observed within overall performances
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 This topic has 16 replies, 9 voices, and was last updated 15 years, 7 months ago by Quainoo.

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October 14, 2006 at 11:29 am #44890
Hello everyone,
When I use Minitab to calculate a capability, I always use the results (in ppm) that are displayed in the “observed performance”.
Can someone explain when the “within” and “overall” performances are supposed to be used ?
Thanks
Vincent
0October 19, 2006 at 8:01 am #145143“Within” perfomance is based on a short term estimate of sigma (for example Rbar/d2) typically derived from a control chart. This will characterise the past and future performance for a STABLE process (i.e. one that does not exhibit any special cause variation) It will not characterise the past or the future performance of an unstable process. Indices derived from sigma are typically Cp and Cpk.
“Overall” performance is based on a global estimate, usually denoted as S, (typically derived using the stdev() calculation in Excel, for example). For a stable process this will be equivalent to the within performance. For an unstable process, the overall performance will characterise the past performance of your process, but not the future performance (UNSTABLE is equivalent to UNPREDICTABLE). Indices derived from S are typically Pp and Ppk.
The best measure of capablility is actually the observed performance in terms of ppm. Everything else depends on assumptions. For example, it is possible to get a process with a Cpk of 0.8 which is 100% conforming, and it is also possible to get a sixsigma process (Cpk=2) which is 100% nonconforming.
Take a look at Don Wheeler’s books “Beyond Capability Confusion” and “Guide to Data Analysis” for more information.
Hope this helps
Phil0October 19, 2006 at 9:35 am #145156Yes!!
If everyone here read Wheeler we might start to see more quality and fewer SS snake oil salesmen.0October 19, 2006 at 10:43 am #145159Philip
You wrote:
“For example, it is possible to get a process with a Cpk of 0.8 which is 100% conforming, and it is also possible to get a sixsigma process (Cpk=2) which is 100% nonconforming.”How?0October 19, 2006 at 11:25 am #145160I agree with Daves. There is no possibility to get Cpk =0,8 with 100% conforming and Cpk=2 with 100% nonconforming. These indicators are depends on specification limits, so situation describd above is impossible (in my opinion).
0October 19, 2006 at 11:46 am #145161Situation 1:
A stable process having a skewed distribution, can give you a capability less than 1.0, whilst having no product outside the specification limits (example given in “Beyond Capability Confusion”) The correct interpretation of a capability index less than 1.0 is that the process MAY be incapable IF the process is symetrical about the mean – only by plotting the control chart, and the histogram can you tell.
Situation 2:
A process which is higly unstable may have very low short term variation, and a mean exactly centered between the spec limits, but still have every result outside the specification limits. The key here is that neither the mean, nor the standard deviation, nor any other summary statistic, will adequately characterise the output of an unstable process, and it is complete nonsense to estimate the capability of an unstable process.0October 19, 2006 at 2:21 pm #145178
Marlon BrandoParticipant@MarlonBrando Include @MarlonBrando in your post and this person will
be notified via email.It is always better (first priority) to have less variation (lower SD),then put your maximum efforts to avhieve the mean (target0.
0October 19, 2006 at 2:26 pm #145181Marlon,
You are right…I agree t you..This is something very basic.
Monk0October 19, 2006 at 2:35 pm #145185Very cryptic …..
How is it relevant to the discussion?
Phil0October 19, 2006 at 2:36 pm #145188
Marlon BrandoParticipant@MarlonBrando Include @MarlonBrando in your post and this person will
be notified via email.Monk
Thank You0October 19, 2006 at 2:38 pm #145190Philip
How is ur sentence related to the discussion?
Just think about it and then reply !0October 19, 2006 at 3:05 pm #145197I still don’t understand the relevance. So all I can do is respond to what I assume Marlon’s comment meant …
The discussion is revolving around the presence or absence of instability in a process. In the presence of instability, SD does not exist, and therefore it is meaningless to try to minimise it. The correct approach to any process problem is
a) eliminate special cause variation (until you have done this you don’t have a process to improve or measure)
b) centre the process on the target (because this is easier than reducing common cause variation)
c) if this still doesn’t give you what you want reduce common cause variation by reengineering your process (dissagregate, stratify, experiment).0October 19, 2006 at 3:44 pm #145199Wheeler is a bit of a snake oil salesman himself.
0October 21, 2006 at 9:26 am #145360Philip,
I have not been able to read the post during the week, I am reading them now.
First of all, thanks very much to you and all other participants for your inputs.
Just a comment about your last post:
Your points 1,2 and 3 make sense to me and I totally agree.
I am not too sure about what you said at the begining of your message:
“In the presence of instability, SD does not exist”
I can see two outcomes of a process being not stable (as far as SD is concerned) :
1) “ups” and “downs” relative to the mean (beyond 3 sigma limits) and as a result, the SD will tend to increase but the distribution may remain normally distributed.
2) High proportion of “ups” or high proportion of “downs” and as a result, the distribution will be skewed to the right or to the left and it is likely that the distribution will no longer be normal.
So, for case 2, the SD is miningless since the distribution is not normal (and this is in agreement with your statement), but for case 1, SD may still exist even though it has increased in magnitude.
What do you think ?
Vincent
0October 22, 2006 at 8:56 pm #145432Hi Vincent.
I think the key is not whether the distribution is normal or not, the key is whether there is a distribution at all.
In the presence of instability, you effectively have different distributions at different times. Whilst you can often model such data using a distribution, and you can estimate parameters of the distribution, what do these estimates, or the model, characterise?
When you estimate the SD for an unstable “process” you are effectively determining the SD of the underlying stable process, plus the inflation due to the instability. This is useful if you want to know if the process is stable, but it is not an estimate of the parameter of a distribution.
Phil0October 23, 2006 at 1:18 am #145439You twerp. I’m sure you’ve never read Wheeler. I’ll bet you still chew the corners of your picture books.
0October 27, 2006 at 7:33 pm #145961Thanks Philip (just read your last post)
Vincent
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