2.008 -spring 20041 2.008 process control. 2.008 -spring 20042 outline 1.optimization 2.statistical...
TRANSCRIPT
2.008 -Spring 2004 1
2.008
Process Control
2.008 -Spring 2004 2
Outline
1. Optimization
2. Statistical Process Control
3. In-Process Control
2.008 -Spring 2004 3
What is quality?
2.008 -Spring 2004 4
Variation: Common and Special Causes
Pieces vary from each other:
But they form a pattern that, if stable, is called a distribution:
Size Size Size Size
Size Size Size
2.008 -Spring 2004 5
Common and Special Causes (cont’d)
Distributions can differ in…
…or any combination of these
Size Size Size
Location Spread Shape
2.008 -Spring 2004 6
Common and Special Causes (cont’d)
If only common causes of variation are present, the output of a process forms a distribution that is stabl
e over time and is predictable:
Size
Prediction
Time
2.008 -Spring 2004 7
Common and Special Causes (cont’d)
IF special causes of variation are present, the process output is not stable over time and is not predictable:
Size
Prediction
Time
2.008 -Spring 2004 8
Variation: Run Chart
Specified diameter = 0.490” +/-0.001”
Dia
met
er o
f p
art
Run numberProcess change
0.4915
0.4910
0.4905
0.4900
0.4895
0.4890
0.4885
0.48800 10 20 30 40 50
2.008 -Spring 2004 9
Process Control
In Control(Special causes eliminated)
Out of Control(Special causes present)
Time
Size
2.008 -Spring 2004 10
Statistical Process Control
1. Detect disturbances (special causes)
2. Take corrective actions
2.008 -Spring 2004 11
Central Limit Theorem
• A large number of independent events have a continuous probability density function that is normal in shape.
• Averaging more samples increases the precision of the estimate of the average.
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Shewhart Control ChartA
vera
ge
(o
f 10
sam
ple
s) D
iam
eter
Run number
Upper control limit
Lower control limit
1.010
0.990
0.995
1.000
1.005
0 10 20 30 40 50 60 70 80 90 100
2.008 -Spring 2004 13
Sampling and Histogram Creation
Wheel of Fortune: Equal probability of outcome 1-
10, P=0.1 Taking 100 random samples, the
resulting histogram would look like this
Taking ∞ random samples, the resulting histogram would look li
ke this
1 2 3 4 5 6 7 8 9 10
Fre
quen
cy
20181614121086420
Outcome
1 2 3 4 5 6 7 8 9 10
Fre
quen
cyOutcome
2.008 -Spring 2004 14
Sampling and Histogram Creation (cont’d)
Wheel of Fortune: Equal probability of outcome 1-
10, P=0.1
Take 10 random samples, calculate their
average, and repeat 100 times, the resulting
histogram would resemble
Take 10 random samples, calculate their average, and repeat ∞ times, the resulting histogram would approach the continuous distributi
on shown
1 2 3 4 5 6 7 8 9 10
Fre
quen
cy
20
18
16
14
12
10
8
6
4
2
0
Outcome
1 2 3 4 5 6 7 8 9 10
2.008 -Spring 2004 15
Uniform DistributionsUnderlying Distribution
Pro
bab
ilit
y d
istr
ibu
tio
n 1.5
1
0.5
0
0 0.5 1
1.5
1
0.5
0
0 0.5 1
Pro
bab
ilit
y d
istr
ibu
tio
n
Distribution resulting from averaging of 2 random values
Distribution resulting from averaging of 3 random values
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0
Distribution resulting from averaging of 10 random values
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0
2.008 -Spring 2004 16
Normal Distribution
Distribution resulting from averaging of 3 random values
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0
Distribution resulting from averaging of 10 random values
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0
Underlying Distribution
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0
Distribution resulting from averaging of 2 random values
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0
2.008 -Spring 2004 17
Precision
Not precise Precise
Not accurate
Accurate
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Shewhart Control Chart
• Upper Control Limit (UCL), Lower Control Limit (LCL)
• Subgroup size ( 5 < n < 20 )
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Shewhart Control Chart (cont’d)
Control chart based on 100 samples
Ave
rag
e (
of
10 s
amp
les)
Dia
met
er
Run number
Upper control limit
Lower control limit
1.010
0.990
0.995
1.000
1.005
0 10 20 30 40 50 60 70 80 90 100
Step disturbance
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Shewhart Control Chart (cont’d)A
vera
ge
(o
f 10
sam
ple
s) D
iam
eter
Run number
Upper control limit
Lower control limit
1.010
0.990
0.995
1.000
1.005
0 10 20 30 40 50 60 70 80 90 100
Step disturbance
Control chart based on average of 10 samples, note the step change that occurs at run 51
2.008 -Spring 2004 21
Control Charts
1. Collection:. Gather data and plot on chart.
2. Control. Calculate control limits from process data, using simple formulae.
. Identify special causes of variation; take local actions to correct.
3. Capability. Quantify common cause variation; take action on the system.
These three phases are repeated for continuing process improvement.
Upper control limit Process average Lower control limit
0 10 20 30 40 50 60 70 80 90 100Run number
2.008 -Spring 2004 22
Setting the Limits
Idea: Points outside the limits will signal that something is wro
ng-an assignable cause. We want limits set so that assignabl
e causes are highlighted, but few random causes are highligh
ted accidentally.
Convention for Control Charts:• Upper control limit (UCL) = x + 3σsg
• Lower control limit (LCL) = x-3σsg
(Where σsg represents the standard deviation of a subgroup of samples)
2.008 -Spring 2004 23
Setting the Limits (cont’d)
Convention for Control Charts (cont’d):
As n increases, the UCL and LCL move closer to the center line, making the control chart more sensitive to shifts in the mean.
processsggroup sub
n/processsubgroup
nx /3UCL process
nx /3LCL process
2.008 -Spring 2004 24
Benefits of Control Charts
Properly used, control charts can:
• Be used by operators for ongoing control of a process
• Help the process perform consistently, predictably, for quality and cost
• Allow the process to achieve:
– Higher quality
– Lower unit cost
– Higher effective capacity
• Provide a common language for discussing process performance
• Distinguish special from common causes of variation; as a guide to local or management action
2.008 -Spring 2004 25
Process Capability
Lower specification limit
Upper specification limit
In Control and Capable(Variation from common cause reduced)
In Control but not Capable(Variation from common cause excessive)
Size
Time
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Process Capability Index
1. Cp = Range/6
2. Cpk
2.008 -Spring 2004 27
Process Capability
• Take an example with:
Mean = .738
Standard deviation, σ = .0725
UCL = 0.900
LCL = 0.500
• Normalizing the specifications:
23.20725.0
738.0900.0UCLUCL
x
Z
28.30725.0
738.0500.0LCLLCL
x
Z
basis) absolutean (on 23.2MIN Z
738.0
x
0.500
LCL
0.900
UCL
28.3LCL
Z
23.2UCLZ
2.008 -Spring 2004 28
Process Capability (cont’d)
• Using the tables of areas under the Normal Curve, the proportions out of specification would be:
PUCL= 0.0129
PLCL = 0.0005
Ptotal = 0.0134
• The Capability Index would be:CPK = Zmin/3 = 2.23 / 3 = 0.74
2.008 -Spring 2004 29
Process Capability (cont’d)
• If this process could be adjusted toward the center of the specification, the proportion of parts falling beyond either or both specification limits might be reduced, even with no change in Standard deviation.
• For example, if we confirmed with control charts a new mean = 0.700, then:
700.0
x
0.500
LCL
0.900
UCL
76.2LCL
Z
76.2UCLZ
76.20725.0
700.0900.0UCL newUCL
x
Z
76.20725.0
700.0500.0LCL newLCL
x
Z
76.2MIN Z
2.008 -Spring 2004 30
Process Capability (cont’d)
• The proportions out of specification would be:PUCL= 0.0029
PLCL = 0.0029
Ptotal = 0.0058
• The Capability Index would be:CPK= Zmin/3 = 2.76 / 3 = 0.92
2.008 -Spring 2004 31
Improving Process Capability
• To improve the chronic performance of the process, concentrate on the common causes that affect all periods. These will usually require management action on the system to correct.
• Chart and analyze the revised process:– Confirm the effectiveness of the system by continued
monitoring of the Control Chart