# Cell Model Assignment

### Cell transmission mechanism

To solve the DAURTN based on the continuous transmission, we build the CTM for the urban rail network (*V*,*A*). For describing the CTM, the cell transmission network is constructed based on cell from the network as follows.

We define each section as a cell chain, and each station as a station cell. For any section , as travel time is fixed, we divide the section into several transmission cells by Δ*T*. The transmission cells of one section compose a cell chain, and passengers flows can be transmitted forward between the transmission cells. Note that travel time may not be exactly divided by Δ*T*, so the time length of the tail cell can be equal to or exceed Δ*T*. The number of cells divided is , where ⌊*x*⌋ is a function of the maximum integer no larger than *x*. Therefore, we denote Cell(*L*_{ld},*i*,*j*) as the *j*th cell in the cell chain of section . Denote *m*(*L*_{ld},*i*) as the number of cells in the *i*th cell chain, and for simplifying the notation, denote Cell(*L*_{ld},*i*,*end*) the last cell in the cell chain of section . Denote *y*_{h}(*L*_{ld},*i*,*j*,*t*_{k}) as the flow of Cell(*L*_{ld},*i*,*j*) traveling to destination *s*^{h} in *t*_{k}. We also denote Cell(*s*^{u}) as the station cell of *s*^{u}, and *y*_{h}(*s*^{u},*t*_{k}) as the flow of Cell(*s*^{u}) traveling to destination *s*^{h} in *t*_{k}.

The transmission relationship between cells is illustrated in **Fig 3**, where a hollow node represents a station cell, a solid node represents a transmission cell, a hollow rectangle represents the corresponding cell chain of a section, and the arrows represent transmission directions. In each time interval, passengers follow the instantaneous dynamic route choice principle [22] and are transmitted between cells. Transmission mechanism between cells is designed and transmission processes of flows between the cells are classified into 4 groups: where ‘⟹’ means the transmission process of flow from the left cell to the right cell.

The flows of station cells and transmission cells at the initial interval *t*_{0} are 0. In *t*_{k}(*k* ≥ 1), the O–D demand {*q*_{uh}(*t*_{k})|*s*^{h} ∈ *S*} inflows into station cell Cell(*s*^{u}). Therefore, the initial value of the variables are *y*_{h}(*s*^{u},*t*_{0}) = 0, *y*_{h}(*s*^{u},*t*_{k}) = *q*_{uh}(*t*_{k}), *y*_{h}(*L*_{ld},*i*,*j*,*t*_{k}) = 0, ∀*h*,*u*,*i*,*j*,*L*_{ld},*k* ≥ 1.

Next, we analyze the transmission mechanism in 3 steps.

Step 1: The transmission processes Cell(*L*_{ld},*i*,*end*) ⟹ Cell(*L*_{ld},*i* + 1,1), and

As the length of time in the tail cell Cell(*L*_{ld},*i*,*end*) is equals to or greater than Δ*T*, only a certain proportion of flow can outflow, and the proportion is . Thus, the outflow of Cell(*L*_{ld},*i*,*end*) in interval *t*_{k} is (9)

Then we can obtain the detained flow of the tail cell (10) where ‘←’ denotes that the value of the right variable is assigned to the left variable.

According to the space priority principle, the outflow *f*_{h}(*L*_{ld},*i*,*end*,*t*_{k}) of tail cell Cell(*L*_{ld},*i*,*end*) has only two choices, i.e., Cell(*L*_{ld},*i* + 1,1) or , and the transmission choice is determined by the instantaneous dynamic route choice principle [22], i.e., the shortest path from platform to destination station *s*^{h} at the current time interval in network (*V*,*A*). If the shortest path passes through , then flow *f*_{h}(*L*_{ld},*i*,*end*,*t*_{k}) is transmitted into cell Cell(*L*_{ld},*i* + 1,1); otherwise, it is transmitted to station cell . The above transmission choice is similar to the all-or-nothing assignment, i.e., the flows follow the shortest path.

We denote the set of destinations to which the shortest path from platform passes through as (11)

When , the flow *f*_{h}(*L*_{ld},*i*,*end*,*t*_{k}) is transmitted from Cell(*L*_{ld},*i*,*end*) to Cell(*L*_{ld},*i* + 1,1). Thus, (12)

When , the flow *f*_{h}(*L*_{ld},*i*,*end*,*t*_{k}) is transmitted from Cell(*L*_{ld},*i*,*end*) to . Note that the average transfer time at station is , so (13)

In order to realize the FCFS in transmission mechanism, we introduce a variable *x*_{h}(*s*^{u},*t*_{v}), 1 ≤ *v* ≤ *N*, which represents the flows arriving at station *s*^{u} in *t*_{v} and detained at the station in *t*_{k}.

Step 2: The transmission process Cell(*L*_{ld},*i*,*j*) ⟹ Cell(*L*_{ld},*i*,*j* + 1)

After Step 1, in the tail cell of the cell chain, there may be some detained flows, and then the flow of the tail cell equals to the detained flows plus the flows from Cell(*L*_{ld},*i*,*end* − 1), so (15)

For other cells in the chain, it is only need to move flows from the forward cell to the backward cell in the chain, that is, (16)

Step 3: The transmission process

According to the principle of the space priority, flows from Cell(*L*_{ld},*i* − 1,*end*) are transmitted to Cell(*L*_{ld},*i*,1) and occupy the capacity of Cell(*L*_{ld},*i*,1) with priority. Thus, the surplus capacity of Cell(*L*_{ld},*i*,1) in *t*_{k} is (17) The flows in *t*_{k}, which are queuing at station *s*^{u} and head to Cell(*L*_{ld},*i*,1), have to compete for the surplus capacity with the FSFC principle.

In order to determine the queuing flow, passengers at station cell Cell(*s*^{u}) first determine which platform to queue. Similar to the method in Step 1, passengers determined the platform by the shortest path from station *s*^{u} to destination station *s*^{h} at the current time interval in network (*V*,*A*). If the shortest path passes through , then flows traveling to destination station *s*^{h} queue on platform . We denote the set of destinations to which the shortest path from station *s*^{u} passes through as (18) Thus, the flow competing for the surplus capacity is .

If , then (19)

If , which means the surplus capacity is insufficient, then there exists , and it makes that According to the FCFS principle, the flow traveling to each destination station can be transmitted, i.e., (20) and a portion of can also be transmitted. According to the equal proportion principle, the proportion of flows transmitted can be calculated by (21) Then (22)

After the above processes, the flow of each cell make a choice by the shortest paths and are all transmitted to the next cell. But the cost of each arc will be changed with the variable flow, so the method of successive average (MSA) is adopted to reach the instantaneous dynamic user optimal state in each time interval. The variables in the above model are updated in MSA.

### An efficient method for solving the shortest path

In the CTM, it is needed to solve the shortest path in *t*_{k} from s ∈ *S* ∪ *S*_{Ω} to *s*^{u} ∈ *S* in network (*V*,*A*). We design a fast method for solving the shortest path as follows.

If the shortest path from *s* ∈ *S* ∪ *S*_{Ω} to *s*^{u} ∈ *S* passes through several transfer stations, then the shortest path can be divided into three segments at most. The first segment of the shortest path is from origin *s* to the first transfer station , and its length is denoted as . The last segment of the shortest path is from the last transfer station to destination *s*^{u}, and its length is . As long as we solve the length of the shortest path between any two transfer stations , we can obtain the cost of the shortest paths in three cases as follows: (23)

If the shortest path from *s* ∈ *S* ∪ *S*_{Ω} to *s*^{u} ∈ *S* does not pass through any transfer station, then it will only use one line and can be solved easily.

The above analysis indicates that the solving method for the shortest path from *s* ∈ *S* ∪ *S*_{Ω} to *s*^{u} ∈ *S* can be decomposed into 3 steps.

Step 1: calculate the shortest path from *s* ∈ *S* ∪ *S*_{Ω} to *s*^{u} ∈ *S* in each network *G*(*L*_{lU},*L*_{lD}), which composed of a pair of opposite directional lines *L*_{lU},*L*_{lD} shown in **Fig 1**.

Step 2: calculate the shortest path between each two transfer stations in the network.

Step 3: calculate all the shortest paths from *s* ∈ *S* ∪ *S*_{Ω} to *s*^{u} ∈ *S*.

In step 1, for any destination , we can structure two subsets of nodes bounded by node , i.e. and , which can form two generated sub-networks of *G*(*L*_{lU},*L*_{lD}). Obviously, the shortest paths from other nodes to *s*^{u} in the two sub-networks are equal to solving the shortest paths in *G*(*L*_{lU},*L*_{lD}). In the former sub-network, there are three cases.

Case 1: solve the shortest paths from nodes to *s*^{u} along the directional line *L*_{lU};

Case 2: solve the shortest paths from

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