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

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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

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Rate of Learning

• Let– Y(u) = time required for the u-th unit

• Then, from the definition of rate of learning,

1002

uYuYL

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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


nd

th

L

1001020


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

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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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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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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.


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)


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)


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.)

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LESSON 2: LEARNING AND EXPERIENCE CURVES