OpenSpecEngineering

Finds the center line and the upper and lower control limits (UCL, LCL) for the common Shewhart control charts. Pick a variables chart (X̄–R, X̄–S or I–MR) for measurements, or an attributes chart (p, np, c or u) for counts of defectives or defects. Quality engineers and operators use it to set up a chart from a baseline run of data, or to check limits that a software package reports [1]–[3].

Equations

X̄–R: x̄̄ ± A₂R̄    R: D₃R̄, D₄R̄    σ = R̄/d₂

X̄–S: x̄̄ ± A₃s̄    S: B₃s̄, B₄s̄    σ = s̄/c₄

I–MR: x̄ ± 2.660·M̄R    MR: 0, 3.267·M̄R

p: p̄ ± 3√(p̄(1−p̄)/n)    np: np̄ ± 3√(np̄(1−p̄))

c: c̄ ± 3√c̄    u: ū ± 3√(ū/n)

x̄̄ is the grand mean of the subgroup means, R̄ the average range, s̄ the average standard deviation and M̄R the average moving range of consecutive readings. n is the subgroup size. p̄ is the average fraction nonconforming, np̄ the average count of nonconforming units per subgroup, c̄ the average defects per inspection unit and ū the average defects per unit. A₂, A₃, B₃, B₄, D₃, D₄, d₂ and c₄ are the standard Shewhart constants for subgroups of 2 to 10 [1], [2]. A lower limit that would fall below zero is set to zero, and a p-chart UCL is capped at 1.

Limits and common mistakes

• Set limits from 20 to 25 subgroups taken while the process is stable. Drop subgroups with known special causes and recompute.
• Control limits are not specification limits. They show what the process does, not what the drawing allows.
• The p and u limits depend on n. If subgroup sizes vary a lot, work out limits for each size.
• The X̄ charts here take subgroups of 2 to 10. Larger subgroups need constants from a full table [1].

Worked example

An operator measures five pins an hour on a grinder. Over 25 hours the grand mean diameter is 12.004 mm and the average range is 0.021 mm. Set up the X̄–R chart.

1. Choose Chart Type X̄–R (mean & range).
2. Enter Subgroup Size (n) 5.
3. Enter Center Line — grand mean x̄̄ / c̄ / ū 12.004 and Average Range R̄ / Std Dev s̄ / Moving Range M̄R 0.021.
4. Press Solve. The Solution shows: X̄ chart UCL: 12.0161
   X̄ chart CL: 12.004
   X̄ chart LCL: 11.9919
   R chart UCL: 0.044394
   R chart CL: 0.021
   R chart LCL: 0
   Estimated σ = R̄ / d2: 0.00902837

Check: for n = 5, A₂ = 0.577, so the limits are 12.004 ± 0.577 × 0.021 = 12.004 ± 0.0121 mm. That is ±3σ/√5 with σ = 0.00903 mm, as it should be.