A two-level iteration approach for modeling and analysis of rapid response process with multiple deteriorating patients


In acute care, a patient’s clinical deterioration is often a precursor to serious and often fatal outcomes. To reduce the severity and frequency of negative outcomes, care providers need to response rapidly by providing quick evaluation, triage, and treatment to patients with declining conditions. However, a provider’s availability to respond can be constrained when multiple patients are deteriorating at the same time. To study the multiple patients rapid response process, we introduce a network model with complex structures, such as split, merge, and parallel. Iterative methods are presented to evaluate the mean decision time (i.e., the average time from the detection of a patient’s declining to a physician’s treatment decision being made). It is shown that such methods lead to convergent results and high accuracy in performance evaluation. Such a model provides a quantitative tool for healthcare professionals to design and improve rapid response systems.

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This work is supported in part by National Science Foundation Grant No. CMMI-1536987 and by National Natural Science Foundation of China Grant No. 71501109.

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Correspondence to Xiaolei Xie.

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Appendix 1: Iteration procedures

Three-patient example: Level-1 iteration procedure

Denote \(\tau _{k,r}^{(j)}\), \(k=1,2,3\), \(r \in X\), as the mean decision time that includes patient k’s waiting time for provider \(r\) during the j-th iteration, \(j=1,2,\ldots\). Let \(p_{k,r}^{(j)}\) be the probability that provider r is treating patient k and there is another request for provider r during the j-th iteration. At the beginning of iteration, assume

$$\begin{aligned} \tau _{k,r}^{(0)}=T_d \quad \text{ and }\quad p_{k,r}^{(0)}=0, \qquad k=1,2,3, \quad r \in X. \end{aligned}$$

First, consider patient 1. During the first iteration, \(\tau _{1,int}^{(1)}\) can be updated as:

$$\begin{aligned} \tau _{1,int}^{(1)} = T_d + (p_{2,int}^{(0)}+p_{3,int}^{(0)})(p_{int} \tau _{int}+ p_{rrt \& int} \tau _{rrt \& int}). \end{aligned}$$

The \(p_{1,int}^{(1)}\) can be updated as:

$$\begin{aligned} p_{1,int}^{(1)} = \frac{p_{int}^2 \tau _{int}+ p_{rrt \& int}^2 \tau _{rrt \& int}}{\tau _{1,int}^{(1)}}. \end{aligned}$$

Next, Consider patient 2. Decision time \(\tau _{2,int}^{(1)}\) and probability \(p_{2,int}^{(1)}\) can be calculated.

$$\begin{aligned} \tau _{2,int}^{(1)}&= T_d + (p_{1,int}^{(1)}+p_{3,int}^{(0)})(p_{int} \tau _{int}+ p_{rrt \& int} \tau _{rrt \& int}),\\ p_{2,int}^{(1)}&= \frac{p_{int}^2 \tau _{int}+ p_{rrt \& int}^2 \tau _{rrt \& int}}{\tau _{2,int}^{(1)}}. \end{aligned}$$

Lastly, consider patient 3, we have

$$\begin{aligned} \tau _{3,int}^{(1)}&= T_d + (p_{1,int}^{(1)}+p_{2,int}^{(1)})(p_{int} \tau _{int}+ p_{rrt \& int} \tau _{rrt \& int}),\\ p_{3,int}^{(1)}&= \frac{p_{int}^2 \tau _{int}+ p_{rrt \& int}^2 \tau _{rrt \& int}}{\tau _{3,int}^{(1)}}. \end{aligned}$$

This completes the update of the intern.

Similar updating process for the resident can be carried out. First, we study patient 1:

$$\begin{aligned} \tau _{1,res}^{(1)}&= T_d + (p_{2,res}^{(0)}+p_{3,res}^{(0)})(p_{res} \tau _{res} +p_{rrt \& res} \tau _{rrt \& res}),\\ p_{1,res}^{(1)}&= \frac{p_{res}^2 \tau _{res}+ p_{rrt \& res}^2 \tau _{rrt \& res}}{\tau _{1,res}^{(1)}}. \end{aligned}$$

Next, consider patient 2:

$$\begin{aligned} \tau _{2,res}^{(1)}&= T_d + (p_{1,res}^{(1)}+p_{3,res}^{(0)})(p_{res} \tau _{res} +p_{rrt \& res} \tau _{rrt \& res}),\\ p_{2,res}^{(1)}&= \frac{p_{res}^2 \tau _{res}+ p_{rrt \& res}^2 \tau _{rrt \& res}}{\tau _{2,res}^{(1)}}. \end{aligned}$$

Then, patient 3 is included:

$$\begin{aligned} \tau _{3,res}^{(1)}&= T_d + (p_{1,res}^{(1)}+p_{2,res}^{(1)})(p_{res} \tau _{res} +p_{rrt \& res} \tau _{rrt \& res}),\\ p_{3,res}^{(1)}&= \frac{p_{res}^2 \tau _{res}+ p_{rrt \& res}^2 \tau _{rrt \& res}}{\tau _{3,res}^{(1)}}. \end{aligned}$$

Similarly, all the rest of providers are updated. Particularly, for the RRT, considering patient 1, we obtain:

$$\begin{aligned} \tau _{1,rrt}^{(1)}&= T_d + (p_{2,rrt}^{(0)}+p_{3,rrt}^{(0)})(p_{rrt} \tau _{rrt} + p_{rrt \& int} \tau _{rrt \& int} + p_{rrt \& res} \tau _{rrt \& res} \\&\quad +\,p_{rrt \& fel}\tau _{rrt \& fel}+ p_{rrt \& atn} \tau _{rrt \& atn}),\\ p_{1,rrt}^{(1)}&= (p_{rrt}^2 \tau _{rrt} + p_{rrt \& int}^2 \tau _{rrt \& int}^2 + p_{rrt \& res}^2 \tau _{rrt \& res} + p_{rrt \& fel}^2 \tau _{rrt \& fel} \\&\quad +\, p_{rrt \& atn}^2 \tau _{rrt \& atn}) / \tau _{1,rrt}^{(1)}. \end{aligned}$$

Regarding patient 2, we have

$$\begin{aligned} \tau _{2,rrt}^{(1)}&= T_d + (p_{1,rrt}^{(1)}+p_{3,rrt}^{(0)})(p_{rrt} \tau _{rrt} + p_{rrt \& int} \tau _{rrt \& int} + p_{rrt \& res} \tau _{rrt \& res} \\&\quad +\, p_{rrt \& fel} \tau _{rrt \& fel} + p_{rrt \& atn} \tau _{rrt \& atn}),\\ p_{2,rrt}^{(1)}&= (p_{rrt}^2 \tau _{rrt} + p_{rrt \& int}^2 \tau _{rrt \& int}^2 + p_{rrt \& res}^2 \tau _{rrt \& res} + p_{rrt \& fel}^2 \tau _{rrt \& fel} \\&\quad +\, p_{rrt \& atn}^2 \tau _{rrt \& atn}) / \tau _{2,rrt}^{(1)}. \end{aligned}$$

Furthermore, parameters of patient 3 are updated:

$$\begin{aligned} \tau _{3,rrt}^{(1)}&= T_d + (p_{1,rrt}^{(1)}+p_{2,rrt}^{(1)})(p_{rrt} \tau _{rrt} + p_{rrt \& int} \tau _{rrt \& int} + p_{rrt \& res} \tau _{rrt \& res} \\&\quad +\, p_{rrt \& fel} \tau _{rrt \& fel} + p_{rrt \& atn} \tau _{rrt \& atn}),\\ p_{3,rrt}^{(1)}&= (p_{rrt}^2 \tau _{rrt} + p_{rrt \& int}^2 \tau _{rrt \& int}^2 + p_{rrt \& res}^2 \tau _{rrt \& res} + p_{rrt \& fel}^2 \tau _{rrt \& fel} \\&\quad +\, p_{rrt \& atn}^2 \tau _{rrt \& atn}) /\tau _{3,rrt}^{(1)}. \end{aligned}$$

Then for the fellow, patients 1 to 3 are considered:

$$\begin{aligned} \tau _{1,fel}^{(1)}&= T_d + (p_{2,fel}^{(0)}+p_{3,fel}^{(0)})(p_{fel} \tau _{fel}+ p_{rrt \& fel} \tau _{rrt \& fel}),\\ p_{1,fel}^{(1)}&= \frac{p_{fel}^2 \tau _{fel}+ p_{rrt \& fel}^2 \tau _{rrt \& fel}}{\tau _{1,fel}^{(1)}},\\ \tau _{2,fel}^{(1)}&= T_d + (p_{1,fel}^{(1)}+p_{3,fel}^{(0)})(p_{fel} \tau _{fel}+ p_{rrt \& fel} \tau _{rrt \& fel}),\\ p_{2,fel}^{(1)}&= \frac{p_{fel}^2 \tau _{fel}+ p_{rrt \& fel}^2 \tau _{rrt \& fel}}{\tau _{2,fel}^{(1)}},\\ \tau _{2,fel}^{(1)}&= T_d + (p_{1,fel}^{(1)}+p_{2,fel}^{(1)})(p_{fel} \tau _{fel}+ p_{rrt \& fel} \tau _{rrt \& fel}),\\ p_{2,fel}^{(1)}&= \frac{p_{fel}^2 \tau _{fel}+ p_{rrt \& fel}^2 \tau _{rrt \& fel}}{\tau _{3,fel}^{(1)}}. \end{aligned}$$

Finally, for the attending, we again address all three patients:

$$\begin{aligned} \tau _{1,atn}^{(1)}&= T_d + (p_{2,atn}^{(0)}+p_{3,atn}^{(0)})(p_{atn} \tau _{atn}+ p_{rrt \& atn} \tau _{rrt \& atn}),\\ p_{1,atn}^{(1)}&= \frac{p_{atn}^2 \tau _{atn}+ p_{rrt \& atn}^2 \tau _{rrt \& atn}}{\tau _{1,atn}^{(1)}},\\ \tau _{2,atn}^{(1)}&= T_d + (p_{1,atn}^{(1)}+p_{3,atn}^{(0)})(p_{atn} \tau _{atn}+ p_{rrt \& atn} \tau _{rrt \& atn}),\\ p_{2,atn}^{(1)}&= \frac{p_{atn}^2 \tau _{atn}+ p_{rrt \& atn}^2 \tau _{rrt \& atn}}{\tau _{2,atn}^{(1)}},\\ \tau _{3,atn}^{(1)}&= T_d + (p_{1,atn}^{(1)}+p_{2,atn}^{(1)})(p_{atn} \tau _{atn}+ p_{rrt \& atn} \tau _{rrt \& atn}),\\ p_{3,atn}^{(1)}&= \frac{p_{atn}^2 \tau _{atn}+ p_{rrt \& atn}^2 \tau _{rrt \& atn}}{\tau _{3,atn}^{(1)}}. \end{aligned}$$

When the first iteration is finished, all the updated parameters will be used for the second iteration to calculate \(\tau _{k,r}^{(2)}\), \(k=1,2,3\), \(r \in X\), and \(p_{k,r}^{(2)}\). The process is repeated until procedure converges. Let \(\delta = 10^{-5}\). When

$$\begin{aligned} | \tau _{i,r}^{(j+1)} - \tau _{i,r}^{(j)} | \le \delta , \qquad | p_{i,r}^{(j+1)} - p_{i,r}^{(j)} | \le \delta , \qquad i=1,2,3, \quad r\in X, \end{aligned}$$

the procedure is convergent, i.e.,

$$\begin{aligned} \lim _{j \rightarrow \infty } \tau _{i,r}^{(j)} = \tau _{i,r}, \qquad \lim _{j \rightarrow \infty } p_{i,r}^{(j)} = p_{i,r}, \qquad i=1,2,3. \end{aligned}$$

In particular, all \(\tau _{i,r}, i=1,2,3,\) are identical and all \(p_{i,r}, i=1,2,3,\) are the same. Then the mean decision time (including waiting time) \({\mathrm {T}}_r\) and provider utilization \(P_r\) can be obtained:

$$\begin{aligned} \tau _{1,r}=\tau _{2,r}=\tau _{3,r}:={\mathrm {T}}_{r},\qquad p_{1,r}=p_{2,r}=p(3,r):=P_{r}. \end{aligned}$$

The mean decision time \(T_{in}\) includes the additional waiting time.

$$\begin{aligned} T_{in} = T_d + \Sigma _{r, r \in X} P_{r} {\mathrm {T}}_{r}. \end{aligned}$$

Three-patient example: Level-2 iteration procedure

Denote \(\rho _k^{(l)}\), \(k=1,2,3\), \(l=1,2,\ldots\), as the percentage of time the patient is in a deteriorating status in iteration j, and \(\lambda _k^{(l)}\), \(k=1,2,3\), \(l=1,2,\ldots\), as the updated mean decision time in iteration j by including the time percentage patient k is declining. When the iteration starts, assume all

$$\begin{aligned} \rho _k^{(0)}=0 \quad \text{ and } \quad \lambda _k^{(0)}=T_{in}, \qquad k=1,2,3. \end{aligned}$$

Considering patient 1, \(\lambda _1^{(1)}\) can be updated as:

$$\begin{aligned} \lambda _1^{(1)} = T_{in}\left[ 1+\rho _1^{(0)}\left( \rho _2^{(0)}+\rho _3^{(0)}\right) \right] . \end{aligned}$$

The time percentage that the first patient is in deteriorating status can be calculated as

$$\begin{aligned} \rho _1^{(1)} = \frac{\lambda _1^{(1)}}{\lambda _1^{(1)}+T_{normal}}. \end{aligned}$$

Next consider patients 2 and 3, where \(\lambda _i^{(1)}\) and \(\rho _i^{(1)}\), \(i=2,3\), can be obtained:

$$\begin{aligned} \lambda _2^{(1)}&= T_{in}\left[ 1+\rho _2^{(0)}\left( \rho _1^{(1)}+\rho _3^{(0)}\right) \right] ,\\ \rho _2^{(1)}&= \frac{\lambda _2^{(1)}}{\lambda _2^{(1)}+T_{normal}},\\ \lambda _3^{(1)}&= T_{in}\left[ 1+\rho _3^{(0)}\left( \rho _1^{(1)}+\rho _2^{(1)}\right) \right] ,\\ \rho _3^{(1)}&= \frac{\lambda _3^{(1)}}{\lambda _3^{(1)}+T_{normal}}. \end{aligned}$$

This finishes the first iteration. Then \(\rho _k^{(1)}\), \(k=1,2,3\), and \(\lambda _k^{(1)}\) are used for the second iteration to evaluate \(\rho _k^{(2)}\) and \(\lambda _k^{(2)}\). The process is repeated until the procedure converges. When the following criteria is met:

$$\begin{aligned} | \lambda _{i}^{(j+1)} - \lambda _{i}^{(l)} | \le \delta ,\qquad | \rho _{i}^{(j+1)} - \rho _{i}^{(l)} | \le \delta , \qquad i=1,2,3, \end{aligned}$$

the procedure is convergent. Again \(\delta = 10^{-5}\). Upon converges, we have

$$\begin{aligned} \lim _{l \rightarrow \infty } \lambda _{i}^{(l)} = \lambda _{i}, \qquad \lim _{l \rightarrow \infty } \rho _{i}^{(l)} = \rho _{i}, \qquad i=1,2,3. \end{aligned}$$

The final mean decision time can be obtained:

$$\begin{aligned} \lambda _{1}=\lambda _{2}=\lambda _3=T_{final}. \end{aligned}$$

General iteration procedure

Procedure 1

(1) Level-1 iteration

Step 1.1 Initialization: Calculate \(p_i\), \(i \in X\), and \(T_d\) using the results in Xie et al. (2012). Set \(j=0\) and

$$\begin{aligned} \tau _{k,i}^{(j)}=p_{k,i}^{(j)}=0. \end{aligned}$$

Step 1.2 Update \(\tau _{k,i}^{(j)}\) and \(p_{k,i}^{(j)}\): For patient 1,

$$\begin{aligned} \tau _{1,int}^{(j+1)}&= T_d + \Sigma _{i=2}^n p_{i,int}^{(j)} (p_{int} \tau _{int}+ p_{rrt \& int} \tau _{rrt \& int}), \\ p_{1,int}^{(j+1)}&= (p_{int}^2 \tau _{int}+ p_{rrt \& int}^2 \tau _{rrt \& int})\big /\tau _{1,int}^{(j+1)}, \\ \tau _{1,res}^{(j+1)}&= T_d + \Sigma _{i=2}^n p_{i,res}^{(j)} (p_{res} \tau _{res}+ p_{rrt \& res} \tau _{rrt \& res}), \\ p_{1,res}^{(j+1)}&= (p_{res}^2 \tau _{res}+ p_{rrt \& res}^2 \tau _{rrt \& res})\big /\tau _{1,res}^{(j+1)}, \\ \tau _{1,rrt}^{(j+1)}&= T_d + \Sigma _{i=2}^n p_{i,rrt}^{(j)} (p_{rrt} \tau _{rrt} + p_{rrt \& int} \tau _{rrt \& int} +p_{rrt \& res} \tau _{rrt \& res} \\&\quad +\, p_{rrt \& fel} \tau _{rrt \& fel}+ p_{rrt \& atn} \tau _{rrt \& atn}), \\ p_{1,rrt}^{(j+1)}&= (p_{rrt}^2 \tau _{rrt} + p_{rrt \& int}^2 \tau _{rrt \& int}^2 + p_{rrt \& res}^2 \tau _{rrt \& res} \\&\quad +\, p_{rrt \& fel}^2 \tau _{rrt \& fel} + p_{rrt \& atn}^2 \tau _{rrt \& atn}) \big / {\tau _{1,rrt}^{(j+1)}}, \\ \tau _{1,fel}^{(j+1)}&= T_d + \Sigma _{i=2}^n p_{i,fel}^{(j)} (p_{fel} \tau _{fel}+ p_{rrt \& fel} \tau _{rrt \& fel}), \\ p_{1,fel}^{(j+1)}&= (p_{fel}^2 \tau _{fel}+ p_{rrt \& fel}^2 \tau _{rrt \& fel})\big /\tau _{1,fel}^{(j+1)}, \\ \tau _{1,atn}^{(j+1)}&= T_d + \Sigma _{i=2}^n p_{i,atn}^{(j)} (p_{atn} \tau _{atn}+ p_{rrt \& atn} \tau _{rrt \& atn}), \\ p_{1,atn}^{(j+1)}&= (p_{atn}^2 \tau _{atn}+ p_{rrt \& atn}^2 \tau _{rrt \& atn})\big /\tau _{1,atn}^{(j+1)}. \end{aligned}$$

For patient \(k=2,\ldots ,m-1\),

$$\begin{aligned} \tau _{k,int}^{(j+1)}&= T_d + \left( \Sigma _{i=1}^{k-1} p_{i,int}^{(j+1)}+\Sigma _{i=k+1}^m p_{i,int}^{(j)}\right) \cdot (p_{int} \tau _{int}+ p_{rrt \& int} \tau _{rrt \& int}), \\ p_{k,int}^{(j+1)}&= (p_{int}^2 \tau _{int}+ p_{rrt \& int}^2 \tau _{rrt \& int})\big /\tau _{k,int}^{(j+1)}, \\ \tau _{k,res}^{(j+1)}&= T_d + \left( \Sigma _{i=1}^{k-1} p_{i,res}^{(j+1)}+\Sigma _{i=k+1}^m p_{i,res}^{(j)}\right) \cdot (p_{res} \tau _{res}+ p_{rrt \& res} \tau _{rrt \& res}), \\ p_{k,res}^{(j+1)}&= (p_{res}^2 \tau _{res}+ p_{rrt \& res}^2 \tau _{rrt \& res})\big /\tau _{k,res}^{(j+1)}, \\ \tau _{k,rrt}^{(j+1)}&= T_d + \left( \Sigma _{i=1}^{k-1} p_{i,rrt}^{(j+1)} +\Sigma _{i=k+1}^m p_{i,rrt}^{(j)}\right) \cdot (p_{rrt} \tau _{rrt} + p_{rrt \& int} \tau _{rrt \& int} \\&\quad +\, p_{rrt \& res} \tau _{rrt \& res}+ p_{rrt \& fel} \tau _{rrt \& fel} + p_{rrt \& atn}\tau _{rrt \& atn}), \\ p_{k,rrt}^{(j+1)}&= (p_{rrt}^2 \tau _{rrt} + p_{rrt \& int}^2 \tau _{rrt \& int}^2 + p_{rrt \& res}^2 \tau _{rrt \& res} + p_{rrt \& fel}^2 \tau _{rrt \& fel} \\&\quad +\, p_{rrt \& atn}^2 \tau _{rrt \& atn}) / \tau _{k,rrt}^{(j+1)}, \\ \tau _{k,fel}^{(j+1)}&= T_d + (\Sigma _{i=1}^{k-1} p_{i,fel}^{(j+1)}+\Sigma _{i=k+1}^m p_{i,fel}^{(j)}) \cdot (p_{fel} \tau _{fel}+ p_{rrt \& fel} \tau _{rrt \& fel}), \\ p_{k,fel}^{(j+1)}&= (p_{fel}^2 \tau _{fel}+ p_{rrt \& fel}^2 \tau _{rrt \& fel})\big /\tau _{k,fel}^{(j+1)}, \\ \tau _{k,atn}^{(j+1)}&= T_d + \left( \Sigma _{i=1}^{k-1} p_{i,atn}^{(j+1)}+\Sigma _{i=k+1}^m p_{i,atn}^{(j)}\right) \cdot (p_{atn} \tau _{atn}+ p_{rrt \& atn} \tau _{rrt \& atn}), \\ p_{k,atn}^{(j+1)}&= (p_{atn}^2 \tau _{atn}+ p_{rrt \& atn}^2 \tau _{rrt \& atn})\big /\tau _{k,atn}^{(j+1)}. \end{aligned}$$

For patient m,

$$\begin{aligned} \tau _{m,int}^{(j+1)}&= T_d + \Sigma _{i=1}^{m-1} p_{i,int}^{(j+1)} (p_{int} \tau _{int} + p_{rrt \& int} \tau _{rrt \& int}), \\ p_{m,int}^{(j+1)}&= \frac{p_{int}^2 \tau _{int}+ p_{rrt \& int}^2 \tau _{rrt \& int}}{\tau _{k,r}^{(j+1)}}, \\ \tau _{m,res}^{(j+1)}&=T_d + \Sigma _{i=1}^{m-1} p_{i,res}^{(j+1)} (p_{res} \tau _{res} + p_{rrt \& res} \tau _{rrt \& res}), \\ p_{m,res}^{(j+1)}&= (p_{res}^2 \tau _{res}+ p_{rrt \& res}^2 \tau _{rrt \& res})\big /\tau _{k,res}^{(j+1)}, \\ \tau _{n,rrt}^{(j+1)}&=T_d + \Sigma _{i=1}^{m-1} p_{i,rrt}^{(j+1)} (p_{rrt} \tau _{rrt} + p_{rrt \& int} \tau _{rrt \& int} + p_{rrt \& res} \tau _{rrt \& res} \\&\quad +\, p_{rrt \& fel} \tau _{rrt \& fel} + p_{rrt \& atn} \tau _{rrt \& atn}), \\ p_{m,r}^{(j+1)}&= (p_{rrt}^2 \tau _{rrt} + p_{rrt \& int}^2 \tau _{rrt \& int}^2 + p_{rrt \& res}^2 \tau _{rrt \& res} + p_{rrt \& fel}^2 \tau _{rrt \& fel} \\&\quad +\, p_{rrt \& atn}^2 \tau _{rrt \& atn}) \big / \tau _{m,rrt}^{(j+1)}, \\ \tau _{m,fel}^{(j+1)}&= T_d + \Sigma _{i=1}^{m-1} p_{i,fel}^{(j+1)} (p_{fel} \tau _{fel}+ p_{rrt \& fel} \tau _{rrt \& fel}), \\ p_{m,fel}^{(j+1)}&= (p_{fel}^2 \tau _{f}+ p_{rrt \& fel}^2 \tau _{rrt \& fel})\big /\tau _{m,fel}^{(j+1)}, \\ \tau _{m,atn}^{(j+1)}&= T_d + \Sigma _{i=1}^{m-1} p_{i,atn}^{(j+1)} (p_{atn} \tau _{atn}+ p_{rrt \& atn} \tau _{rrt \& atn}), \\ p_{m,atn}^{(j+1)}&= (p_{atn}^2 \tau _{atn}+ p_{rrt \& atn}^2 \tau _{rrt \& atn})\big /\tau _{m,atn}^{(j+1)}. \end{aligned}$$

Step 1.3 Iteration: Set \(j=j+1\). If the terminating criteria is not met, go back to Step 1.2. Let \(\delta = 10^{-5}\), the Level-1 iteration is finished if

$$\begin{aligned} | \tau _{i,r}^{(j+1)} - \tau _{i,r}^{(j)} | \le \delta ,\qquad | p_{i,r}^{(j+1)} - p_{i,r}^{(j)} | \le \delta ,\qquad i=1,2,\ldots ,m. \end{aligned}$$

Step 1.4 Termination: If the stopping conditions are met, set

$$\begin{aligned} \tau _{i,r}^{(j+1)}&={\mathrm {T}}_{r}, \qquad p_{i,r}^{(j+1)}=P_{r}, \qquad i=1,\ldots ,m, \\ T_{in}&= T_d + \Sigma _{r, r \in X} P_{r} {\mathrm {T}}_{r}. \end{aligned}$$

(2) Level-2 iteration

Step 2.1 Initialization: Set \(l=0\) and

$$\begin{aligned} \rho _1^{(l)}=0,\qquad \lambda _1^{(l)} = T_{in}. \end{aligned}$$

Step 2.2 Update \(\rho _k^{(l)}\) and \(\lambda _k^{(l)}\): For patient 1,

$$\begin{aligned} \lambda _1^{(l+1)}&= T_{in}(1+\rho _1^{(l)} \Sigma _{i=2}^m \rho _i^{(l)}), \\ \rho _1^{(l+1)}&= \frac{\lambda _1^{(l+1)}}{\lambda _1^{(l+1)}+T_{normal}}. \end{aligned}$$

For patient \(k=2,\ldots ,m-1\),

$$\begin{aligned} \lambda _k^{(l+1)}&= T_{in}(1+\rho _k^{(l)} (\Sigma _{i=1}^{k-1} \rho _i^{(l+1)}+\Sigma _{i=k+1}^m \rho _{i}^{(l)})), \\ \rho _k^{(l+1)}&= \frac{\lambda _k^{(l+1)}}{\lambda _k^{(l+1)}+T_{normal}}. \end{aligned}$$

For patient m,

$$\begin{aligned} \lambda _m^{(l+1)}&= T_{in}(1+\rho _k^{(l)} \Sigma _{i=1}^{m-1} \rho _i^{(l+1)}), \\ \rho _m^{(l+1)}&= \frac{\lambda _k^{(l+1)}}{\lambda _m^{(l+1)}+T_{normal}}. \end{aligned}$$

Step 2.3 Iteration: Set \(l=l+1\). If the terminating criteria is not met, go back to Step 2.2.

$$\begin{aligned} | \lambda _{i}^{(l+1)} - \lambda _{i}^{(l)} | \le \delta , \qquad | \rho _{i}^{(l+1)} - \rho _{i}^{(l)} | \le \delta , \quad i=1,\ldots ,m. \end{aligned}$$

Step 2.4 Termination: If the terminating condition is met, set

$$\begin{aligned} \lambda _{i}^{(l+1)}=\lambda _{i}, \quad \rho _{i}^{(l+1)}=\rho _{i}, \quad \lambda _{i}=T_{final},\quad i=1,\ldots ,m. \end{aligned}$$

Appendix 2: Proofs

To prove Proposition 1, Lemmas 1 and 2 are needed.

Lemma 1

Under assumptions (1)–(5), when\(m=2\), if\(p_{2,r}^{(j)} > p_{2,r}^{(j-1)}\), \(r \in X\), \(j=1,2,\ldots\), then\(\tau _{1,r}^{(j+1)} > \tau _{1,r}^{(j)}\), \(p_{1,r}^{(j+1)} < p_{1,r}^{(j)}\), \(\tau _{2,r}^{(j+1)} < \tau _{2,r}^{(j)}\), \(p_{2,r}^{(j+1)} > p_{2,r}^{(j)}\).

Lemma 2

Under assumptions (1)–(5), when\(m=2\), the sequences\(p_{1,r}^{(j)}\) and\(\tau _{2,r}^{(j)}\) are monotonically decreasing, while the sequences\(p_{2,r}^{(j)}\) and\(\tau _{1,r}^{(j)}\) are monotonically increasing.

Proof of Lemma 1

From all the equations related to the update of \(\tau _{i,r}^{(j)}\) and \(p_{i,r}^{(j)}\), which are from (4) to (6), define \(C_{1,r}\) and \(C_{2,r}\) as constants related to resource r, \(r \in X\). We have

$$\begin{aligned} C_{1,r} &= {} \left\{ \begin{array}{ll} p_{int} \tau _{int}+ p_{rrt \& int} \tau _{rrt \& int}, &{}\text{ if } r=int, \\ \\ p_{res} \tau _{res}+ p_{rrt \& res} \tau _{rrt \& res} &{}\text{ if } r=res, \\ \\ p_{rrt} \tau _{rrt} + p_{rrt \& int} \tau _{rrt \& int} &{}\\ + p_{rrt \& res} \tau _{rrt \& res}\\ + p_{rrt \& fel} \tau _{rrt \& fel}+ p_{rrt \& atn} \tau _{rrt \& atn} &{}\text{ if } r=rrt, \\ \\ p_{fel} \tau _{fel}+ p_{rrt \& fel} \tau _{rrt \& fel} &{}\text{ if } r=fel, \\ \\ p_{atn} \tau _{atn}+ p_{rrt \& atn} \tau _{rrt \& atn} &{}\text{ if } r=atn. \end{array}\right. \\ C_{2,r} &= {} \left\{ \begin{array}{ll} p_{int}^2 \tau _{int}+ p_{rrt \& int}^2 \tau _{rrt \& int}, &{}\text{ if } r=int, \\ \\ p_{res}^2 \tau _{res}+ p_{rrt \& res}^2 \tau _{rrt \& res} &{}\text{ if } r=res, \\ \\ p_{rrt}^2 \tau _{rrt} + p_{rrt \& int}^2 \tau _{rrt \& int}^2 \\ + p_{rrt \& res}^2 \tau _{rrt \& res}\\ + p_{rrt \& fel}^2 \tau _{rrt \& fel}+ p_{rrt \& atn}^2 \tau _{rrt \& atn} &{}\text{ if } r=rrt, \\ \\ p_{fel}^2 \tau _{fel}+ p_{rrt \& fel}^2 \tau _{rrt \& fel} &{}\text{ if } r=fel, \\ \\ p_{atn}^2 \tau _{atn}+ p_{rrt \& atn}^2 \tau _{rrt \& atn} &{}\text{ if } r=atn. \end{array}\right. \end{aligned}$$

For iteration j, if \(p_{2,r}^{(j)} > p_{2,r}^{(j-1)}\), then for patient 1:

$$\begin{aligned} \tau _{1,r}^{(j)} &= {} T_d + p_{2,r}^{(j-1)} C_{1,r} < T_d + p_{2,r}^{(j)} C_{1,r} = \tau _{1,r}^{(j+1)}, \end{aligned}$$
$$\begin{aligned} p_{1,r}^{(j)} &= {} \frac{C_{1,r}}{\tau _{1,r}^{(j)}} > \frac{C_{1,r}}{\tau _{1,r}^{(j+1)}} = p_{1,r}^{(j+1)}. \end{aligned}$$

This leads to, for patient 2,

$$\begin{aligned} \tau _{2,r}^{(j)} &= T_d + p_{1,r}^{(j)} C_{1,r} > T_d + p_{1,r}^{(j+1)} C_{1,r} = \tau _{2,r}^{(j+1)}, \end{aligned}$$
$$\begin{aligned} p_{2,r}^{(j)} &= {} \frac{C_{2,r}}{\tau _{2,r}^{(j)}} < \frac{C_{2,r}}{\tau _{2,r}^{(j+1)}} = p_{2,r}^{(j+1)}. \end{aligned}$$

The obtained results in the above four inequations complete the proof. \(\square\)

Proof of Lemma 2

Induction is used for the proof of the lemma.

Initial Step: When \(j=1\), since \(p_{2,r}^{(0)}=0\), from Eq. (14), we have

$$\begin{aligned} p_{2,r}^{(1)} > p_{2,r}^{(0)}=0. \end{aligned}$$

Then, from Lemma 1, we obtain

$$\begin{aligned} \tau _{1,r}^{(2)}> \tau _{1,r}^{(1)},\qquad p_{1,r}^{(2)}< p_{1,r}^{(1)},\qquad \tau _{2,r}^{2)} < \tau _{2,r}^{(1)},\qquad p_{2,r}^{(2)} > p_{2,r}^{(1)}. \end{aligned}$$

The base case is proved.

Inductive Step: Assume when \(j=k\), we have

$$\begin{aligned} \tau _{1,r}^{(k+1)}> \tau _{1,r}^{(k)},\qquad p_{1,r}^{(k+1)}< p_{1,r}^{(k)},\qquad \tau _{2,r}^{(k+1)} < \tau _{2,r}^{(k)},\qquad p_{2,r}^{(k+1)} > p_{2,r}^{(k)}. \end{aligned}$$

From Lemma 1, this leads to

$$\begin{aligned} \tau _{1,r}^{(k+2)}> \tau _{1,r}^{(k+1)},\qquad p_{1,r}^{(k+2)}< p_{1,r}^{(k+1)},\qquad \tau _{2,r}^{(k+2)} < \tau _{2,r}^{(k+1)},\qquad p_{2,r}^{(k+2)} > p_{2,r}^{(k+1)}. \end{aligned}$$

Thus, the case of \(j=k+1\) also holds.

By induction, we obtain that, when \(m=2\), the sequences \(p_{1,r}^{(j)}\) and \(\tau _{2,r}^{(j)}\) are monotonically decreasing, while the sequences \(p_{2,int}^{(j)}\) and \(\tau _{1,int}^{(j)}\) are monotonically increasing, \(r\in X\), \(j=1,2,\ldots\). \(\square\)

Proof of Proposition 1

From Lemma 2, we obtain the monotonicity of decreasing sequences \(p_{1,r}^{(j)}\) and \(\tau _{2,r}^{(j)}\) and increasing sequences \(p_{2,int}^{(j)}\) and \(\tau _{1,int}^{(j)}\), \(r\in X\), \(j=1,2,\ldots\). Next we show that the sequences \(\tau _{i,r}^{(j)}\) and \(p_{i,r}^{(j)}\), are bounded from above and below. For \(p_{i,r}^{(j)}\)s, from Eqs. (12) and (14), we have

$$\begin{aligned} 0< p_{i,r}^{(j)} < 1. \end{aligned}$$

For \(\tau _{i,r}^{(j)}\)s, from Eqs. (11) and (13), since \(0<p_{i,r}^{(j)}<1\), we obtain

$$\begin{aligned} T_d< \tau _{i,r}^{(j)} < T_d + C_{i,r}. \end{aligned}$$

Since the sequences \(\tau _{i,r}^{(j)}\) and \(p_{i,r}^{(j)}\), \(r \in X\); \(j=1,2,\ldots\), are monotonic and bounded from above and below, they are convergent. Thus, Level-1 iteration is convergent. \(\square\)

To prove Proposition 2, Lemma 3 is needed.

Lemma 3

Under assumptions (1)–(5), if\(\rho _{i}^{(l)} > \rho _{i}^{(l-1)}\), \(i=1,\ldots ,m\), \(l=1,2,\ldots\), then\(\rho _{i}^{(l+1)} > \rho _{i}^{(l)}\).

Proof of Lemma 3

From Eq. (8), we obtain

$$\begin{aligned} \lambda _1^{(l+1)} = T_{in}(1+\rho _1^{(l)} \Sigma _{i=2}^m \rho _i^{(l)}) > T_{in}(1+\rho _1^{(l-1)} \Sigma _{i=2}^m \rho _i^{(l-1)}) = \lambda _1^{(l)}. \end{aligned}$$

This implies that

$$\begin{aligned} \rho _1^{(l+1)} &= {} \frac{\lambda _1^{(l+1)}}{\lambda _1^{(l+1)}+T_{normal}}=\frac{1}{1+\frac{T_{normal}}{\lambda _1^{(l+1)}}}>\frac{1}{1+\frac{T_{normal}}{\lambda _1^{(l)}}}=\rho _1^{(l)}. \end{aligned}$$

When \(2 \le k \le m-1\), from (9), we have

$$\begin{aligned} \lambda _k^{(l+1)} &= T_{in}(1+\rho _k^{(l)} \left( \Sigma _{i=1}^{k-1} \rho _i^{(l+1)}+\Sigma _{i=k+1}^{m} \rho _i^{(l)}\right) \\ &> T_{in}(1+\rho _1^{(l-1)} \left( \Sigma _{i=1}^{k-1} \rho _i^{(l)}+\Sigma _{i=k+1}^{m} \rho _i^{(l-1)}\right) \\ &= {} \lambda _k^{(l)},\\ \rho _k^{(l+1)}&= \frac{\lambda _k^{(l+1)}}{\lambda _k^{(l+1)}+ T_{normal}}>\frac{1}{1+\frac{T_{normal}}{\lambda _k^{(l)}}}=\rho _k^{(l)}. \end{aligned}$$

Finally, for \(k=m\), from (10), it follows that

$$\begin{aligned} \lambda _m^{(l+1)}&= T_{in}\left( 1+\rho _m^{(l)} \Sigma _{i=1}^{m-1} \rho _i^{(l+1)} \right)> T_{in}\left( 1+\rho _m^{(l-1)} \Sigma _{i=1}^{m-1} \rho _i^{(l)}\right) =\lambda _m^{(l)},\\ \rho _m^{(l+1)} &= \frac{\lambda _m^{(l+1)}}{\lambda _m^{(l+1)}+T_{normal}}> \frac{1}{1+\frac{T_{normal}}{\lambda _m^{(l-1)}}}=\rho _m^{(l)}. \end{aligned}$$

The arguments follow directly. \(\square\)

Proof of Proposition 2

First we prove that the sequences \(\lambda _{i}^{(l)}\) and \(\rho _{i}^{(l)}\), \(i=1,2,\ldots ,m\); \(l=1,2,\ldots\), are monotonically increasing using mathematical induction.

Initial Step: When \(l=1\), since \(\rho _{i}^{(0)}=0\), from Lemma 3,

$$\begin{aligned} \rho _{i}^{(1)}>\rho _{i}^{(0)}=0. \end{aligned}$$

This leads to

$$\begin{aligned} \rho _{i}^{(2)}> \rho _{i}^{(1)}, \qquad \lambda _{i}^{(2)} > \lambda _{i}^{(1)}. \end{aligned}$$

The base case is proved.

Inductive Step: Assume when \(l=k\), we have

$$\begin{aligned} \lambda _{i}^{(k)}> \lambda _{i}^{(k-1)},\qquad \rho _{i}^{(k)}>\rho _{i}^{(k-1)}, \qquad i=1,2,\ldots ,m. \end{aligned}$$

Then from Lemma 3, we have

$$\begin{aligned} \lambda _{i}^{(k+1)}> \lambda _{i}^{(k)}, \qquad \rho _{i}^{(k+1)}>\rho _{i}^{(k)}. \end{aligned}$$

Therefore, the case where \(l=k+1\) also holds. Then, the sequences \(\lambda _{i}^{(l)}\) and \(\rho _{i}^{(l)}\), \(i=1,2,\ldots ,m\); \(l=1,2,\ldots\), are monotonically increasing.

For boundedness, it is clear that \(\rho _{i}^{(l)}\)s are bounded between 0 and 1 from Eqs. (8), (9), and (10), while \(\lambda _{i}^{(l)}\)s are also bounded according to equations (8) and (9).

Since the sequences \(\lambda _{i}^{(l)}\) and \(\rho _{i}^{(l)}\), \(i=1,2,\ldots ,m\), are both monotonic and bounded from above and below, they are convergent. \(\square\)

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Zeng, Z., Fan, Z., Xie, X. et al. A two-level iteration approach for modeling and analysis of rapid response process with multiple deteriorating patients. Flex Serv Manuf J 32, 35–71 (2020). https://doi.org/10.1007/s10696-019-09347-6

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  • Rapid response
  • Decision time
  • Mean waiting time
  • Multiple patients
  • Patient deterioration
  • Iterations