Question 6 of 8: Control Criterion for a Purchased Watch Gear, and its Relation to the Specification

Question 6: Control Criterion for a Purchased Watch Gear, and its Relation to the Specification (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Twenty-five subgroups of $n=5$ consecutive pieces, with grand average $\bar{\bar{X}}=0.125$ in and average range $\bar{R}=0.002$ in on the key dimension. The specification limits are not stated.

Find. The control criterion itself, that is the control limits for the mean and the range; the basis on which that criterion may legitimately be compared with the drawing specification; and the courses of action open to the vendor if the two are incompatible.

Approach. Build the $\bar{X}$ and $R$ charts from the two statistics supplied, convert the average range into an estimate of the process standard deviation, and use that to express the process capability in the same units as the specification before comparing them.

1. Set the criterion for the range first. Control of the mean is meaningless while the spread is unstable, so the $R$ chart is established first: $$UCL_{R}=D_{4}\bar{R}=2.114(0.002)=0.004228\ \text{in},\qquad LCL_{R}=D_{3}\bar{R}=0$$ A subgroup range above 0.00423 in is evidence that the within-subgroup variability has increased — a worn or loose cutter, a changed operator, mixed material.
2. Set the criterion for the mean. With the spread in control, the limits on the subgroup average are placed three standard errors either side of the grand average, which the $A_{2}$ factor delivers directly from the average range: $$UCL_{\bar{X}}=\bar{\bar{X}}+A_{2}\bar{R}=0.125+0.577(0.002)=\boxed{0.12615\ \text{in}}$$ $$LCL_{\bar{X}}=\bar{\bar{X}}-A_{2}\bar{R}=0.125-0.577(0.002)=\boxed{0.12385\ \text{in}}$$ The operating rule is then the ordinary Shewhart one: plot the mean and range of every sample of five, and investigate whenever a point falls outside a limit, or when the points show a run of seven on one side of the centre line, a trend, or any other non-random pattern.
3. Convert the average range into a process standard deviation. The control limits above apply to averages of five, so they cannot be compared with a specification, which applies to individual gears. The bridge between the two is $$\hat{\sigma}=\frac{\bar{R}}{d_{2}}=\frac{0.002}{2.326}=0.00086\ \text{in}$$ from which the standard error used in the chart is $\hat{\sigma}/\sqrt{5}=0.000385$ in, and three of those is 0.00115 in — which reproduces $A_{2}\bar{R}$ and confirms the limits.
4. Express the process capability in specification units. The spread of individual pieces from a stable process is $$6\hat{\sigma}=6(0.00086)=\boxed{0.00516\ \text{in}}$$ so the natural tolerance of the process runs from $0.125-0.00258=0.12242$ in to $0.125+0.00258=0.12758$ in. This band, not the control limits, is what must be set against the drawing.

How the criterion compares with the specification. The comparison is legitimate only between the natural tolerance and the specification, and it has three possible outcomes. If the specification band is wider than 0.00516 in and the process is centred on the nominal, the process is capable: a specification of $0.125\pm0.003$ in, for example, gives a capability index $C_{p}=0.006/0.00516=1.16$, and the vendor may run to the control chart and ship without screening. If the specification band is comparable with 0.00516 in — say $0.125\pm0.0025$ in, giving $C_{p}=0.97$ — the process is marginal and will produce a few thousand parts per million outside tolerance even while perfectly in control. If the specification is tighter still, say $0.125\pm0.002$ in with $C_{p}=0.78$, the process is simply incapable, and a chart showing perfect control will nevertheless be accompanied by rejected lots. That last case is the one the question is pointing at: control and capability are different properties, and a process can possess either without the other.

Alternatives if the criterion and the specification are incompatible. Four courses of action are open, in the order a vendor should consider them. First, centre the process: if the grand average is off nominal, part of the non-conformance is free to remove, since re-setting the machine costs nothing in variability and the capability index $C_{pk}$ rises immediately. Second, reduce the variability so that the natural tolerance fits inside the specification — better tooling and fixturing, closer control of material and temperature, refurbishing or replacing the machine, or moving the operation to a more precise process such as grinding rather than hobbing. This is the only permanent answer and it is the one statistical process control exists to support, because the chart identifies which sources of variation are assignable and worth chasing. Third, re-examine the specification with the horologist: tolerances are often inherited rather than engineered, and if the functional requirement genuinely permits a wider band the drawing should be changed rather than the process punished. Fourth, if none of these is available in the time required, screen — sort every piece by 100 percent inspection or automatic gauging, or grade the output into selective-assembly classes — accepting that screening adds cost, is never perfectly effective, and is a containment measure rather than a solution. Changing to a different vendor or process is the last resort, and it is the one the horologist will impose if the vendor does not act.