CN106097226A - City Routine Transit Network Design method based on Hierarchical Programming - Google Patents

Abstract

The invention discloses a kind of City Routine Transit Network Design method based on Hierarchical Programming, the present invention goes out row mode according to passenger and gauze carries out distinguishing hierarchy, and according to " Hierarchical Programming " result, searches for the optional circuit of each level successively, and carry out demand assignment；With demand assignment structure as foundation, build one with departure frequency as variable, with passenger and network operator's cost minimization as object function, with the City Routine Transit Network Design model that the restrictive conditions such as line length, departure frequency, transport capacity, number of, lines are constraint；Use simulated annealing that model is solved, finally can determine rail network structure and circuit departure frequency, it is achieved the once one-tenth net of " Hierarchical Programming " Transit Network Design model simultaneously.Also corresponding to traffic zone in the present invention multiple set out or arrive website, more identical with the actual facilities of traffic zone periphery website, the result of calculation of model is more stable simultaneously.

Description

City Routine Transit Network Design method based on Hierarchical Programming

Technical field

The present invention relates to public transport field, be specifically related to a kind of City Routine Transit Network Design based on Hierarchical Programming Method.

Background technology

Transit villages design is the key component in public transport development plan, and it is to resident's daily life, work Trip or even the development of urban economy Community health have very important effect.Regular public traffic is the weight of urban public tranlport system Want ingredient, thus, the research to City Routine Transit Network Design has important practical significance.City Routine public transport line The design of net mainly includes determination and the calculating of circuit departure frequency of public bus network structure.Conventional Transit Network Design research In merely by a traffic zone corresponding to single website, do not correspond with the actual facilities of traffic zone periphery website, can Bigger error can be caused；Domestic existing " Hierarchical Programming " theory shortage is corresponding can determine that public bus network is tied simultaneously Structure and the method for departure frequency；And existing " Hierarchical Programming " line net design model lacks the description that passenger changes to behavior；This Outward, in existing model solution result, the website repetitive rate on circuit is higher, and cannot realize synchronization determine rail network structure and send out Car frequency.

Summary of the invention

In order to overcome shortcoming that prior art exists with not enough, the present invention provides a kind of City Routine based on Hierarchical Programming Transit Network Design method.

The present invention adopts the following technical scheme that

As it is shown in figure 1, regular public traffic gauze method for designing based on " Hierarchical Programming ", comprise the following steps:

Known road network is simplified and processes by S1, increases Centroid and distribution node.

S2 uses dijkstra's algorithm to solve the shortest path between each distribution node OD pair, then uses K shortest path to candidate Circuit expands, and initially screens solving the shortest path obtained according to length constraint, will meet condition Short-term road, as initial candidate circuit, stores to set A.

Expand, particularly as follows: the K shortest path tried to achieve is screened according to line length constraints, will meet The K shortest path of constraints is defined as candidate line, and stores in our candidate line set.

As in figure 2 it is shown, the dijkstra's algorithm basic ideas used are:

(1) initialize.Selected search starting point s, makes d (s)=0, prd (s)=0, simultaneously permanent label n node and interim mark Number node is respectively as follows: P=Φ,For other all nodes: p (i)=-1, d (i)=∞ i=1,2 ... N；Incite somebody to action Point s is labeled as permanent label n, and is transferred in permanent label n set P, makes k=1, and starts Shortest Path Searching.

(2) search permanent label n.Search smporary label setIn in permanent label n set P labelling point closest Node i, and the label of node i is become permanent label n, evenMake p (i)=1, and by node i from Smporary label setIn be transferred in permanent label n set P；K=k+1.

(3) data in road network are updated.To any other nodes j in road network, if (i, j) ∈ A (i), and d (j) > d (i) +C_ij, then d (j)=d (i)+C is made_ij, the more distance between new node j and starting point s, and make prd (j)=i.

(4) if all nodes are the most labeled in road network, i.e. k=N, then Shortest Path Searching is complete；If k < N, then return Step (2), continues the search procedure of permanent label n.

Wherein, the number of nodes during N is road network；I represents that the arbitrary node in road network is numbered, i=1,2 ... N；A (i) table Show the section being connected in road network；(i, j) represents node i and the section being connected between node j, and arbitrarily (i, j) ∈ A (i)； C_ijRepresent section (i, length j)；P is the set storing permanent label n node；For storing the set of smporary label node；s Represent the starting point of shortest path, the starting point i.e. searched for；D (i) represents the starting point s length to the shortest path of node i；P (i) is auxiliary Helping label variable, if p (i)=-1, then node i is smporary label, if p (i)=1, then node i is permanent label n；Prd (j) is Auxiliary variable, the numbering of the previous node that labelling shortest path is connected with node j, on shortest path, node chases after Track.

As it is shown on figure 3, the basic ideas of employing K shortest path algorithm are:

(1) shortest path between 1-N is tried to achieve according to Dijkstra methodAnd by it Store to set A, and make k=1, Num=N_k；

(2) judge whether k≤K sets up, if setting up, then make k=k+1, t=1 and perform step (3), otherwise, calculate knot Bundle；

(3) inspection A^K-1With A^j, j=1,2 ..., front t the node of k-1 is the most identical, if identical, then makes d_tq=∞, Wherein, q represents A^jIn the t+1 node；

(4) Q is tried to achieve according to Dijkstra method_tShortest path between-N, during noting this Shortest Path Searching, should arrange Except A^K-1Front t node；

(5) by node 1 to node Q_tBetween path store to(Part I), node Q_tRoad between node N Footpath store to(Part I)；

(6) askWillWithMerge, and willStore to set B, and make t=t+1；

(7) t is judged > N_k-1Whether-1 set up, if setting up, then performs step (8), otherwise, return step (3)；

(8) path that in search set B, length is the shortest, this path of labelling is A^K, and stored to set A, other Path remains in set B.

Wherein, i is the node serial number in road network, i=1,2 ... N, wherein 1 is starting point, and N is terminal；N_kRepresent kth bar Nodes on short path；1-i-... j, i ≠ j ≠ ... ≠ 1 represents from node 1 to the path of node j, and this path is through node I ...；Represent the i-th node on kth bar shortest path； Represent the kth bar shortest path between 1-j, whereinRepresent the 2nd on path, 3 ..., N respectively_k-1 joint Point；Represent the A of t node i on kth article shortest path^k-1Set, i.e.With A^k-1From starting point 1 starts to the t node i the most identical, and the node between the t+1 node N to terminal differs；RepresentFirst Point, i.e. the K shortest path of candidateStarting point 1 to the t node i between path, the node in this set all withIn phase With；RepresentPart II, i.e. the K shortest path of candidateThe t+1 node i N to terminal between path, this collection In closing in addition to terminal N, withWithout other same node point；K represents the serial number of shortest path, k=1,2,3 ..., K；d_ijTable Show the distance in section between node i-j.

S3 determines the line set of initial solution, public transport network is carried out distinguishing hierarchy, at level by the row mode that goes out of passenger In partition process, number of transfer and long walking number of times by passenger carry out distinguishing hierarchy, concrete distinguishing hierarchy such as Fig. 6.

S4, according to distinguishing hierarchy result, searches for each optional circuit of level passenger, by always the going out of passenger on each candidate line The row time carries out demand assignment.Considering under practical situation, passenger is generally biased towards the circuit trip selecting the travel time shorter, only Retain the travel time circuit less than or equal to minimum Trip Costs 1.3 times.After circuit has screened, further according to the trip of each circuit Time, the trip requirements between OD is allocated.

1) the first level (0 transfer 0 walking)

The generally central point (Centroid) that also non-selected traffic zone is set of website, but in the traffic zone divided Periphery arranges bus station, and therefore, the traveler in community, for realizing trip purpose, need to produce point (trip herein from trip Produce point, attraction refers both to center of housing estate point starting point) walk to periphery bus station.In like manner, after passenger getting off car, the most also need step Walk to trip purpose ground.Therefore, this goes out under row mode, and the computing formula of total travel time of each circuit is as follows:

${\mathrm{TT}}_{kij}=t_{i,ik}^{access}+t_{ik,k}^{wait}+t_{ik,k,kj}^{invt}+t_{kj,j}^{access}---\left(1\right)$

Wherein: TT_kijFor traffic zone OD to (i, j) between, kth bar 0 changes to total travel time of 0 walking circuit；For The walking time of origin cell i to a corresponding ik that gets on the bus；Represent the walking time of the kj community j to terminal that gets off； Wait the average Waiting time of circuit k at the website ik that gets on the bus for passenger；For the time of occupant ride public bus network k, i.e. car Slave site ik is to the running time of site k j.Calculating formula is respectively as follows:

$t_{i,ik}^{access}=\frac{d_{i,ik}}{v_w},t_{kj,j}^{access}=\frac{d_{kj,j}}{v_w},t_{ik,k}^{wait}=aw\frac{1}{f_{kij}},t_{ik,k,kj}^{invt}=\frac{d_{ik,kj}}{v_v}---\left(2\right)$

Wherein, the circuit of occupant ride under k represents this pattern；d_i,ikRepresent the distance between node i and node i k, d_kj,j、 d_ik,kjImplication be similar to；v_wRepresent walking speed；v_vRepresent running speed；Aw is Waiting time conversion coefficient, generally takes 0.5； f_kijRepresent that OD is to (i, j) in the departure frequency of kth candidate line of this level.

Total travel time of all candidate lines between each OD pair of this level is calculated, further according to the travel time according to formula (1) Being allocated demand OD demand, concrete distribution formula is as follows:

${\left(q_{ij}\right)}_{r_k}=\frac{1/{\mathrm{TT}}_{kij}}{\underset{r_m\&Element;R_{ij}}{\&Sigma;}(1/{\mathrm{TT}}_{mij})}\&CenterDot;q_{ij}---\left(3\right)$

Wherein, r_kRepresent (i, j) between kth bar candidate line；(q_ij)_rkRepresent that OD is to (i, trip requirements j) is assigned to Demand on kth bar candidate line therebetween；q_ijRepresent (i, j) between total trip requirements；R_ijFor in this level (i, j) between full The candidate line set of foot condition.

After the search of ground floor secondary line, to this Level Search to can the traffic zone OD of row line to being marked, Avoid repeat search.

2) the second level

If the road network of the first level cannot meet (i, j) between trip requirements, then for realizing trip purpose, passenger need transfer Or walking (select 1 transfer 0 walking trip pattern or 0 transfer 1 walking go out row mode), i.e. in the road network of the second level search for (i, j) between candidate line, and carry out assignment of traffic.Note, in the search procedure of the second level candidate line, should be got rid of Circuit on level, i.e. straightforward line, therefore in the candidate line search procedure of this level, tackle searching of line sentence Disconnected, it is to avoid the repeat search of circuit.

(1) 1 transfer 0 walking

This goes out under row mode, and the whole trip process of passenger is decomposed into following 4 processes: circuit is walked in origin cell 1 get on the bus website, take circuit 1, take circuit 2, circuit 2 is got off and is a little walked to terminal community.Simultaneously as this goes out row mode Under, passenger's trip needs carry out 1 transfer, therefore in the calculating of total travel time, penalties at 1 transfer should be increased.Concrete Total travel time calculating formula is as follows:

${\mathrm{TT}}_{kij}^1={\mathrm{TT}}_{it\left(n_1\right)t^{\&prime;}j}{t}_{i,it}^{access}+t_{it,t}^{wait}+t_{it,t,n_1}^{invt}+t_{n_1,t^{\&prime;}}^{wait}+t_{transfer\_penalty}+t_{n_1,t^{\&prime;},{\mathrm{jt}}^{\&prime;}}^{invt}+t_{{\mathrm{jt}}^{\&prime;},j}^{access}---\left(4\right)$

Wherein,Represent the second level small area OD (i, j) between total travel time of kth bar circuit；T represents that passenger takes The circuit 1 taken advantage of；It represents getting on the bus a little of circuit 1；n₁Represent transfer website, be circuit 1 get off a little, be again the upper of circuit 2 Che Dian；T ' represents the circuit 2 of occupant ride；Jt ' represents getting off a little of circuit 2；t_{transfer_penalty}Represent penalties at transfer；Its His symbol due to similar, be not stated otherwise herein.

(2) 0 transfer 1 walkings

This goes out under row mode, and passenger is without transfer, but need to carry out 1 vice-minister's walking, and it may be for starting point to getting on the bus a little Long walking or the long walking a little located to terminal of getting off.It should be noted that in the search procedure of this pattern, searching for long walking Time, need to first get rid of the situation that terminus community is adjacent.Therefore, the travel time under this pattern calculates a point following two situation.

When starting point needs long walking:

${\mathrm{TT}}_{kij}^1={\mathrm{TT}}_{{\mathrm{ii}}^{N}kj}=t_{i,i^0}^{access}+t_{i^0,i^N}^{access}+t_{i^N,k}^{wait}+t_{i^N,k,kj}^{invt}+t_{kj,j}^{access}---\left(5\right)$

Wherein, i⁰Represent the node (distribution node of cell i) that passenger is connected with road network from starting point center of housing estate point；i^NTable Show the terminal of the long walking of starting point, be also getting on the bus a little of occupant ride circuit；K is the circuit of occupant ride；Kj is debarkation stop Point；For node i⁰To node i^NWalking time；Other symbol implications are with similar.

When destination county needs long walking:

${\mathrm{TT}}_{kij}^1={\mathrm{TT}}_{{\mathrm{ikj}}^{N}j}=t_{i,ik}^{access}+t_{ik,k}^{wait}+t_{ik,k}^{invt}+t_{kj,j^N}^{access}+t_{j^N,j}^{access}---\left(6\right)$

Wherein, ik is for getting on the bus a little；K is the circuit of occupant ride；Kj represents and gets off a little；j^NRepresent the long walking of destination county eventually Point (distribution node of community j)；For node kj to node j^NWalking time；Other symbol implications are with similar.

In conjunction with result of calculation, according to following demand assignment formula, to this level searched up to OD pair, community between Demand is allocated, if this level still cannot search minizone up to circuit, then continue the search of next level circuit.

${\left(q_{ij}\right)}_{r_k}=\frac{1/{\mathrm{TT}}_{kij}^1}{\underset{r_m\&Element;R_{ij}^1}{\&Sigma;}(1/{\mathrm{TT}}_{it\left(n_1\right)t^{\&prime;}j})+\underset{r_m\&Element;R_{ij}^2}{\&Sigma;}(1/{\mathrm{TT}}_{{\mathrm{ii}}^{N}kj})+\underset{r_m\&Element;R_{ij}^3}{\&Sigma;}(1/{\mathrm{TT}}_{{\mathrm{ikj}}^{N}j})}\&CenterDot;q_{ij}---\left(7\right)$

Wherein,Represent the travel time of the second level candidate line；Represent respectively in the second level 1 transfer 0 walking, 0 transfer 1 walking (starting point director's walking), 0 transfer 1 walking (terminal director's walking) go out the candidate under row mode Line set；Other symbol implications are with above.

This Level Search of labelling is to can OD pair, all traffic zones of row line.

3) third level

This level is number of transfer and the road network layer that long walking number of times summation is 2, if first and second level all cannot search Between traffic zone up to circuit, then passenger must go out row mode by 2 transfers, 2 walkings or 1 walking of 1 transfer, Trip purpose could be realized.

(1) 2 transfer 0 walking

In this pattern, passenger need to change to through 2 times could be during starting point cell i reaches home community j, i.e. this trip Passenger need to take altogether 3 circuits (circuit 1, circuit 2, circuit 3).Wherein, circuit 2 both without starting point community, also without Terminal community, but it has a common website (transfer point) respectively with circuit 1, circuit 3, and 3 line combinations get up to constitute OD Travel route between pair.In this research, it is assumed that k₁、k₂、k₃Respectively represent passenger trip during once take circuit 1, circuit 2, circuit 3；ik₁For getting on the bus a little of circuit 1；n₁For point of getting off, first transfer point of passenger of circuit 1, also it is the upper of circuit 2 Che Dian；n₂For point of getting off, second transfer point of passenger of circuit 2, also it is getting on the bus a little of circuit 3；k₃J is getting off of circuit 3 Point.Concrete calculating formula is as follows:

$\begin{array}{c}{\mathrm{TT}}_{kij}^2={\mathrm{TT}}_{{\mathrm{in}}_1{n}_2,j}=t_{i,{\mathrm{ik}}_1}^{access}+t_{{\mathrm{ik}}_1,k_1}^{wait}+t_{{\mathrm{ik}}_1,k_1,n_1}^{invt}+t_{n_1,k_2}^{wait}+t_{transfer\_penalty}\\ +t_{n_1,k_2,n_2}^{invt}+t_{n_2,k_3}^{wait}+t_{transfer\_penalty}+t_{n_2,k_3,k_{3}j}^{invt}+t_{k_{3}j,j}^{access}\end{array}---\left(8\right)$

Wherein,Represent third level small area OD (i, j) between total travel time of kth bar circuit；Other related symbol Implication with the most above.

It should be noted that and this pattern needs change to 2 times, therefore, need in the calculating of travel time at plus 2 transfers Penalties.

(2) 0 transfer 2 walkings

Similar as above, before the search of the travel route of this pattern starts, should first get rid of: terminus cell i, j are adjacent The situation of community；The situation of a community (neighbor cell is identical) only it is spaced between terminus cell i, j.Then, search terminus Constitute OD pair of the neighbor cell of community up to circuit, and with the public transport line that circuit is occupant ride under this pattern searched Road k.It is assumed that i⁰Distribution node (node being connected with road network) for cell i；Ik be circuit k get on the bus a little, i.e. this node is little The distribution node point of district's i adjacent node；Kj is getting off a little of circuit k；j^NDistribution node for j neighbor cell, community.Total trip The calculating formula of time is:

${\mathrm{TT}}_{kij}^2={\mathrm{TT}}_{{\mathrm{ii}}^0{j}^{N}j}=t_{i,i^0}^{access}+t_{i^0,ik}^{access}+t_{ik,k}^{wait}+t_{ik,k,kj}^{invt}+t_{kj,j^N}^{access}+t_{j^N,j}^{access}---\left(9\right)$

Wherein,Represent the long walking spent time at terminus community respectively；Other symbols in formula Implication is with above.

(3) 1 transfer 1 walkings

Long gait processes in this pattern is similar with 0 transfer 1 walking trip pattern, it may occur however that at starting point, destination county Or between transfer website.Herein for simplifying the search of candidate line, in this goes out row mode, only consider long walking occur in starting point or The situation of destination county.Therefore, total travel time of this pattern calculates and also should be divided into two kinds of situations.

When starting point needs long walking:

${\mathrm{TT}}_{kij}^2={\mathrm{TT}}_{{\mathrm{ii}}^N{n}_{1}j}=t_{i,i^0}^{access}+t_{i^0,i^N}^{access}+t_{i^N,k_1}^{wait}+t_{i^N,k_1,n_1}^{invt}+t_{n_1,k_2}^{wait}+t_{n_1,k_2,k_{2}j}^{invt}+t_{k_{2}j,j}^{access}---\left(10\right)$

Wherein, i⁰Represent the node (distribution node of cell i) that passenger is connected with road network from starting point center of housing estate point；i^NTable Show the terminal of the long walking of starting point, be also occupant ride circuit k₁Get on the bus a little；k₁Circuit 1 for occupant ride；n₁For circuit k₁Get off a little, be also the transfer website of passenger；k₂Circuit 2 for occupant ride；k₂J is circuit k₂Get-off stop；Other symbols Number implication is with similar.

When destination county needs long walking:

${\mathrm{TT}}_{kij}^2={\mathrm{TT}}_{{\mathrm{in}}_1{j}^{N}j}=t_{i,{\mathrm{ik}}_1}^{access}+t_{{\mathrm{ik}}_1,k_1}^{wait}+t_{{\mathrm{ik}}_1,k_1,n_1}^{invt}+t_{n_1,k_2}^{wait}+t_{n_1,k_2,k_{2}j}^{invt}+t_{k_{2}j,j^N}^{access}+t_{j^N,j}^{access}---\left(11\right)$

Wherein, k₁Circuit 1 for occupant ride；ik₁For circuit k₁Get on the bus a little；n₁For circuit k₁Get off a little, also take advantage of The transfer website of visitor；k₂Circuit 2 for occupant ride；；k₂J is circuit k₂Get-off stop；j^NFor the long walking of destination county passenger Terminal, this node is the distribution node of community j；Other symbol implications are with similar.

In sum, the demand assignment formula of this level is as follows:

${\left(q_{ij}\right)}_{r_k}=\frac{1/{\mathrm{TT}}_{kij}^2}{\underset{r_m\&Element;R_{ij}^1}{\&Sigma;}(1/{\mathrm{TT}}_{{\mathrm{in}}_1{n}_{2}j})+\underset{r_m\&Element;R_{ij}^2}{\&Sigma;}(1/{\mathrm{TT}}_{{\mathrm{ii}}^0{j}^{N}kj})+\underset{r_m\&Element;R_{ij}^3}{\&Sigma;}(1/{\mathrm{TT}}_{{\mathrm{ii}}^N{n}_{1}j})+\underset{r_m\&Element;R_{ij}^4}{\&Sigma;}(1/{\mathrm{TT}}_{{\mathrm{in}}_1{j}^{N}j})}\&CenterDot;q_{ij}---\left(12\right)$

In above formula,Represent third level small area OD (i, j) between total travel time of kth bar circuit；Other correlative symbol Number implication with the most above.

S5, so that the Trip Costs of passenger and enterprise operation cost minimization are turned to target, sets up object function, wherein, and passenger Trip Costs include waiting for bus cost and transfer cost three part in cost, car；

Min TC=C_W+C_R+C_I+C_o

Wherein, C_WIt is the cost of waiting for bus of passenger, C_RFor the transfer cost of passenger, C_IIt is cost in the car of passenger, C_oVehicle is transported Battalion's cost；Distinguishing hierarchy in comprehensive S4 and the calculating of each level related data, the calculating formula that can obtain all kinds of cost is as follows:

$C_W=\underset{k\&Element;A}{\&Sigma;}{\mathrm{\&alpha;}}_k\&CenterDot;t_k^{wait}\&CenterDot;q_k\&CenterDot;(V_W^{NR}{\mathrm{\&beta;}}_{k1}+V_W^R{\mathrm{\&beta;}}_{k2})---\left(13\right)$

$C_I=\underset{k\&Element;A}{\&Sigma;}{\mathrm{\&alpha;}}_k\&CenterDot;(t_k^{invt}+N_k\&CenterDot;t_d)\&CenterDot;q_k\&CenterDot;V_I---\left(14\right)$

$C_R=\underset{k\&Element;A}{\&Sigma;}{\mathrm{\&alpha;}}_k\&CenterDot;{\mathrm{\&beta;}}_{kr}\&CenterDot;t_k^{transfer}\&CenterDot;q_k\&CenterDot;V_W^R---\left(15\right)$

$C_O=\underset{k\&Element;A}{\&Sigma;}{\mathrm{\&alpha;}}_k\&CenterDot;\frac{{\mathrm{dis}}_k}{v_v}\&CenterDot;f_k\&CenterDot;c---\left(16\right)$

Wherein: A is candidate line set；f_kFrequency for circuit k；α_kFor auxiliary variable, if f_k≠ 0, then take 1, otherwise, Take 0；q_kFor the flow being assigned on circuit k；β_k1、β_k2It is auxiliary variable, if current search is to go out row mode containing change to In circuit, then β_k1=0, β_k2=1, otherwise, β_k1=1, β_k2=0；Represent non-transfer mode, transfer mode respectively Under passenger's unit Waiting time cost；Represent the time of vehicle operation of circuit k；N_kStop on circuit k, vehicle is operating Pull in and count；t_dFor vehicle website (without head and the tail website) average stopping time；V_I: time cost in passenger's flat vehicle； β_krFor auxiliary variable, if current search is transfer circuit, then β to circuit k_kr=1, otherwise β_kr=0；For transfer circuit k's The average transfer time of average Waiting time, i.e. passenger；dis_kLength for circuit k；v_vTravel speed for vehicle；_cFor vehicle Travel the required unit operation cost paid of 1 km.

In sum, object function expression formula is:

$\begin{array}{c}TC=C_W+C_R+C_I+C_O\\ =\underset{k\&Element;A}{\mathrm{\&Sigma;}}{\mathrm{\&alpha;}}_k\&CenterDot;t_k^{wait}\&CenterDot;q_k\&CenterDot;(V_W^{NR}{\mathrm{\&beta;}}_{k1}+V_W^R{\mathrm{\&beta;}}_{k2})+\underset{k\&Element;A}{\&Sigma;}{\mathrm{\&alpha;}}_k\&CenterDot;{\mathrm{\&beta;}}_{kr}\&CenterDot;t_k^{transfer}\&CenterDot;q_k\&CenterDot;V_W^R+\underset{k\&Element;A}{\&Sigma;}{\mathrm{\&alpha;}}_k\&CenterDot;(t_k^{invt}+N_k\&CenterDot;t_d)\&CenterDot;q_k\&CenterDot;V_I\\ +\underset{k\&Element;A}{\&Sigma;}{\mathrm{\&alpha;}}_k\&CenterDot;\frac{{\mathrm{dis}}_k}{v_v}\&CenterDot;f_k\&CenterDot;c\end{array}$

S6 sets relevant constraint, and described constraints includes:

(1) line length l_ij

200≤l_ij≤3000

(2) frequency f_k

Same circuit bi-directional frequency is equal, and circuit departure frequency need to meet maximum, the constraint of minimum departure frequency (in example Greatly, minimum departure frequency take 30,0 respectively), meet the restriction of integer condition simultaneously, it may be assumed that

f_k=f_k+1K=1,3,5 ...

30≥f_k≥0

f_k is integer

(3) transport capacity constraint

Can the service that transport capacity constraint is gauze and is provided meet all trip requirements in road network.In practical study In, generally in units of circuit, transport capacity retrained and check, i.e. the transport capacity of circuit can all meet on circuit Maximum section flow.

q_kmax≤f_k·cap·load

Wherein, q_kmaxFor the maximum stream flow on circuit k, the statement of this parameter is the passenger flow demand on circuit, and equation is right While be the transport capacity of available transportation service, i.e. circuit in circuit, the meaning of this constraints is the transport energy of circuit Power must be able to meet passenger flow demand；Cap is vehicle capacity；Load is the maximum load factor allowed.

(4) maximum number of, lines

Obtaining more excellent solution for solving, traffic planners is maximum operating line number in arranging road network generally according to Fleet size The constraint of amount.The formulation of this constraint plan of arranging an order according to class and grade driver has significant effect.The addition of this constraints, adds optimum The practicality of gauze solution.

N_R≤N_Rmax

Wherein, N_RRepresent the number of, lines in road network；N_RmaxRepresent the maximum number of, lines in the road network solution allowed.

S7, by above-mentioned object function and constraints, arranges the nonlinear mathematics programming that can obtain Transit Network Design as follows Model:

Min TC=C_W+C_R+C_I+C_O

s.t.200≤l_ij≤3000

f_k=f_k+1K=1,3,5 ...

30≥f_k≥0

f_k is integer

q_kmax≤f_k·cap·load

N_R≤N_Rmax

S8 uses simulated annealing to solve above-mentioned nonlinear mathematics programming model, determines the line in public transport network Line structure and departure frequency.The concrete solution procedure of algorithm is as follows:

S8.1 determines the set A of all shortest paths, makes quantity n=0 of initial candidate circuit.

S8.2 determines maximum iteration time L calculated at initial temperature T, global cycle number of times G, each temperature, makes n=n+2, G=0.

S8.3 randomly chooses n bar initial candidate circuit in shortest path set A, and the frequency of initial candidate circuit is taken maximum Frequency values, t=T, l=0.

S8.4 according to the current circuit selected and frequency values thereof, by distinguishing hierarchy successively search between OD pair, each community up to Circuit, is deposited the travel route searched and related data (average Waiting time, website number, time of vehicle operation etc.) thereof by layer Storage；Further according to Level Search result, OD demand is allocated, and calculating target function TC (S₀)。

S8.5 checks level Search Results, if all having up to circuit between OD pair, all communities, if meeting condition, then continuing Following step；Otherwise, return S8.3.

S8.6 solves the largest cumulative flow of each circuit, and tests the transport capacity of each circuit, if meeting transport Capacity consistency, then record current target function value and optimal solution, i.e. TC=TC (S₀), S=S₀；Otherwise, return S8.3.

S8.7, according to the largest cumulative flow of each circuit tried to achieve, recalculates the frequency of candidate line, by newly try to achieve Frequency carries out Level Search and assignment of traffic again to gauze, and calculates the object function TC (S of correspondence_g) and optimal solution S_g；Relatively TC(S_g) and TC, if TC is (S_g) < TC then makes TC=TC (S_g), S=S_g, l=l+1.

S8.8t=t/l, if t < ε or l < L, returns S8.4；Otherwise, continue following step.

If S8.9 is g > G, then export current optimal solution, algorithm terminates；Otherwise, if g≤G, g=g+1 return S8.3.

If S8.10 is n > N_RmaxThen exporting current optimal solution, algorithm terminates；Otherwise, return S8.2.

Meaning of parameters in Fig. 1 is: g, l, k are auxiliary variable；T represents current temperature；T is initial temperature；ε represents mould Intend minimum temperature given in annealing algorithm；S_g、TC(S_g), Δ t represent solution that kth time iteration obtains, totle drilling cost, cost respectively Increment；R^betterRepresent and solve the relatively shortest path set obtained；G, L, K represent that given auxiliary parameter g, the maximum of l, k take respectively Value, represents under the maximum iteration time at the maximum iteration time of simulated annealing, each temperature, each number of, lines respectively Candidate line collection maximum search number of times；The line set that R obtains after representing adjustment.

The present invention goes out row mode according to passenger and gauze carries out distinguishing hierarchy, and according to " Hierarchical Programming " result, searches successively The optional circuit of each level of rope, and carry out demand assignment；With demand assignment structure as foundation, build one with departure frequency for becoming Amount, with passenger and network operator's cost minimization as object function, limits with line length, departure frequency, transport capacity, number of, lines etc. Condition processed is the City Routine Transit Network Design model of constraint；Use simulated annealing that model is solved, finally may be used Determine rail network structure and circuit departure frequency, it is achieved " once becoming net " of " Hierarchical Programming " Transit Network Design model simultaneously；This Traffic zone is also corresponded to by invention multiple set out or arrive website, with the actual facilities of traffic zone periphery website more Coincideing, the result of calculation of model is more stable simultaneously.

Beneficial effects of the present invention:

Present invention achieves " Hierarchical Programming " and " once becoming net " of Transit Network Design；The model proposed in the present invention is examined Consider the transfer behavior of passenger, effectively overcomed the problem that in gauze, on circuit, website repetitive rate is high, it is achieved that synchronized to determine Line construction in public transport network and the departure frequency of correspondence, and result of calculation is more stable.

Accompanying drawing explanation

Fig. 1 is the workflow diagram of the present invention；

Fig. 2 is the flow chart of dijkstra's algorithm in the present invention；

Fig. 3 is the flow chart of K shortest path algorithm in the present invention；

Fig. 4 is the road network structure figure of the embodiment of the present invention；

Fig. 5 is the road network structure figure after the embodiment of the present invention processes；

Fig. 6 is the gauze distinguishing hierarchy schematic diagram of the embodiment of the present invention；

Fig. 7 is the number of, lines variation diagram in embodiment of the present invention solving result；

Fig. 8 is this variation diagram of system synthesis in embodiment of the present invention solving result.

Detailed description of the invention

Below in conjunction with embodiment and accompanying drawing, the present invention is described in further detail, but embodiments of the present invention are not It is limited to this.

Embodiment

Fig. 4 provides a road network being made up of (two-way) 7 traffic zones, 15 nodes (crossover node), 44 sections. For ease of research, need to process former road network (increase dummy node), the road network after process is by 7 traffic zones, 15 joints Point (crossover node) constitutes 82 sections (two-way 41 sections) and constitutes, and concrete road network structure figure is as shown in Figure 5.In road network respectively Passenger's the beginning and the end demand between traffic zone, the related data of node and each traffic zone is listed in table 1～table 4 respectively.Public transport The travel speed of car is 30km/h.It is ready that the passenger's ratio arrived at by transfer is 0.05.Time valency in passenger car Value 30 (unit/people h) is reluctant that the outer time value 50 (unit/people h) of passenger car of transfer is ready the outer time valency of car of transfer passenger Value 70 (unit/people h).Vehicle capacity 70 (people/car)；The bus stop time is 0.02h every time.Passenger vehicle operating cost mainly includes combustion Oil consumption, pass cost, personnel's wage etc., add tire consume expense, personal expenditures, enterprise's operating clerical cost etc., equivalent list Position operation totle drilling cost is that 30 (unit/car km) calculates.Public bus network length is limited between 200m-3000m, and line frequency values limits System is between 0-30, and maximum load factor takes 1.3.

Table 1 passenger's OD demand

Table 2 traffic zone related data

The traffic zone ownership situation that table 3 distribution node is corresponding

Table 4 crossover node and the coordinate of distribution node

Based on data above, model can synchronize to determine the line construction in optimal gauze and departure frequency.Ask according to model The departure frequency obtained determines the circuit facilities in road network between each traffic zone.Model calculate gained road network in circuit and Departure frequency is shown in Table 5.If the departure frequency that circuit is corresponding is 0, then it is not provided with this circuit.Through 300 independent calculating, this example selects Select one group of result of wherein target function value minimum as route arrangement result.From result of calculation, optimal gauze needs altogether 4 circuits (unidirectional, two-way 8) are set, are specifically shown in Table 5.

Table 5 route arrangement result

The optimal solution of model solution gained is: totle drilling cost 15538 yuan, wherein, cost 2945 yuan of waiting for bus, change to cost 0 yuan, Time cost 5153 yuan in car, operation cost 7440 yuan.

In this optimal solution, total transfer cost of passenger is 0, and i.e. all passengers in road network are all without transfer, can accept out Through service between row OD point.Owing to, under this solution throughway, searching for first, second and third level successively, and once at current level When searching up to travel route, it, to being marked, in other follow-up Level Searchs, is no longer searched by Ji Dui community OD Rope, which greatly improves the service level of the gauze tried to achieve.

Investigate 300 independent result of calculations, it appeared that the circuit sum in each result of calculation and system synthesis basis Change is little, and result of calculation is the most stable.Choose 10 the trial result (being shown in Table 6), the stability to result of calculation in this example It is analyzed as shown in Figure 7.

Table 6 solving result stability analysis

Note: in table, number of, lines is the quantity of one-way line.

As shown in Figure 8, can find according to result of calculation repeatedly, Transit Network Design model solution based on " Hierarchical Programming " On circuit sum and system synthesis change originally, fluctuate less, i.e. the solving result of model is more stable.

Above-described embodiment is the present invention preferably embodiment, but embodiments of the present invention are not by described embodiment Limit, the change made under other any spirit without departing from the present invention and principle, modify, substitute, combine, simplify, All should be the substitute mode of equivalence, within being included in protection scope of the present invention.

Claims (8)

1. a City Routine Transit Network Design method based on Hierarchical Programming, it is characterised in that comprise the steps:

Known public transport network is simplified and processes by S1, particular by increasing Centroid and distribution node, makes each hand over The corresponding multiple starting points in logical community or arrival website；

S2 uses dijkstra's algorithm to solve the shortest path between each distribution node OD pair, and applies K shortest path algorithm to candidate Line set expands, and initially screens solving the shortest path obtained according to length constraint, will meet condition Generic line as initial candidate circuit, store in set A；

S3 determines the set of initial candidate circuit, public transport network is carried out distinguishing hierarchy and flow divides by the row mode that goes out of passenger Joining, during distinguishing hierarchy, number of transfer and long walking number of times by passenger carry out distinguishing hierarchy；

S4, according to distinguishing hierarchy result, searches for each optional circuit of level passenger, by when always going on a journey of passenger on each candidate line Between carry out demand assignment；

S5 turns to target with Trip Costs and the enterprise operation cost minimization of passenger, sets up object function, wherein, and the trip of passenger Cost includes waiting for bus cost and transfer cost three part in cost, car；

Min TC=C_W+C_R+C_I+C_o, wherein, C_WIt is the cost of waiting for bus of passenger, C_RFor the transfer cost of passenger, C_IIt it is the car of passenger Interior cost, C_oVehicle operating cost；

S6 sets relevant constraint, and described constraints includes:

(1) line length l_ijConstraint, 200≤l_ij≤ 3000, unit is rice；

(2) departure frequency f_k, the two-way departure frequency of same circuit is equal, is constrained to:

f_k=f_k+1K=1,3,5 ...；

30≥f_k≥0；

f_kis integer；

(3) transport capacity constraint,

q_kmax≤f_k·cap·load

Wherein, q_kmaxIt it is the maximum stream flow on circuit k；Cap is vehicle capacity；Load is the maximum load factor allowed；

(4) maximum number of, lines

N_R≤N_Rmax

Wherein, N_RRepresent the number of, lines in road network；N_RmaxRepresent the maximum number of, lines in the road network solution allowed；

S7, by above-mentioned object function and constraints, arranges the nonlinear mathematics programming mould that can obtain Transit Network Design as follows Type:

Min TC=C_W+C_R+C_I+C_O

s.t.200≤l_ij≤3000

f_k=f_k+1K=1,3,5 ...

30≥f_k≥0

f_k is integer

q_kmax≤f_k·cap·load

N_R≤N_Rmax

S8 uses simulated annealing to solve above-mentioned nonlinear mathematics programming model, determines the circuit knot in public transport network Structure and departure frequency.

City Routine Transit Network Design method the most according to claim 1, it is characterised in that the optional circuit tool of passenger Body refers to passenger's travel time circuit less than or equal to minimum Trip Costs 1.3 times.

City Routine Transit Network Design method the most according to claim 1, it is characterised in that concrete the solving of described S8 Process is:

S8.1 determines the set A of all shortest paths, makes quantity n=0 of initial candidate circuit；

S8.2 determines maximum iteration time L calculated at initial temperature T, global cycle number of times G, each temperature, makes n=n+2, g= 0；

S8.3 randomly chooses n bar initial candidate circuit in set of minimal paths A, and the frequency of initial candidate circuit takes maximum frequency Rate value, t=T, l=0；

S8.4 according to the current circuit selected and frequency values thereof, by distinguishing hierarchy successively search between OD pair, each community up to line Road, is stored the travel route searched and related data thereof by layer；Further according to Level Search result, OD demand is allocated, And calculating target function TC (S₀)；

S8.5 checks level Search Results, if all have up to circuit between OD pair, all communities, if meeting condition, then continues following Step；Otherwise, return S8.3；

S8.6 solves the largest cumulative flow of each circuit, and tests the transport capacity of each circuit, if meeting transport capacity Constraint, then record current target function value and optimal solution, i.e. TC=TC (S₀), S=S₀；Otherwise, return S8.3；

S8.7, according to the largest cumulative flow of each circuit tried to achieve, recalculates the frequency of candidate line, by the frequency newly tried to achieve Again gauze is carried out Level Search and assignment of traffic, and calculates the object function TC (S of correspondence_g) and optimal solution S_g；Relatively TC (S_g) and TC, if TC is (S_g) < TC then makes TC=TC (S_g), S=S_g, l=l+1；

S8.8t=t/l, if t < ε or l < L, returns S8.4；Otherwise, continue following step；

If S8.9 is g > G, then export current optimal solution, algorithm terminates；Otherwise, if g≤G, g=g+1 return S8.3；

If S8.10 is n > N_RmaxThen exporting current optimal solution, algorithm terminates；Otherwise, return S8.2；

Wherein g, l, k are auxiliary variable；T represents current temperature；T is initial temperature；ε represents in simulated annealing given Minimum temperature；S_g、TC(S_g), Δ t represent solution, totle drilling cost and the cost increase that kth time iteration obtains respectively；R^betterExpression solves The relatively shortest path set obtained；G, L, K represent given auxiliary parameter g, the maximum occurrences of l, k respectively, represent that simulation is moved back respectively Maximum iteration time at the fire maximum iteration time of algorithm, each temperature, the candidate line collection under each number of, lines are maximum Searching times；The line set that R obtains after representing adjustment.

City Routine Transit Network Design method the most according to claim 3, it is characterised in that described related data includes Average Waiting time, website number and time of vehicle operation.

City Routine Transit Network Design method the most according to claim 1, it is characterised in that described object function, tool Body is

Min TC=C_W+C_R+C_I+C_o

$C_W=\underset{k\&Element;A}{\mathrm{\&Sigma;}}{\mathrm{\&alpha;}}_k\&CenterDot;t_k^{wait}\&CenterDot;q_k\&CenterDot;(V_W^{NR}{\mathrm{\&beta;}}_{k1}+V_W^R{\mathrm{\&beta;}}_{k2})$

$C_I=\underset{k\&Element;A}{\mathrm{\&Sigma;}}{\mathrm{\&alpha;}}_k\&CenterDot;(t_k^{invt}+N_k\&CenterDot;t_d)\&CenterDot;q_k\&CenterDot;V_I$

$C_R=\underset{k\&Element;A}{\&Sigma;}{\mathrm{\&alpha;}}_k\&CenterDot;{\mathrm{\&beta;}}_{kr}\&CenterDot;t_k^{transfer}\&CenterDot;q_k\&CenterDot;V_W^R$

$C_O=\underset{k\&Element;A}{\&Sigma;}{\mathrm{\&alpha;}}_k\&CenterDot;\frac{{\mathrm{dis}}_k}{v_v}\&CenterDot;f_k\&CenterDot;c$

Wherein: A is candidate line set；f_kFrequency for circuit k；α_kFor auxiliary variable, if f_k≠ 0, then take 1, otherwise, take 0；q_k For the flow being assigned on circuit k；β_k1、β_k2It is auxiliary variable, if the line gone out in row mode for containing transfer of current search Road, then β_k1=0, β_k2=1, otherwise, β_k1=1, β_k2=0；Represent taking advantage of under non-transfer mode, transfer mode respectively Visitor's unit Waiting time cost；Represent the time of vehicle operation of circuit k；N_kFor vehicle operating stop website on circuit k Number；t_dFor vehicle at website, and the average stopping time without head and the tail website；V_IRepresent time cost in passenger's flat vehicle；β_kr For auxiliary variable, if current search is transfer circuit, then β to circuit k_kr=1, otherwise β_kr=0；For putting down of transfer circuit k All average transfer times of Waiting time, i.e. passenger；dis_kLength for circuit k；v_vTravel speed for vehicle；C is vehicle row Sail the required unit operation cost paid of 1 km.

City Routine Transit Network Design method the most according to claim 1, it is characterised in that according to passenger in described S3 The row mode that goes out public transport network is carried out distinguishing hierarchy and assignment of traffic is as follows:

S3-1 searches for the first level, and i.e. 0 changes to 0 Walking Mode,

This goes out under row mode, and the computing formula of total travel time of each circuit is as follows:

${\mathrm{TT}}_{kij}=t_{i,ik}^{access}+t_{ik,k}^{wait}+t_{ik,k,kj}^{invt}+t_{kj,j}^{access}$

Wherein,Wherein, passenger under k represents this pattern The circuit taken；d_i,ikRepresent the distance between node i and node i k, v_wRepresent walking speed；v_vRepresent running speed；Aw is Car time scale factor, generally takes 0.5；f_kijRepresent that OD is to (i, j) in the departure frequency of kth candidate line of this level；

According to total travel time of all candidate lines between each OD pair of this level, OD demand is allocated, specific as follows: ${\left(q_{ij}\right)}_{r_k}=\frac{1/{\mathrm{TT}}_{kij}}{\underset{r_m\&Element;R_{ij}}{\&Sigma;}(1/{\mathrm{TT}}_{mij})}\&CenterDot;q_{ij}$

Wherein, r_kRepresent (i, j) between kth bar candidate line；Represent that OD is to (i, trip requirements j) is assigned to kth therebetween Demand on bar candidate line；q_ijRepresent (i, j) between total trip requirements；R_ijFor in this level (i, j) between meet condition Candidate line set；

After the search of ground floor secondary line, to this Level Search to can the traffic zone OD of row line to being marked, it is to avoid Repeat search；

S3-2 searches for second layer secondary line, and described second level includes 1 transfer 0 Walking Mode and 0 transfer 1 Walking Mode；

When 1 transfer 0 Walking Mode, passenger's trip has only to carry out 1 transfer, its concrete total travel time computing formula For:

${\mathrm{TT}}_{kij}^1={\mathrm{TT}}_{it\left(n_1\right)t^{\&prime;}j}=t_{i,it}^{access}+t_{it,t}^{wait}+t_{it,t,n_1}^{invt}+t_{n_1,t^{\&prime;}}^{wait}+t_{transfer\_penalty}+t_{n_1,t^{\&prime;},{\mathrm{jt}}^{\&prime;}}^{invt}+t_{{\mathrm{jt}}^{\&prime;},j}^{access}$

Wherein,Represent the second level small area OD (i, j) between total travel time of kth bar circuit；T represents occupant ride Circuit 1；It represents getting on the bus a little of circuit 1；n₁Represent transfer website, be circuit 1 get off a little, be again getting on the bus a little of circuit 2； T ' represents the circuit 2 of occupant ride；Jt ' represents getting off a little of circuit 2；t_{transfer_penalty}Represent penalties at transfer；

When 0 transfer 1 walking, passenger is without transfer, but need to carry out 1 vice-minister's walking and be specifically divided into two kinds of situations:

When starting point to the long walking got on the bus a little, its travel time is:

${\mathrm{TT}}_{kij}^1={\mathrm{TT}}_{{\mathrm{ii}}^{N}kj}=t_{i,i^0}^{access}+t_{i^0,i^N}^{access}+t_{i^N,k}^{wait}+t_{i^N,k,kj}^{invt}+t_{kj,j}^{access}$

Wherein, i⁰The distribution node of the cell i that expression passenger is connected with road network from starting point center of housing estate point；i^NRepresent starting point The terminal of long walking, is also getting on the bus a little of occupant ride circuit；K is the circuit of occupant ride；Kj is get-off stop；For joint Point i⁰To node i^NWalking time；

The long walking a little located to terminal when getting off, its total travel time is:

${\mathrm{TT}}_{kij}^1={\mathrm{TT}}_{{\mathrm{ikj}}^{N}j}=t_{i,ik}^{access}+t_{ik,k}^{wait}+t_{ik,k}^{invt}+t_{kj,j^N}^{access}+t_{j^N,j}^{access}$

Wherein,

Ik is for getting on the bus a little；K is the circuit of occupant ride；Kj represents and gets off a little；j^NRepresent the long walking terminal i.e. community j of destination county Distribution node；For node kj to node j^NWalking time；

According to the travel time, carry out demand assignment, according to equation below:

${\left(q_{ij}\right)}_{r_k}=\frac{1/{\mathrm{TT}}_{kij}^1}{\underset{r_m\&Element;R_{ij}^1}{\&Sigma;}(1/{\mathrm{TT}}_{it\left(n_1\right)t^{\&prime;}j})+\underset{r_m\&Element;R_{ij}^2}{\&Sigma;}(1/{\mathrm{TT}}_{{\mathrm{ii}}^{N}kj})+\underset{r_m\&Element;R_{ij}^3}{\&Sigma;}(1/{\mathrm{TT}}_{{\mathrm{ikj}}^{N}j})}\&CenterDot;q_{ij}$

Wherein,Represent the travel time of the second level candidate line；Represent in the second level that 1 changes respectively Take advantage of the candidate line set under 0 walking, starting point director's walking, terminal director's walking trip pattern；

What labelling met that the second layer time searches can all traffic OD pair of row line；

The road network layer that S3-3 third level specially takes advantage of number of times and long walking number of times summation to be 2, if first and second level all cannot be searched Rope between traffic zone up to circuit, then passenger must be by 2 transfers, 2 walkings or the trip of 1 walking of 1 transfer Pattern, could realize trip purpose；

Under (1) 2 transfer 0 Walking Mode, total travel time is:

$\begin{array}{c}{\mathrm{TT}}_{kij}^2={\mathrm{TT}}_{{\mathrm{in}}_1{n}_{2}j}=t_{i,{\mathrm{ik}}_1}^{access}+t_{{\mathrm{ik}}_1,k_1}^{wait}+t_{{\mathrm{ik}}_1,k_1,n_1}^{invt}+t_{n_1,k_2}^{wait}+t_{transfer\_penalty}\\ +t_{n_1,k_2,n_2}^{invt}+t_{n_2,k_3}^{wait}+t_{transfer\_penalty}+t_{n_2,k_3,k_{3}j}^{invt}+t_{k_{3}j,j}^{access}\end{array}$

Wherein,Represent third level small area OD_(i,j)Between total travel time of kth bar circuit；

(2) 0 transfer 2 walkings:

It is assumed that i⁰Distribution node for cell i；Ik be circuit k get on the bus a little, i.e. this node be cell i adjacent node distribution joint Point point；Kj is getting off a little of circuit k；j^NFor the distribution node of j neighbor cell, community, the calculating formula of total travel time is:

${\mathrm{TT}}_{kij}^2={\mathrm{TT}}_{{\mathrm{ii}}^0{j}^{N}j}=t_{i,i^0}^{access}+t_{i^0,ik}^{access}+t_{ik,k}^{wait}+t_{ik,k,kj}^{invt}+t_{kj,j^N}^{access}+t_{j^N,j}^{access}$

Wherein,Represent the long walking spent time at terminus community respectively；In formula, the implication of other symbols is same Above；

(3) 1 transfer 1 walkings:

In this goes out row mode, only consider that long walking occurs the situation at beginning or end, therefore, when always going on a journey of this pattern Between calculate and also should be divided into two kinds of situations.

When starting point needs long walking:

${\mathrm{TT}}_{kij}^2={\mathrm{TT}}_{{\mathrm{ii}}^N{n}_{1}j}=t_{i,i^0}^{access}+t_{i^0,i^N}^{access}+t_{i^N,k_1}^{wait}+t_{i^N,k_1,n_1}^{invt}+t_{n_1,k_2}^{wait}+t_{n_1,k_2,k_{2}j}^{invt}+t_{k_{2}j,j}^{access}$

Wherein, i⁰Represent the node that passenger is connected with road network from starting point center of housing estate point；i^NRepresent the end of the long walking of starting point Point, is also occupant ride circuit k₁Get on the bus a little；k₁Circuit 1 for occupant ride；n₁For circuit k₁Get off a little, be also passenger Transfer website；k₂Circuit 2 for occupant ride；k₂J is circuit k₂Get-off stop；

When destination county needs long walking:

${\mathrm{TT}}_{kij}^2={\mathrm{TT}}_{{\mathrm{in}}_1{j}^{N}j}=t_{i,{\mathrm{ik}}_1}^{access}+t_{{\mathrm{ik}}_1,k_1}^{wait}+t_{{\mathrm{ik}}_1,k_1,n_1}^{invt}+t_{n_1,k_2}^{wait}+t_{n_1,k_2,k_{2}j}^{invt}+t_{k_{2}j,j^N}^{access}+t_{j^N,j}^{access}$

Wherein, k₁Circuit 1 for occupant ride；ik₁For circuit k₁Get on the bus a little；n₁For circuit k₁Get off a little, be also passenger Transfer website；k₂Circuit 2 for occupant ride；k₂J is circuit k₂Get-off stop；j^NFor the terminal of the long walking of destination county passenger, This node is the distribution node of community j；

In sum, the demand assignment formula of this level is as follows:

${\left(q_{ij}\right)}_{r_k}=\frac{1/{\mathrm{TT}}_{kij}^2}{\underset{r_m\&Element;R_{ij}^1}{\&Sigma;}(1/{\mathrm{TT}}_{{\mathrm{in}}_1{n}_{2}j})+\underset{r_m\&Element;R_{ij}^2}{\&Sigma;}(1/{\mathrm{TT}}_{{\mathrm{ii}}^0{j}^{N}kj})+\underset{r_m\&Element;R_{ij}^3}{\&Sigma;}(1/{\mathrm{TT}}_{{\mathrm{ii}}^N{n}_{1}j})+\underset{r_m\&Element;R_{ij}^4}{\&Sigma;}(1/{\mathrm{TT}}_{{\mathrm{in}}_1{j}^{N}j})}\&CenterDot;q_{ij}$

In above formula,Represent third level small area OD_(i,j)Between total travel time of kth bar circuit.

City Routine Transit Network Design method the most according to claim 6, it is characterised in that in the second level, search During the long walking of rope, to first get rid of starting point and the adjacent situation of terminal.

City Routine Transit Network Design method the most according to claim 1, it is characterised in that described Centroid represents The center of traffic zone, distribution node the setting out or arriving website as this traffic zone that each community is corresponding.