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Capability Formula

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  • June 5, 2003 at 11:54 am #32447

    David Condron
    ★ 10 Years ★
    Participant

    Hi there,
    Can anyone tell me what this formula relates to ? We are reviewing Cpk data from a MFG process. The formula behind the calculation for the Cpk is as follows….
    Cpk = ((1-abs(x-bar – nominal)/(tol/2) x Cp
    Anyone ever seen this for calculating Cpk.
    It would be nice if any help was thrown this way.
     

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    June 5, 2003 at 12:31 pm #86693

    Robert Butler
    ★ 20 Years ★
    Participant

    It looks like a derivation of the kt factor which expresses the loss in potential capability when a target not equal to M (where M in this case is nominal) is specified for the process average.Cp* = Cp(1-kt) where kt = |T-M|/(tolerance/2)  in the case you are describing T = xbar and M = nominalpp. 160 of Bothe Measuring Process Capability has more information

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    June 5, 2003 at 12:32 pm #86694

    Gabriel
    ★ 20 Years ★
    Participant

    It’s simple algebra, just instead of “nominal” I would say “center of the tolerance”.Let X=Xbar, T=USL-LSL, C=(USL+LSL)/2, s=sigmaUSL-X=(C+T/2)-X=T/2*(1-(X-C)/(T/2))X-LSL=X-(C-T/2)=T/2*(1-(C-X)/(T/2))Note that if X is closer to USL, then (X-C) is possitive, and if it is closer to LSL, then (C-X) is possitive. ThenMIN(USL-X;X-LSL)=T/2*(1-|X-C|/(T/2)), thenCpk=MIN(USL-X;X-LSL)/3s=T/2*(1-|X-C|/(T/2))/3s=T*(1-|X-C|/(T/2))/6s=(T/6s)*(1-|X-C|/(T/2))=(1-|X-C|/(T/2)*Cp

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    June 5, 2003 at 12:56 pm #86696

    David Condron
    ★ 10 Years ★
    Participant

    Is ABS therefore as refered to in original note meaning absolute measurement i.e Nominal ?
    What has the specified tolerance got to do with the calculation, should it not be the STD DEV based on the individual points taken from the process under study ? 
    Which book is the one you refer to ? Is the Ford Manual ?

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    June 5, 2003 at 1:46 pm #86700

    David Condron
    ★ 10 Years ★
    Participant

    Thanks for the reply. Unfortunately, we cannot correlate with the original formula.See previous mail to the other guys re: clarification.Is the industry std (or more recognised) not the correct and fundamental formula which we should all be adhering to….Cpk = min{Cpl, Cpu}As…..Cpl = x-bar – LSL/3 sigma’s ; Cpu = USL – x-bar/3 sigma’sShould we not be following this one, every book i am searching thru’ shows this one outlined above ????

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    June 5, 2003 at 2:24 pm #86702

    Robert Butler
    ★ 20 Years ★
    Participant

    In the reference I cited, “abs” means the absolute value of the quantity T-M.  The measure
    C*p = Cp(1-kt)  with kt = |T-M|
    is a measure of the potential capability of the system.  In Bothe’s book all potential measurements are denoted with an ‘*’.  The use of potential capability measurements is driven by your goals.  Bothe indicates that you would use C*p in those cases where the goal is to center the process at some target T which is not equal to the middle of the tolerance.
     The choice of capability index should be driven by the questions that you want to ask of the process.  Cpk is well and good when the process output is normal, your specs are symmetric with respect to your process, etc.  If you have unilateral tolerances, your output is not normal, etc. then you need to identify a capability index that is appropriate for your situation. 
      The fact that someone has chosen to express a process capability in terms of C*p could, of course, be due to ignorance.  Given the “esoteric” nature of C*p I would be more willing to believe that the decision to use this index was driven by an understanding of the process and the need to properly express its capabilities.

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    June 5, 2003 at 4:39 pm #86711

    Gabriel
    ★ 20 Years ★
    Participant

    David, I have a problem. I’m trying to calculate the average between 2 values, A and B. I was used to the typical formula:
    average=(A+B)/2
    But now I found another one:
    average=A+(B-A)/2
    and yet a third one:
    average=B+(A-B)/2
    Which is the correct one? Shouldn’t we be all following the first one that it’s in all books?
    It’s a mathematical equality. They are exactly the same thing. Try and you will get exactly (not more or less) the same figure with any of them.
    Now, reread my post and you will find, in the last line, a long chain of equalities that goes from the formula “in all books” to the “new strange” formula.

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    June 5, 2003 at 4:41 pm #86712

    None
    ★ 20 Years ★
    Participant

    None of your formulas can be the same as the average, mathematically if you subtract

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