Question 10.P.6: Process capability. Determine which of these three processes......
Process capability. Determine which of these three processes are capable:
| Process | Mean | Standard Deviation |
Lower Spec |
Upper Spec |
| 1 | 7.5 | .10 | 7.0 | 8.0 |
| 2 | 4.6 | .12 | 4.3 | 4.9 |
| 3 | 6.0 | .14 | 5.5 | 6.7 |
Notice that the means of the first two processes are exactly in the center of their upper and lower specs. Hence, the C_{p} index (Formula 10–10) is appropriate. However, the third process is not centered, so C_{pk} (Formula 10–11) is appropriate.
Process capability index, C_{p}\,=\,\frac{\mathrm{Specification~width~}}{\mathrm{Process~width}} \\ = \frac{\mathrm{Upper~specification~} -\,\mathrm{Lower~specification}}{6\sigma \mathrm{~of~the~process}} (10-10)
{\frac{\mathrm{Upper\;specification}-\mathrm{Process~Mean}}{3\sigma}} (10-11)
For Processes 1 and 2: C_{p}\,=\,{\frac{\mathrm{Upper~spec~}\!\!-\,\mathrm{Lower~spec}}{6\sigma}}
In order to be capable, C_{p} must be at least 1.33.
Process 1: C_{p}\;=\;{\frac{8.0-7.0}{6(.10)}}\;=\;1.67\;({\mathrm{capable}})
Process 2: C_{p}\,=\,{\frac{4.9-4.3}{6(.12)}}\,=\,.83\,{\mathrm{(not~capable)}}
For Process 3, C_{pk} must be at least 1.33. It is the lesser of these two:
{\frac{\mathrm{Upper\;spec}-\mathrm{Mean}}{3\sigma}}={\frac{6.7-6.0}{3(.14)}}= 1.67\\ \frac{\mathrm{{Mean}}-\,{\mathrm{Lower~spec}}}{3\sigma}=\frac{6.0-5.5}{3(.14)}=1.19\,\,(\mathrm{{not~capable}})