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A two-stage intervened decision system with multi-state decision units and dynamic system configuration

  • S.I. : Reliability Modeling with Applications Based on Big Data
  • Published:
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Abstract

This paper develops the performability and cost–benefit models for a two-stage intervened decision system with majority voting rule and binary input and output. The decision process of the system contains two stages: an inspection stage (stage 1) and a result submission stage (stage 2). During the first stage, each decision unit in the system will have multiple states and a supervisor will come to visit each unit and check its state for at most twice. The supervisor will conduct the first visit to each unit for certain. However, the behavior of the second visit to each unit will be determined by its state during the first visit. In addition, each decision unit may be removed from the system given certain states during each visit. Therefore the structure of system may change during the decision process. The units which are not removed during the first stage can submit the result at any time during the second stage. However, the performance of each remaining unit will be determined by the ending state of the first stage. Moreover, in order to improve the efficiency of the decision process, a check point is added to the second stage. The performability and cost–benefit models for this dynamic system are developed by considering the distribution of states at the end of the first stage. A three-step method will be proposed for model optimization. Some numerical examples for the three-step method will be presented. The proposed intervened decision system in this paper can be applied in many contexts such as financial investment, paper submission review and proposal evaluation, credit evaluation and loan application and product release and recall.

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Acknowledgements

We would like to appreciate the guest editor and anonymous reviewers for their valuable suggestions and thorough review, which have significantly improved this paper.

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Authors and Affiliations

  1. Department of Industrial and Systems Engineering, Rutgers University, Piscataway, NJ, 08854, USA

    Tingnan Lin & Hoang Pham

Authors
  1. Tingnan Lin
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  2. Hoang Pham
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Correspondence to Tingnan Lin.

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Appendix

Appendix

The distribution of states at the end of stage 1 can be calculated following the procedure below:

Scenario 1 \( T_{{I_{1} }} \le \frac{1}{2}t_{0} \), in which major mistakes can be fixed during the first visit.

Define \( P_{H} \triangleq P\left( {T_{{I_{1} }} \le \frac{1}{2}t_{0} } \right) = H_{1} \left( {\frac{1}{2}t_{0} } \right) = \mathop \int \limits_{0}^{{\frac{1}{2}t_{0} }} h_{1} \left( {y_{1} } \right)dy_{1} \).

Given that \( T_{{I_{1} }} = y_{1} \in \left( {0,\frac{1}{2}t_{0} } \right] \),

$$ \begin{aligned} S_{{0|y_{1} ,H}} & \triangleq P\left( {X\left( {T_{{I_{1} }} } \right) = 0 |T_{{I_{1} }} \le \frac{1}{2}t_{0} ,T_{{I_{1} }} = y_{1} } \right) = e^{{ - \mathop \int \limits_{0}^{{y_{1} }} \lambda \left( t \right)dt}} \\ S_{{i|y_{1} ,H}} & \triangleq P\left( {X\left( {T_{{I_{1} }} } \right) = i |T_{{I_{1} }} \le \frac{1}{2}t_{0} ,T_{{I_{1} }} = y_{1} } \right) \\ & = \alpha_{i} \left( {1 - e^{{ - \mathop \int \limits_{0}^{{y_{1} }} \lambda \left( t \right)dt}} } \right)\quad \left( {i = 1, 2, 3} \right). \\ \end{aligned} $$

(1-1) If \( X\left( {T_{{I_{1} }} } \right) = 3 \), the agent has made critical mistakes and it will be removed from the system for sure. In this way,

$$ \begin{aligned} P_{{34|y_{1} ,H}} & \triangleq P\left( {X\left( {t_{0} } \right) = 4 |T_{{I_{1} }} \le \frac{1}{2}t_{0} ,T_{{I_{1} }} = y_{1} ,X\left( {T_{{I_{1} }} } \right) = 3} \right) = 1 \\ P_{{3k|y_{1} ,H}} & \triangleq P\left( {X\left( {t_{0} } \right) = k |T_{{I_{1} }} \le \frac{1}{2}t_{0} ,T_{{I_{1} }} = y_{1} ,X\left( {T_{{I_{1} }} } \right) = 3} \right) = 0 \quad \left( {k = 1, 2, 3} \right). \\ \end{aligned} $$

(1-2) If \( X\left( {T_{{I_{1} }} } \right) = 2 \), major mistakes are found during the first visit and the supervisor will come for the second visit for sure after the problems are fixed.

Given that \( T_{{I_{2} }} = y_{2} \in \left( {y_{1} ,t_{0} } \right) \),

$$ \begin{aligned} S_{{20|y_{1} y_{2} ,H}} & \triangleq P\left( {X\left( {T_{{I_{2} }} } \right) = 0 |T_{{I_{1} }} \le \frac{1}{2}t_{0} ,T_{{I_{1} }} = y_{1} ,X\left( {T_{{I_{1} }} } \right) = 2,T_{{I_{2} }} = y_{2} } \right) = e^{{ - \mathop \int \limits_{{y_{1} }}^{{y_{2} }} \lambda_{2} \left( t \right)dt}} \\ S_{{2j|y_{1} y_{2} ,H}} & \triangleq P\left( {X\left( {T_{{I_{2} }} } \right) = j |T_{{I_{1} }} \le \frac{1}{2}t_{0} ,T_{{I_{1} }} = y_{1} ,X\left( {T_{{I_{1} }} } \right) = 2,T_{{I_{2} }} = y_{2} } \right) \\ & = \alpha_{j} \left( {1 - e^{{ - \mathop \int \limits_{{y_{1} }}^{{y_{2} }} \lambda_{2} \left( t \right)dt}} } \right) \quad\left( {j = 1, 2, 3} \right). \\ \end{aligned} $$

If the result of the second visit is 0, the state at \( t_{0} \) can be 0, 1, 2 or 3:

$$ \begin{aligned} P_{{200|y_{1} y_{2} ,H}} & \triangleq P\left( {X\left( {t_{0} } \right) = 0 |T_{{I_{1} }} \le \frac{1}{2}t_{0} ,T_{{I_{1} }} = y_{1} ,} \right. \\ & \quad \left. {X\left( {T_{{I_{1} }} } \right) = 2,T_{{I_{2} }} = y_{2} ,X\left( {T_{{I_{2} }} } \right) = 0} \right) = e^{{ - \mathop \int \limits_{{y_{2} }}^{{t_{0} }} \lambda_{20} \left( t \right)dt}} \\ P_{{20k|y_{1} y_{2} ,H}} & \triangleq P\left( {X\left( {t_{0} } \right) = k |T_{{I_{1} }} \le \frac{1}{2}t_{0} ,T_{{I_{1} }} = y_{1} ,X\left( {T_{{I_{1} }} } \right) = 2,T_{{I_{2} }} = y_{2} ,} \right. \\ & \quad \left. {X\left( {T_{{I_{2} }} } \right) = 0} \right) = \alpha_{k} \left( {1 - e^{{ - \mathop \int \limits_{{y_{2} }}^{{t_{0} }} \lambda_{20} \left( t \right)dt}} } \right)\quad \left( {k = 1, 2, 3} \right). \\ \end{aligned} $$

If the result of the second visit is not 0, the agent will be removed from the system:

$$ \begin{aligned} P_{{2j4|y_{1} y_{2} ,H}} & \triangleq P\left( {X\left( {t_{0} } \right) = 4 |T_{{I_{1} }} \le \frac{1}{2}t_{0} ,T_{{I_{1} }} = y_{1} ,X\left( {T_{{I_{1} }} } \right) = 2,} \right. \\ & \quad \left. {T_{{I_{2} }} = y_{2} ,X\left( {T_{{I_{2} }} } \right) = j} \right) = 1 \quad \left( {j = 1, 2, 3} \right). \\ \end{aligned} $$

Un-condition on the result of the second visit:

$$ \begin{aligned} P_{{2k|y_{1} y_{2} ,H}} & \triangleq P\left( {X\left( {t_{0} } \right) = k |T_{{I_{1} }} \le \frac{1}{2}t_{0} ,T_{{I_{1} }} = y_{1} ,} \right. \\ & \quad \left. {X\left( {T_{{I_{1} }} } \right) = 2,T_{{I_{2} }} = y_{2} } \right) = P_{{20k|y_{1} y_{2} ,H}} S_{{20|y_{1} y_{2} ,H}} \left( {k = 0, 1, 2, 3} \right) \\ P_{{24|y_{1} y_{2} ,H}} & \triangleq P\left( {X\left( {t_{0} } \right) = 4 |T_{{I_{1} }} \le \frac{1}{2}t_{0} ,T_{{I_{1} }} = y_{1} ,} \right. \\ & \quad \left. {X\left( {T_{{I_{1} }} } \right) = 2,T_{{I_{2} }} = y_{2} } \right) = \mathop \sum \limits_{j = 1}^{3} S_{{2j|y_{1} y_{2} ,H}} = 1 - e^{{ - \mathop \int \limits_{{y_{1} }}^{{y_{2} }} \lambda_{2} \left( t \right)dt}} . \\ \end{aligned} $$

Un-condition on the time of the second visit:

$$ \begin{aligned} P_{{2k|y_{1} ,H}} & \triangleq P\left( {X\left( {t_{0} } \right) = k |T_{{I_{1} }} \le \frac{1}{2}t_{0} ,T_{{I_{1} }} = y_{1} ,X\left( {T_{{I_{1} }} } \right) = 2} \right) \\ & = \mathop \int \limits_{{y_{1} }}^{{t_{0} }} P_{{2k|y_{1} y_{2} ,H}} h_{2} \left( {y_{2} } \right)dy_{2}\quad \left( {k = 0, 1, 2, 3, 4} \right). \\ \end{aligned} $$

(1-3) If \( X\left( {T_{{I_{1} }} } \right) = 1 \), minor mistakes are found during the first visit and the supervisor will come for the second visit for sure.

Given that \( T_{{I_{2} }} = y_{2} \in \left( {y_{1} ,t_{0} } \right) \),

$$ \begin{aligned} S_{{10|y_{1} y_{2} ,H}} & \triangleq P\left( {X\left( {T_{{I_{2} }} } \right) = 0 |T_{{I_{1} }} \le \frac{1}{2}t_{0} ,T_{{I_{1} }} = y_{1} ,X\left( {T_{{I_{1} }} } \right) = 1,T_{{I_{2} }} = y_{2} } \right) = e^{{ - \mathop \int \limits_{{y_{1} }}^{{y_{2} }} \lambda_{1} \left( t \right)dt}} \\ S_{{1j|y_{1} y_{2} ,H}} & \triangleq P\left( {X\left( {T_{{I_{2} }} } \right) = j |T_{{I_{1} }} \le \frac{1}{2}t_{0} ,T_{{I_{1} }} = y_{1} ,} \right. \\ & \quad \left. {X\left( {T_{{I_{1} }} } \right) = 1,T_{{I_{2} }} = y_{2} } \right) = \alpha_{j} \left( {1 - e^{{ - \mathop \int \limits_{{y_{1} }}^{{y_{2} }} \lambda_{1} \left( t \right)dt}} } \right)\quad \left( {j = 1, 2, 3} \right). \\ \end{aligned} $$

If the result of the second visit is 0 or 1, the state at \( t_{0} \) can be 0, 1, 2 or 3:

$$ \begin{aligned} P_{{1j0|y_{1} y_{2} ,H}} & \triangleq P\left( {X\left( {t_{0} } \right) = 0 |T_{{I_{1} }} \le \frac{1}{2}t_{0} ,T_{{I_{1} }} = y_{1} ,X\left( {T_{{I_{1} }} } \right) = 1,} \right. \\ & \quad \left. {T_{{I_{2} }} = y_{2} ,X\left( {T_{{I_{2} }} } \right) = j} \right) = e^{{ - \mathop \int \limits_{{y_{2} }}^{{t_{0} }} \lambda_{1j} \left( t \right)dt}} \left( {j = 0, 1} \right) \\ P_{{1jk|y_{1} y_{2} ,H}} & \triangleq P\left( {X\left( {t_{0} } \right) = k |T_{{I_{1} }} \le \frac{1}{2}t_{0} ,T_{{I_{1} }} = y_{1} ,X\left( {T_{{I_{1} }} } \right) = 1,T_{{I_{2} }} = y_{2} ,X\left( {T_{{I_{2} }} } \right) = j} \right) \\ & = \alpha_{k} \left( {1 - e^{{ - \mathop \int \limits_{{y_{2} }}^{{t_{0} }} \lambda_{1j} \left( t \right)dt}} } \right)\quad \left( {j = 0, 1;\,\,k = 0, 1, 2, 3} \right). \\ \end{aligned} $$

If the result of the second visit is 2 or 3, the agent will be removed from the system:

$$ \begin{aligned} P_{{1j4|y_{1} y_{2} ,H}} & \triangleq P\left( {X\left( {t_{0} } \right) = 4 |T_{{I_{1} }} \le \frac{1}{2}t_{0} ,T_{{I_{1} }} = y_{1} ,} \right. \\ & \quad \left. {X\left( {T_{{I_{1} }} } \right) = 1,T_{{I_{2} }} = y_{2} ,X\left( {T_{{I_{2} }} } \right) = j} \right) = 1\quad \left( {j = 2, 3} \right). \\ \end{aligned} $$

Un-condition on the result of the second visit:

$$ \begin{aligned} P_{{1k|y_{1} y_{2} ,H}} & \triangleq P\left( {X\left( {t_{0} } \right) = k |T_{{I_{1} }} \le \frac{1}{2}t_{0} ,T_{{I_{1} }} = y_{1} ,X\left( {T_{{I_{1} }} } \right) = 1,T_{{I_{2} }} = y_{2} } \right) \\ & = P_{{10k|y_{1} y_{2} ,H}} S_{{10|y_{1} y_{2} ,H}} + P_{{11k|y_{1} y_{2} ,H}} S_{{11|y_{1} y_{2} ,H}} \left( {k = 0, 1, 2, 3} \right) \\ P_{{14|y_{1} y_{2} ,H}} & \triangleq P\left( {X\left( {t_{0} } \right) = 4 |T_{{I_{1} }} \le \frac{1}{2}t_{0} ,T_{{I_{1} }} = y_{1} ,X\left( {T_{{I_{1} }} } \right) = 1,T_{{I_{2} }} = y_{2} } \right) \\ & = S_{{12|y_{1} y_{2} ,H}} + S_{{13|y_{1} y_{2} ,H}} \\ \end{aligned} $$

Un-condition on the time of the second visit:

$$ \begin{aligned} P_{{1k|y_{1} ,H}} & \triangleq P\left( {X\left( {t_{0} } \right) = k |T_{{I_{1} }} \le \frac{1}{2}t_{0} ,T_{{I_{1} }} = y_{1} ,X\left( {T_{{I_{1} }} } \right) = 1} \right) \\ & = \mathop \int \limits_{{y_{1} }}^{{t_{0} }} P_{{1k|y_{1} y_{2} ,H}} h_{2} \left( {y_{2} } \right)dy_{2} \quad \left( {k = 0, 1, 2, 3, 4} \right). \\ \end{aligned} $$

(1-4) If \( X\left( {T_{{I_{1} }} } \right) = 0 \), no mistake is found during the first visit and the supervisor will come to visit this agent again with probability \( q \).

(1-4-1) The supervisor comes to visit the agent for the second time.

Given that \( T_{{I_{2} }} = y_{2} \in \left( {y_{1} ,t_{0} } \right) \),

$$ \begin{aligned} S_{{00|y_{1} y_{2} ,H,q}} & \triangleq P\left( {X\left( {T_{{I_{2} }} } \right) = 0 |T_{{I_{1} }} \le \frac{1}{2}t_{0} ,T_{{I_{1} }} = y_{1} ,} \right. \\ & \quad \left. {X\left( {T_{{I_{1} }} } \right) = 0,y_{1} < T_{{I_{2} }} < t_{0} ,T_{{I_{2} }} = y_{2} } \right) = e^{{ - \mathop \int \limits_{{y_{1} }}^{{y_{2} }} \lambda_{0} \left( t \right)dt}} \\ S_{{0j|y_{1} y_{2} ,H,q}} & \triangleq P\left( {X\left( {T_{{I_{2} }} } \right) = j |T_{{I_{1} }} \le \frac{1}{2}t_{0} ,T_{{I_{1} }} = y_{1} ,X\left( {T_{{I_{1} }} } \right) = 0,y_{1} < T_{{I_{2} }} < t_{0} ,T_{{I_{2} }} = y_{2} } \right) \\ & = \alpha_{j} \left( {1 - e^{{ - \mathop \int \limits_{{y_{1} }}^{{y_{2} }} \lambda_{0} \left( t \right)dt}} } \right)\quad \left( {j = 1, 2, 3} \right). \\ \end{aligned} $$

If the result of the second visit is 0 or 1, the state at \( t_{0} \) can be 0, 1, 2, 3:

$$ \begin{aligned} P_{{0j0|y_{1} y_{2} ,H,q}} & \triangleq P\left( {X\left( {t_{0} } \right) = 0 |T_{{I_{1} }} \le \frac{1}{2}t_{0} ,T_{{I_{1} }} = y_{1} ,X\left( {T_{{I_{1} }} } \right) = 0,} \right. \\ & \quad \left. {y_{1} < T_{{I_{2} }} < t_{0} ,T_{{I_{2} }} = y_{2} ,X\left( {T_{{I_{2} }} } \right) = j} \right) = e^{{ - \mathop \int \limits_{{y_{2} }}^{{t_{0} }} \lambda_{0j} \left( t \right)dt}} \left( {j = 0, 1} \right) \\ P_{{0jk|y_{1} y_{2} ,H,q}} & \triangleq P\left( {X\left( {t_{0} } \right) = k |T_{{I_{1} }} \le \frac{1}{2}t_{0} ,T_{{I_{1} }} = y_{1} ,X\left( {T_{{I_{1} }} } \right) = 0,} \right. \\ & \quad \left. {y_{1} < T_{{I_{2} }} < t_{0} ,T_{{I_{2} }} = y_{2} ,X\left( {T_{{I_{2} }} } \right) = j} \right) \\ & = \alpha_{k} \left( {1 - e^{{ - \mathop \int \limits_{{y_{2} }}^{{t_{0} }} \lambda_{0j} \left( t \right)dt}} } \right)\quad \left( {j = 0, 1;\,\,k = 1, 2, 3} \right). \\ \end{aligned} $$

If the result of the second visit is 2 or 3, the agent will be removed from the system:

$$ \begin{aligned} P_{{0j4|y_{1} y_{2} ,H,q}} & \triangleq P\left( {X\left( {t_{0} } \right) = 4 |T_{{I_{1} }} \le \frac{1}{2}t_{0} ,T_{{I_{1} }} = y_{1} ,X\left( {T_{{I_{1} }} } \right) = 0,} \right. \\ & \quad \left. {y_{1} < T_{{I_{2} }} < t_{0} ,T_{{I_{2} }} = y_{2} ,X\left( {T_{{I_{2} }} } \right) = j} \right) = 1\,\, \left( {j = 2, 3} \right). \\ \end{aligned} $$

Un-condition on the result of the second visit:

$$ \begin{aligned} P_{{0k|y_{1} y_{2} ,H,q}} & \triangleq P\left( {X\left( {t_{0} } \right) = k |T_{{I_{1} }} \le \frac{1}{2}t_{0} ,T_{{I_{1} }} = y_{1} ,X\left( {T_{{I_{1} }} } \right) = 0,y_{1} < T_{{I_{2} }} < t_{0} ,T_{{I_{2} }} = y_{2} } \right) \\ & = P_{{00k|y_{1} y_{2} ,H,q}} S_{{00|y_{1} y_{2} ,H,q}} + P_{{01k|y_{1} y_{2} ,H,q}} S_{{01|y_{1} y_{2} ,H,q}}\,\, \left( {k = 0, 1, 2, 3} \right) \\ P_{{04|y_{1} y_{2} ,H,q}} & \triangleq P\left( {X\left( {t_{0} } \right) = 4 |T_{{I_{1} }} \le \frac{1}{2}t_{0} ,T_{{I_{1} }} = y_{1} ,X\left( {T_{{I_{1} }} } \right) = 0,y_{1} < T_{{I_{2} }} < t_{0} ,T_{{I_{2} }} = y_{2} } \right) \\ & = S_{{02|y_{1} y_{2} ,H,q}} + S_{{03|y_{1} y_{2} ,H,q}} \\ \end{aligned} $$

Un-condition on the time of the second visit:

$$ \begin{aligned} P_{{0k|y_{1} ,H,q}} & \triangleq P\left( {X\left( {t_{0} } \right) = k |T_{{I_{1} }} \le \frac{1}{2}t_{0} ,T_{{I_{1} }} = y_{1} ,X\left( {T_{{I_{1} }} } \right) = 0,y_{1} < T_{{I_{2} }} < t_{0} } \right) \\ & = \mathop \int \limits_{{y_{1} }}^{{t_{0} }} P_{{0k|y_{1} y_{2} ,H,q}} h_{2} \left( {y_{2} } \right)dy_{2} \quad\left( {k = 0, 1, 2, 3, 4} \right). \\ \end{aligned} $$

(1-4-2) There is no second visit.

This case can be represented as \( T_{{I_{2} }} > t_{0} \) and this agent will remain in system for sure:

$$ \begin{aligned} P_{{00|y_{1} ,H,1 - q}} & \triangleq P\left( {X\left( {t_{0} } \right) = 0 |T_{{I_{1} }} \le \frac{1}{2}t_{0} ,T_{{I_{1} }} = y_{1} ,X\left( {T_{{I_{1} }} } \right) = 0,T_{{I_{2} }} > t_{0} } \right) = e^{{ - \mathop \int \limits_{{y_{1} }}^{{t_{0} }} \lambda_{0} \left( t \right)dt}} \\ P_{{0k|y_{1} ,H,1 - q}} & \triangleq P\left( {X\left( {t_{0} } \right) = k |T_{{I_{1} }} \le \frac{1}{2}t_{0} ,T_{{I_{1} }} = y_{1} ,X\left( {T_{{I_{1} }} } \right) = 0,T_{{I_{2} }} > t_{0} } \right) \\ & = \alpha_{k} \left( {1 - e^{{ - \mathop \int \limits_{{y_{1} }}^{{t_{0} }} \lambda_{0} \left( t \right)dt}} } \right)\,\,\left( {k = 1, 2, 3} \right) \\ P_{{04|y_{1} ,H,1 - q}} & \triangleq P\left( {X\left( {t_{0} } \right) = 4 |T_{{I_{1} }} \le \frac{1}{2}t_{0} ,T_{{I_{1} }} = y_{1} ,X\left( {T_{{I_{1} }} } \right) = 0,T_{{I_{2} }} > t_{0} } \right) = 0. \\ \end{aligned} $$

Un-condition on the existence of the second visit by combining (1-4-1) and (1-4-2).

$$ \begin{aligned} P_{{0k|y_{1} ,H}} & \triangleq P\left( {X\left( {t_{0} } \right) = k |T_{{I_{1} }} \le \frac{1}{2}t_{0} ,T_{{I_{1} }} = y_{1} ,X\left( {T_{{I_{1} }} } \right) = 0} \right) \\ & = qP_{{0k|y_{1} ,H,q}} + \left( {1 - q} \right)P_{{0k|y_{1} ,H,1 - q}} \quad\left( {k = 0, 1, 2, 3, 4} \right). \\ \end{aligned} $$

Combine situations (1-1)–(1-4) and un-condition on the result of the first visit:

$$ \begin{aligned} P_{{k|y_{1} ,H}} & \triangleq P\left( {X\left( {t_{0} } \right) = k |T_{{I_{1} }} \le \frac{1}{2}t_{0} ,T_{{I_{1} }} = y_{1} } \right) \\ & = \mathop \sum \limits_{i = 0}^{3} (P_{{ik|y_{1} ,H}} S_{{i|y_{1} ,H}} )\quad \left( {k = 0, 1, 2, 3, 4} \right). \\ \end{aligned} $$

Un-condition on the time for the first visit:

$$ \begin{aligned} P_{k|H} & \triangleq P\left( {X\left( {t_{0} } \right) = k |T_{{I_{1} }} \le \frac{1}{2}t_{0} } \right) \\ & = \frac{1}{{H_{1} \left( {\frac{1}{2}t_{0} } \right)}}\mathop \int \limits_{0}^{{\frac{1}{2}t_{0} }} P_{{k|y_{1} ,H}} h_{1} \left( {y_{1} } \right)dy_{1} \quad\left( {k = 0, 1, 2, 3, 4} \right). \\ \end{aligned} $$

Scenario 2 \( T_{{I_{1} }} > \frac{1}{2}t_{0} \), in which major mistakes cannot be fixed during the first visit.

Define \( P_{1 - H} \triangleq P\left( {T_{{I_{1} }} > \frac{1}{2}t_{0} } \right) = 1 - H_{1} \left( {\frac{1}{2}t_{0} } \right) = \mathop \int \limits_{{\frac{1}{2}t_{0} }}^{{t_{0} }} h_{1} \left( {y_{1} } \right)dy_{1} \).

Given that \( T_{{I_{1} }} = y_{1} \in \left( {\frac{1}{2}t_{0} ,t_{0} } \right) \),

$$ \begin{aligned} S_{{i|y_{1} ,1 - H}} & \triangleq P\left( {X\left( {T_{{I_{1} }} } \right) = i |T_{{I_{1} }} > \frac{1}{2}t_{0} ,T_{{I_{1} }} = y_{1} } \right) \\ & = S_{{i|y_{1} ,H}} \quad\left( {i = 0, 1, 2, 3} \right). \\ \end{aligned} $$

Given \( X\left( {T_{{I_{1} }} } \right) = 0, 1, 3 \), the calculation results are the same with case (1-4), (1-3) and (1-1) respectively, that is,

$$ \begin{aligned} P_{{ik|y_{1} ,1 - H}} & \triangleq P\left( {X\left( {t_{0} } \right) = k |T_{{I_{1} }} > \frac{1}{2}t_{0} ,T_{{I_{1} }} = y_{1} ,X\left( {T_{{I_{1} }} } \right) = i} \right) \\ & = P_{{ik|y_{1} ,H}} \quad\left( {i = 0, 1, 3;\,\,k = 0, 1, 2, 3, 4} \right). \\ \end{aligned} $$

If \( X\left( {T_{{I_{1} }} } \right) = 2 \), major mistakes are detected during the first visit and the agent will be removed from the system, which is different from case (1-2) in Scenario 1. In this way,

$$ \begin{aligned} P_{{24|y_{1} ,1 - H}} & \triangleq P\left( {X\left( {t_{0} } \right) = 4 |T_{{I_{1} }} > \frac{1}{2}t_{0} ,T_{{I_{1} }} = y_{1} ,X\left( {T_{{I_{1} }} } \right) = 2} \right) = 1 \\ P_{{2k|y_{1} ,1 - H}} & \triangleq P\left( {X\left( {t_{0} } \right) = k |T_{{I_{1} }} > \frac{1}{2}t_{0} ,T_{{I_{1} }} = y_{1} ,X\left( {T_{{I_{1} }} } \right) = 2} \right) = 0\quad \left( {k = 0, 1, 2, 3} \right). \\ \end{aligned} $$

Un-condition on the result of the first visit:

$$ P_{{k|y_{1} ,1 - H}} \triangleq P\left( {X\left( {t_{0} } \right) = k |T_{{I_{1} }} > \frac{1}{2}t_{0} ,T_{{I_{1} }} = y_{1} } \right) = \mathop \sum \limits_{i = 0}^{3} (P_{{ik|y_{1} ,1 - H}} S_{{i|y_{1} ,1 - H}} )\quad\left( {k = 0, 1, 2, 3, 4} \right). $$

Un-condition on the time for the first visit:

$$ P_{k|1 - H} \triangleq P\left( {X\left( {t_{0} } \right) = k |T_{{I_{1} }} > \frac{1}{2}t_{0} } \right) = \frac{1}{{1 - H_{1} \left( {\frac{1}{2}t_{0} } \right)}}\mathop \int \limits_{{\frac{1}{2}t_{0} }}^{{t_{0} }} P_{{k|y_{1} ,1 - H}} h_{1} \left( {y_{1} } \right)dy_{1}\quad \left( {k = 0, 1, 2, 3, 4} \right). $$

Finally, combining the results of the two scenarios above, the distribution of states for each agent can be expressed as:

$$ P_{k} \left( {t_{0} } \right) = P\left( {X\left( {t_{0} } \right) = k} \right) = P_{H} P_{k|H} + P_{1 - H} P_{k|1 - H} \quad \left( {k = 0, 1, 2, 3, 4} \right). $$

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Lin, T., Pham, H. A two-stage intervened decision system with multi-state decision units and dynamic system configuration. Ann Oper Res 311, 255–277 (2022). https://doi.org/10.1007/s10479-019-03334-8

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