# Cpk higher than Cp - Explain the relationship and why this happens?

F

#### freeda

I'm a little embarrassed to ask this question, but it is one that is asked of me and I can't explain it.
We did a capability study on the straightness of some shafts. Our sample size was 200. We have a Cp value that is less than the Cpk.
Why does this happen?
I already know that the process is not capable. That is why we are doing the study, to find out just how uncapable it is.
We had two points out of tolerance and one on the dead high, the Cp was .95 and the Cpk was 1.00.
I haven't been able to find anything in any of my SPC books that explain what causes this.

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

K

#### Ken K.

What was the mean of your process?
How are you calculating your values? (Excel, MINITAB, JMP, other package, by hand)
What is your estimate of sigma (though it shouldn't matter when comparing Cp & Cpk)

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

A

#### Al Dyer

Freeda,

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...

[This message has been edited by Al Dyer (edited 23 March 2001).]

F

#### freeda

I use CimWorks VisualSPC program to calculate the data.
The Mean is .00236
The Max is .007
The Min is .001
The USL is .005, since it is a unilateral tolerance (straightness), the LSL is NoLoSpec (this is what you put as your LSL in VisualSPC).

D

#### Don Winton

freeda,

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.

Regards,

Don

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

[This message has been edited by dWizard (edited 24 March 2001).]

K

#### Ken K.

Oh, this is a one-sided specification.

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.

This is one of those cases where people really need to understand their software rather than blindly use it (no offence).

F

#### freeda

I already know what the formulas are. CimWorks provides that information to their customers. And I don't believe that I "blindly" use the software.
I was only looking for a good, simple explanation of why.
I tried to explain to the powers above that you can't look at Cp on a unilateral tolerance, but I couldn't come up with a good enough explanation for them.
Thanks to all that contributed, they now understand.

[This message has been edited by freeda (edited 26 March 2001).]

A

#### ASanchez

Ken

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.

Alex

A

#### Al Dyer

Ken,

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!

ASD...

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