Mathematical modeling - bus dispatch optimization

This paper identifies peak periods by establishing a profit threshold model, comprehensively considers bus line resource allocation and passenger waiting time, establishes a multi-objective optimization model, optimizes the bus dispatching plan through an artificial immune algorithm, and predicts passenger flow by establishing a gradient boosting tree model. Thus predicting "peak" and "off-peak" periods.

Considering that citizens' travel is not "even", "peak" needs to be characterized as a period when there are more passengers. In order to facilitate processing, we will divide a period of time $The\; difference\; between\; the\; total\; operating\; revenue\; and\; the\; total\; operating\; cost\; within$$\Delta T$ is not less than a given "threshold" period. The key to the problem is how to establish the relationship between time, operating income and cost, and the selection of thresholds. This group established a net profit threshold model and introduced the concept of traffic index as the basis for threshold selection, thereby reasonably defining "peak" and "flat peak" ” basis for division. \par

Net profit threshold model

Taking into account the cost of operating a bus, there are many factors, including variable costs over a period of time $\tilde{C}$——Fuel consumption, wear factor, etc., and fixed cost $\overline{C}$——Driver wages and car purchase expenses. Here we consider that in order to distinguish the peak period or the off-peak period, this group first divides the departure time into a number of close intervals, and observes whether the net profit of each divided interval reaches the threshold: Ω = [ T
$$Oh=\left[T_0,T_1\right]\cup \left[T_1,T_2\right]\cup \cdots \cup \left[T_{N-1},T_N\right]$$
The length of the divided interval is:$\Delta T=T_n-T_{n-1}$
Book $The\; driving\; time\; of\; flight\; i$ is recorded as:$[t_i,t_i+t_{total}]$ , so each shift is in the sampling division area$[T_n,T_{n+1}$The cost consumed on$]$$c_n^i$：
$$c_n^i=\{\begin{array}{l}\left[\min \left(T_{n+1},t_{i+1}\right)-\max \left(T_n,t_i\right)\right]\cdot v\cdot f{T}_n⩽t_{i+1}and{T}_{n+1}⩾t_i\\ 0T_n>t_{i+1}or{T}_{n+1}<t_i\end{array}$$
Among them, Part IIThe starting time of bus$i$$t_i$：
$$t_i=T_0+\left(i-1\right)\Delta t_{i-1}$$
Then the sampling division area $[T_n,T_{n+1}$The total variable cost consumed on$]$
$$D{\tilde{C}}_N=\sum _{i=1}^M{c}_N^i$$

Therefore, the sampling division area $[T_n,T_{n+1}]$ The total cost CCconsumed$C$ is:
$$\Delta C=D\tilde{C}+D\overline{C}$$
Considering that the influencing factor of income is passenger arrival rate, let $The\; arrival\; rate\; of\; passengers\; at\; j$ stations over time is$r_j$, assuming $r_j$Obeying the uniform distribution (person/minute), the total income in the sampling divided area $ \left[ T_{n},T_{n+1} \right]$ is: Δ I = P ⋅ ∑ Δ t Δ t ⋅
$$\Delta I\_=P\cdot \sum _{\Delta t}\Delta t\cdot r_j$$

Model establishment for question 2

Assume that the passenger arrival rate follows a uniform distribution:
$$r_j\sim U\left(t_{j-1},t_j\right)$$
The average waiting time of passengers is:
$$\bar{t}=\frac{t_i-t_{i-1}}{2}$$
Book $The\; departure\; time\; of\; vehicle\; i$ is$T_i$, No. $Car\; i$ arrives (or leaves)jjThe time at station$j$$T_i^j$则:
$$T_i^j=\{\begin{array}{l}{T}_i+{\sum }_{n=2}^j{d}_{n}j>1\\ T_{i}j=1\end{array}$$
Taking into account the customer's time interests and service satisfaction, the waiting timeout rate $WT$ represents the convenience of the client:
$$WT=\frac{\text{The total number of people waiting for more than}15}{\text{The total number of people boarding the bus during the dispatch period}}$$
即为：
$$WT=\frac{\sum _{T_i^j}\sum _{h=1}^{h_{ij}}\left\{W_i^j\left(h\right)|W{T}_{ij}\left(h\right)⩾15\right\}}{\sum _{T_i^j}{P}_i\left(T_i^j\right)}$$
In the formula $W_i^j\left(h\right)$ Performance$Car\; i$ arrives at$When\; station\; j$ , hhhas already been waited for on this station.$h$ bus, but the number of people who have not yet boarded the bus, assuming that the waiting time of these people is:
$$W{T}_i^j\left(h\right)=T_i^j-T_{i-h}^j,W_i^j\left(0\right)=U_{ij}$$
In the formula $U_i^j$Expressed in $T_{i-1}^j$到$T_i^j$During the time period, the number of new people waiting for the bus at the station:
$$W{W}_i^j\left(0\right)={\int }_{T_{i-1}^j}^{T_i^j}{u}_i\left(t\right)dt$$
设$h_{ij}$represents the $k$ cars arrive$At\; station\; j$ , the number of waiting vehicles for the longest waiting passengers at that station is then$Car\; i$ leavesjjThe number of passengers on the bus at station$j$
$$P_i\left(T_i^j\right)=\{\begin{array}{ll}{a}_{ij}+{\sum }_{r=0}^{h_{ij}}{W}_i^j\left(r\right),& h_{ij}^{\ast }=0,\\ M,& h_{ij}^{\ast }>0.\end{array}$$
inside $a_{ij}$Part $Car\; i$ arrives at$After\; station\; j$ , after the passengers get off the bus, the number of remaining passengers on the bus:
$$a_{ij}=\max \left\{\left(P_k\left(T_i^{j-1}\right)-D_j\right)\text{,}0\right\}$$
where$D_j$For in jjThe number of passengers getting off at stop$j\; .$
Assume$M$ is the maximum number of people the vehicle can carry, then the number of people that can be accommodated on the vehicle at this time is recorded as:
$$b_{ij}=M-a_{ij}$$
Based on the queuing principle of first come first boarding, at $k+1$ car arrives atjjWhen station$j$$h_{ij}^{\ast }$:
$$h_{ij}^{\ast }=\max \left\{h|\sum _{r=0}^{h_{ij}}{W}_{ij}\left(r\right)⩽b_{ij}\right\},$$