iSixSigma

PPM equivalent for multiple characteristics

Six Sigma – iSixSigma › Forums › Old Forums › General › PPM equivalent for multiple characteristics

Viewing 4 posts - 1 through 4 (of 4 total)
  • Author
    Posts
  • January 21, 2010 at 3:37 pm #53160

    Sinnicks
    ★ 10 Years ★
    Participant

    Can anyone tell me how to calculate an equivalent part PPM for something having 4 characteristics each having a 600PPM?
    I read something about Spk (Cpk but for multiple characteristics) but couldn’t find much about it.
    Thanks 

    0
    January 22, 2010 at 7:47 am #188576

    Remi
    ★ 10 Years ★
    Participant

    Hi Mark,
    as a first approximation you could sum them up.
    More correct is to compensate for products that have more than one characteristic failing. If the characteristics occur independently of each other you get the following calculation (if not a similar formula can be derived but you have to count the multiple occurences).
    Pi = P (char i failed). Then P(reject) = P( at least one char failed)= P1+P2 +P3+P4 – SUM PiPj + Sum PiPjPk – P1P2P3P4.
    Note that if P1= 600 ppm =P2 then P1P2 =0.36 ppm so ‘forgetting’ to count the products that have more than one characteristic is in your calculation resolution.
    Remi

    0
    January 23, 2010 at 2:04 pm #188613

    Eric Maass
    ★ 10 Years ★
    Participant

    Mark,Remi is right. Alternatively, if you are more comfortable with
    probability theory, you could estimate the yield for
    each of the 6 characteristics as [1e6 – 600 ppm]*1e-
    6 – .9994, and – like Remi said – if yo assume they
    are independent, you could multiply them…or, since
    each of the 6 characteristics have 600 ppm, you
    would have
    [(1e6 – 600ppm)*1e-6]^6 = .9994^6 = .9964 which
    corresponds to 3600 ppm (equivalent to Remi’s answer
    of 600 ppm x 6). There is a slight difference in
    these two calculations a couple decimal points
    further out…but the bigger question is whether the
    6 types of failures due to different characteristics
    are independent.Best regards,
    Eric

    0
    January 25, 2010 at 12:56 pm #188647

    Jonathon Andell
    ★ 10 Years ★
    Participant

    I agree with both respondents thus far. For the relatively low PPM of your operation you will find that both methods result in nearly the same outcomes. Processes with higher PPM will show divergence between the two methods, at which point Eric’s method would be advised.As Eric pointed out, this works when the defect types occur more or less independently. If they were to track one another closely, then you’d need to make some adjustments.

    0
  • Author
    Posts
Viewing 4 posts - 1 through 4 (of 4 total)

The forum ‘General’ is closed to new topics and replies.