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,
Jeff

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....

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 !
:bigwave:

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.

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...

Yes, it is due to unilateral tolerances, and also, try PpK for your process.

Al...

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 :rolleyes:

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? :confused:

Yep, I think you have it FWIW. "Sigma hat" is estimated process SD which is rbar/d2 -- looks like they just forgot to explain that.

sigma = R-bar / d2

Cpl = (x-double bar minus LSL) / (3 *sigma)

or you can use autocorrelation factor (usefull for Individuals and moving range).

sigma = MR-bar / d2 * (1 /(1-r^2)^0.5)

but as howste said "the problem comes in when there is a physical limit which causes the data to be skewed"

In my opinion, use Cpmk (or the performance equivalent).

read this article, I tink is great (It's a must see).

http://www.scausa.com/capaqa.pdf

You can also transform the variable or use non-parametrics (median, percentile)

Thank you Mike S. and Darius. I appreciate your help. I will review the attachment Darius.

Now that I can confidently calculate the Cpl, I only now need to figure how to reduce the variability...:rolleyes:

Small steps though.

Originally posted by Darius
sigma = R-bar / d2

Cpl = (x-double bar minus LSL) / (3 *sigma)

or you can use autocorrelation factor (usefull for Individuals and moving range).

sigma = MR-bar / d2 * (1 /(1-r^2)^0.5)

but as howste said "the problem comes in when there is a physical limit which causes the data to be skewed"

In my opinion, use Cpmk (or the performance equivalent).

read this article, I tink is great (It's a must see).

http://www.scausa.com/capaqa.pdf

You can also transform the variable or use non-parametrics (median, percentile)
Darius,

I'm confused. I scanned part of your attachment and do not understand section 4. How can there be a Cpk calculated for a one-sided (upper-spec limit only)?

It says Cpk is unsuitable for a one-sided tolerance, which I agree with, but IMO it gives the wrong reason. It says example A and B both give the same Cpk while the parts in process B are much better. But Cpk -- by definition (Cpk = min of Cpl and Cpu) -- cannot exist for a one-sided tolerance -- it is intended for a 2-sided tolerance. You would use Cpu which would show that the parts in example B are much better than the parts in example A.

Statistics is difficult enough for me -- seeing stuff like this makes it even more difficult. JMO.

It said "The use of Cpk is not suitable for an S-type tolerance" but it sould said "Why the use of Cpk is not suitable for an S-type tolerance", I didn't wrote the article, but the ideas have great insight. I take it, as one of the few articles about that problem in an example way (easy to understand the point of view of the writer).

I my self found many cases where the Cpk (for one sided tolerance), gets better farter from the target because of the variation reduction (but the Cpmk is worse), even cases where all the data (from that set of data) is farter that the maximum of the set nearer from the target.

In most of the books or articles, said that in S-type tolerance, just use the one that can be calculated saying Cpk = Cpu or Cpk = Cpl depending of the spec limit that you have. Maybe could be a different way to tell the same thing: "Use the Cpu" or "Cpk = Cpu". I don't tink that matter, it's just nomenclature.

:confused: I tink that the calculus of capability-performance is important in order to find where we are, many practitioners said Cpk can be only calculated in stable state, others don't care about stability, others that said that you can not use any index at all because of the non normality of the data, and at the last group, like me, tink it as a mean to find if something change, agreed that if the process is not stable the values tend to fluctuate a lot, but I we chart the values (Cpk, ppk, Cpmk of different periods) in a control chart, as Wheeler said in one article on quality magazine the values out of the chart show a change in the process.

If Quality is get nearer from target and with less variation (IN THAT ORDER, Cpmk is better that Cpk because it take the target in account (Cpk take target as in the middle of the specs and in one sided spec the "target" of Cpk is lost).

:thedeal: