PPM equivalent for multiple characteristics
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 This topic has 3 replies, 4 voices, and was last updated 12 years, 6 months ago by Jonathon Andell.

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January 21, 2010 at 3:37 pm #53160
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.
Thanks0January 22, 2010 at 7:47 am #188576Hi 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.
Remi0January 23, 2010 at 2:04 pm #188613
Eric MaassParticipant@poetengineer Include @poetengineer in your post and this person will
be notified via email.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)*1e6]^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,
Eric0January 25, 2010 at 12:56 pm #188647
Jonathon AndellParticipant@JonathonAndell Include @JonathonAndell in your post and this person will
be notified via email.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.
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