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observed within overall performances

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  • #44890

    Quainoo
    Member

    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
     
     
     

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    #145143

    Philip
    Participant

    “Within” perfomance is based on a short term estimate of sigma (for example R-bar/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 six-sigma process (Cpk=2) which is 100% non-conforming. 
    Take a look at Don Wheeler’s books “Beyond Capability Confusion” and “Guide to Data Analysis” for more information.
    Hope this helps
    Phil

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    #145156

    mand
    Member

    Yes!!
    If everyone here read Wheeler we might start to see more quality and fewer SS snake oil salesmen.

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    #145159

    DaveS
    Participant

    Philip-
    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 six-sigma process (Cpk=2) which is 100% non-conforming.”How?

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    #145160

    Maciek
    Participant

    I 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).

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    #145161

    Philip
    Participant

    Situation 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.

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    #145178

    Marlon Brando
    Participant

    It  is  always  better (first  priority)  to  have  less  variation (lower SD),then  put  your  maximum  efforts  to  avhieve  the  mean (target0. 

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    #145181

    Monk
    Participant

    Marlon,
    You are right…I agree t you..This is something very basic.
    Monk

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    #145185

    Philip
    Participant

    Very cryptic …..
    How is it relevant to the discussion?
    Phil

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    #145188

    Marlon Brando
    Participant

    Monk
    Thank You

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    #145190

    Monk
    Participant

    Philip
    How is ur sentence related to the discussion?
    Just think about it and then reply !

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    #145197

    Philip
    Participant

    I 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 re-engineering your process (dissagregate, stratify, experiment).

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    #145199

    Mikel
    Member

    Wheeler is a bit of a snake oil salesman himself.

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    #145360

    Quainoo
    Member

    Philip,
    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
     

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    #145432

    Philip
    Participant

    Hi 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.
    Phil

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    #145439

    Len L
    Participant

    You twerp.  I’m sure you’ve never read Wheeler.  I’ll bet you still chew the corners of your picture books.

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    #145961

    Quainoo
    Member

    Thanks Philip (just read your last post)
    Vincent
     

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