McGraw-Hill/Irwin Copyright © 2009 by The McGraw-Hill Companies, Inc. All Rights Reserved.

Supplement 7

Learning Curves

7S-2

LC estimates (SLOPE):

• Every doubling of repetitions results in a constant percentage decrease in the time per repetition Typical decreases range from 10 to 20 percent 10% of Improvement Rate = 90% of Learning Curve Slope = 90% of Learning Percent

7S-3

LC estimates (SLOPE):

• T2n = Learning Percent x Tn

T22 = Learning Percent x T11

T11 = T22 / Learning Percent

Learning Percent = T22 / T11

Tn = Time for nth unit

7S-4

Learning

7S-5

The Learning Effect

7S-6

Learning Curves: On a Log-Log Graph

7S-7

The Learning Effect

7S-8

Learning Illustrated

• Each time cumulative output doubles, the time per unit for that amount should be approximately equal to the previous time multiplied by the learning percentage.

• If the first unit of a process took 100 hours and the learning rate is 90%:

Unit Unit Time (hours)1 = 100

2 .90(100) = 90

4 .90(90) = 81

8 .90(81) = 72.9

16 .90(72.9) = 65.61

32 .90(65.61) = 59.049

7S-9

Unit Times: Formula Approach

logarithm natural for the standsln percentage rate learning

2lnln

unitfirst for Time unitth for Time

where

1

1

r

rb

TnT

nTT

n

bn

7S-10

Example: Formula Approach

• If the learning rate is 90%, and the first unit took 100 hours to complete, how long would it take to complete the 25th unit?

Or, Check table 7s.1 (PAGE 255) T25 = T1 x ( ) .90, 25

T25 = 100 x .613 = 61.3 hours

ln.90.15200ln 2

25 100 25 100 2561.3068 hours

T

7S-11

Unit Times: Learning Factor Approach

• The learning factor approach uses a table that shows two things for selected learning percentages:– Unit value for the number of repetitions (unit number)

– Cumulative value, which enables us to compute the total time required to complete a given number of units.

factor Unit time1 TTn

factor timeTotal1TTn

7S-12

Example: Learning Factor Approach

• If the learning rate is 90%, and the first unit took 100 hours to complete, how long would it take to complete the 25th unit?

• How long would it take to complete the first 25 units?

hours 3.61 613.10025

T

hours 3.771,1

713.71100251

T

7S-13

eg : learning rate is 80%, T1st Jet = 400 Days

a. Estimate the expected number of labor days of direct labor for the 20th jet.

• T20 = T1 x ( ) .80, 20

T20 = 400 x .381 = 152.4 Days

b. Estimate the expected number of labor days of direct labor for all 20 jets.

• T1-20 = T1 x ( ) .80, 1-20

T1-20 = 400 x 10.485 = 4194 Days

Avg Days/Jet = 4194 days / 20jets = 209.7 days/jet

7S-14

Eg : learning rate is 80, T1st Jet = 400 Days

c. If the company expects a contribution to overhead & profit of $150/day on top of a labor cost $200/day, what should be the Total Price Quote?

• T1-20 = 4194 Days Charge per day=150+200=350

Total price quote = $350 x 4194days = $1467900d. If the company expects a contribution to overhead &

profit of $150/day on top of a labor cost $200/day, what should be the Unit Price Quote?

• Unit price quote = $1467900 / 20jets = $73395

7S-15

Eg : learning rate is 80, T1st Jet = 400 Days

e. Estimate the expected number of labor days of direct labor for jet 10 through jet 15.

• T1-15 – T1-9 = T1 x ( ).80, 1-15 – T1 x ( ).80, 1-9

= 400 x 8.511 – 400 x 5.839

= 400 x (8.511 – 5.839) = 1068.8 Days

7S-16

Eg : learning rate is 80, T1st Jet = 400 Days

f. If the total labor cost of 20 jets is $1467900, what should be the total labor cost for 30 jets and the average cost per unit?

• T1-20 = 4194 Days • $1467900 /4194days = $350/day T1-30 = 400 x 14.020 = 5608 Days Total labor cost =$350x5608days=$1962800• Avg cost/unit =$1962800/30Jets=$65426.70