Supplement 7
description
Transcript of Supplement 7
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