Transcript Six Sigma and Process Capability Indices

SIX SIGMA AND CALCULATION OF
PROCESS CAPABILITY INDICES:
SOME RECOMMENDATIONS
• P.B. Dhanish
• Department of
Mechanical
Engineering
• [email protected]
Overview
• What is Six Sigma?
• The sample size problem
• The distribution problem
• The control problem
• Conclusion
What is Six Sigma?
• Disciplined quality improvement
• AIM: Near elimination of defects
• NUMERICALLY: 3.4 DPMO!
Defect levels at various sigmas
Sigma
6 sigma
5 sigma
4 sigma
3 sigma
2 sigma
1 sigma
Table 1
Defects per million
3.4 defects per million
230 defects per million
6200 defects per million
67,000 defects per million
310,000 defects per million
700,000 defects per million
What is sigma?
• Statistics:
• The process standard deviation
Then, six sigma?
• Specification limits should be at
• +/- six sigma
Then, how many defects?
• Assuming
Normal distribution
• The area under the normal curve
beyond +/-six sigma
:Fraction non-conforming
• Multiply by 1,000,000 to get DPMO
• If the process is centred on target,
0.002 DPMO
Nonconforming
Nonconforming
Target
8
6
4
2
0
-2
-4
-6
-8
USL
LSL
Such perfect centering
• Not possible in practice
• Allow +/-1.5 sigma shift
• Then the defect level will be 3.4DPMO
8
6
4
2
0
-2
-4
-6
-8
Nonconforming
Nonconforming
USL
LSL
Shift
Target
The sample size problem
•
•
•
•
•
•
In practice, true sigma is unknowable
Sample standard deviation
From a finite number of samples
Sampling error in sigma level
Single value not meaningful
Hence: Give Confidence Limits
If x is normally distributed
• an upper 100(1-α)% confidence limit
for σ is (n  1) s 2
 12 ,n 1
• and an upper 100(1- α)% confidence
limit for μ is
s
x  t , n 1
n
Assuming that mean and
sigma are independent
• to calculate an upper 100(1-α)%
confidence limit for the proportion
nonconforming p,
• construct 100(1-α)1/2 % confidence
limits for each parameter separately
• using these two values, determine p
For example,
• If n=25, USL=6, LSL=-6, x =0, and s=1,
a 97.47% upper confidence limit for σ is
(25  1)1
 1.3898
12 .43
2
and a 97.47% upper confidence limit for μ
is
1
0  2.0577
 0.4115
25
Then the proportion
nonconforming:
 6  0.4115 
 (6)  0.4115 
p  1 
  

 1.3898 
 1.3898

 1  0.999971  1.98396X10
-6
 30.967 X 10
6
= 30.967ppm
Alternative: Determine
sample size for the
required confidence level
Table 2
Parameters
USL=+6
LSL=-6
Sample mean=0
Sample standard deviation=1
Upper confidence limit=0.95
Sample size
20
30
40
45
50
DPMO
79.17
15.01
5.191
3.459
2.441
Recommendation 1
• Do NOT give a single value for the
sigma level of your process
• Instead,
Give confidence limits
OR
the confidence level
The distribution problem
• Above calculations utilise:
• the tail end of the normal distribution
• Does any process in nature match the
values in the tail?
The values in the tail:
Sigma value
-7
-6.5
-6
-5.5
-5
-4.5
-4
-3.5
-3
Table 3
Value of the standard normal
distribution
9.13472E-12
2.66956E-10
6.07588E-09
1.07698E-07
1.48672E-06
1.59837E-05
0.00013383
0.000872683
0.004431848
To check the correctness,
• We need millions of samples!
• No shift in the process during this
production!
• Hence impossible to verify
• the exact values of the defect levels,
say 3.4 or 5 DPMO, may not have
practical significance
Not normally distributed:
• Surface Finish
• Circularity
• Runout
• Hence, take care
Cramer (1945): Lippman:
• everybody believes in the law of errors,
the experimenters because they think it
is a mathematical theorem,
the mathematicians because they think
it is an experimental fact
Recommendation 2
• Verify that the distribution is normal
• Otherwise, utilise the appropriate
distribution
• Realise that the exact values of low
defect levels are meaningless
The Control Problem
• To claim that future performance would
be similar,
the process should be stable
OR
in statistical control
Deming (1986):
• One sees much wrong practice in
connection with capability of the
process. It is totally wrong to take any
number of pieces such as 8, 20, 50 or
100, measure them with calipers or
other instruments, and take 6 standard
deviations of these measurements as
the capability of the process
Consider thirty
observations:
-1.6, -1.2, -1.9, -0.6, -1.6, -1.4,
-0.5, -0.9, -0.2, -0.7, 0.2, -0.5,
0.3, -0.4, 0.5, -0.3, 0.4, -0.2,
0.8, 0.6, 0, 1.2, 2, 0.5,
0.9, 0.8, 0.1, 1.4, 0.6, 1.7
• Mean 0
• Standard deviation 1
Frequency
Histogram:
7
6
5
4
3
2
1
0
-2
-1.5
-1
-0.5
0
0.5
Observation
1
1.5
2
An excellent process?
• If specification limits are +/-3
• Wait,
Just plot a run chart for the given
measurements
The process is drifting!
3
Observation
2
1
0
-1 0
5
10
15
20
-2
-3
Sample number
25
30
No capability can be
ascribed!
• QS9000: Distinguishes between Cpk
and Ppk
• An undisturbed process: Shouldn’t it be
in control?
• Very very unlikely
• An SPC program necessary
Nelson (2001):
• Getting a process in statistical control is
not improvement (though it may be
thought of as improvement of the
operation), getting a process in
statistical control only reveals the
process, and after a process is in
statistical control, improving it can
begin.
Another pitfall:
• Samples taken too infrequently
• This way, any process can be made to
appear in control!
• A common cause for failure of SPC
Recommendation 3:
• Ensure that the process is stable or in
statistical control
Any questions?
Conclusion:
Industries while claiming Six Sigma,
should
1. reveal the confidence level of their
sigma calculation
2. The process distribution utilised
3. How process stability was verified