posted 23 March 2001 02:36 PM               

BTW Marc,
I just wanted to say thanks for providing this site. And thanks to all that contribute to it. It has really helped me a lot.
Thanks

posted 23 March 2001 02:58 PM              

It can only be one of two things (I think):

1. The mean of your process is outside the process spec. limits, causing the MIN[(mean-LSL),(USL-mean)] to be larger than (USL-LSL)/2

posted 23 March 2001 03:26 PM               

I'm sure you will get more definitive explanations from this forum but here I go:

Cp = Rating of how much of the specification limits are being used.

Cpk = Rating of how close your process is to the expected/targeted value.

The higher the CP, the less of the spec. limits being used. The higher the Cpk the closer you are to the target.

You could be running a high Cp (good) but the process is running too close to a spec. limit which would give you a higher Cp than Cpk.

This will probably lead to the old discussion as to how a process can be in control but not be capable.

Input from experts hopefully forthcoming!

ASD...

posted 24 March 2001 09:28 AM               

You failed to provide a value for dispersion, but I will try to help.

The distribution appears to be skewed right. With a mean of 0.00236 and 3 of 200 out or near the USL (more I am sure), this appears to be the case. However, without the data or values for skewness or kurtosis, I cannot know for sure.

If you are willing to provide the raw data (Excel or Tab Delimited Text), I may be able to provide a more complete analysis for you.

One other thing:

Calculation of predictable process capability indices is dependent on the statistical control of the process. If the process is not in statistical control, then the results of the study are subject to fluctuate unpredictably. The statistical control of the process can be studied using control charts. If the distribution is skewed right, the state of statistical control is in question.

------------------
I was better but I got over it.

posted 24 March 2001 12:19 PM              

To my knowledge, the Cp is not defined in this situation. You can't compare the process variation to the tolerance width if the tolerance is one-sided.

With an upper spec you would use the CPU as an estimate of Cpk (defined as the minimum of [CPL,CPU]), but Cp isn't defined.

CPU = (USL-mean)/3sigma

I can't find anything on the web about the formulas that CimWorks VisualSPC uses for one-sided Cp & Cpk. I'd strongly suggest you understand these formulas before using the software's output. Maybe call the supplier and ask for an explanation.

posted 27 March 2001 11:54 AM               

My point of view is that because the Cp is a comparison of the tolerance and your process variation it is not feasible to compare one sided specification to your process variation for example if you have a spec 0.25 max you physically could have parts from 0 to 0.25, but if you have an spec of 0.25 min you could physlcally have pieces from 0.25 to infinite.
What I'm telling you is not in any book is only my point of view.

posted 27 March 2001 06:33 PM               

In the case of say, runout, the spec might be Max = .025.

USL (.025) - LSL (.000) = .025

CP = .025 / 6 sigma (r-bar-d2)

Same as a bilateral spec.

I won't go into a unilateral spec. that calls for a minimum specification as I have never run across one. I'll leave that to the experts!

posted 28 March 2001 08:58 AM              

C'pu = the maximum of either (USL - target/3 Sigma short term) or USL-mean/3 Sigma short term) where Sigma short term = the short-term process standard deviation, which reflects only within-subgroup variation.

Hope this helps.

Ben