Question 4 of 6: p/np/c/u and Demerit-Point Charts, the $u_i=0$ Policy, and a u Chart Build

Question 4: p/np/c/u and Demerit-Point Charts, the $u_i=0$ Policy, and a u Chart Build (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-point charts

$p$ chart plots the fraction nonconforming in a sample of $n$ units, each simply classified conforming/nonconforming (go/no-go), modeled Binomial$(n,p)$; used whenever the inspection result per unit is binary and sample size can vary. $np$ chart plots the raw count of nonconforming units (rather than the fraction) for a constant sample size $n$ — statistically equivalent to a $p$ chart but sometimes preferred operationally because whole numbers are easier for operators to read on the shop floor. $c$ chart plots the count of nonconformities (not nonconforming units — one unit can carry several nonconformities) found in a constant-size inspection unit, modeled Poisson$(c)$; used when a single unit can have multiple distinct defects and the opportunity for defects (unit size) never changes. $u$ chart plots the average nonconformities per inspection unit, $u=c/n$, when the number of inspection units per sample varies — the natural generalization of the $c$ chart to a variable-size sample, with sample-size-dependent control limits. Demerit-point chart weights different nonconformity classes by severity (e.g. critical/major/minor demerit weights $w_1>w_2>w_3$) before summing, $D=\sum w_iC_i$, so that a single serious defect is not treated as equal in importance to a cosmetic blemish — used whenever nonconformities differ substantially in their impact on fitness for use, which a plain count-based $c$/$u$ chart cannot express.

(b) $LCL=0$, $u_i=0$: when to stop the process

A single observation of $u_i=0$ sits exactly at the boundary $LCL=0$; it is not a control-limit violation and, under a Poisson-type process with a moderate in-control mean, an occasional zero count is an expected, normal outcome — so the process should not be stopped for one isolated $u_i=0$. The situation calls for stopping and searching for an assignable cause only when zero (or unusually low) counts occur far more often than the underlying model predicts — most concretely, a run of several consecutive $u_i=0$ points (a Western-Electric-style run rule), because that pattern is itself statistically improbable under a stable Poisson process with the estimated mean. Such a run should always be investigated, because it can represent either a genuine, desirable process improvement (worth locking in and re-basing the chart on) or a false signal (inspection not actually being performed, a measurement-system change, or nonconformities being mis-recorded) — both possibilities require action, just of different kinds.

(c) u chart for disk-drive assembly nonconformities (variable daily sample size)

Given. Final-inspection data for 10 days; the number of assemblies inspected varies day to day.

Find. The appropriate control chart with 3-sigma limits, revised if necessary, and the in-control mean number of nonconformities per assembly.

Approach. The number of assemblies inspected varies day to day (2 or 4), so a plain $c$ chart (constant inspection-unit count) is not appropriate — a $u$ chart (nonconformities per assembly, with sample-size-dependent limits) is required, treating 1 inspection unit = 1 assembly.

1. Daily rate. $u_i=c_i/n_i$.

   Day	1	2	3	4	5	6	7	8	9	10
   $u_i$	6.0	8.0	5.0	7.5	9.0	5.0	10.0	6.0	7.5	7.0

2. Centre line. $$\bar u=\frac{\sum c_i}{\sum n_i}=\frac{195}{28}=6.964$$
3. Variable 3-sigma limits, $UCL_i/LCL_i=\bar u\pm3\sqrt{\bar u/n_i}$. For $n_i=2$: $UCL=6.964+3\sqrt{6.964/2}=12.56$, $LCL=6.964-3\sqrt{6.964/2}=1.37$. For $n_i=4$: $UCL=6.964+3\sqrt{6.964/4}=10.92$, $LCL=6.964-3\sqrt{6.964/4}=3.01$.
4. Check for out-of-control points. Comparing every $u_i$ against its own day's limits (1.37–12.56 for the $n_i=2$ days, 3.01–10.92 for the $n_i=4$ days), all 10 points fall inside their limits — the largest is $u_5=9.0<12.56$ ($n_i=2$) and the smallest is $u_3=u_6=5.0>3.01$ ($n_i=4$'s own minimum comparator is day 4/8/10 at 7.5/6.0/7.0, all above 3.01). $$\boxed{\text{No point is out of control} \Rightarrow \text{no revision of the limits is needed}}$$
5. In-control mean nonconformities per assembly. Since 1 inspection unit = 1 assembly here, $\bar u$ is already the rate per assembly: $$\boxed{\hat c_{\text{assembly}}=\bar u=6.96\text{ nonconformities/assembly}}$$