is defined as follows:
1002
item th- produce to required Timeitem th- produce to required Time
uuL
L
3
Rate of Learning
• Let– Y(u) = time required for the u-th unit
• Then, from the definition of rate of learning,
1002
uYuYL
4
Rate of Learning
• For example, using rate of learning,
• Using rate of learning,
• Using rate of learning,
,1u
10012
item produce to required Timeitem produce to required Time
st
nd
L
,2u
10024
item produce to required Timeitem produce to required Time
nd
th
L
1001020
item produce to required Timeitem produce to required Time
th
th
L
,10u
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Rate of Learning
• Rewriting the definition of rate of learning,
• Hence,
• If we know the time required for the first unit, Y(1), we can find the time required for the 2nd, 4th, 8th, ….. units using the above equation iteratively.
1002
uYuYL
uYLuY100
2
6
Rate of Learning
• The time required for the 1st unit = Y(1) [notation]• The time required for the 2nd unit,
• The time required for the 4th unit,
• The time required for the 8th unit,
1100
2 YLY
2100
4 YLY
4100
8 YLY
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Rate of Learning
• An example of 80% learning rate: Suppose that it requires 100 hours to produce the first unit. Then, the 2nd unit requires (0.80)(100)=80 hours. The 4th units requires (0.80)(80)=64 hours, and so on.Unit Number of Hours Required1st unit 100 hours2nd unit (0.80)100=80 hours4th unit (0.80)(80)=64 hours8th unit (0.80)(64)=51.2 hours
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• In general, for any unit u, not necessarily 1, 2, 4, 8, …, the time required can be obtained from the learning curve equation.
• The learning curve is of the formY(u) = au-b
Where, a and b are parameters.a = time required for the first unitb = - ln (L) / ln (2), where L is the
rate of learning, 0.80 for 80% learning, 0.90 for 90% learning, etc.
Learning Curve
Here, ln = natural log. A review on logarithms follows.
Pro
cess
ing
time
per u
nit,
Y(u)
Units produced, u
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Learning Curve
Suppose that
– a = 18 hours– Learning rate = 80%– What is time for the 9th unit?
Y(u) = au-b
= auln(L)/ln(2)
Y(9) =
Here, ln = natural log. A review on logarithms follows.
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Logarithms (Review)
• Recall that if then, . • Here, p is the base. • If the base is e, “ln” (natural log) replaces “log”. So,
• Here, e is a constant:
qp x qx plog
qq elogln
....71828.2
...241
61
21
11
11
...!4
1!3
1!2
1!1
1!0
1
11
e
nnLim
en
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Logarithms (Review)
• A scientific calculator usually contains 2 buttons:• log x provides log10 x , logarithmic value of some
number x with base 10 • ln x provides loge x , logarithmic value of some
number x with base e =2.71828…• To get a logarithmic value with a base other than 10 or e,
use the following formula:
pq
pqq
pq
pqq
e
ep
p
lnln
logloglog
loglog
logloglog
10
10
or,
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• Recall, that learning curve is of the formY(u) = au-b
Where, a and b are parameters.• If we observe the time required to
produce various units, we can estimate parameters a and b along with the rate of learning L.
• The relationship between u and Y(u), as shown on the left, is not linear. But, the relationship between ln(u) and ln(Y(u)) is linear.
Pro
cess
ing
time
per u
nit,
Y(u)
Units produced, u
Estimating Parameter Values
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Estimating Parameter Values
Y(u) = au-b (Learning Curve)
or, ln(Y(u)) = ln(au-b) (Take logarithm on both sides)
or, ln(Y(u)) = ln(a)+ln(u-b)or, ln(Y(u)) = ln(a) - bln(u)
This equation has the form of a straight line y = c + mx (straight line, with slope m and intercept c)
Thus, a plot of ln(u) vs ln(Y(u)) fits a straight line
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Estimating Parameter Values
ln(Y(u)) = ln(a) - bln(u) (Learning Curve)y = c +mx (straight line)
Notice thatIntercept = ln(a) Hence, a = eintercept
Slope = - b Hence, b = -slope
Finally, Since, b = - ln (L) / ln (2), we have L = eslope*ln(2)
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• It’s an important fact that the relationship between ln(u) and ln(Y(u)) is linear. Because if the relationship between two variables is linear, we can fit a straight line that provides the relationship.
• The slope and intercept of the straight line are obtained by using linear regression on ln(u) and ln(Y(u)).
• The slope and intercept can then be used to get paratmeters a and b and rate of learning L.
Estimating Parameter Values
ln(Y
(u))
ln(u)
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• We estimate parameters as follows:• Step 1: Given a set of u and Y(u)
values, compute the set of ln(u) and ln(Y(u)) values.
• Step 2: Using linear regression on ln(u) and ln(Y(u)), compute slope, m and intercept, c of the straight line that best fits the set of ln(u) and ln(Y(u)) values.
ln(Y
(u))
ln(u)
Estimating Parameter Values
c
1m
• Step 3: Compute a, b and L using the following formula: – a = eintercept = ec
– b = -slope = -m– L = eslope*ln(2) = em*ln(2)
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• An interpretation of the intercept, c:– ec is an estimate of the time
required for the first unit denoted by a or Y (1).
• An interpretation of the slope, m:– em*ln(2) is an estimate of the rate
of learning, L.– Learning is demonstrated by
the negative slope. – If the slope is less, then the
line is steeper, L is less and the learning is faster.
ln(Y
(u))
ln(u)
Estimating Parameter Values
c
1m
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Estimating Parameter Values: Example
Consider the text example:
Cumulative Number of Number of Hours RequiredUnits Produced For the Next Unit
u Y (u )10 9.2225 4.85
100 3.8250 2.44500 1.7
1000 1.035000 0.6
10000 0.5
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Relationship Between u and Y(u)A plot of u vs Y(u) is not linear.
u vs Y(u)
0123456789
10
0 2000 4000 6000 8000 10000
u
Y(u)
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A plot of ln(u) vs ln(Y(u)) is linear. Hence, linear regression is used on ln(u) and ln(Y(u)).
Relationship Between ln(u) and ln(Y(u))
ln(u ) vs ln(Y (u )
-1
0
1
2
3
0 5 10ln(u )
ln(Y
(u))
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Step 1
Step 1: Compute the logarithmic values.
The Excel function for computing natural logarithms is LN()e.g., if a value of u is in B6, formula for ln(u) is =LN(B6)
u Y (u ) ln(u ) ln(Y (u ))10 9.2225 4.85
100 3.8250 2.44500 1.71000 1.035000 0.6
10000 0.5
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Step 2 by HandStep 2: Compute
i x y xy x ^2ln(u ) ln(y (u )
1 2.302585093 2.221375042 3.218875825 1.57897873 4.605170186 1.335001074 5.521460918 0.891998045 6.214608098 0.530628256 6.907755279 0.02955887 8.517193191 -0.51082568 9.210340372 -0.6931472
SumAverage
yxxxyyx and ,,,, 2
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Step 3 by Hand
Step 3: Compute the slope and intercept:
)
2
11
2
111
xy
xxn
yxyxn
n
ii
n
ii
n
ii
n
ii
n
iii
slope(Intercept
Slope
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Steps 2 and 3 by Excel
• If Excel is used, steps 2 and 3 can be replaced by a single step. Two built-in Excel functions provides slope and intercept as shown below:
• Suppose thatln(u) values are in column A rows 18-25ln(Y(u)) values are in column B rows 18-25
• Excel formulae forIntercept is INTERCEPT(B18:B25,A18:A25)Slope is SLOPE(B18:B25,A18:A25)
• Thus, intercept = 3.1301, and slope = -0.42276.
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Step 4
Step 4: Compute the parameters a, b, L
Suppose that the values of intercept and slope are in cells B29 and B30 respectively
Parameter Formula Excel formula Valuea eintercept = EXP(B29)b -slope = -B30L eslope*ln(2) = EXP(B30*ln(2))
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READING AND EXERCISES
Lesson 2
Reading: Section 1.10, pp. 32-38 (4th Ed.), pp. 29-36 (5th Ed.)
Exercises: 29, 30, 33, pp. 37-38 (4th Ed.), pp. 35-36 (5th Ed.)