Production Line Capability After a Process Change
A turning cell produces a pin with a specified diameter of 25.00 ± 0.05 mm. Before a change of tool grade, 50 consecutive parts gave a mean of 25.010 mm and a standard deviation of 0.012 mm. After the change, 50 parts gave a mean of 25.028 mm and a standard deviation of 0.018 mm. The line runs 480 minutes per shift with a 30-minute planned break, and produces 120 parts per shift. Scrap rate rose from 0.5% to 3.2% after the change, and the cell now occasionally stops for tool breakage. The production manager wants to revert to the old tool grade; the quality engineer wants to keep the new grade and adjust the process.
1. Calculate the process capability before and after the change, and interpret the results.
Read the worked answer
Specification limits: USL = 25.05 mm, LSL = 24.95 mm, tolerance width = 0.10 mm.
Before the change:
- Cp = (USL − LSL) / (6σ) = 0.10 / (6 × 0.012) = 0.10 / 0.072 = 1.39
- Cpk = min[(USL − x̄)/(3σ), (x̄ − LSL)/(3σ)]
- Upper: (25.05 − 25.010)/(3 × 0.012) = 0.040/0.036 = 1.11
- Lower: (25.010 − 24.95)/(0.036) = 0.060/0.036 = 1.67
- Cpk = 1.11
After the change:
- Cp = 0.10 / (6 × 0.018) = 0.10 / 0.108 = 0.93
- Cpk = min[(25.05 − 25.028)/(3 × 0.018), (25.028 − 24.95)/(0.054)]
- Upper: 0.022/0.054 = 0.41
- Lower: 0.078/0.054 = 1.44
- Cpk = 0.41
Interpretation. Before the change the process was marginally capable (Cpk 1.11, below the usual 1.33 target) but centred reasonably. After the change the process is not capable: Cp = 0.93 means even a perfectly centred process would produce out-of-spec parts, because the natural spread (6σ = 0.108 mm) exceeds the 0.10 mm tolerance. The mean has also drifted upward by 0.018 mm, pushing Cpk down to 0.41. The upper specification limit is the binding constraint: the mean is only 0.022 mm below USL, i.e. about 1.2 standard deviations, so roughly 12% of parts would exceed USL on a normal distribution - consistent with the observed 3.2% scrap plus rework.
Conclusion: The new tool grade increased both the mean and the variation. Reverting is one option, but the more informative question is why the variation rose - tool wear rate, rigidity, or coolant. Simply reverting may mask a fixable problem.
2. Using the shift data, assess the productivity impact and recommend an action plan.
Read the worked answer
Productivity calculation
- Available time per shift = 480 − 30 = 450 min.
- At 120 parts/shift, cycle time = 450/120 = 3.75 min/part.
- Good parts before: 120 × 0.995 = 119.4 per shift.
- Good parts after: 120 × 0.968 = 116.2 per shift.
- Loss = 3.2 good parts per shift, about 2.7% of output. Over three shifts, about 9.6 parts/day lost, plus the tool-breakage downtime which is not yet quantified.
Action plan
- Quantify the tool-breakage downtime. Log stoppages for two weeks. If mean time between failures has fallen, the new grade may be running outside its recommended cutting speed or feed.
- Re-centre the process. Adjust the tool offset to bring the mean from 25.028 mm back to 25.000 mm. This alone raises Cpk from 0.41 to about 0.93 (still below 1.33 because Cp = 0.93).
- Reduce variation. Investigate the cause of the higher σ: check tool holder rigidity, spindle runout, coolant flow, and whether the new grade wears faster (which would cause a drift within the tool life). If wear is the cause, shorten the tool-change interval and re-centre after each change.
- Only then decide on the tool grade. If, after re-centring and variation reduction, Cpk ≥ 1.33, keep the new grade (it may offer longer life or better surface finish). If not, revert.
- Set up control charts. Use X-bar and R charts with the new process; the shift in mean would have been caught within a few subgroups. Establish a reaction plan for out-of-control points.
The decision should be data-driven: the new grade is not inherently unsuitable, but the process is currently incapable and must be corrected before either accepting or rejecting the change.