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  • #37937

    Tim
    Member

    Can anyone provide me with a formula to calculate the minimum tolerance needed to maintain a 1.33 Cpk  on an assembly length given that it is derived from the sum of lengths of 3 component parts.  The 3 components have length tolerances of +/- 0.075mm & assume capability of 1.33 is maintained on each independent component.
    Any help greatly appreciated.
    Tim
     
     

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    #112875

    Loehr
    Member

    Hi Tim,
    Assuming the average is centered at the middle of the tolerance, sigma for the length of the first component (sigma1) is found as follows (given a tolerance of  +- .075 mm and a Cpk of 1.33, which means 8sigma1 = USL – LSL);
    8sigma1 = USL – LSL
    8 sigma1 = .075 – -.075
    8 sigma1 = .150
    sigma1 = .15 / 8
    The sigmas for the second component, sigma2, and the third component, sigma3, are also .15 / 8.
    The standard deviation for the combined length, sigmaTOTAL, is found as follows, asuming the component lengths are independent;
    sigmaTOTAL = sqrt(sigma1^2 + sigma2^2 + sigma3^2)
    sigmaTOTAL = sqrt[(.15/8)^2 + (.15/8)^2 + (.15/8)^2]
    sigmaTOTAL = sqrt[3*(.15/8)^2] = sqrt(3)*(.15/8)
    Thus, to achieve a Cpk of 1.33 for the combined length of all three components(assuming the overall process average is centered at the middle of the tolerance), the +- tolerance for the combined length is;
    +-Tolerance = 4*sqrt(3)*(.15/8)
    Hope this helps (and I hope this wasn’t a homework/exam problem that you were supposed to figure out on your own!).

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    #113439

    Da Agent
    Participant

    Ross,
    How did you get to the formula of (8sigma1 = USL – LSL)?

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    #113449

    Loehr
    Member

    Hi,
    The goal was to center the process at the middle of the tolerance and get a Cpk index of 1.33.  If the average is centered in the middle of the tolerance, then mu – LSL will equal USL – mu, both of which will equal 1/2 (USL – LSL), where mu is the process average and USL – LSL is the width of the tolerance.
    If we let 8sigma = USL – LSL, then 3sigma = 3/8 (USL – LSL).  Now we plug these quantities into the Cpk formula.
    Cpk = Minimum[ (mu -LSL) / (3sigma) , (USL – mu) / (3sigma)]
    Cpk =  Minimum[ {1/2 (USL – LSL)} / {3/8 (USL – LSL)} , {1/2 (USL – LSL)} / {3/8 (USL – LSL)}]
    Cpk = Minimum[ (1/2) / (3/8) , (1/2) / (3/8)]
    Cpk = Minimum[ 8/6 , 8/6 ] = 1.33
    That’s the long way, the short way to figure how wide the tolerance must be (in terms of standard deviations), is to multiply the Cpk goal by 6.  For the above example, 6 (1.33) = 8.  Thus, to hit a Cpk goal of 1.33 and have the process average centered at the middle of the tolerance, 8sigma must equal the tolerance.
    If the Cpk goal had been 1.67, then 10sigma would have to equal USL – LSL, since 6 (1.67) = 10.
    Hope this helps.

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