6.15 Capability Studies, Attributes Capability, Yield Metrics, and Long-Term Capability

Designing and conducting a capability study

A capability index computed on unsuitable data is worse than no index, because it is believed. Four prerequisites, in order:

1. Validated measurement system. If gage R&R consumes 30% of the tolerance, the observed spread is substantially measurement noise and the capability index is understated.
2. Statistical stability. Capability is a prediction about future output, which is only defensible if the process is in control. Plot a control chart first; if special causes are present, remove them before computing capability.
3. Verified distribution. Standard capability formulas assume normality. Test it, and if the data is non-normal, transform or use a percentile method.
4. Rational sampling plan. The sample must span the sources of variation the study is meant to describe.

Sampling plan design

The distinction is the whole point. A short-term study deliberately excludes between-setup and between-lot variation to estimate the process's inherent potential. A long-term study deliberately includes them to estimate what the customer will actually receive.

Report with the study: the sample size, the period covered, the stratification included, the normality result, the MSA result, and the control chart evidence. A capability number without that context cannot be evaluated.

Natural process limits versus specification limits

This distinction is tested repeatedly and confused constantly.

The two are completely independent. A process can be in perfect statistical control and produce 100% defective output, if its natural limits sit entirely outside the specification. Conversely, a process can be wildly out of control and still produce nothing outside specification if the tolerance is wide.

Two errors follow from confusing them, and both are serious:

• Putting specification limits on a control chart. Control limits describe what the process does; specification limits describe what is required. Charting subgroup means against specification limits is meaningless, because the distribution of means is narrower than the distribution of individuals by a factor of $\sqrt{n}$.
• Adjusting the process whenever a unit is outside specification. This is tampering: reacting to common-cause variation increases variation.

Yield and defect metrics

Definitions first. A unit is the item produced; a defect is any single non-conformance; a defective is a unit with one or more defects; an opportunity is a chance for a defect on a unit.

$DPU = \frac{\text{Defects}}{\text{Units}} \qquad DPO = \frac{\text{Defects}}{\text{Units} \times \text{Opportunities per unit}} \qquad DPMO = DPO \times 10^6$

$\text{Percent defective} = \frac{\text{Defective units}}{\text{Total units}} \times 100 \qquad PPM = \text{Percent defective} \times 10^4$

First pass yield and rolled throughput yield

First pass yield (FPY) for a step is the proportion of units passing that step with no rework. Rolled throughput yield (RTY) is the probability that a unit passes every step with no rework anywhere:

$RTY = \prod_{i=1}^{k} FPY_i$

RTY is always lower than final yield, because final yield counts reworked units as good. The gap between them is the size of the hidden factory.

Worked example, four steps:

$RTY = 0.950 \times 0.949 \times 0.970 \times 0.990 = 0.8658$

So 86.6% of applications pass cleanly all the way through, even though final yield after rework may be 99%+. The normalized yield -- the average per-step yield -- is $\sqrt[4]{0.8658} = 0.9646$, useful for comparing processes with different numbers of steps.

When defects follow a Poisson model, yield relates to DPU as:

$Y = e^{-DPU} \qquad \Longleftrightarrow \qquad DPU = -\ln(Y)$

With $RTY = 0.8658$, total $DPU = -\ln(0.8658) = 0.1441$ defects per unit across the whole process.

Capability for attributes data

There is no Cp or Cpk for attributes data, because there is no continuous distribution and no specification width. Capability is expressed as a process sigma level derived from the defect rate.

Procedure:

1. Compute DPMO (or the proportion defective, if working at the unit level).
2. Convert the corresponding proportion to a $Z$ value from the standard normal table. This is $Z_{LT}$, the long-term sigma level.
3. Add 1.5 to report the conventional short-term equivalent: $Z_{ST} = Z_{LT} + 1.5$.

Worked example. A process produces 6,210 defects per million opportunities.

• $DPO = 0.006210$, so the yield is 0.993790.
• The $Z$ value with 0.006210 in the upper tail is $Z_{LT} = 2.50$.
• $Z_{ST} = 2.50 + 1.5 = 4.0$, so the process is reported as a "4 sigma" process.

The step from 3 to 4 sigma removes roughly 90% of the defects; the step from 5 to 6 sigma removes another 98.5% of what remains. This is why capability improvement gets progressively harder and why the business case for the last increment must be made on consequence, not on defect count.

Short-term and long-term capability

$Z_{shift} = Z_{ST} - Z_{LT} \approx 1.5$

The 1.5 sigma shift is an empirical convention originating in Motorola's observation that process means drift over time. It is a default, not a law: where enough long-term data exists, compute the actual shift as $Z_{ST} - Z_{LT}$ rather than assuming 1.5.

Interpreting the gap between the two index sets is diagnostic:

When only short-term data is available, the convention is to subtract 1.5 from the short-term $Z$ to estimate long-term performance. When only long-term data is available, add 1.5 to state the short-term equivalent. Say which convention you used; reporting a "6 sigma process" without stating whether it is short- or long-term is ambiguous by a factor of roughly 20,000 in defect rate.