A passenger pick-up and drop-off scheduling method for customized bus systems between urban areas

Abstract

The present invention discloses a passenger pick-up and drop-off scheduling method for a customized bus system between urban areas, the steps of which are as follows: all stops are divided into a starting point area containing n starting points and a terminal area containing m terminal points, and a travel node sequence set Y={y 1 ,y 2 ,…,y n+m}; According to the inverse of the distance of each sequence, find the probability of each sequence being selected, sequence y i The probability of being selected is #imgabs0# and then based on the random number in the interval [0.0,1.1], the determination is to use the sequence y i nodes, or randomly add nodes to generate a new travel node sequence y 0 , and then iterate continuously until convergence; the present invention converts the path weight into a probability distribution, and on the basis of continuously passing on the advantageous subsequent nodes, adopts a 10% mutation probability for each node, which can quickly iterate and evolve; since the data exchange with the memory is reduced, the running speed can be improved; the iterative evolution of the present invention has strong cohesion and can conveniently realize parallel computing.

Description

Technical Field

[0001] The invention relates to road traffic, and in particular to a passenger pick-up and drop-off scheduling method for a customized public transportation system between urban areas. Background Art

[0002] For commuting between urban areas, bus companies have launched customized bus systems between urban areas, which can provide services close to carpooling at prices based on bus transfers; while ensuring the completion of pick-up and drop-off tasks, optimizing the shortest path is a core element for bus companies to reduce operating costs; the closest prior art to the present invention is the invention patent "A Customized Bus System and Dispatch Method between Urban Areas" with application number 202210086214.X. The customized bus passenger pick-up and drop-off path selection method proposed in the invention can quickly complete relatively optimal path selection, but it cannot continuously optimize, especially when there are a large number of passengers to be picked up and dropped off, the selected path may be significantly different from the optimal path. Summary of the invention

[0003] Purpose of the invention: The purpose of the present invention is to provide a passenger pick-up and drop-off scheduling method for a customized bus system between urban areas, so that a better route can be found within a reasonable time when there are many starting and ending points for picking up and dropping off passengers.

[0004] Technical solution: The passenger pick-up and drop-off scheduling method of the inter-city customized public transportation system described in the present invention comprises the following steps:

[0005] (1) Let the bus departure point be v 0 , the starting points of all k passengers assigned to this bus are added to the set Vo, that is, Vo = {v 1 ,v 2 ,…,v n}, n is the starting number of passengers on this bus, v 1 ~v n is the starting point of k passengers, n≤k, and the destinations of all passengers on this bus are added to the set Vd, that is, Vd= {v n+1 ,v n+2 ,…,v n+m},v n+1 ~v n+m is the destination of k passengers, m≤k, and Vo and Vd are combined into a stop point set V=Vo∪Vd={v 1 ,v 2 ,…,v n ,v n+1 ,v n+2 ,…,v n+m};

[0006] (2) Based on the static map data, list the OD matrix D of Vo→Vo (Vo→Vo)=(d i,j ) n×n , where i∈Vo,j∈Vo; list the OD matrix D of Vo→Vd (Vo→Vd) =(d i,j ) n×m , where i∈Vo,j∈Vd; list the OD matrix D of Vd→Vd (Vd→Vd) =(d i,j ) m×m , where i∈Vd, j∈Vd, the above d i,j is the distance traveled by bus from starting point i to end point j. The distances mentioned below are all distances traveled by bus;

[0007] (3) Construct the initial travel node sequence set Y = {y 1 ,y 2 ,…,y n+m}, where y 1 ~y n+m They are all travel node sequences, and their construction method can be created first Here ← means assigning the value on the right to the variable on the left, and then filling the set Y until Y contains n+m different sequences:

[0008] In step (3), the steps of filling the set Y are:

[0009] (3.1) Create a sequence Create a process set Temp←Vo and create an intermediate variable i←v 0 ;

[0010] (3.2) According to the distance from i to each node in the Temp set, obtain the distance from i to any node v in Temp j Probability where v j ∈Temp, From i to v j distance, |Temp| is the number of elements in the set Temp, p←generates a random number in the interval [0.0,1.0], and Distribution in Temp, get the next node v t The method is Temp in v 1 The condition for a node to become the next node of i is a random number v j The condition for a node to become the next node of i is a random number

[0011] (3.3)Temp←(Temp-{v t}), y 0 ←(y 0 ⊙{v t}), ⊙ means to connect the sequences before and after it in order, i←v t , if |Temp|>1, return to step (3.2), otherwise y 0 ←(y 0 ⊙Temp);

[0012] (3.4)i←(x|x∈Temp), Temp←Vd, which means that the last node of the starting point set is assigned to i, and the end point set is assigned to Temp;

[0013] (3.5) According to the distance from i to each node in the Temp set, obtain the distance from i to any node v in Temp j Probability where v j ∈Temp, From i to v j distance, p←generates a random number in the interval [0.0,1.0], based on the value of p and Distribution in Temp, get the next node v t ;

[0014] (3.6)Temp←(Temp-{v t}), y 0 ←(y 0 ⊙{v t}), i←v t , if |Temp|>1, return to step (3.5), otherwise y 0 ←(y 0 ⊙Temp);

[0015] (3.7) If Y does not contain the sequence y 0 , then Y←(Y∪{y 0});

[0016] (3.8) If |Y|<(n+m), return to step (3.1);

[0017] (4) Find the row node sequence set Y = {y 1 ,y 2 ,…,y n+m}, if y i = {v 1 ,v 2 ,…,v n ,v n+1 ,v n+2 ,…,v n+m}, sequence y i Total distance

[0018] (5) According to the inverse of the distance of each sequence, find the probability of each sequence being selected, sequence y i The probability of being selected is Generate a new travel node sequence y 0 ;

[0019] In step (5), a new travel node sequence y is generated 0 The steps are:

[0020] (5.1) Create a sequence Create a process set Temp←Vo and create an intermediate variable i←v 0 ;

[0021] (5.2) p←generates a random number in the interval [0.0,1.1]. If p≤1.0, select the corresponding sequence y according to the probability of each sequence being selected in Y. j , in the sequence y j In the loop traversal method, the first node v belonging to the set Temp after node i is found t ∈Temp joins the sequence y 0 In, that is, y 0 ←(y 0 ⊙{v t});

[0022] (5.3) If p>1.0, according to the distance from i to each node in the Temp set, obtain the distance from i to any node v in Temp. j Probability where v j ∈Temp, p←regenerate random numbers in the interval [0.0,1.0], according to the value of p and Distribution in Temp, get the next node v t , the node v t ∈Temp joins the sequence y 0 In, that is, y 0 ←(y 0 ⊙{v t});

[0023] (5.4) v t Delete from Temp, Temp←(Temp-{v t}), the current node points to v t , i←v t , if |Temp|>1, return to step (5.2), otherwise y 0 ←(y 0 ⊙Temp);

[0024] (5.5)i←(x|x∈Temp), Temp←Vd;

[0025] (5.6) Generate a random number p in the interval [0.0, 1.1]. If p ≤ 1.0, select the corresponding sequence y according to the probability of each sequence in Y being selected. j In the sequence y j Using a loop traversal method, add the first node v belonging to the set Temp after node i to the sequence y t ∈Temp to the sequence y 0 That is, y 0 ←(y 0 ⊙{v t});

[0026] (5.7) If p > 1.0, obtain the probability from node i to any node v in the Temp set according to the distance from node i to each node in the Temp set j where v ∈Temp, p ← regenerate a random number in the interval [0.0, 1.0]. According to the value of p and j the distribution in Temp, obtain the next node v Add the node v t ∈Temp to the sequence y t That is, y 0 ←(y 0 ⊙{v 0}); t});

[0027] (5.8) Temp ← (Temp - {v t}), i ← v t , if |Temp| > 1, return to execute step (5.6). Otherwise, y 0 ←(y 0 ⊙Temp);

[0028] (6) Calculate the total distance of the travel node sequence y 0 using the method in step (4) Find the sequence with the maximum distance in the travel node sequence set Y If In the set Y, replace y 0 with y i , that is, y i ←y 0 , clear the counter count, that is, count ← 0; if Increment the counter count by one, that is, count ← (count + 1);

[0029] (7) If count < C × (n + m), where C is a preset constant integer, which can be set as C ∈ [1, 10], return to step (4). The larger the value of C, the longer the operation time, and the corresponding result may be better.

[0030] (8) Find the sequence with the shortest distance in the set of travel node sequences Y At this time, y i This is the optimal travel node sequence.

[0031] A computer storage medium stores a computer program, which, when executed by a processor, implements the above-mentioned passenger pick-up and drop-off scheduling method for a customized public transportation system between urban areas.

[0032] A computer device includes a memory, a processor, and a computer program stored in the memory and executable on the processor. When the processor executes the computer program, the above-mentioned method for dispatching passengers for a customized public transportation system between urban areas is implemented.

[0033] Beneficial effects: Compared with the prior art, the present invention has the following advantages:

[0034] 1. Since the optimal path selection belongs to the full permutation problem, when the number of stop nodes is n, the calculation amount of the optimal solution is n! The present invention draws on the idea of ​​genetic algorithm, combines the probability distribution converted from the path weight, and can adopt a 10% mutation probability for each node on the basis of continuously passing on the advantage subsequent nodes, which can quickly iterate and evolve;

[0035] 2. The core reason why traditional genetic algorithms are not used for path selection problems is that information exchange between parents is difficult to implement in path selection algorithms. The present invention implements downward inheritance in the form of subsequent nodes, solving the problem of offspring generation in genetic algorithms;

[0036] 3. The current mainstream path selection algorithm is generally based on the path weight in the OD matrix. This method is mainly based on the node sequence, which avoids the repeated changes of the OD matrix in the traditional algorithm, resulting in frequent reading and writing of the memory and reducing the operation speed. In order to achieve rapid evolution, the present invention increases the amount of calculation, but due to the reduction of data exchange with the memory, it can improve the operation speed;

[0037] 4. The iterative evolution of the present invention has strong cohesion and can easily realize parallel computing. It can not only utilize cloud computing resources, but also make full use of local multi-core systems to realize multi-path simultaneous evolution. BRIEF DESCRIPTION OF THE DRAWINGS

[0038] Figure 1 is a flow chart of the steps of the present invention;

[0039] Figure 2 For subprocess Figure 1 -Build the initial node;

[0040] Figure 3 For subprocess Figure 2 - Generate a new node. DETAILED DESCRIPTION

[0041] The technical solution of the present invention is further described below in conjunction with the accompanying drawings.

[0042] like Figures 1 to 3 As shown, a passenger pick-up and drop-off scheduling method for a customized bus system between urban areas includes the following steps:

[0043] 1. Passengers submit travel information to the backend system, including the earliest boarding time, the latest alighting time, the starting point, and the destination. The backend assigns a group of passengers to a customized bus based on the latest alighting time interval, the starting area, and the destination area. The bus needs to pick up passengers at 50 locations in the starting area and drop off passengers at 40 locations in the destination area.

[0044] 2. The bus leaves from the parking point, and the parking point is taken as the starting point, recorded as v 0 , add the 50 locations in the starting area to the set Vo, that is, Vo = {v 1 ,v 2 ,…,v 50}, add the 40 locations in the destination area to the set Vd, that is, Vd = {v 51 ,v 52 ,…,v 90}, merge Vo and Vd into a stop point set V = Vo ∪ Vd = {v 1 ,v 2 ,…,v 50 ,v 51 ,…,v 90};

[0045] 3. Based on the static map data, the OD matrices of Vo→Vo, Vo→Vd, and Vd→Vd are listed. The OD matrix of Vo→Vo is shown in Table 1, where the rows are the starting points, the columns are the end points, and the values ​​are the distance traveled by the bus in meters;

[0046] Table 1 OD matrix of Vo→Vo

[0047]

[0048]

[0049] 4. Create Create a sequence Create a process set Temp←Vo and create an intermediate variable i←v 0 ;

[0050] 5. According to the distance from i to each node in the Temp set, obtain the distance from i to any node v in Temp j Probability If i to v are required 8 The probability of i to v is , then j = 8, |Temp| = 50 is substituted into the above formula. 8 The distance is 2456m, that is If the sum of the distances from i to all nodes in the Temp set is 100000, that is We can find the values ​​from i to v 8 Probability p←generates a random number in the interval [0.0,1.0], based on the value of p and Distribution in Temp, get the next node v t , at this time, the vth j The condition for a node to become the next node of i is a random number If i to v 1 Probability i to v 2 Probability The random number p = 0.03100, at this time the random number p∈(0.01986,0.04020], so the next node v of i t =v 2 ;

[0051] 6. V t Join 0 In the sequence, y 0 = {v 2}, delete v in Temp t , at this time |Temp|=49, i←v 2 , return to repeat steps 5 and 6 until |Temp|=1. If |Temp|=1, Temp={v 28}, then v 28 Add to y 0 The end of the sequence, y 0 = {v 2 ,……,v 28},i←v 28 , at this time v 28 for y 0 The last node in the starting area of ​​the sequence, Temp←Vd, puts the node in the ending area into the Temp set;

[0052] 7. According to the distance from i to each node in the Temp set, obtain the distance from i to any node v in Temp j Probability where v j ∈Temp, From i to v j distance, p←generates a random number in the interval [0.0,1.0], based on the value of p and Distribution in Temp, get the next node v t ;

[0053] 8. Temp←(Temp-{v t}), y 0 ←(y 0 ⊙{v t}), i←v t , if |Temp|>1, return to step 7, otherwise y 0 ←(y 0 ⊙Temp), if Y does not contain sequence y 0 , then Y←(Y∪{y 0});

[0054] 9. If |Y|<(n+m), that is, the number of sequences in the travel node sequence set Y is less than 90, return to step 4;

[0055] 10. Find the row node sequence set Y = {y 1 ,y 2 ,…,y n+m}, if y i = {v 1 ,v 2 ,…,v n ,v n+1 ,v n+2 ,…,v n+m}, sequence y i Total distance If y 1 = {v 2 ,v 1 ,v 4 ,v 3 ,…,v 68 ,v 82},but If v 0 to v 2 The distance is 2034m. Referring to Table 1, we get

[0056] 11. According to the inverse of the distance of each sequence, find the probability of each sequence being selected, sequence y i The probability of being selected is For example

[0057] 12. Generate a new travel node sequence y according to the following method 0 :

[0058] (12.1) Creating a sequence Create a process set Temp←Vo and create an intermediate variable i←v 0 ;

[0059] (12.2) p←generates a random number in the interval [0.0,1.1]. If p≤1.0, select the corresponding sequence y according to the probability of each sequence being selected in Y. j , in the sequence y j In the loop traversal method, the first node v belonging to the set Temp after node i is found t ∈Temp joins the sequence y 0 In, that is, y 0 ←(y 0 ⊙{v t}), if the random number p = 0.1325, and Then the nodes after i follow y 7 Select, if y 7 = {v 5 ,v 12 ,…}, at this time i=v 0 , that is, the newly created y 0 In the sequence v 0 The first node after that is v 5 , that is, y 0 = {v 5}, i←v 5 ,Temp←(Temp-{v 5}); Generate random number p=0.5368 again, if Then the nodes after i follow y 47 Select, if y 47 = {v 25 ,v 12 ,…,v 5 ,v 55 ,…}, and y 47 i=v 5 The node after is taken as y 0 In the sequence v 5 The node after that, at this time y 47 Medium 5 All the nodes behind belong to Vd, and no node belongs to Temp = Vo, so the loop traversal method is used to traverse from y 47 Start searching from the head and find the next node is v 25 , that is, y 0 = {v 5 ,v 25}, i←v 25 , Temp←(Temp-{v 25}); Generate a random number p = 1.0368 again. Since p>1.0, a mutation occurs at this time and no longer changes from the existing y 1 ~y 90 Instead of getting the next node from i, we get the distance from i to any node v in Temp according to the distance from i to each node in Temp. j Probability where v j ∈Temp, by generating a random number p in the interval [0.0,1.0] again, determine the node after i. The method is the same as in step 5. If the next node obtained randomly is v 38 , then y 0 = {v 5 ,v 25 ,v 38}, i←v 38 , Temp←(Temp-{v 38}); Repeat this step until only one element remains in Temp;

[0060] (12.3) Add the remaining node in Temp to y 0 At this time, all nodes in the starting area Vo have been added to y 0 If the only node left in Temp is v 21 , then i←v 21 , Temp←Vd, that is, setting the current node to v 21 , the subsequent node set is Vd;

[0061] (12.4) Using a method similar to step (12.2), add all nodes in Vd to y 0 In the above example, find the row node sequence y 0 Total distance If the travel node sequence set Y = {y 1 ,y 2 ,…,y 90 The travel node sequence with the largest total distance in} is because Therefore, in Y, y 0 Replace y with the value 1 The value of y 1 ←y 0 , and clear the counter count; if the obtained travel node sequence y 0 Total distance because Then y 0 For an invalid travel node sequence, add one to the counter count, that is, count←(count+1);

[0062] 13. Determine the value of count. Here, it is determined that the preset constant integer C = 2. Then, if count < C×(n + m), that is, count < 180, continue to return to step twelve until count ≥ 180;

[0063] 14. After completing the above operations, the sequence in the set Y of travel node sequences with the minimum distance y at this time i is the relatively optimal travel node sequence.

Claims

1. A passenger pick-up and drop-off scheduling method for a customized public transportation system between urban areas, characterized in that: Including the following steps: (1) Let the starting point of the bus be v0, and the starting points of all k passengers assigned to this bus be added to the set Vo, that is, Vo = {v1, v2, …, v n }, n is the starting number of passengers on this bus, v1~v n is the starting point of k passengers, n≤k, and the destinations of all passengers on this bus are added to the set Vd, that is, Vd={v n+1 ,v n+2 ,…,v n+m },v n+1 ~v n+m is the destination of k passengers, m≤k, and Vo and Vd are combined into a set of stops V=Vo∪Vd={v1,v2,…,v n ,v n+1 ,v n+2 ,…,v n+m }; (2) Based on the static map data, list the OD matrix D of Vo→Vo (Vo→Vo) =(d i,j ) n×n , where i∈Vo,j∈Vo; list the OD matrix D of Vo→Vd (Vo→Vd) =(d i,j ) n×m , where i∈Vo,j∈Vd; list the OD matrix D of Vd→Vd (Vd→Vd) =(d i,j ) m×m , where i∈Vd, j∈Vd, the above d i,j is the distance traveled by bus from starting point i to end point j. The distances mentioned below are all distances traveled by bus; (3) Construct the initial travel node sequence set Y = {y1, y2, …, y m+m }, where y1~y n+m They are all travel node sequences, and their construction method can be created first Here ← means assigning the value on the right to the variable on the left, and then filling the set Y until Y contains n+m different sequences: (4) Find the set of row node sequences Y = {y1, y2, …, y n+m }, if y i ={v1,v2,…,v n ,v n+1 ,v n+2 ,…,v n+m }, sequence y i Total distance (5) According to the inverse of the distance of each sequence, find the probability of each sequence being selected, sequence y i The probability of being selected is Generate a new travel node sequence y0; (6) Calculate the total distance of the trip node sequence y0 according to the method in step (4) Find the sequence with the largest distance in the travel node sequence set Y like Replace y with y0 in the set Y i , that is, y i ←y0, clear the counter count, that is, count←0; if The counter count is incremented by one, i.e. count←(count+1); (7) If count < C×(n + m), where C is a preset constant integer and can be set as C ∈ [1, 10], return to step (4). The larger the value of C, the longer the operation time and the better the corresponding result may be; (8) Find the sequence with the shortest distance in the set of travel node sequences Y At this time, y i That is the optimal travel node sequence; In the said step (3), the step of filling set Y is: (3.1) Create a sequence Create a process set Temp←Vo and create an intermediate variable i←v0; (3.2) According to the distance from i to each node in the Temp set, obtain the distance from i to any node v in Temp j Probability where v j ∈Temp, From i to v j distance, |Temp| is the number of elements in the set Temp, p←generates a random number in the interval [0.0,1.0], and Distribution in Temp, get the next node v t The method is that the condition for the v1 node in Temp to become the next node of i is a random number v j The condition for a node to become the next node of i is a random number (3.3)Temp←(Temp-{v t }), y0←(y0⊙{v t }), ⊙ means to connect the sequences before and after it in order, i←v t , if |Temp|>1, return to step (3.2), otherwise y0←(y0⊙Temp); (3.4) i ← (x|x ∈ Temp), Temp ← Vd, which means assigning the last node of the starting point set to i and assigning the ending point set to Temp; (3.5) According to the distance from i to each node in the Temp set, obtain the distance from i to any node v in Temp j Probability where v j ∈Temp, From i to v j distance, p←generates a random number in the interval [0.0,1.0], based on the value of p and Distribution in Temp, get the next node v t ; (3.6)Temp←(Temp-{v t }), y0←(y0⊙{v t }), i←v t , if |Temp|>1, return to step (3.5), otherwise y0←(y0⊙Temp); (3.7) If sequence y0 is not included in Y, then Y ← (Y ∪ {y0}); (3.8) If |Y| < (n + m), return to execute step (3.1); In the said step (5), the step of generating a new travel node sequence y0 is: (5.1) Create a sequence Create a process set Temp←Vo and create an intermediate variable i←v0; (5.2) p←generates a random number in the interval [0.0,1.1]. If p≤1.0, select the corresponding sequence y according to the probability of each sequence being selected in Y. j , in the sequence y j In the loop traversal method, the first node v belonging to the set Yemp after node i is found t ∈Temp is added to the sequence y0, that is, y0←(y0⊙{v t }); (5.3) If p>1.0, according to the distance from i to each node in the Temp set, obtain the distance from i to any node v in Temp. j Probability where v j ∈Temp, p←regenerate random numbers in the interval [0.0,1.0], according to the value of p and Distribution in Temp, get the next node v t , the node v t ∈Temp is added to the sequence y0, that is, y0←(y0⊙{v t }); (5.4) v t Delete from Temp, Temp←(Temp-{v t }), the current node points to v t , i←v t , if |Temp|>1, return to step (5.2), otherwise y0←(y0⊙Temp); (5.5) i ← (x|x ∈ Temp), Temp ← Vd; (5.6) p←generates a random number in the interval [0.0,1.1]. If p≤1.0, select the corresponding sequence y according to the probability of each sequence being selected in Y. j , in the sequence y j In the loop traversal method, the first node v belonging to the set Temp after node i is found t ∈Temp is added to the sequence y0, that is, y0←(y0⊙{v t }); (5.7) If p>1.0, according to the distance from i to each node in the Temp set, obtain the distance from i to any node v in Temp. j Probability where v j ∈Temp, p←regenerate random numbers in the interval [0.0,1.0], according to the value of p and Distribution in Temp, get the next node v t , the node v t ∈Temp is added to the sequence y0, that is, y0←(y0⊙{v t }); (5.8)Temp←(Temp-{v t }), i←v t If |Temp|>1, return to step (5.6), otherwise y0←(y0⊙Temp).

2. A computer storage medium having a computer program stored thereon, characterized in that: When the computer program is executed by a processor, it implements a method for passenger pick-up and drop-off scheduling of an inter-city customized bus system as described in claim 1.

3. A computer device comprising a memory, a processor and a computer program stored in the memory and executable on the processor, characterized in that: When the processor executes the computer program, it implements a method for passenger pick-up and drop-off scheduling of an inter-city customized bus system as described in claim 1.

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