Economic design of two-stage control charts with skewed and dependent measurements

Abstract

In many instances, the cost is high to monitor primary quality characteristic called performance variable, but it could be more economical to monitor its surrogate. To cover asymmetric processes in an alternating fashion of two-stage charting design using either performance variable or surrogate variable, both process variables are modeled by a skew normal distribution, respectively. The proposed two-stage control charts are constructed with an economic viewpoint using Markov chain approach. Two algorithms are provided to implement the proposed charting method. The application of the proposed charting method and its advantages over the existing methods are presented through an illustrating example.

Appendices

Appendices

1.1 Appendix 1: the proof of Theorem 3

• (a) Since X _τ , X _τ + 1, ⋯, are iid SN(ξ ₃, σ ₂, λ ₁) performance variables in an out-of-control state, it can be shown that the process mean of the performance variables is μ ^X₁  = ξ ₃ + μ ^X₀ and the process standard deviation is σ ^X₀ . Using Eqs. (4) and (5), we can show that

  $$ \begin{array}{c}\hfill {P}_r\left({\mathrm{LCL}}_{\overline{X}}<\overline{X}<{\mathrm{UCL}}_{\overline{X}}\Big|{\mu}^X={\mu}_1^X,{\sigma}^X={\sigma}_0^X\right)\hfill \\ {}\hfill ={P}_r\left({\mathrm{LCL}}_{\overline{X}}<{\mu}_1^X+\left({\gamma}_1{\displaystyle {\overline{Z}}_X}-{\gamma}_2\right){\sigma}_0^X<{\mathrm{UCL}}_{\overline{X}}\right)\hfill \\ {}\hfill ={P}_r\left(-\frac{\xi_3}{\gamma_1{\sigma}_0^X}+{\displaystyle {\overline{z}}_{\lambda_1,{n}_X,p/2}}<{\displaystyle {\overline{Z}}_X}\right.\kern0.5em \left.<-\frac{\xi_3}{\gamma_1{\sigma}_0^X}+{\displaystyle {\overline{z}}_{\lambda_1,{n}_X,1-p/2}}\right).\hfill \end{array} $$

  (21)

• (b) Replacing $\left({\mathrm{LCL}}_{\overline{X}},{\mathrm{UCL}}_{\overline{X}}\right)$ by $\left({\mathrm{LWL}}_{\overline{X}},{\mathrm{UWL}}_{\overline{X}}\right)$ and p by p ₂, respectively. Theorem 3(b) can be proved using an analog proof process to that of (a).

1.2 Appendix 2: The transition probabilities p _ij in P

$$ \begin{array}{c}\hfill {p}_{11}= \Pr \left\{\left.\overline{Y}\kern0.5em \in {I}_Y\right|{\mu}^Y={\mu}_0^Y,{\sigma}^Y={\sigma}_0^Y\right\}{e}^{-\theta hy}\hfill \\ {}\hfill {p}_{12}=\left[1- \Pr \left\{\left.\overline{Y}\kern0.5em \in {I}_Y\right|{\mu}^Y={\mu}_0^Y,{\sigma}^Y={\sigma}_0^Y\right\}\right]{e}^{-\theta {h}_X}\hfill \\ {}\hfill {p}_{13}= \Pr \left\{\left.\overline{Y}\in {I}_Y\right|{\mu}^Y={\mu}_0^Y,{\sigma}^Y={\sigma}_0^Y\right\}\left(1-{e}^{-\theta hy}\right)\hfill \\ {}\hfill {p}_{14}=\left[1-{P}_r\left\{\left.\overline{Y}\in {I}_Y\right|{\mu}^Y={\mu}_0^Y,{\sigma}^Y={\sigma}_0^Y\right\}\right]\times \left(1-{e}^{-\theta {h}_X}\right)\hfill \\ {}\hfill {p}_{21}={P}_r\left\{\left.\overline{X}\kern0.5em \in {I}_{X_W}\right|\overline{X}\in {I}_{X_A},{\mu}^X={\mu}_0^X,{\sigma}^X={\sigma}_0^X\right\}\times {e}^{-\theta {h}_Y}\hfill \\ {}\hfill {p}_{22}={P}_r\left\{\left.\overline{X}\in {I}_{X_{WA}}\right|\overline{X}\in {I}_{X_A},{\mu}^X={\mu}_0^X,{\sigma}^X={\sigma}_0^X\right\}\times {e}^{-\theta {h}_X}\hfill \\ {}\hfill {p}_{23}={P}_r\left\{\left.\overline{X}\in {I}_{X_W}\right|\overline{X}\in {I}_{X_A},{\mu}^X={\mu}_0^X,{\sigma}^X={\sigma}_0^X\right\}\times \left(1-{e}^{-\theta {h}_Y}\right)\hfill \\ {}\hfill {p}_{24}={P}_r\left\{\left.\overline{X}\in {I}_{X_{WA}}\right|\overline{X}\in {I}_{X_A},{\mu}^X={\mu}_0^X,{\sigma}^X={\sigma}_0^X\right\}\times \left(1-{e}^{-\theta {h}_X}\right)\hfill \\ {}\hfill {p}_{33}={P}_r\left\{\left.\overline{Y}\in {I}_Y\right|{\mu}^Y={\mu}^Y,{\sigma}^Y={\sigma}_0^Y\right\}\hfill \\ {}\hfill {p}_{34}=1-{p}_{33}\hfill \\ {}\hfill {p}_{43}={P}_r\left\{\left.\overline{X}\in {I}_{Xw}\right|\overline{X}\in {I}_{X_A},{\mu}^X={\mu}_1^X,{\sigma}^X={\sigma}_0^X\right\}\hfill \\ {}\hfill {p}_{44}={P}_r\left\{\overline{X}\in \left.{I}_{X_{WA}}\right|\overline{X}\in {I}_{X_A},{\mu}^X={\mu}_1^X,{\sigma}^X={\sigma}_0^X\right\}\hfill \\ {}\hfill {p}_{45}=1-{P}_r\left\{\left.\overline{X}\in {I}_{X_A}\right|{\mu}^X={\mu}_1^X,{\sigma}^X={\sigma}_0^X\right\}\hfill \\ {}\hfill {p}_{55}=1\hfill \end{array} $$