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23-Ind-A5 Quality Planning, Control, and Assurance · May 2017

Question 2 of 6: Process Variation, Run Length, and the X̄-R Control Chart

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

National Exams, May 2017. Closed-book examination. Any five of the six questions constitute a complete paper; all six are answered in full below. Relevant statistical tables (cumulative standard normal distribution, factors for constructing variables control charts, the F distribution, and a blank Weibull probability chart) are reproduced/applied from the paper's own attached appendices.

Reference texts: Montgomery, Introduction to Statistical Quality Control (8th ed.) — Ch. 1–2 (quality philosophy, cost of quality, Six Sigma/DMAIC), Ch. 5–6 (variables control charts: X̄-R, X̄-S, process capability), Ch. 7 (attributes charts: p, np, c, u, demerit systems), Ch. 9 (CUSUM and EWMA control charts), Ch. 8 & 13 (reliability, life testing and Weibull analysis; designed experiments and factorial designs).

Question 2: Process Variation, Run Length, and the X̄-R Control Chart (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) Common vs. special-cause variation; statistical control; charting processes with a trend; run length and ARL

Production processes exhibit two types of variation. Common-cause (chance) variation is the aggregate of many small, inherent sources — material, machine, environment, operator — that are always present; it is random, stable, and describable by a fixed probability distribution. Special-cause (assignable) variation arises from a specific, identifiable event (a tool wearing out, a bad lot of raw material, an untrained operator) that shifts the process away from its stable distribution. A process is said to be in statistical control when only common-cause variation is present, so that the process mean and variance are constant over time and all future output can be predicted from the historical distribution — operationally, this is the state in which every point on the control chart falls within the control limits with no non-random patterns.

Standard Shewhart charts assume a stationary mean and can give a very poor (often misleadingly "in control") picture when the process exhibits a genuine natural trend (e.g. gradual tool wear steadily shrinking a bore diameter). Two adaptations handle this: (1) fit and subtract the deterministic trend first, then chart the residuals on an ordinary X̄ chart (residuals are approximately stationary if the trend model is adequate); or (2) use a chart designed for trending/autocorrelated data, such as a regression control chart with control limits around the fitted trend line, or an EWMA/CUSUM tuned as an engineering-process-control adjustment. Example: a chart of bore diameter vs. tool-change cycle shows a steady downward drift of about 0.001 mm per part; plotting the residuals from a fitted regression line on a standard X̄ chart correctly separates the expected wear trend from a genuine special-cause jump.

The run length (RL) is the number of samples plotted before a chart first signals (a point outside the control limits, or a run-rule violation). The average run length (ARL) is its expectation; for a chart with false-alarm/detection probability $p$ per sample, $RL$ is geometric with $E[RL]=1/p$. The average time to signal (ATS) converts this to real time: $ATS=ARL\times h$, where $h$ is the sampling interval. Sample size effect: increasing $n$ narrows the control limits (they scale as $1/\sqrt n$ around the centre line) for a FIXED shift size, so a given real shift produces a larger standardized deviation and is detected faster — ARL decreases (chart becomes more sensitive) as $n$ increases, at the cost of more inspection per sample.

(b) X̄-R chart: control limits and in-control process parameters

Given. $m=40$ samples of size $n=5$; $\sum\bar X_i=18{,}740$; $\sum R_i=680$. From the paper's own Appendix VI (Factors for Constructing Variables Control Charts), for $n=5$: $A_2=0.577$, $D_3=0$, $D_4=2.115$, $d_2=2.326$.

Find. The X̄ and R chart control limits and the in-control estimates $\hat\mu$, $\hat\sigma$.

Approach. Compute the grand average $\bar{\bar X}$ and average range $\bar R$ from the given sums, then apply the standard $A_2$/$D_3$/$D_4$ factor relations for the control limits and $\bar R/d_2$ for the process standard deviation.

  1. Grand average and average range. $$\bar{\bar X}=\frac{\sum \bar X_i}{m}=\frac{18{,}740}{40}=\boxed{468.5},\qquad \bar R=\frac{\sum R_i}{m}=\frac{680}{40}=\boxed{17.0}.$$
  2. X̄ chart limits. Substituting into $UCL,LCL=\bar{\bar X}\pm A_2\bar R$: $$UCL_{\bar X}=468.5+0.577(17.0)=\boxed{478.31},\qquad LCL_{\bar X}=468.5-0.577(17.0)=\boxed{458.69}.$$
  3. R chart limits. Substituting into $UCL_R=D_4\bar R$, $LCL_R=D_3\bar R$: $$UCL_R=2.115(17.0)=\boxed{35.96},\qquad LCL_R=0(17.0)=\boxed{0}.$$
  4. In-control process parameters. The centre lines themselves are the point estimates of the true process mean and standard deviation: $$\hat\mu_0=\bar{\bar X}=\boxed{468.5},\qquad \hat\sigma_0=\frac{\bar R}{d_2}=\frac{17.0}{2.326}=\boxed{7.309}.$$
LSL=445USL=495μ₀=468.5Tensile strength — process distribution vs. specification (Q2c)
Fig. 2.1 — The fitted in-control process ($\hat\mu_0=468.5$, $\hat\sigma_0=7.309$) plotted against the tensile-strength specification $470\pm25$ used in part (c).
QuantityValue
Grand average $\bar{\bar X}$468.5
Average range $\bar R$17.0
X̄ chart limitsLCL = 458.69, UCL = 478.31
R chart limitsLCL = 0, UCL = 35.96
In-control mean $\hat\mu_0$468.5
In-control std. dev. $\hat\sigma_0$7.309
Check: The estimates $\hat\mu_0$, $\hat\sigma_0$ are valid only if all 40 subgroups are themselves in statistical control (no points beyond the just-computed limits, no non-random patterns) — i.e. the 40-sample baseline is assumed free of assignable causes. We also assume the tensile-strength measurements within each subgroup are approximately normally distributed and independent, which is what justifies using the $d_2$-based estimator $\hat\sigma_0=\bar R/d_2$ for the common-cause standard deviation.

(c) Process capability and a chart re-designed for a specified ARL

Given. From part (b): $\hat\mu_0=468.5$, $\hat\sigma_0=7.309$. Specification $470\pm25$, i.e. $LSL=445$, $USL=495$. The chart is to be re-designed with 3-sigma limits so that $ARL\le5$ when the mean shifts from $\mu_0$ to $\mu_0+1.2\sigma$.

Find. $C_p$, $C_{pk}$, the process fraction nonconforming, and the sample size $n$ (with resulting control limits) that meets the ARL target.

Approach. Compute $C_p$/$C_{pk}$ from the spec width and $\hat\sigma_0$; get the nonconforming fraction from the standard normal CDF at each specification limit; then scan integer $n$ in the ARL formula for a $k=3$-sigma chart until the target shift's ARL first drops to 5 or below.

  1. Process capability. $$C_p=\frac{USL-LSL}{6\hat\sigma_0}=\frac{495-445}{6(7.309)}=\boxed{1.140},$$ $$C_{pk}=\min\!\left(\frac{USL-\hat\mu_0}{3\hat\sigma_0},\ \frac{\hat\mu_0-LSL}{3\hat\sigma_0}\right)=\min\!\left(\frac{26.5}{21.93},\ \frac{23.5}{21.93}\right)=\min(1.209,\,1.072)=\boxed{1.072}.$$
  2. Fraction nonconforming. Standardizing each limit, $Z_{USL}=(495-468.5)/7.309=3.626$ and $Z_{LSL}=(445-468.5)/7.309=-3.215$: $$\hat p=P(Z>3.626)+P(Z<-3.215)=0.000144+0.000651=\boxed{0.000795}\ (\approx0.0795\%,\ 795\text{ ppm}).$$
  3. ARL of an $n$-sample, 3-sigma X̄ chart at shift $\delta\sigma$. The chance a shifted process still plots inside the limits is $\beta=\Phi\!\left(k-\delta\sqrt n\right)-\Phi\!\left(-k-\delta\sqrt n\right)$, giving $ARL=1/(1-\beta)$. Scanning $n=1,2,3,\dots$ at $k=3$, $\delta=1.2$: $$n=3:\ ARL=5.61;\qquad n=4:\ ARL=3.65.$$ The smallest sample size that meets $ARL\le5$ is $\boxed{n=4}$.
  4. Control limits for the re-designed chart. With $n=4$: $$UCL_{\bar X}=468.5+\frac{3(7.309)}{\sqrt4}=\boxed{479.46},\qquad LCL_{\bar X}=468.5-\frac{3(7.309)}{\sqrt4}=\boxed{457.54}.$$
QuantityValue
$C_p$1.140
$C_{pk}$1.072
Process fraction nonconforming0.0795% (795 ppm)
Required sample size for $ARL\le5$ at a $1.2\sigma$ shift$n=4$
Re-designed X̄ chart limits ($n=4$)LCL = 457.54, UCL = 479.46