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Emergency evacuation problem for a multi-source and multi-destination transportation network: mathematical model and case study

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Abstract

Disasters such as earthquake or tsunami can easily take the lives of thousands of people and millions worth of property in a fleeting moment. A successful emergency evacuation plan is critical in response to disasters. In this paper, we seek to investigate the multi-source, multi-destination evacuation problem. First, we construct a mixed integer linear programming model. Second, based on K shortest paths and user equilibrium, we propose a novel algorithm (hereafter KPUE), whose complexity is polynomial in the numbers of nodes and evacuees. Finally, we demonstrate the effectiveness of algorithm KPUE by a real evacuation network in Shanghai, China. The numerical examples show that the average computation time of the proposed algorithm is 95% less than that of IBM ILOG CPLEX solver and the optimality gap is no more than 5%.

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Acknowledgements

The first author’s work was partially supported by the National Natural Science Foundation of China (Grant Nos. 71201093, 71571111), the Innovation Method Fund of China (Grant No. 2018IM020200), and the Fundamental Research Funds of Shandong University (Grant No. 2018JC055). The corresponding and third author’s work was partially supported by the National Natural Science Foundation of China (Grant Nos. 71871063, 71371052). The authors also would like to thank the Qilu Young Scholars and Tang Scholars of Shandong University for financial and technical supports.

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Correspondence to Yingxue Zhao.

Appendices

Proof

Proof of Proposition 1

We define \(T_a^*\) as the optimal solution value and \(N_a^{k,*}\) as the corresponding number of evacuees that are allocated to path \(p_a^k\). According to the definition of \(T_a\), we have \(T_a=\max _{1 \le k \le \hat{K}(a)}\{t_a^k(D_0)-1+\left\lceil \frac{N_a^{k}}{f_a^k} \right\rceil \}\). Hence, we have

$$\begin{aligned} \begin{aligned} T_a^* \ge t_a^k(D_0)-1+\left\lceil \frac{N_a^{k,*}}{f_a^k} \right\rceil ; \ \\ T_a^* \cdot f_a^k \ge (t_a^k(D_0)-1)f_a^k+N_a^{k,*}. \end{aligned} \end{aligned}$$

\(\square \)

Summing over k, we can get \(T_a^* \cdot \sum _{k=1}^{\hat{K}(a)}{f_a^k} \ge \sum _{k=1}^{\hat{K}(a)}(t_a^k(D_0)-1)f_a^k+{ ICS}_a\). We have

$$\begin{aligned} \begin{aligned} T_a^* \ge \frac{\sum _{k=1}^{\hat{K}(a)}{\left( t_a^k(D_0)-1\right) f_a^k}+{ ICS}_a}{\sum _{k=1}^{\hat{K}(a)}{f_a^k}}. \end{aligned} \end{aligned}$$

We denote

$$\begin{aligned} \begin{aligned} \overline{T}_a&=\frac{\sum _{k=1}^{\hat{K}(a)}{(t_a^k(D_0)-1)f_a^k}+{ ICS}_a}{\sum _{k=1}^{\hat{K}(a)}{f_a^k}}; \\ \overline{N}_a^k&=f_a^k \cdot \overline{T}_a-(t_a^k(D_0)-1)\cdot f_a^k \\&=f_a^k \cdot \left( \overline{T}_a-t_a^k(D_0)+1\right) . \end{aligned} \end{aligned}$$

In the following, we proceed the analysis according two cases: (a) \(\overline{T}_a\) is integral; (b) \(\overline{T}_a\) is fractional.

Case (a) If \(\overline{T}_a\) is integral, \(\overline{N}_a^k\) can be evacuated in \(\overline{T}_a\), and a total of \(\sum _{k=1}^{\hat{K}(a)}{\overline{N}_a^k}={ ICS}_a\) people can be evacuated.

Case (b) If \(\overline{T}_a\) is fractional, denote \(\hat{N}_a^k=\left\lceil \overline{N}_a^k \right\rceil \), and \(\hat{N}_a^k\) people can be evacuated in \(\left\lceil \overline{T}_a \right\rceil \).

Because \(\left\lceil \frac{\overline{N}_a^k}{f_a^k}\right\rceil \ge \frac{\overline{N}_a^k}{f_a^k}\), we can get \(\left\lceil \frac{\overline{N}_a^k}{f_a^k} \right\rceil \cdot f_a^k \ge \overline{N}_a^k\). Hence, we have \(\left\lceil \frac{\overline{N}_a^k}{f_a^k} \right\rceil \cdot f_a^k \ge \left\lceil \overline{N}_a^k\right\rceil =\hat{N}_a^k\) and \(\left\lceil \frac{\overline{N}_a^k}{f_a^k} \right\rceil \ge \frac{\hat{N}_a^k}{f_a^k}\). So, we also have

figure j

Because \(\overline{N}_a^k \le \hat{N}_a^k\), we have

figure k

According to (A.1) and (A.2), we can get \(\left\lceil \frac{\overline{N}_a^k}{f_a^k} \right\rceil = \left\lceil \frac{\hat{N}_a^k}{f_a^k} \right\rceil \). Hence, sending \(\hat{N}_a^k\) people to super destination node \(D_0\) need \(\left\lceil \frac{\overline{N}_a^k}{f_a^k} \right\rceil +t_a^k(D_0)-1=\left\lceil \overline{T}_a \right\rceil \). All \(\sum _{k=1}^{\hat{K}(a)}{N_a^k}> { ICS}_a\) people can be evacuated in \(\left\lceil \overline{T}_a \right\rceil \).

To sum up case (a) and (b), \(T_a^*=\left\lceil \frac{\sum _{k=1}^{\hat{K}(a)}{(t_a^k(D_0)-1)f_a^k}+{ ICS}_a}{\sum _{k=1}^{\hat{K}(a)}{f_a^k}}\right\rceil \) is the optimal solution.

Data and evacuation networks of the literature (Chalmet et al. 1982; Lu et al. 2005)

In Sect. 5, we present different scenarios for numerical study. The evacuation network of Scenario S01 and S02 are presented in Fig. 6. The number of evacuees at source nodes [N1, N2, N3] from S01 are [10, 5, 15], while the number of evacuees at source nodes [N1, N2, N3] from S02 are [220, 240, 450]. The evacuation network of Scenarios S11, S12, S13, S14, S15 and S16 are presented in Fig. 1. The number of evacuees at source nodes [N1, N2, N3] from S11 are [220, 240, 450]. The number of evacuees at source nodes [N1, N2, N3, N4] from scenarios S12, S13, S14 and S15 are [150, 350, 305, 105], [300, 700, 600, 200], [500, 1150, 1050, 300], [850, 1800, 1550, 400], respectively. The number of evacuees at source nodes [N1, N2, N3, N4, N5, N6, N7, N8] from scenario S16 are [100, 100, 200, 100, 110, 100, 100, 100]. The evacuation network of Scenario S21 are presented in Fig. 7. The number of evacuees at source nodes [N1, N2, N3] from S21 are [220, 240, 450].

Fig. 6
figure 6

Evacuation network from Lu et al. (2005)

Fig. 7
figure 7

Evacuation network from Chalmet et al. (1982)

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Zhang, J., Liu, Y., Zhao, Y. et al. Emergency evacuation problem for a multi-source and multi-destination transportation network: mathematical model and case study. Ann Oper Res 291, 1153–1181 (2020). https://doi.org/10.1007/s10479-018-3102-x

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