Question 4 of 6: Attribute Control Charts and a $u$-Chart for Disk-Drive Assembly Nonconformities

Question 4: Attribute Control Charts and a $u$-Chart for Disk-Drive Assembly Nonconformities (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.

(a) $p$, $np$, $c$, $u$, and demerit charts

The $p$-chart plots the fraction of nonconforming UNITS in a sample ($\hat p=x/n$, a binomial-based proportion); because it plots a proportion, it naturally accommodates a sample size $n$ that VARIES from period to period (its limits $\bar p\pm3\sqrt{\bar p(1-\bar p)/n}$ simply widen or narrow with $n$). The $np$-chart plots the raw COUNT of nonconforming units instead of the fraction; it is the same underlying binomial model but requires a CONSTANT sample size, since a raw count is only comparable from sample to sample if the opportunity to accumulate it does not change — it is often preferred on the shop floor because operators find a plain count more intuitive than a proportion. The $c$-chart plots the COUNT of NONCONFORMITIES (not nonconforming units — one unit can have several) observed in an inspection unit of CONSTANT size, modelled as Poisson; the $u$-chart plots nonconformities PER UNIT ($u=c/n$), the Poisson analogue of the $p$-chart, and likewise accommodates a VARYING inspection-unit size or a varying number of units per sample — exactly the situation in part (c), where the number of assemblies inspected changes from day to day. The demerit chart extends the $c$/$u$ idea by weighting different classes of nonconformity by severity (e.g. critical, major, minor nonconformities weighted 100, 50, 10 points respectively) and plotting the total (or average) weighted "demerit" score per unit, since a chart that simply counts nonconformities treats a cosmetic scratch and a safety-critical defect as equally serious — the demerit chart restores that distinction.

(b) Why a $c$-chart's inspection unit is often several physical parts, and when zero $u_i$ signals a problem

The $c$-chart's Poisson model behaves well, and its lower control limit is informative, only when the average count $\lambda=c$ is not too close to zero. If a single physical part rarely has more than zero or one nonconformity, the resulting $c$-chart would sit at a very low average with an $LCL$ that is uselessly pinned at zero and an $UCL$ so close to the mean that the chart has almost no ability to discriminate a real increase from ordinary Poisson noise. Bundling several physical units (e.g. 5 parts) into ONE inspection unit raises the expected count per inspection unit to a workable magnitude (often several nonconformities) while keeping the same, CONSTANT "area of opportunity" every sample — a requirement of the $c$-chart, as distinct from the $u$-chart, which instead lets the unit size vary and rescales by dividing by $n$.

When $LCL=0$ for a $u$-chart and a particular sample's $u_i=0$: this should not, by itself, trigger a search for an assignable cause. A count of zero nonconformities is entirely consistent with in-control operation at the estimated rate $\bar u$ (indeed, under the Poisson model it is simply the most likely single outcome whenever $\lambda$ is modest), and it sits ON the boundary of, not beyond, the control limits — there is no statistical signal. The situation calling for investigation is instead: a point falling ABOVE the $UCL$ (a genuine, statistically unlikely EXCESS of nonconformities, indicating the process has gotten worse), or a nonrandom PATTERN such as an implausibly long run of zeros that is itself statistically improbable under the fitted $\bar u$ (which would suggest inspectors are failing to detect/record real nonconformities, or a gaming of the count, rather than genuine process improvement). A single isolated $u_i=0$, with no such pattern, is simply good news, not a signal to stop the line.

(c) $u$-chart for disk-drive assembly nonconformities

Given. Nonconformities and assemblies inspected over 10 days, sample size varying day to day (table above); total nonconformities $\sum c_i=182$ over $\sum n_i=30$ assemblies.

Find. The appropriate 3-sigma control chart, whether the process is in control (revising if necessary), and the in-control mean nonconformities per assembly.

Approach. Since the number of assemblies inspected varies by day, the $u$-chart (not $c$-chart) applies; compute $u_i=c_i/n_i$ per day, the centre line $\bar u$ from the pooled totals, and DAY-SPECIFIC control limits $\bar u\pm3\sqrt{\bar u/n_i}$ that widen or narrow with each day's own $n_i$.

1. Centre line. $$\bar u=\frac{\sum_{i=1}^{10}c_i}{\sum_{i=1}^{10}n_i}=\frac{182}{30}=\boxed{6.067}\ \text{nonconformities/assembly}$$
2. Per-day $u_i$ and control limits. $u_i=c_i/n_i$ for each day, and $UCL_i,LCL_i=\bar u\pm3\sqrt{\bar u/n_i}$ (the "3 sigma limits" requested), evaluated at each distinct sample size present ($n_i=2,3,4$):

   $n_i$	$UCL_i=\bar u+3\sqrt{\bar u/n_i}$	$LCL_i$
   2	$6.067+3\sqrt{6.067/2}=11.29$	$6.067-3\sqrt{6.067/2}=0.84$
   3	$6.067+3\sqrt{6.067/3}=10.33$	$6.067-3\sqrt{6.067/3}=1.80$
   4	$6.067+3\sqrt{6.067/4}=9.76$	$6.067-3\sqrt{6.067/4}=2.37$

3. Plot and check for out-of-control points. Computing $u_i$ for every day (4.00, 5.25, 5.00, 7.50, 6.00, 5.00, 6.67, 6.00, 7.50, 6.50) and comparing each against its OWN day-specific limits from step 2 (see figure), every single point falls between its $LCL_i$ and $UCL_i$ — there is no point beyond either limit and no nonrandom pattern (runs, trend) evident in the sequence.

4. Conclusion — no revision necessary. Because no point signals out of control, the trial limits computed in step 2 ARE the final limits (there is nothing to remove and re-average), and the centre line from step 1 is already the best in-control estimate: $$\hat\lambda=\bar u=\boxed{6.067\ \text{nonconformities per assembly}}$$