Reading an X-bar/R chart when the gauge itself eats 30% of the tolerance
A shaft OD has a specification of 25.000 ± 0.050 mm. The shop monitors it with an X-bar and R chart, subgroups of n = 5, limits frozen from 25 subgroups: grand mean 25.012 mm and mean range 0.024 mm. The chart constants are A2 = 0.577, D4 = 2.114 and d2 = 2.326. On the most recent shift one subgroup averaged 25.031 mm; the R chart stayed in control throughout. In parallel, a gauge repeatability and reproducibility study on the same feature (10 parts, 3 operators, 3 trials, digital micrometer) gives GRR = 0.030 mm expressed as 6σ. The micrometer is in calibration and reads to 0.001 mm. At the morning meeting the quality engineer proposes, on the basis of a Cpk of about 1.2, either tightening the drawing tolerance or budgeting for a new grinder. The machine has a fixed tool offset set at the start of each shift only, and the last three shifts all averaged slightly above nominal.
1. Compute the control limits, Cp and Cpk from the chart data, and interpret them together with the single out-of-control subgroup.
Read the worked answer
Limits: UCL = 25.012 + 0.577 × 0.024 = 25.012 + 0.0138 = 25.0258; LCL = 25.012 − 0.0138 = 24.9982. R chart: UCL_R = 2.114 × 0.024 = 0.0507, LCL_R = 0.
Process sigma from within-subgroup range: σ = R-bar/d2 = 0.024/2.326 = 0.01032 mm.
Cp = (USL − LSL)/(6σ) = 0.100/(6 × 0.01032) = 0.100/0.0619 = 1.62. Cpk = min(25.050 − 25.012, 25.012 − 24.950)/(3 × 0.01032) = 0.038/0.0310 = 1.23.
Interpretation: Cp 1.62 says the spread is capable; Cpk 1.23 says the process is off-centre by 0.012 mm, about 1.2σ. The gap between Cp and Cpk, not the absolute value, is the actionable part.
The subgroup at 25.031 is 0.019 mm above the grand mean, i.e. 0.019/(0.01032/√5) = 4.1 standard errors, with an in-control R chart. That is a genuine mean shift, not a spread change, and needs an assignable cause (a new lot of bar stock, coolant temperature, a partially worn wheel, or a tool offset keyed in wrongly) before any limit is recalculated.
2. The GRR figure of 0.030 mm has been handed to you. Reassess Cp and Cpk, and judge whether the gauge is fit for this chart.
Read the worked answer
GRR is 6σ_gauge = 0.030 mm, so σ_gauge = 0.005 mm. As a fraction of tolerance, %GRR = 0.030/0.100 = 30%, sitting at the boundary where a gauge is normally judged unacceptable for product decision-making and certainly for tight SPC limits.
The chart's within-subgroup sigma of 0.01032 contains gauge error, so the true process sigma is smaller: σ_actual = √(0.01032² − 0.005²) = √(1.065 × 10⁻⁴ − 2.5 × 10⁻⁵) = √(8.15 × 10⁻⁵) = 0.00903 mm.
Corrected capability: Cp = 0.100/(6 × 0.00903) = 1.85 Cpk = 0.038/(3 × 0.00903) = 1.40.
So the gauge inflates the apparent spread and depresses Cp by roughly 0.2; the reported 1.62 and 1.23 understate the process.
The gauge is still marginally adequate for detecting the 25.031 subgroup: with true σ = 0.00903 the standard error of a subgroup mean is 0.00404, and the deviation of 0.019 mm is 4.7 standard errors. The signal survives the measurement noise. But a 30% GRR makes the R chart insensitive and inflates every capability number, so it must be fixed before any capability figure is quoted to a customer.
3. Given all of the above, what should the plant actually do, and why are the two proposals on the table the wrong first moves?
Read the worked answer
Both proposals are premature:
- Tightening the drawing tolerance would raise cost and make the same process look worse; the process is capable on spread (corrected Cp 1.85), so the tolerance is not the constraint.
- A new grinder addresses a spread problem that does not exist. Grinding variation is around 0.009 mm; the loss of capability is 0.012 mm of mean offset.
Correct sequence:
- Fix the measurement system first. Replace the hand-held micrometer with a comparator stand, a master setting ring and a spring-loaded gauge, or at minimum a fixture and a single trained operator. Target %GRR below 10%, or at worst below 30% with the limits recalculated from the improved data. Until then every Cp/Cpk and every control limit is biased.
- Then find the assignable cause of the 25.031 subgroup. The R chart is stable, so something shifted the mean on one shift; check the bar stock lot certificate, the coolant and machine thermal state, and whether the offset was keyed correctly.
- Then revalidate and recentre. The last three shifts averaged above nominal, which is a systematic offset, not noise. Shifting the mean from 25.012 to 25.000 gives Cpk = 0.050/(3 × 0.00903) = 1.85, matching Cp, at the cost of one offset change per shift.
Only after steps 1–3, if capability still falls short, is capital spend justified. Doing it in this order costs a fixture and a day of study instead of a grinder.