Process capability: Cp and Cpk

Indices comparing the specification width with the process's natural spread (Cp) and accounting for how well the process is centred (Cpk).

C_p = \frac{\text{USL} - \text{LSL}}{6\sigma_{\text{within}}}, \qquad C_{pk} = \frac{\min(\text{USL} - \hat\mu, \; \hat\mu - \text{LSL})}{3\sigma_{\text{within}}}

Capability indices (short-term)

$\text{USL}, \text{LSL}$: upper and lower specification limits (customer requirements)
$\hat\mu$: process mean (grand mean of the data)
$\sigma_{\text{within}}$: within-subgroup sd: $\bar R/d_2$ (default), $\bar S/c_4$ , or $\overline{MR}/d_2(2)$ for individual values

P_p = \frac{\text{USL} - \text{LSL}}{6s}, \qquad P_{pk} = \frac{\min(\text{USL} - \hat\mu, \; \hat\mu - \text{LSL})}{3s}

Performance indices (overall sd)

$s$: overall sample standard deviation of all values ( $n - 1$ )

When to use

After a process has been shown to be in statistical control, to judge whether it can meet specifications.

Assumptions

The process is stable (in control) — otherwise no single σ describes it.
Approximately normal data: the "6σ covers almost everything" logic relies on it.
With one specification limit only the one-sided index (CPU or CPL) exists, and Cpk is that index.

How to interpret

Cp ≥ 1.33 is a common minimum; Cp = 1 means the specification width equals 6σ.
Cpk ≤ Cp always; Cpk much smaller than Cp means the process is off-centre — recentring could help without reducing variation.
Pp/Ppk use the overall spread; a big gap from Cp/Cpk points to drift between subgroups.

Common mistakes

Computing indices for an unstable process.
Treating a point estimate from 20 values as precise — capability indices have wide sampling variability.
Mixing up specification limits and control limits.

Example

The 20 measurements of the X̄–R example (mean 10.035, $\sigma_{\text{within}} = \bar R/d_2 = 0.243$ ), specifications 9.4 to 10.8: $C_p = 1.4/(6 \times 0.243) = 0.96$ ; $C_{pk} = (10.035 - 9.4)/(3 \times 0.243) = 0.87$ . Not capable by the 1.33 standard, and slightly closer to the lower limit. Here $s = 0.208$ gives $P_p = 1.12$ , $P_{pk} = 1.02$ — with so few subgroups the two σ estimates differ by chance.