Question 2 of 6: Trend Charts, EPC vs. SPC and EWMA, and X̄/R Chart Design

Question 2: Trend Charts, EPC vs. SPC and EWMA, and X̄/R Chart Design (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) Charting trend processes; EPC vs. SPC; the rationale for EWMA in both

An ordinary Shewhart $\bar X$ chart assumes the underlying process mean is constant (in control at a fixed target) between assignable-cause events; a "trend" process such as tool wear violates that assumption by design — the mean drifts monotonically and predictably as the tool wears, so a fixed centre line and fixed 3-sigma limits will signal a false "out of control" alarm almost immediately as the natural, expected drift carries the mean past the limits, even though nothing abnormal has happened. The traditional-chart fix is to chart against a MOVING target rather than a fixed one: fit (or specify from engineering knowledge) the expected trend line for the mean over time, and plot each sample's DEVIATION from the value the trend line predicts at that time, with control limits computed for those residuals exactly as for an ordinary characteristic. Any residual outside the (now-appropriately-centred) limits signals a genuine departure from the EXPECTED wear pattern — an abnormally fast or slow drift, a step change, or a new source of variation — while the expected wear itself no longer trips the chart.

Engineering Process Control (EPC) and Statistical Process Control (SPC) represent two different philosophies for handling process variation. SPC assumes the process, once brought into a state of statistical control, should be left alone: only assignable-cause variation should trigger an adjustment (a control chart signal), because any adjustment made in response to ordinary common-cause noise ("tampering") actually increases output variability, per Deming's funnel experiment. EPC (also called automatic/feedback process control) instead assumes the process is subject to a continuously drifting disturbance (exactly a trend process such as tool wear, or a raw-material property that wanders) that CANNOT economically be eliminated at its source, and compensates for it continuously by adjusting a manipulable process variable (e.g. a machine offset) via a feedback or feedforward control law, so the output stays near target even though the underlying disturbance is never removed. In short: SPC seeks to detect and eliminate the CAUSE of variation and otherwise not touch the process; EPC accepts an inevitable, structured disturbance and continuously COMPENSATES for its effect on the output.

The EWMA (exponentially weighted moving average) statistic, $z_t=\lambda x_t+(1-\lambda)z_{t-1}$, is used for both purposes because of what that recursion actually computes. As an SPC tool, $z_t$ is a weighted average that discounts old observations exponentially, so it accumulates evidence of a small, sustained shift across several consecutive samples, giving it far greater sensitivity to SMALL mean shifts than a memoryless Shewhart chart (which only looks at the current sample) — exactly the shift-detection improvement SPC wants. As an EPC tool, the identical recursion IS the classical exponentially-weighted forecast of the process level one step ahead, and is mathematically equivalent to the update law of a discrete integral (PI-type) feedback controller: $z_t$ is precisely the quantity an EPC scheme would compute as its best current estimate of "where the process is heading" and adjust against. Because the same statistic simultaneously serves as (i) a sensitive detector of sustained shifts for a control chart and (ii) the natural one-step-ahead predictor a compensating controller needs, EWMA is the standard bridge used to combine SPC monitoring with EPC compensation on the same trending process: the EPC loop uses $z_t$ to keep the mean on target, while an EWMA (or similar) chart of the RESIDUALS after compensation continues to watch for genuine assignable causes the controller was never designed to compensate for.

(b) X̄/R chart control limits, process mean/σ estimates, and probability of missing a shift to 25

Given. $n=4$ per sample, 30 samples, $\bar{\bar X}=20$, $\bar R=4.56$. Control-chart factors for $n=4$ (Appendix VI): $A_2=0.729$, $D_3=0$, $D_4=2.282$, $d_2=2.059$.

Find. The 3-sigma limits for the $\bar X$ and $R$ charts; $\hat\mu$ and $\hat\sigma$; and $P(\text{miss the shift to }\mu_1=25\text{ on BOTH of the next 2 samples})$.

1. $\bar X$-chart limits. $$CL_{\bar X}=\bar{\bar X}=20,\qquad UCL_{\bar X}=\bar{\bar X}+A_2\bar R=20+0.729(4.56)=\boxed{23.324},\qquad LCL_{\bar X}=20-0.729(4.56)=\boxed{16.676}.$$
2. $R$-chart limits. $$CL_R=\bar R=4.56,\qquad UCL_R=D_4\bar R=2.282(4.56)=\boxed{10.406},\qquad LCL_R=D_3\bar R=0(4.56)=\boxed{0}.$$
3. Process mean and standard deviation estimates. The centre line of the $\bar X$ chart is itself the unbiased estimate of the process mean, and $\bar R/d_2$ is the standard unbiased estimator of $\sigma$ from the average range: $$\hat\mu=\bar{\bar X}=\boxed{20},\qquad \hat\sigma=\frac{\bar R}{d_2}=\frac{4.56}{2.059}=\boxed{2.215}.$$
4. Probability of not detecting a shift to $\mu_1=25$ on one sample. After the shift, $\bar X\sim N\!\left(25,\ (\hat\sigma/\sqrt n)^2\right)$ with $\hat\sigma/\sqrt n=2.215/2=1.107$. Not detecting means $\bar X$ still falls inside the (unchanged) control limits: $$Z_U=\frac{23.324-25}{1.107}=-1.513,\qquad Z_L=\frac{16.676-25}{1.107}=-7.517.$$ $$P(\text{miss}\mid 1\text{ sample})=\Phi(Z_U)-\Phi(Z_L)=\Phi(-1.513)-\Phi(-7.517)=0.0651-0.0000=\boxed{0.0651}.$$
5. Probability of missing the shift on BOTH of the next two samples. Successive samples are independent, so $$P(\text{miss both of 2 subsequent samples})=\big[P(\text{miss}\mid 1)\big]^2=(0.0651)^2=\boxed{0.00423\ (0.42\%)}.$$

The chart is quite effective against this size of shift ($\delta=(25-20)/2.215=2.26\hat\sigma$ in the mean): roughly 93.5% of the time it flags the very first sample after the shift, and the probability of running two consecutive samples without a signal is under half a percent.

(c) Minimum sample size $n$ so ARL $\le 4$ for a shift to 23

Given. $\hat\mu_0=20$, $\hat\sigma=2.215$ (from part (b)); shift target $\mu_1=23$; required average run length $ARL\le 4$; the $\bar X$ chart keeps its standard $\pm3\hat\sigma/\sqrt n$ limits about $\mu_0=20$ for each candidate $n$.

Find. The minimum future sample size $n$.

Approach. $ARL=1/p$, where $p$ is the probability a single sample signals once the mean has shifted to $\mu_1$; require $p\ge 1/ARL=0.25$ and search over integer $n$, recomputing the $\pm3\hat\sigma/\sqrt n$ limits and $p(n)$ each time.

1. Detection probability as a function of $n$. With limits $UCL(n)=20+3\hat\sigma/\sqrt n$, $LCL(n)=20-3\hat\sigma/\sqrt n$, after the shift $\bar X\sim N(23,(\hat\sigma/\sqrt n)^2)$, so $$p(n)=1-\Big[\Phi\!\Big(\tfrac{UCL(n)-23}{\hat\sigma/\sqrt n}\Big)-\Phi\!\Big(\tfrac{LCL(n)-23}{\hat\sigma/\sqrt n}\Big)\Big].$$
2. Search over $n$. $$n=2:\ UCL=24.70,\ LCL=15.30,\ p=0.139,\ ARL=1/0.139=7.19\ (\text{fails, }ARL>4).$$ $$n=3:\ UCL=23.84,\ LCL=16.16,\ p=0.257,\ ARL=1/0.257=\boxed{3.90\le4}\ (\text{passes}).$$
3. Minimum sample size. $n=2$ gives $ARL=7.19>4$ and $n=3$ gives $ARL=3.90\le4$, so the minimum sample size satisfying the requirement is $$\boxed{n=3}.$$