Developing Control Charts for Monitoring Time Interval Between Nonconforming Items in High-Quality Processes

Abstract

High-quality processes with very low rate of nonconforming items can be found in many industries nowadays. For these processes, the traditional Shewhart control charts such as p-chart, u-chart, and np-chart are not applicable. Due to low fraction of nonconforming, it is better to establish the control chart to monitor the time between successive defectives of the process. In fact, time between successive defectives of high-quality processes usually follows exponential distribution. Due to the fact that the exponential distribution is highly skewed, some transformation techniques should be applied to help developing the control charts with unbiased in-control average run length (ARL). In this chapter, two different approaches have been applied using Weibull transformation and Cornish–Fisher expansion to develop unbiased ARL control charts in such a way that the probability of false alarm is at acceptable value.

Appendix

The moment generating the function of the exponential distribution is $\phi \left( t \right) = \frac{\lambda }{\lambda - t}\left( {\lambda > 0} \right)$, where $\lambda = {\raise0.7ex\hbox{$1$} \!\mathord{\left/ {\vphantom {1 \theta }}\right.\kern-0pt} \!\lower0.7ex\hbox{$\theta $}}$. So, the first four ordinary moments can be computed as $\mu_{1}^{\prime } = E\left( X \right) = \frac{1}{\lambda },\mu_{2}^{\prime } = E\left( {X^{2} } \right) = \frac{2}{{\lambda^{2} }},\mu_{3}^{\prime } = E\left( {X^{3} } \right) = \frac{6}{{\lambda^{3} }},\mu_{4}^{\prime } = E\left( {X^{4} } \right) = \frac{24}{{\lambda^{4} }}$.

Also, the central moments of the exponential distribution can be calculated as $\mu_{h} = E\left( {X - \mu } \right)^{h} = \int_{{D_{x} }} {\left( {x - \mu } \right)^{h} \cdot f\left( x \right) {\text{d}}x}$. And hence, $\mu_{1} = 0\,\mu_{2} = \mu_{2}^{\prime } - \mu^{2} = \sigma_{x}^{2} = \frac{1}{{\lambda^{2} }}, \mu_{3} = \mu_{3}^{\prime } - 3\mu \mu_{2}^{\prime } + 2\mu^{3} = \frac{2}{{\lambda^{3} }}, \mu_{4} = \mu_{4}^{\prime } - 4\mu \mu_{3}^{\prime } + 6\mu^{2} \mu_{2}^{\prime } - 3\mu^{4} = \frac{9}{{\lambda^{4} }}.$

The cumulants can then be obtained from the cumulants generating function (logarithm of the moments generating function) with the central moments to be standardized. The results are $K_{1} = \frac{\mu }{\sigma } = 1, K_{2} = \frac{{\mu_{2} }}{{\sigma^{2} }} = 1, K_{3} = \frac{{\mu_{3} }}{{\sigma^{3} }} = 2, K_{4} = \frac{{\mu_{4} }}{{\sigma^{4} }} - 3\left( {\frac{{\mu_{2} }}{{\sigma^{2} }}} \right)^{2} = 6.$