## 2 Model Formulation
### Platform Exit Width Range
#### 2.2.1 Passenger Flow Assignment Based on Path Utility
Constraint (4) requires that the total passenger flow of a node pair must be equal to the sum of passenger flows on all related paths. Inequality (5) regulates the threshold of passenger flow on a link. Theorem 1: _The optimal solution of the passenger flow assignment model proposed in this study is equivalent to the passenger selection based on path utility._  
Proof of Theorem 1: Model (3) is a convex optimization problem with linear constraints, transforming constraints (4) and (5) into Lagrange multipliers and incorporating them into model (3) to construct model (6) [43]. \[ \begin{array}{l}\text{min}Z=\sum\limits_{(r,s)}\sum\limits_{m\in W_{rs}}U_{rs }^{m}.f_{rs}^{m}+\frac{1}{\theta}\sum\limits_{(r,s)}\sum\limits_{m\in W_{rs}} f_{rs}^{m}(\ln f_{rs}^{m}-1)+\\ \sum\limits_{(r,s)}\pi_{rs}.(\sum\limits_{m\in W_{rs}}f_{rs}^{m}-q_{rs})+\sum \limits_{p}\pi_{p}.(\sum\limits_{(r,s)}\sum\limits_{m\in W_{rs}}\delta_{p}^{m }.f_{rs}^{m}-\sigma_{p}^{T})\end{array}  \tag{6}\]  
For any \(f_{rs}^{m},\forall(r,s),\forall m\),  
\[\frac{\partial Z}{\partial f_{rs}^{m}}=U_{rs}^{m}+\frac{1}{\theta}.\ln f_{rs}^ {m}+\pi_{rs}+\sum\limits_{p}\pi_{p}.\delta_{p}^{m} \tag{7}\]  
Let \(\frac{\partial Z}{\partial f_{rs}^{m}}=0\); then,  
\[f_{rs}^{m}=\exp(-U_{rs}^{m}-\pi_{rs}-\sum\limits_{p}\pi_{p}.\delta_{p}^{m}) \tag{8}\]Because \(\pi_{rs},\pi_{p}>0\), \(\sum\limits_{(r,s)}\sum\limits_{m\in A}\delta_{p}^{m}f_{rs}^{m}-\sigma_{p}^{T}=0\), and \(\sum\limits_{m\in A}f_{rs}^{m}-q_{rs}=0\), Equation (8) is just the KKT (Karush-Kuhn-Tucker) condition of model (3). Meanwhile, it is notable that the probability of passenger selection can be expressed as  
\[ \begin{array}{l}\Pr{b}_{rs}^{m}=\frac{f_{rs}^{m}}{\sum\limits_{m\in W_{rs}}f_ {rs}^{m}}=\frac{\exp(-U_{rs}^{m}-\pi_{rs}-\pi_{p}\cdot\delta_{p}^{m})}{\sum \limits_{m\in W_{rs}}\exp(-U_{rs}^{m}-\pi_{rs}-\pi_{p}\cdot\delta_{p}^{m})}\\ =\frac{\exp(\tilde{U}_{rs}^{m})}{\sum\limits_{m\in W_{rs}}\exp( \tilde{U}_{rs}^{m})}\end{array}  \tag{9}\]  
\(\square\)  
It is obvious that Formulas (2) and (9) are equivalent. Therefore, Theorem 1 has been proven. By using Theorem 1, we can directly solve the passenger flow assignment problem through calculating the path utilities to simplify solving the original problem. \[f(p,T)=\sum\limits_{(r,s)}\sum\limits_{m\in W_{rs}}\delta_{p}^{m}\cdot f_{rs}^ {m} \tag{10}\]  
Constraints (11)-(17) ensure that the proposed model conforms to the application scenario in this paper, where \(c_{i}\) represents the number of passengers disembarking from the train on track \(i\), which follows a normal distribution, \(c_{i}\sim N(\mu_{i},\sigma_{i}^{2})\). The gap between the cumulative probability of passenger flow that satisfies a normal distribution and 100% is defined as the risk probability. Thus, \(\Phi_{c_{i}}^{-1}(90\%)\) means the number of passengers corresponding to the cumulative probability of 90%, and the risk probability is 10%. The risk probability quantifies the risk of passengers falling off the platform from a probabilistic perspective. \[q_{A}^{T}=\sum\limits_{i}q_{Ai}^{T}=\sum\limits_{i}\frac{l_{AB}}{l_{AB}+l_{CB }+l_{B}}\cdot\Phi_{c_{i}}^{-1}(90\%),T=1 \tag{11}\]  
\[q_{C}^{T}=\sum\limits_{i}q_{Ci}^{T}=\sum\limits_{i}\frac{l_{CB}}{l_{AB}+l_{CB }+l_{B}}\cdot\Phi_{c_{i}}^{-1}(90\%),T=1 \tag{12}\]  
\[q_{B}^{T}=\sum\limits_{i}q_{Ci}^{T}=\sum\limits_{i}\frac{l_{B}}{l_{AB}+l_{CB }+l_{B}}\cdot\Phi_{c_{i}}^{-1}(90\%),T=1 \tag{13}\]  
\[q_{AB}^{T}=q_{A}^{T}=\sum\limits_{m\in\omega_{AB}}f_{AB}^{m},T=1 \tag{14}\]  
\[q_{CB}^{T}=q_{C}^{T}=\sum\limits_{m\in\omega_{CB}}f_{CB}^{m},T=1 \tag{15}\]  
\[q_{B}^{T}=q_{BD}^{T}+q_{BE}^{T},T=1 \tag{16}\]  
For period 2 (\(T=2\)), only passengers at node B proceed to exit D or E. Constraint (17) indicates that the passengers departing from node B in the second period are equal to the arrived passengers from nodes A and C in the first period. \[q_{BD}^{T}+q_{BE}^{T}=q_{AB}^{T-1}+q_{CB}^{T-1},T=2 \tag{17}\]