Fundamental Changes in Theory of Duncan’s Model for Economic Design of Control Charts in the Presence of Multiple Assignable Causes

Abstract

One of the most widely used tools for quality engineers in quickly detecting assignable causes and monitoring processes is control charts. Duncan (in J Am Stat Assoc 51(274):228–242, 1956), in order to improve product quality and reduce the economic costs of the quality cycle, presented the first economic design of the $\bar{X}$ control chart in the presence of multiple assignable causes. In his model and all the economic designs derived from it, it is assumed that after the occurrence of an assignable cause, no other assignable cause occurs until the correct alarm is issued. This assumption is unrealistic and impractical in production and service processes. Therefore, in this paper, we present a realistic and practical economic design in the presence of multiple assignable causes for the $\bar{X}$ control chart under the Weibull shock model in industry. The numerical results of our model show well that in the previous models, the average cost per unit time of the quality cycle is severely underestimated compared to the actual value. Therefore, it is suggested that in order to eliminate the shortcomings of the previous economic design in the presence of multiple assignable causes, in future research, they should be redesigned based on our proposed model.

Appendix

Proof of Theorem 1

• 1. From Eq. (2), $w_{j} = j^{\frac{1}{k}} h_{1}$ is obtained. According to Fig. 1, the random variable X represents the time elapsed from the beginning of the quality cycle to the issuance of the true alarm. In order to summarize, we define the steps of proving the following equations:

  $$ \lambda = \mathop \sum \limits_{i = 1}^{m} \lambda_{i} ,\quad \lambda_{ - i} = \lambda - \lambda_{i} $$

  (25)
  $$ \left\{ {{\mathbf{T}}_{{{\mathbf{ - i}}}} > X} \right\} = \{ T_{1} > X, \ldots ,T_{i - 1} > X, T_{i + 1} > X , \ldots ,T_{m} > X\} $$

  (26)

  Clearly with respect to Eq. (26) the occurrence of the cause of the ith documentable cause before the issuance of the true alarm and the absence of other attributable causes until the issuance of the true alarm ({I_i}) with the occurrence $\{ T_{i} \le X ,T_{ - i} > X\}$ is equivalent, so using the law we have the total probability:

  $$ \begin{aligned} P\left( {I_{i} } \right) & = P\left( {T_{i} \le X ,{\mathbf{T}}_{{{\mathbf{ - i}}}} > X} \right) \\ & = \mathop \sum \limits_{j = 1}^{\infty } P(T_{i} \le w_{j} ,{\mathbf{T}}_{{{\mathbf{ - i}}}} > w_{j} ,X = w_{j} ) \\ & = \mathop \sum \limits_{j = 1}^{\infty } \mathop \sum \limits_{l = 1}^{j} P\left( {w_{l - 1} < T_{i} \le w_{l} ,{\mathbf{T}}_{{{\mathbf{ - i}}}} > w_{j} ,X = w_{j} } \right) \\ \end{aligned} $$

  (27)

  According to the definition of variables as well as assumptions we have:

  $$ P\left( {X = w_{j} |w_{l - 1} < T_{i} \le w_{l} ,{\mathbf{T}}_{{{\mathbf{ - i}}}} > w_{j} } \right) = \beta_{i}^{j - l} (1 - \beta_{i} ) $$

  (28)
  $$ P\left( {w_{l - 1} < T_{i} \le w_{l} } \right) = \left( {1 - {\text{e}}^{{ - \lambda_{i} h^{k} }} } \right) \cdot {\text{e}}^{{ - \lambda_{i} \left( {l - 1} \right)h_{1}^{k} }} $$

  (29)
  $$ P\left( {{\mathbf{T}}_{{{\mathbf{ - i}}}} > w_{j} |w_{l - 1} < T_{i} \le w_{l} } \right) = {\text{e}}^{{ - \lambda_{ - i} jh_{1}^{k} }} $$

  (30)

  From Eqs. (28) to (30) we have:

  $$ \begin{aligned} P\left( {I_{i} } \right) & = \mathop \sum \limits_{j = 1}^{\infty } \mathop \sum \limits_{l = 1}^{j} \beta_{i}^{j - l} (1 - \beta_{i} )\left( {1 - {\text{e}}^{{ - \lambda_{i} h_{1}^{k} }} } \right).{\text{e}}^{{ - \lambda_{i} \left( {l - 1} \right)h_{1}^{k} }} {\text{e}}^{{ - \lambda_{ - i} jh_{1}^{k} }} \\ & = \mathop \sum \limits_{j = 1}^{\infty } (1 - \beta_{i} )(1 - {\text{e}}^{{ - \lambda_{i} h_{1}^{k} }} )\frac{{\left( {{\text{e}}^{{ - \lambda h_{1}^{k} }} } \right)^{j} - \left( {\beta_{i} {\text{e}}^{{ - \lambda_{ - i} h_{1}^{k} }} } \right)^{j} }}{{{\text{e}}^{{ - \lambda_{i} h_{1}^{k} }} - \beta_{i} }} \\ & = \frac{{\left( {1 - \beta_{i} } \right)\left( {1 - {\text{e}}^{{ - \lambda_{i} h_{1}^{k} }} } \right){\text{e}}^{{ - \lambda_{ - i} h_{1}^{k} }} }}{{(1 - e^{{ - \lambda h_{1}^{k} }} )(1 - \beta_{i} e^{{ - \lambda_{ - i} h_{1}^{k} }} )}} \\ \end{aligned} $$

  (31)

  Clearly $\left\{ {I = 1} \right\} = \bigcup\nolimits_{i = 1}^{m} {I_{i} }$ and since we have $I_{i} \cap I_{j} = \emptyset$ for every i ≠ j, therefore, $P\left( {I = 1} \right) = \sum\nolimits_{i = 1}^{m} P \left( {I_{i} } \right)$.

• 2. Considering that for $i = 1, 2, \ldots ,m$, $\lim\nolimits_{{h \to 0^{ + } }} P\left( {I_{i} } \right) = \frac{0}{0}$, using the hopital rule, it can be easily shown that $\lim\nolimits_{{h \to 0^{ + } }} P\left( {I_{i} } \right) = \lambda_{i} /\lambda$ and therefore $\lim\nolimits_{{h \to 0^{ + } }} P\left( I \right) = 1$.

• 3. Clearly for $i = 1, 2, \ldots ,m$, $\lim\nolimits_{L \to + \infty } \beta_{i} = 1$ resulting in $\lim\nolimits_{L \to + \infty } P\left( I \right) = 0$.

Proof of Lemma 1

Given that it is assumed that the time of taking and interpreting the sample is negligible, then the true alarm will occur in one of the sampling times, i.e. $w_{1}$, $w_{2}$, etc. Therefore, according to the Eqs. (28) to (30), we have

$$ \begin{aligned} E\left( {X|I = 1} \right) & = \sum\limits_{j = 1}^{\infty } {w_{j} } P\left( {X = w_{j} |I = 1} \right) = \sum\limits_{j = 1}^{\infty } {\sum\limits_{i = 1}^{m} {w_{j} } } \frac{{P(T_{i} \le w_{j} ,T_{ - i} > w_{j} ,X = w_{j} )}}{P(I = 1)} \\ & = \sum\limits_{i = 1}^{m} {\sum\limits_{j = 1}^{\infty } {\left[ {w_{j} \frac{{(1 - \beta_{i} )(1 - {\text{e}}^{{ - \lambda_{i} h_{1}^{k} }} )}}{P(I = 1)}\frac{{\left( {{\text{e}}^{{ - \lambda h_{1}^{k} }} } \right)^{j} - \left( {\beta_{i} {\text{e}}^{{ - \lambda_{ - i} h_{1}^{k} }} } \right)^{j} }}{{{\text{e}}^{{ - \lambda_{i} h_{1}^{k} }} - \beta_{i} }}} \right]} } . \\ \end{aligned} $$

Proof of Lemma 2

Clearly if $T = \min (T_{1} , \ldots ,T_{m} )$ then $X_{1} = T$ and we have:

$$ E\left( {X_{1} |I = 1} \right) = \mathop \int \limits_{0}^{\infty } tf\left( {t|I = 1} \right){\text{d}}t = \mathop \sum \limits_{j = 1}^{\infty } \mathop \int \limits_{{w_{j - 1} }}^{{w_{j} }} tf\left( {t|I = 1} \right){\text{d}}x $$

Therefore, it suffices to obtain $f(t|I = 1)$ for $w_{j - 1} < t < w_{j}$ as follows:

$$ \begin{aligned} f\left( {t|I = 1} \right){\text{d}}t & \approx P\left( {t < T < t + {\text{d}}t|I = 1} \right) \\ & = \mathop \sum \limits_{l = 1}^{\infty } \frac{{P\left( {t < T < t + {\text{d}}t,I = 1,X = w_{l} } \right)}}{{P\left( {I = 1} \right)}} \\ & = \mathop \sum \limits_{l = j}^{\infty } \frac{{P\left( {t < T < t + {\text{d}}t,I = 1,X = w_{l} } \right)}}{{P\left( {I = 1} \right)}} \\ & = \mathop \sum \limits_{l = j}^{\infty } \frac{{P\left( {t < T < t + {\text{d}}t,\bigcup\nolimits_{i = 1}^{m} {\left\{ {T_{i} < w_{l} ,T_{ - i} > w_{l} ,X = w_{l} } \right\}} } \right)}}{{P\left( {I = 1} \right)}} \\ & = \mathop \sum \limits_{l = j}^{\infty } \mathop \sum \limits_{i = 1}^{m} \frac{{P\left( {t < T_{i} < t + {\text{d}}t,T_{i} < w_{l} ,T_{ - i} > w_{l} ,X = w_{l} } \right)}}{{P\left( {I = 1} \right)}} \\ & = \mathop \sum \limits_{i = 1}^{m} \mathop \sum \limits_{l = j}^{\infty } \frac{{\beta_{i}^{l - j} \left( {1 - \beta_{i} } \right) P\left( {t < T_{i} < t + {\text{d}}t} \right)P\left( {T_{ - i} > w_{l} } \right)}}{{P\left( {I = 1} \right)}} \\ \end{aligned} $$

Now if we have ${\text{d}}t\, \to \,0$ then,

$$ \begin{aligned} f\left( {t|I = 1} \right) & = \mathop \sum \limits_{i = 1}^{m} \mathop \sum \limits_{l = j}^{\infty } \frac{{\beta_{i}^{l - j} \left( {1 - \beta_{i} } \right) f_{{T_{i} }} \left( t \right)P\left( {T_{ - i} > w_{l} } \right)}}{{P\left( {I = 1} \right)}} \\ & = \mathop \sum \limits_{i = 1}^{m} \mathop \sum \limits_{l = j}^{\infty } \frac{{\beta_{i}^{l - j} \left( {1 - \beta_{i} } \right) {\text{e}}^{{ - \lambda_{ - i} lh_{1}^{k} }} f_{{T_{i} }} \left( t \right)}}{{P\left( {I = 1} \right)}} \\ & = \mathop \sum \limits_{i = 1}^{m} \left[ {\frac{1}{P(I = 1)}.\frac{{\left( {1 - \beta_{i} } \right) {\text{e}}^{{ - \lambda_{ - i} jh_{1}^{k} }} f_{{T_{i} }} (t)}}{{1 - \beta_{i} {\text{e}}^{{ - \lambda_{ - i} h_{1}^{k} }} }}} \right]. \\ \end{aligned} $$

Therefore,

$$ \begin{aligned} E\left( {X_{1} |I = 1} \right) & = \mathop \sum \limits_{j = 1}^{\infty } \mathop \int \limits_{{w_{j - 1} }}^{{w_{j} }} tf\left( {t|I = 1} \right){\text{d}}t \\ & = \mathop \sum \limits_{i = 1}^{m} \mathop \sum \limits_{j = 1}^{\infty } \left[ {\frac{{1 - \beta_{i} }}{{P\left( {I = 1} \right)}} \times \frac{{{\text{e}}^{{ - \lambda_{ - i} jh^{k} }} }}{{1 - \beta_{i} {\text{e}}^{{ - \lambda_{ - i} h^{k} }} }} \times \mathop \int \limits_{{w_{j - 1} }}^{{w_{j} }} t.k\lambda_{i} t^{k - 1} {\text{e}}^{{ - \lambda_{i} t^{k} }} {\text{d}}t} \right] \\ & = \mathop \sum \limits_{i = 1}^{m} \mathop \sum \limits_{j = 1}^{\infty } \left[ {\frac{{1 - \beta_{i} }}{P(I = 1)} \times \frac{{{\text{e}}^{{ - \lambda_{ - i} jh_{1}^{k} }} }}{{1 - \beta_{i} {\text{e}}^{{ - \lambda_{ - i} h_{1}^{k} }} }} \times \frac{{\Gamma \left( {\frac{k + 1}{k},\lambda_{i} (j - 1)h_{1}^{k} } \right) - \Gamma \left( {\frac{k + 1}{k},\lambda_{i} jh_{1}^{k} } \right)}}{{\lambda_{i}^{1/k} }}} \right], \\ \end{aligned} $$

where the above relation $\Gamma \left( {s,x} \right) = \mathop \int \nolimits_{x}^{\infty } t^{s - 1} {\text{e}}^{ - t} {\text{d}}t$ and $f_{{T_{i} }} \left( t \right) = k\lambda_{i} t^{k - 1} {\text{e}}^{{ - \lambda_{i} t^{k} }}$.

Proof of Lemma 3

If $Z_{2i}$ is the correction time of the ith attributable cause, then the average process correction time provided that only one attributable cause occurs until the true alarm is issued after identifying the attributable cause ($E(X_{4} |I = 1)$) Is equal to:

$$ \begin{aligned} E\left( {X_{4} |I = 1} \right) & = \mathop \sum \limits_{i = 1}^{m} Z_{2i} P\left( {X_{4} = Z_{2i} |I = 1} \right) \\ & = \mathop \sum \limits_{i = 1}^{m} \mathop \sum \limits_{j = 1}^{\infty } Z_{2i} \frac{{P(T_{i} \le w_{j} ,T_{ - i} > w_{j} ,X = w_{j} )}}{P(I = 1)} \\ & = \mathop \sum \limits_{j = 1}^{\infty } \mathop \sum \limits_{i = 1}^{m} Z_{2i} \frac{{(1 - \beta_{i} )(1 - {\text{e}}^{{ - \lambda_{i} h_{1}^{k} }} )}}{P(I = 1)}\frac{{\left( {{\text{e}}^{{ - \lambda h_{1}^{k} }} } \right)^{j} - \left( {\beta_{i} {\text{e}}^{{ - \lambda_{ - i} h_{1}^{k} }} } \right)^{j} }}{{{\text{e}}^{{ - \lambda_{i} h_{1}^{k} }} - \beta_{i} }} \\ & = \mathop \sum \limits_{i = 1}^{m} Z_{2i} \left[ {\frac{{(1 - \beta_{i} )(1 - {\text{e}}^{{ - \lambda_{i} h_{1}^{k} }} )}}{{P(I = 1) ({\text{e}}^{{ - \lambda_{i} h_{1}^{k} }} - \beta_{i} )}} \times \left( {\frac{{{\text{e}}^{{ - \lambda h_{1}^{k} }} }}{{1 - {\text{e}}^{{ - \lambda h_{1}^{k} }} }} - \frac{{\beta_{i} {\text{e}}^{{ - \lambda_{ - i} h_{1}^{k} }} }}{{1 - \beta_{i} {\text{e}}^{{ - \lambda_{ - i} h_{1}^{k} }} }}} \right)} \right] \\ & = \mathop \sum \limits_{i = 1}^{m} \left[ {Z_{2i} \frac{{P(I_{i} )}}{P(I = 1)}} \right]. \\ \end{aligned} $$

Proof of Lemma 4

If $N_{{{\text{TST}}}}$ is equal to the number of samples taken from the process to the true alarm and $N_{{{\text{INC}}}}$ is equal to the number of samples taken from the process in-control state, it is clearly $N_{{{\text{OC}}}} = N_{{{\text{TST}}}} - N_{{{\text{INC}}}}$. Therefore, $E\left( {N_{{{\text{OC}}}} |I = 1} \right) = E\left( {N_{{{\text{TST}}}} |I = 1} \right) - E(N_{{{\text{INC}}}} |I = 1)$, where,

$$ \begin{aligned} E\left( {N_{{{\text{TST}}}} |I = 1} \right) & = \mathop \sum \limits_{j = 1}^{\infty } j .\frac{{P(N_{{{\text{TST}}}} = j,I = 1)}}{P(I = 1)} \\ & = \mathop \sum \limits_{j = 1}^{\infty } j .\frac{{P\left( {\bigcup\nolimits_{i = 1}^{m} {\left\{ {T_{i} \le w_{j} ,T_{ - i} > w_{j} ,X = w_{j} } \right\}} } \right)}}{P(I = 1)} \\ & = \mathop \sum \limits_{j = 1}^{\infty } \mathop \sum \limits_{i = 1}^{m} j .\frac{{P(T_{i} \le w_{j} ,T_{ - i} > w_{j} ,X = w_{j} )}}{P(I = 1)} \\ & = \mathop \sum \limits_{j = 1}^{\infty } \mathop \sum \limits_{i = 1}^{m} j .\frac{{(1 - \beta_{i} )(1 - {\text{e}}^{{ - \lambda_{i} h_{1}^{k} }} )}}{P(I = 1)}\frac{{\left( {{\text{e}}^{{ - \lambda h_{1}^{k} }} } \right)^{j} - \left( {\beta_{i} {\text{e}}^{{ - \lambda_{ - i} h_{1}^{k} }} } \right)^{j} }}{{{\text{e}}^{{ - \lambda_{i} h_{1}^{k} }} - \beta_{i} }}. \\ & = \mathop \sum \limits_{i = 1}^{m} \left[ {\frac{{(1 - \beta_{i} )(1 - {\text{e}}^{{ - \lambda_{i} h_{1}^{k} }} )}}{{P(I = 1) ({\text{e}}^{{ - \lambda_{i} h_{1}^{k} }} - \beta_{i} )}}\left[ {\frac{{{\text{e}}^{{ - \lambda h_{1}^{k} }} }}{{\left( {1 - {\text{e}}^{{ - \lambda h_{1}^{k} }} } \right)^{2} }} - \frac{{\beta_{i} {\text{e}}^{{ - \lambda_{ - i} h_{1}^{k} }} }}{{\left( {1 - \beta_{i} {\text{e}}^{{ - \lambda_{ - i} h_{1}^{k} }} } \right)^{2} }}} \right]} \right] \\ & = \mathop \sum \limits_{i = 1}^{m} \left[ {\frac{{P(I_{i} )}}{P(I) }\left[ {\frac{{{\text{e}}^{{ - \lambda h_{1}^{k} }} }}{{1 - {\text{e}}^{{ - \lambda h_{1}^{k} }} }} + \frac{1}{{1 - \beta_{i} {\text{e}}^{{ - \lambda_{ - i} h_{1}^{k} }} }}} \right]} \right], \\ \end{aligned} $$

and

$$ \begin{aligned} E(N_{{{\text{INC}}}} |I = 1) & = \mathop \sum \limits_{j = 1}^{\infty } j .\frac{{P(N_{{{\text{INC}}}} = j,I = 1)}}{P(I = 1)} \\ & = \mathop \sum \limits_{j = 1}^{\infty } \mathop \sum \limits_{l = 1}^{\infty } j .\frac{{P\left( {w_{j} < T < w_{j + 1} ,\bigcup\nolimits_{i = 1}^{m} {\left\{ {T_{i} \le w_{l} ,T_{ - i} > w_{l} ,X = w_{l} } \right\}} } \right)}}{P(I = 1)} \\ & = \mathop \sum \limits_{j = 1}^{\infty } \mathop \sum \limits_{l = j + 1}^{\infty } j .\frac{{P\left( {\bigcup\nolimits_{i = 1}^{m} {\left\{ {w_{j} < T_{i} < w_{j + 1} ,T_{i} \le w_{l} ,T_{ - i} > w_{l} ,X = w_{l} } \right\}} } \right)}}{P(I = 1)} \\ & = \mathop \sum \limits_{i = 1}^{m} \mathop \sum \limits_{j = 1}^{\infty } \mathop \sum \limits_{l = j + 1}^{\infty } j .\frac{{P(w_{j} < T_{i} < w_{j + 1} ,T_{ - i} > w_{l} ,X = w_{l} )}}{P(I = 1)} \\ & = \mathop \sum \limits_{i = 1}^{m} \mathop \sum \limits_{j = 1}^{\infty } \mathop \sum \limits_{l = j + 1}^{\infty } j .\left( {1 - \beta_{i} } \right)\beta_{i}^{l - j - 1} \frac{{P\left( {w_{j} < T_{i} < w_{j + 1} } \right)P(T_{ - i} > w_{l} )}}{P(I = 1)} \\ & = \mathop \sum \limits_{i = 1}^{m} \left[ {\frac{{\left( {1 - \beta_{i} } \right)\left( {1 - {\text{e}}^{{ - \lambda_{i} h_{1}^{k} }} } \right){\text{e}}^{{ - \lambda_{ - i} h_{1}^{k} }} }}{{P\left( {I = 1} \right) (1 - \beta_{i} {\text{e}}^{{ - \lambda_{ - i} h_{1}^{k} }} )}} \times \frac{{{\text{e}}^{{ - \lambda h_{1}^{k} }} }}{{\left( {1 - {\text{e}}^{{ - \lambda h_{1}^{k} }} } \right)^{2} }}} \right] \\ & = \frac{{{\text{e}}^{{ - \lambda h_{1}^{k} }} }}{{1 - {\text{e}}^{{ - \lambda h_{1}^{k} }} }}. \\ \end{aligned} $$