# Calculating capability indices with one specification

The following formula for Cpk is easily found in most statistics books, as well as in software products such as SQCpack.

Cpk = Zmin / 3

Zmin = smaller of Zupper or Zlower

Zupper = [(USL – Mean) / Estimated sigma*]

Zlower = [(Mean – LSL) / Estimated sigma*]

Estimated sigma = average range / d2

And, we’ve all learned that generally speaking, the higher the Cpk, the better the product or process that you are measuring. That is, as the process improves, Cpk climbs.

What is not apparent, however, is how to calculate Cpk when you have only one specification or tolerance. For example, how do you calculate Cpk when you have an upper tolerance and no lower tolerance?

When faced with a missing specification, you could consider:

- Not calculating Cpk since you don’t have all of the variables. Entering in an arbitrary specification.

- Ignoring the missing specification and calculating Cpk on the only Z value.

An example may help to illustrate the outcome of each option. Let’s assume you are making plastic pellets and your customer has specified that the pellets should have a low amount of moisture. The lower the moisture content, the better. No more than .5 is allowed. If the product has too much moisture, it will cause manufacturing problems. The process is in statistical control.

It is not likely your customer would be happy if you went with option A and decided not to calculate a Cpk.

Going with option B, you might argue that the lower specification limit (LSL) is 0 since it is impossible to have a moisture level below 0. So with USL at .5 and LSL at 0, Cpk is calculated as follows:

If USL = .5, X-bar = .0025, and estimated sigma = .15, then:

Zupper = [(.5 - .0025) / .15] = 3.316,

Zlower = [(.0025 – 0) / .15] = .01667 and

Zmin = .01667

Cpk = .01667 /3 = .005

Your customer will probably not be happy with a Cpk of .005 and this number is not representative of the process.

Example C assumes that the lower specification is missing. Since you do not have a LSL, Zlower is missing or non-existent. Zmin therefore becomes Zupper and Cpk is Zupper/3.

Zupper = 3.316 (from above)

Cpk = 3.316 / 3 = 1.10.

A Cpk of 1.10 is more realistic than .005 for the data given in this example and is representative of the process.

As this example illustrates, setting the lower specification equal to 0 results in a lower Cpk. In fact, as the process improves (moisture content decreases) the Cpk will decrease. When the process improves, Cpk should increase. Therefore, when you only have one specification, you should enter only that specification, and treat the other specification as missing.

An interesting debate (well, about as interesting as statistics gets) occurs with what to do with Cp (or Pp). Most textbooks show Cp as the difference between both specifications (USL – LSL) divided by 6 sigma. Because only one specification exists, some suggest that Cp can not be calculated. Another suggestion is to look at ~ ½ of the Cp. For example, instead of evaluating [(USL – Mean) + (Mean – LSL)] / 6*sigma, instead think of Cp as (USL – Mean) / 3*sigma or (Mean – LSL) / 3*sigma. You might note that when you only have one specification, this becomes the same formula as Cpk.