Capability and Control on a Beverage Fill Line
A beverage bottler fills 500 mL bottles on a single line. The fill-volume specification is 495–505 mL. An X-bar and R chart based on 25 subgroups of n = 5 shows X-double-bar = 500.2 mL and R-bar = 4.0 mL. The most recent subgroup mean is 503.5 mL. Control-chart constants for n = 5 are A2 = 0.577, D4 = 2.114, D3 = 0, and d2 = 2.326. The customer's purchase specification requires Cpk ≥ 1.33. Production wants to widen the specification to 494–506 mL and continue running. The quality manager asks you to assess the process before the next production meeting.
1. Compute Cp and Cpk. Does the process meet the customer requirement?
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Estimate the process standard deviation from the range chart:
- σ̂ = R-bar / d2 = 4.0 / 2.326 = 1.72 mL
- Cp = (USL − LSL) / (6σ̂) = (505 − 495) / (6 × 1.72) = 10 / 10.32 = 0.969
- Cpk = min(USL − X̿, X̿ − LSL) / (3σ̂) = min(505 − 500.2, 500.2 − 495) / (3 × 1.72) = min(4.8, 5.2) / 5.16 = 0.930
Both indices are below the required 1.33. The process is not capable. Cpk is slightly lower than Cp because the mean is about 0.2 mL above the 500 mL target, but even a perfectly centered process would have Cp = 0.969, so centering alone will not solve the problem.
2. Is the latest subgroup a signal? What should you do before changing the process or the specification?
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The X-bar chart limits are:
- UCL = X̿ + A2 × R-bar = 500.2 + 0.577 × 4.0 = 502.51 mL
- LCL = 500.2 − 2.31 = 497.89 mL
The latest subgroup mean of 503.5 mL is above the UCL, so it is a special-cause signal. Do not respond by widening the specification; that merely hides the variation and does not improve capability. First contain suspect product since the last in-control subgroup, investigate likely assignable causes (fill nozzle wear, pressure fluctuation, product viscosity, measurement error), and restore control. Only after the process is stable and special causes are removed should you reassess capability. Widening the specification to 494–506 would also violate the customer's drawing unless formally approved, and it does not change Cp or Cpk.
3. If the process were centered exactly at 500 mL, what standard deviation would be needed to achieve Cpk = 1.33? Comment on the centering-versus-variation decision.
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Centered at 500 mL, Cpk = (USL − μ) / (3σ) = 5 / (3σ). Setting this equal to 1.33:
- σ_required = 5 / (3 × 1.33) = 1.253 mL
Current σ̂ is 1.72 mL, so the standard deviation must fall by about 27%. Centering the process to 500 mL would raise Cpk only to about 0.969, still far below 1.33. The dominant problem is common-cause variation, not just an off-center mean. A defensible plan is to reduce variation first (nozzle maintenance, fill-time control, incoming viscosity control), then fine-tune the mean; simply adjusting the mean or widening specifications would not meet the customer requirement.