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  #1  
Old 12th June 2003, 02:06 PM
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Let Me Help You Calculating Cp and Cpk for Product with only LSL (Lower Specification Limit)

Does anyone have any insight into how I can calculate the Cp and Cpk values for characteristics that only have a lower spec limit? Cp wants the tolerance divided by 6s and Cpk is the average of the process average subtracted from the LSL divided by 3s.

When I calculate Cp in this manner I get that Cp=2.94 and Cpk=0.52. I believe the Cpk value, but Cp is hard to believe because there are so many points that are out-of-spec in the process. Is the Cp this way due to no tolerance. Is there another way to calculate.

Does anyone else have any experience that would help me.

Thanks,
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Old 12th June 2003, 02:29 PM
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There are statistics gurus here who can probably answer better, but if memory serves me you cannot have a Cpk for a single-sided lower limit spec. -- you can only have Cpl which is (mean - lower spec. limit) / 3 sigma.

I think this is right but will wait to see if someone can back me up on this....
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Old 12th June 2003, 03:35 PM
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Cool C p & k

Cpk is the lower of:

X-double bar-LSL / 3sigma (R-bar/d2)

or

USL-x-double bar / 3 sigma (R-bar/d2)

It cannot be used for unilateral tolerances. It's one of a few misconceptions: I think a couple more are:

-- Being estimated sigma, Cpk cannot be condfidently used to predict the process without at least ~ 1,000 subgroups of data
-- Widening spec limits improves Cpk but does nothing to improve the quality of the product
--I can submit a population of parts illustrating a Cpk if >2.0 (2 dPPB) and some of the parts will not function
--Cpk is only accurate if the process illustrates a state of control

Anyway, just rambling on now.........I am bored. Have a day and good luck !
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Old 12th June 2003, 04:41 PM
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Cpk is also defined as | X double bar - NSL / 3 sigma |

Where NSL is nearest spec. limit. If there is only one specification limit, why wouldn't this be valid? We are using Cpk to help to understand the capability of the process to meet the specification requirements.

I think the problem comes in when there is a physical limit which causes the data to be skewed (non-normal). Looking at perpendicularity, it's physically impossible to have a value less than zero. The data will tend to be skewed away from zero. If this is the case, other methods must be used to determine the process capability.

Cp really is meaningless though, unless you have a tolerance range. When you calculated Cp, you must have put some number in for an upper limit just to be able to get a value other than infinity.
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Old 12th June 2003, 05:43 PM
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howste,

I believe that by definition (according to a cheat sheet from long ago I found) Cpk can only be gotten from a 2-sided tolerance -- as nobox says (with a few tweeks):

Cpk is the lower of:

Cpl = (X-double bar-LSL) / 3 sigma

or

Cpu = (USL- X double bar) / 3 sigma

But, IMO, for a 1-sided tolerance, Cpl or Cpu should tell you as much as Cpk for a 2-sided tolerance. Again, I'm no stat's guru, but this is as I remember it and what my old notes say. I didn't consult my Jurans...
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Old 12th June 2003, 06:05 PM
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Yes, it is due to unilateral tolerances, and also, try PpK for your process.

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Old 13th June 2003, 04:39 AM
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Definitely not a stats guru, but I would agree that you would use Cpl to see how capable your process is of making parts to the lower spec limit. ( or cpu for upper spec limit ) I have used these in PPAP submissions in the past & never had any problems with either customers or auditors.

Attached file may help in explaining
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Last edited by Teknow; 13th June 2003 at 05:11 AM.
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Old 13th June 2003, 11:07 AM
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First, thank to all for your insight, I really appreciate the help. Let me see if I have this.

I cannot use Cp because it uses a tolerance, and in my case I only have the LSL to work with. So I must use Cpl and determine how capable my process is of holding to that LSL.

Cpl = (x-double bar minus LSL) / (3 X (R-bar / d2))

However, I am not a stat guru and AIAG SPC manual says that Cpl is calculated as:

Cpl = (x-double bar minus LSL) / 3ô R-bar/d2

Is this the same calculation?
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