New Results for Two-Sided CUSUM-Shewhart Control Charts

Abstract

Already Yashchin (IBM J Res Dev 29(4):377–391, 1985), and of course Lucas (J Qual Technol 14(2):51–59, 1982) 3 years earlier, studied CUSUM chart supplemented by Shewhart limits. Interestingly, Yashchin proposed to calibrate the detecting scheme via P _∞ (RL > K) ≥ 1 − α for the run length (stopping time) RL in the in-control case. Calculating the RL distribution or related quantities such as the ARL (Average Run Length) are slightly complicated numerical tasks. Similarly to Capizzi and Masarotto (Stat Comput 20(1):23–33, 2010) who utilized Clenshaw-Curtis quadrature to tackle the ARL integral equation, we deploy less common numerical techniques such as collocation to determine the ARL. Note that the two-sided CUSUM chart consisting of two one-sided charts leads to a more demanding numerical problem than the single two-sided EWMA chart.

Appendices

Appendix 1: Collocation Design for More Than r = 2 Intervals

Here we provide the generalization from r = 2, dealt with in Sect. 2, to general r ∈{2, 3, …}. To start with the first interval, 0 ≤ s ≤ h − ε, we simply state that the shape of the collocation design does not change for greater r. For the succeeding intervals, h − (r − m + 1)ε < s ≤ h − (r − m)ε with m = 2, …, r − 1, the following structure is utilized.

$$\displaystyle \begin{aligned} \sum_{j=1}^N c_{mj} T_{mj} (s) & = 1 + \Phi(k-s) \mathcal{L}(0) \\ & \quad + \sum_{j=1}^N c_{1j} \int_0^{h -(r-1)\varepsilon} \varphi(z+k-s) T_{1j}(z) \, dz \\ & \quad + \sum_{t=2}^m \sum_{j=1}^N c_{tj} \int_{h -(r-t+1)\varepsilon}^{h -(r-t)\varepsilon} \varphi(z+k-s) T_{tj}(z) \, dz \\ & \quad + \sum_{j=1}^N c_{m+1,j} \int_{h -(r-m+1)\varepsilon}^{\varepsilon+s} \varphi(z+k-s) T_{m+1,j}(z) \, dz \,. \end{aligned} $$

Note that these equations are not present for r = 2. However, the last interval, h − ε < s ≤ h, is considered for all r ≥ 2. The general structure of the corresponding collocation equation is similar to the above one (now with m = r) except for the upper limit of the last integral where ε + s has to be replaced by h.

Appendix 2: Two-Sided CUSUM Chart

Starting with case (i), s ⁺ + s ⁻≤ 2k, and re-writing the corresponding integral equation results in:

$$\displaystyle \begin{aligned} \mathcal{L}(s^+,s^-) & = \Phi(k - s^+) \mathcal{L}(0,0) + \int_{k - s^+}^{h + k - s^+} \varphi(x) \mathcal{L}(s^+ + x-k, 0) \,dx \\ & \, + \Phi(k - s^-) \mathcal{L}(0,0) + \int_{k - s^-}^{h + k - s^-} \varphi(-x) \mathcal{L}(0, s^- + x-k) \,dx \\ & \, + 1 - \mathcal{L}(0,0) \,. \end{aligned} $$

Setting s ⁺ or s ⁻ to zero in (6) yields:

$$\displaystyle \begin{aligned} \mathcal{L}(0,0) & = \frac{\mathcal{L}^+(0)\mathcal{L}^-(0)}{\mathcal{L}^+(0) + \mathcal{L}^-(0)} \,, \\ \mathcal{L}(s^+,0) & = \frac{\mathcal{L}^+(s^+)\mathcal{L}^-(0)}{\mathcal{L}^+(0) + \mathcal{L}^-(0)} \;,\;\; \mathcal{L}(0,s^-) = \frac{\mathcal{L}^+(0)\mathcal{L}^-(s^-)}{\mathcal{L}^+(0) + \mathcal{L}^-(0)} \end{aligned} $$

Using this, the first line of the integral equation’s right-hand side changes to

$$\displaystyle \begin{aligned} \frac{\mathcal{L}^-(0)}{\mathcal{L}^+(0) + \mathcal{L}^-(0)} \left( \Phi(k - s^+)\mathcal{L}^+(0) + \int_{k - s^+}^{h + k - s^+} \varphi(x) \mathcal{L}^+(s^+ + x-k) \,dx \right) \end{aligned}$$

Substituting x = z + k − s in (2) while replacing the upper limit by h results in

$$\displaystyle \begin{aligned} \mathcal{L}(s) - 1 = \Phi(k-s) \mathcal{L}(0) + \int_{k-s}^{h+k-s} \varphi(x) \mathcal{L}(s+x-k) \, dx \,, {} \end{aligned} $$

(7)

so that the line under analysis simplifies heavily to

$$\displaystyle \begin{aligned} \frac{\mathcal{L}^-(0) (\mathcal{L}^+(s^+) - 1)}{\mathcal{L}^+(0) + \mathcal{L}^-(0)} \,. \end{aligned}$$

In a similar way we treat the second line ending in

$$\displaystyle \begin{aligned} \frac{\mathcal{L}^+(0) (\mathcal{L}^-(s^-) - 1)}{\mathcal{L}^+(0) + \mathcal{L}^-(0)} \,. \end{aligned}$$

For the second line we made use of φ(−x) = φ(x) in the in-control case (δ = 0), while for δ≠0 we have to change the sign of δ, hence φ _δ (−x) = φ _−δ(x). All together resembles (the “1” consumes the disturbing parts of the above two ratios)

$$\displaystyle \begin{aligned} \frac{\mathcal{L}^+(s^+) \mathcal{L}^-(0) + \mathcal{L}^-(s^-) \mathcal{L}^+(0) - \mathcal{L}^+(0)\mathcal{L}^-(0)}{\mathcal{L}^+(0) + \mathcal{L}^-(0)} \quad \text{ which confirms (\ref{eq:L2L1}).} \end{aligned}$$

Turning to case (ii), 2k < s ⁺ + s ⁻≤ h + 2k, we recall the shape of the related integral equation:

$$\displaystyle \begin{aligned} \mathcal{L}(s^+,s^-) & = 1 \,+ \int_{s^- - k}^{h + k - s^+} \varphi(x) \mathcal{L}(s^+ + x-k, 0) \,dx \\ & \qquad + \int_{k - s^+}^{s^- - k} \varphi(x) \mathcal{L}(s^+ + x-k, s^- - x-k) \, dx \\ & \qquad + \int_{-h - k + s^-}^{k - s^+} \varphi(x) \mathcal{L}(0, s^- - x-k) \,dx \,. \end{aligned} $$

First we plug (6) into and transform the second line

$$\displaystyle \begin{aligned} & \int_{k - s^+}^{s^- - k} \varphi(x) \mathcal{L}(s^+ + x-k, s^- - x-k) \, dx \\ &\quad = \frac{\mathcal{L}^-(0)}{\mathcal{L}^+(0) + \mathcal{L}^-(0)} \int_{k - s^+}^{s^- - k} \varphi(x) \mathcal{L}^+(s^+ + x-k) \, dx \\ & \quad + \frac{\mathcal{L}^+(0)}{\mathcal{L}^+(0) + \mathcal{L}^-(0)} \int_{k - s^+}^{s^- - k} \varphi(x) \mathcal{L}^-(s^- - x-k) \, dx \\ & \quad - \frac{\mathcal{L}^-(0) \mathcal{L}^+(0)}{\mathcal{L}^+(0) + \mathcal{L}^-(0)} \int_{k - s^+}^{s^- - k} \varphi(x) \, dx \,, \end{aligned} $$

with the last integral subsequently reduced to Φ(s ⁻− k) − Φ(k − s ⁺). We rewrite the first line as for case (i) and merge, borrowing $\mathcal {L}^-(0) / \big (\mathcal {L}^+(0) + \mathcal {L}^-(0)\big )$,

$$\displaystyle \begin{aligned} & \frac{\mathcal{L}^-(0)}{\mathcal{L}^+(0) + \mathcal{L}^-(0)} \\ & + \frac{\mathcal{L}^-(0)}{\mathcal{L}^+(0) + \mathcal{L}^-(0)} \Phi(k - s^+) \mathcal{L}^+(0) \\ & + \frac{\mathcal{L}^-(0)}{\mathcal{L}^+(0) + \mathcal{L}^-(0)} \int_{k - s^+}^{s^- - k} \varphi(x) \mathcal{L}^+(s^+ + x-k) \, dx \\ & + \frac{\mathcal{L}^-(0)}{\mathcal{L}^+(0) + \mathcal{L}^-(0)} \int_{s^- - k}^{h + k - s^+} \varphi(x) \mathcal{L}^+(s^+ + x-k) \,dx \end{aligned} $$

to get

$$\displaystyle \begin{aligned} \frac{\mathcal{L}^+(s^+)\mathcal{L}^-(0)}{\mathcal{L}^+(0) + \mathcal{L}^-(0)} \end{aligned}$$

by applying again (7). Exploiting Φ(s ⁻− k) = 1 − Φ(k − s ⁻) we proceed in a similar way with the third line by collecting after transforming both integrals as in the first case

$$\displaystyle \begin{aligned} & \frac{\mathcal{L}^+(0)}{\mathcal{L}^+(0) + \mathcal{L}^-(0)} \\ & + \frac{\mathcal{L}^+(0)}{\mathcal{L}^+(0) + \mathcal{L}^-(0)} \Phi(k - s^-) \mathcal{L}^-(0) \\\noalign{} & + \frac{\mathcal{L}^+(0)}{\mathcal{L}^+(0) + \mathcal{L}^-(0)} \int_{s^- - k}^{k - s^+} \varphi(x) \mathcal{L}^-(s^- + x-k) \, dx \\ & + \frac{\mathcal{L}^+(0)}{\mathcal{L}^+(0) + \mathcal{L}^-(0)} \int_{k - s^+}^{h + k - s^-} \varphi(x) \mathcal{L}^-(s^- + x-k) \,dx \end{aligned} $$

which results in

$$\displaystyle \begin{aligned} \frac{\mathcal{L}^-(s^-)\mathcal{L}^+(0)}{\mathcal{L}^+(0) + \mathcal{L}^-(0)} \,. \end{aligned}$$

The two “borrowed” terms

$$\displaystyle \begin{aligned} -\frac{\mathcal{L}^-(0)}{\mathcal{L}^+(0) + \mathcal{L}^-(0)} - \frac{\mathcal{L}^+(0)}{\mathcal{L}^+(0) + \mathcal{L}^-(0)} = -1 \end{aligned} $$

are compensated with the 1 on the right-hand side of the original equation. The last remaining term forms together with the two others

$$\displaystyle \begin{aligned} \frac{\mathcal{L}^+(s^+) \mathcal{L}^-(0) + \mathcal{L}^-(s^-) \mathcal{L}^+(0) - \mathcal{L}^+(0)\mathcal{L}^-(0)}{\mathcal{L}^+(0) + \mathcal{L}^-(0)} \,. \end{aligned}$$

This completes the proof.