Mathematical modeling -- bus scheduling optimization

This paper identifies the peak period of Gaoping by establishing a profit threshold model , Comprehensively consider the allocation of bus line resources and passenger waiting time , Establish a multi-objective optimization model , Optimize the bus scheduling scheme through artificial immune algorithm , Predict the passenger flow by establishing a gradient lifting tree model , So as to predict “ peak ” and “ Flat peak ” period .

Considering that citizens' travel is not “ uniform ” Of ,“ peak ” The need for representation refers to the period when there are more passengers , For the convenience of handling , We will spend some time $\Delta T$ The difference between the total operating revenue and the total operating cost within the is not less than a given “ threshold ” Time period of . The key to the problem is how to establish the relationship between time and operation revenue and cost, as well as the choice of threshold , This group establishes a net profit threshold model , The concept of traffic index is introduced as the basis for threshold selection , Thus, it reasonably defines “ peak ” And “ Flat peak ” The division basis of .\par

Net profit threshold model

Considering the cost of operating buses, there are many factors , Including variable costs over a period of time $\tilde{C}$—— Fuel consumption 、 Wear factor, etc , And fixed costs $\overline{C}$—— Driver's salary and car purchase expenses , Here we consider to distinguish peak period or peak period , This group first divides the departure time into several tight intervals , Observe whether the net profit of each division reaches the threshold ：
$$\Omega =\left[T_0,T_1\right]\cup \left[T_1,T_2\right]\cup \cdots \cup \left[T_{N-1},T_N\right]$$
The division interval length is ：$\Delta T=T_n-T_{n-1}$
Set the first $i$ The travel time of the shift is recorded as :$[t_i,t_i+t_{total}]$, Therefore, each shift is divided into areas by sampling $\left[T_n,T_{n+1}\right]$ The cost consumed on is $c_n^i$：
$$c_n^i=\{\begin{array}{l}\left[\min \left(T_{n+1},t_{i+1}\right)-\max \left(T_n,t_i\right)\right]\cdot v\cdot f{T}_n⩽t_{i+1}and{T}_{n+1}⩾t_i\\ 0T_n>t_{i+1}or{T}_{n+1}<t_i\end{array}$$
among , The first $i$ The starting time of the bus is $t_i$：
$$t_i=T_0+\left(i-1\right)\Delta t_{i-1}$$
Then sample and divide the area $\left[T_n,T_{n+1}\right]$ The total cost of variable costs consumed on is ：
$$\Delta {\tilde{C}}_N=\sum _{i=1}^M{c}_N^i$$

therefore , Sample and divide the area $[T_n,T_{n+1}]$ Total cost consumed on $C$ by ：
$$\Delta C=\Delta \tilde{C}+\Delta \overline{C}$$
Considering that the influencing factor of income is the arrival rate of passengers , Set the first $j$ The arrival rate of passengers at stations over time is $r_j$, hypothesis $r_j$ Subject to uniform distribution ( people / minute ), Then sample and divide the area $ \left[ T_{n},T_{n+1} \right]$ The total income on the is ：
$$\Delta I=P\cdot \sum _{\Delta t}\Delta t\cdot r_j$$

problem 2 Model building

Suppose the arrival rate of passengers follows a uniform distribution, that is ：
$$r_j\sim U\left(t_{j-1},t_j\right)$$
The average waiting time of passengers is ：
$$\bar{t}=\frac{t_i-t_{i-1}}{2}$$
Set the first $i$ The departure time of each train is $T_i$, The first $i$ Cars arrive （ Or leave ） The first $j$ The time of standing is ：$T_i^j$ be :
$$T_i^j=\{\begin{array}{l}{T}_i+{\sum }_{n=2}^j{d}_{n}j>1\\ T_{i}j=1\end{array}$$
Considering the time interests of customers 、 Service satisfaction , Introduce waiting overtime rate $WT$ Show the convenience of the client :
$$WT=\frac{\text{Waiting time exceeds}15\text{The total number of people in minutes}}{\text{The total number of people on board during the dispatching period}}$$
That is to say ：
$$WT=\frac{\sum _{T_i^j}\sum _{h=1}^{h_{ij}}\left\{W_i^j\left(h\right)|W{T}_{ij}\left(h\right)⩾15\right\}}{\sum _{T_i^j}{P}_i\left(T_i^j\right)}$$
In style $W_i^j(h)$ Show the third $i$ The car arrives at $j$ When standing , The station has been waiting $h$ Vehicles , The number of people who still haven't got on the bus , Let these people have waited for :
$$W{T}_i^j\left(h\right)=T_i^j-T_{i-h}^j,W_i^j\left(0\right)=U_{ij}$$
In style $U_i^j$ It means that $T_{i-1}^j$ To $T_i^j$ Within the time frame , The number of new people waiting at the station :
$$W{W}_i^j\left(0\right)={\int }_{T_{i-1}^j}^{T_i^j}{u}_i\left(t\right)dt$$
set up $h_{ij}$ It means the first one $k$ Cars arrive $j$ When standing , The number of waiting vehicles for the longest waiting passengers at the station , Is the first $i$ The car left the $j$ The number of passengers on the train at the station can be expressed as ：
$$P_i\left(T_i^j\right)=\{\begin{array}{ll}{a}_{ij}+{\sum }_{r=0}^{h_{ij}}{W}_i^j\left(r\right),& h_{ij}^{\ast }=0,\\ M,& h_{ij}^{\ast }>0.\end{array}$$
among $a_{ij}$ For the first time $i$ The car arrives at $j$ Post station , Wait for the passengers to get off , The number of remaining passengers on the train :
$$a_{ij}=\max \left\{\left(P_k\left(T_i^{j-1}\right)-D_j\right)\text{,}0\right\}$$
In style $D_j$ For in the $j$ Number of passengers getting off at the station .
set up $M$ It is the maximum number of people carried by the vehicle , Then the number of passengers that can be accommodated at this time is recorded as ：
$$b_{ij}=M-a_{ij}$$
Based on the queuing principle of first come, first get on , In the $k+1$ The car arrives at $j$ Standing time , The maximum number of times you still have to wait for a vehicle $h_{ij}^{\ast }$:
$$h_{ij}^{\ast }=\max \left\{h|\sum _{r=0}^{h_{ij}}{W}_{ij}\left(r\right)⩽b_{ij}\right\},$$