# Synthetic Optimization Model and Algorithm for Railway Freight Center Station Location and Wagon Flow Organization Problem

Authors: Xing-cai Liu; Shi-wei He; Rui Song; Hao-dong Li; Long Wang; Yang Sun
Journal: Mathematical Problems in Engineering
 (2014)

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## Abstract

The railway freight center stations location and wagon flow organization in railway transport are interconnected, and each of them is complicated in a large-scale rail network. In this paper, a two-stage method is proposed to optimize railway freight center stations location and wagon flow organization together. The location model is present with the objective to minimize the operation cost and fixed construction cost. Then, the second model of wagon flow organization is proposed to decide the optimal train service between different freight center stations. The location of the stations is the output of the first model. A heuristic algorithm that combined tabu search (TS) with adaptive clonal selection algorithm (ACSA) is proposed to solve those two models. The numerical results show the proposed solution method is effective.


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## Body

## 1. Introduction

In the past decade, China railway has invested many railway freight center stations, which are equipped comprehensive transportation facilities and logistics facilities for the purpose of centralized transportation. At the same time, most of the stations with small transport demand were closed. Those center stations can help to centralize transport demand and gain economic of scale, while the transport demand of closed stations may be lost, if the transport cost of the railway is higher than the other transport modes such as road and water transport. In order to attract the transport demand, railway must improve the wagon flow organization to provide service with high level.The center station location and wagon flow organization are interacted and interconnected. In order to organize products with high level of service, the amount of goods must meet the train size limitation. Otherwise, long waiting time may be caused. At the same time, the efficiency of wagon flow organization can decide the operation efficiency of stations. For example, the limited storage capacity of station will be occupied, if there is no suitable train to serve the demand.In order to solve the complicated station location and wagon flow organization problem, a two-stage method is proposed to optimize station location and wagon flow organization together. This paper is organized as follows. Section2 briefly reviews the relevant literature. Section 3 introduces the optimal models in a two-stage method. In Section 4, a heuristic algorithm is proposed to solve the models. Finally, a numerical example is provided to illustrate the application of the models and algorithm.


## 2. Literature Review

The location of railway freight center stations is similar to hub location problem, which is among the most critical management decisions. Many models have been proposed such as set covering problem (SCP),p-center problem, andp-median problem [1–5].The SCP model was presented by Caprara et al. [5] and solved by a heuristic method based on Lagrangian relaxation. Campbell [6] came up with formulations for the multiple and single allocationp-hub median problem with two heuristic methods. Besides, Campbell et al. [7] concluded that there are two types of hub location problem: single allocation and multiple allocation. Skorin-Kapov et al. [8] studied the uncapacitated multiple and single allocationp-hub median problems. O’Kelly [9] analyzed the models for both single-hub and two-hub problems. And O’Kelly [10] presented a quadratic integer program for the hub location. The model is linearized and solved by a heuristic algorithm. Furthermore, O’Kelly and Bryan [11] proposed a reliable model by considering the scale economies effect of traffic. Racunica and Wynter [12] studied the location problem in a hub-and-spoke network, aiming to increase the share of rail in intermodal transport.Recently, fuzzy theory and dynamic environment were introduced into the location problem. Batanović et al.’s study [13] concerned the uncertain parameters in maximum covering location problem, which were modeled by fuzzy set. Sáez-Aguado and Trandafir [14] improved thep-median model by considering a coverage constraint. A dynamic uncapacitated hub location problem was present in Contreras et al.’s paper [15]. Correia et al. [16] proposed four mixed-integer linear programming formulations in order to study the extension of classical capacitated single-allocation hub location problem.For the wagon flow organization problem, Bodin et al. [17] developed a nonlinear, mixed-integer programming model for the railroad blocking problem. Martinelli and Teng [18] applied neural network to solve the railway operation problem. Some research described the railway blocking problem with network theory. Newton et al. [19] took the railroad blocking problem as a network design problem and developed a column generation and branch-and-bound algorithm to solve it. Similarly, Barnhart et al. [20] proposed a Lagrangian relaxation approach for optimizing railway blocking problem. Fukasawa et al. [21] considered both the loaded and empty cars in the network and proposed an integer multicommodity flow model for the problem. Woxenius [22] provided six principles of rail operation and applied them into intermodal rail freight transport network system. Jeong et al. [23] formulated a linear integer programming model in a hub-and-spoke network. Jha et al. [24] compared arc-based with path-based formulations of the block-to-train assignment problem and proved that the latter is smaller in scale. Keaton [25] presented a mixed-integer programming for railroad blocking problem.In China, Wu [26] studied the organization scheme for through trains. Lin et al. [27] formulated a nonlinear 0-1 through train formation model considering the different cost parameters and block capacity constraint. Nonlinear 0-1 programming models were established in Cao et al.’s study [28, 29] to determine the freight train scheduling plan based on analyzing the logistics system costs.Most of the research in literature focused either on the location of stations or the wagon flow organization, while the influence of the location on wagon flow organization should be considered. In this paper, a two-stage method is used to solve this problem.


## 3. The Two-Stage Programming

### 3.1. Model of Optimal Railway Freight Center Station Location Problem

As the location of railway freight center stations is similar to the hub location problem, customer specifies the ordinary railway freight station, while service point specifies the candidate railway freight center station.


#### 3.1.1. Decision Variables

The objective of this model is to find the optimal location of service points and the assignment between customer and service point. The location decision and assignment are treated as decision variables. Those arexij equals 1 if customer i is assigned to service point j. Otherwise, it equals 0.yj equals 1 if service point is located at candidate service point j. Otherwise, it equals 0.


#### 3.1.2. Objective Function

Customers hope that the service points are located close to themselves, so as to send the goods to the station quickly and cheaply, while the planners want to maximize the railway coverage with limited center stations, so as to reduce the total investment. Both the transport cost from costumers to service points and construction costs of stations are considered by the following objective function:(1)min⁡⁡Z=μ1∑i∈I∑j∈Jλdijqi′xij+μ2∑j∈JCjyj,
whereλ is unit cost to transport the goods from customer to service point;μ1 and μ2 are weight of transport cost and construction cost in objective function. The values are defined in advance. And μ1+μ2=1, μ1≥0, μ2≥0;dij is distance between customer i and candidate service point j;qi′ is transport demand of customer i;Cj is fixed cost to construct a service point at candidate service point j;I is set of customers, i∈I;J is set of candidate service points, j∈J.


#### 3.1.3. Constraints

The coverage constraint of a service point is considered. The whole distance which is greater than the preestablished coverage distance at a service point should not exceed a previously chosen value. This constraint is related to the maximum covering location problem (MCLP), whereas the investment of service point may change the transport demand. The model also takes this situation into account.(i)
Each customer must be assigned to one service point(2)∑j∈Jxij=1∀i∈I.
(ii)
Candidate service pointj cannot serve any customer, if j is not chosen as a service point
(3)xij≤yj∀i∈I,j∈J.
(iii)
The total number of chosen service points should be constrained(4)∑j∈Jyj≤p.
p is the maximum number of the stations.
(iv)
Sum of the distance which is greater than coverage distance DC at a service point should not exceedδ. Both DC and δ are prespecified
(5)∑i∈Ilijxij≤δ∀j∈J.
l
i
j is defined as follows:
(6)lij={dijdij>DC0otherwise.
(v)
The goods serviced by pointj cannot exceed its capacity Capj. 													(7)∑i∈Iqi′xij≤Capj∀j∈J.
q
i
′ is defined as follows:
(8)qi′={εqipointiisnotaservicepointqiotherwise∀i∈I,where ε is a discount coefficient of demand at point i if i is not a service point, which describes the change of transport demand. qi is the expected transport demand at point i.


#### 3.1.4. Mathematical Model

The location model of railway freight center stations can be stated as(M-I)(9)min⁡Z=μ1∑i∈I∑j∈Jλdijqi′xij+μ2∑j∈JCjyjs.t.∑j∈Jxij=1∀i∈Is.t.xij≤yj∀i∈I,j∈Js.t.∑j∈Jyj≤ps.t.∑i∈Ilijxij≤δ∀j∈Js.t.∑i∈Iqi′xij≤Capj∀j∈Js.t.xij,yj∈{0,1}∀i∈I,j∈J.


### 3.2. Model of Wagon Flow Organization Problem

The outputs of M-I include the location of railway freight center stations and the transport demand of each station. Based on the average loaded weight of a wagon, the transport demand can be turned into wagon flows. Those are the input data of wagon flow organization.


#### 3.2.1. An Example of Wagon Flow Organization Problem

The wagon flow organization problem can be illustrated by a simple network (see Figure1), which has three technical stations, two freight center stations, and three shipments (N12, N14, and N15).Figure 1
A simple rail network.There are 12 combinations for routing all the shipments on potential train services (see Figure2). Arcs in the network specify the available train connections. The OD pair will be reclassified at the technical stations in the itinerary if it is not served by through train. If transport demand between two adjacent stations is positive, the train service must be provided. So shipment N12 has only one service strategy that is served by the train service 1→2.Twelve combinations for routing all the shipments on potential train services.
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
(i)
(j)
(k)
(l)Strategy 1 (see Figure 2(a)). The OD pairs N15, N14, and N12 are consolidated in train service 1→2 at the origin station 1. In this case, N15 is reclassified at technical stations 2, 3, and 4. N14 is reclassified at technical stations 2 and 3.Strategy 2 (see Figure 2(b)). There are two through train services 1→4 and 1→5. In this case, both N15 and N14 are directly shipped to the destination.Strategy 3 (see Figure 2(c)). There are two through train services 1→3 and 1→4. In this case, N14 is directly shipped to the destination. While N15 is directly shipped to technical station 3, and N15 still needs to be reclassified at technical stations 3 and 4.Strategy 4 (see Figure 2(d)). Strategy 3 and Strategy 4 share the common train service set, but the train services are different. N15 is directly shipped to technical station 4 and reclassified there. N14 is directly shipped to technical station 3 and reclassified there.Strategy 5 (see Figure 2(e)). There is one through train service 1→4. In this case, N15 is directly shipped to the technical station 4 and reclassified there. N14 and N12 are consolidated in train service 1→2. N14 is reclassified at technical stations 2 and 3.Strategy 6 (see Figure 2(f)). Strategy 5 and Strategy 6 share the common train service set, but the train services are different. N14 is directly shipped to technical station 4 by train service 1→4. N15 and N12 are consolidated in train service 1→2. N15 is reclassified at technical stations 2, 3, and 4.Strategy 7 (see Figure 2(g)). There is one through train service 1→3. In this case, N15 is directly shipped to the technical station 3 and reclassified at technical stations 3 and 4. N14 and N12 are consolidated in train service 1→2. N14 is reclassified at technical stations 2 and 3.Strategy 8 (see Figure 2(h)). Strategy 7 and Strategy 8 share the common train service set, but the train services are different. N14 is directly shipped to technical station 3 by train service 1→3 and reclassified there. N15 and N12 are consolidated in train service 1→2. N15 is reclassified at technical stations 2, 3, and 4.Strategy 9 (see Figure 2(i)). There is one through train service 1→3. In this case, N15 and N14 are directly consolidated together and shipped to the technical station 3 by train service 1→3. N14 is reclassified at technical station 3. N15 is reclassified at technical station 3 and 4.Strategy 10 (see Figure 2(j)). There is one through train service 1→4. In this case, N15 and N14 are consolidated together and directly shipped to the technical station 4 by train service 1→4. N15 is reclassified there.Strategy 11 (see Figure 2(k)). There are two through train services 1→3 and 1→5. In this case, N15 is directly shipped to destination by train service 1→5. N14 is directly shipped to the technical station 3 by train service 1→3 and reclassified there.Strategy 12 (see Figure 2(l)). There is one through train service 1→5. In this case, N15 is directly shipped to destination by train service 1→5. N14 and N12 are consolidated in train service 1→2. N14 is reclassified at technical station 2 and 3.


#### 3.2.2. Analysis of Wagon Flow Organization Problem

Among the twelve strategies above, there are two extreme ones, that is, Strategy 1 and Strategy 2.(1)
In Strategy 1, train connection services are only provided between adjacent stations. In this way, the number of through train services reaches minimum; that is, the total waiting time at the dispatching station reaches minimum. However, each shipment will be reclassified at each technical station on its itinerary. Thus the reclassification fee and time consumption will increase significantly. The increasing number of reclassified wagons may cause congestion and time delay at some technical stations.(2)
In Strategy 2, transport demands are all delivered to their destinations without reclassification, which will reduce the reclassification fee and time consumption. However, this strategy may be unworkable. Such strategy requires enough classification tracks to store outbound trains, and single OD pair must be large enough to dispatch a through train service. Waiting time at the dispatching station will increase if the OD pair is small.The solution of wagon flow organization problem is a tradeoff between Strategy 1 and Strategy 2, aiming at minimizing the total shipping and handling cost for all shipments. The time consumption is used to weigh the cost.


#### 3.2.3. Decision Variables

From the example above, it can be concluded that there are three train services for a wagon flow: the first one is shipped to the destination by nonstop shipping scheme at the origin station, the second one is served by nonthrough shipment, and the third one is shipped to a technical station in the itinerary by nonstop shipping scheme consolidated with other OD pairs. These variables are(10)xst={1Nstisshippedbynonstopscheme0otherwisexstk={1Nstisshippedtofirsttechnicalstationkintheitineraryandreclassifiedthere0otherwisemmmmmmmnmmmmmnmmm∀k∈V(s)ystk={1Nstisshippedtotcehnicalstationkandreclassifiedthere0otherwisemmmmmmmmmmmmmmmnmmm∀k∈P(s),
whereNst is number of wagon flow from origin s to destination t;s is origin station;t is destination station;V(s) is set of technical stations next to the origin station s, k∈V(s);P(s) is set of technical stations in the itinerary of wagon flow Nst; exclude the technical station next to the origin station, P(s)∩V(s)=∅, k∈P(s).


#### 3.2.4. Objective Function

The optimal goal is to obtain the wagon flow organization plan with minimum time consumption, which means the level of service. The travel time is not considered for it is identical. In order to simplify the problem, the following assumptions are proposed: (1) the supply of empty wagons is not considered, (2) the storage capacity of the origin and destination station are not considered, (3) the shipping routes for all the service are given, and (4) the train formation plans (TFP) of the technical stations in the itinerary are known. This information can be obtained previously.Time consumption of three train services forNst is as follows.(i)
For the nonstop service, the service time equals the loading and unloading times As there is no intermediate service such as reclassification(11)F1=Nstwstxst,wherewst is loading and unloading time at the origin and destination for a wagon served by nonstop shipping scheme.(ii)
Nonthrough shipment cost, which includes the reclassified time in the itinerary and loading and unloading time(12)F2=∑k∈V(s)Nstw--stxstk+∑k∈V(s)∑g∈G(k,t)Nsttgxstk,wherew--st is loading and unloading time at the origin and destination for a wagon shipped by non-through shipment;tg is reclassified time at technical station g;G(k,t) is set of technical stations in the itinerary after Nst has been reclassified at technical station k, and g∈G(k,t).(iii)
The wagon flow is shipped to a technical station in the itinerary by nonstop shipping scheme. This item includes the reclassified time consumption in the itinerary and time consumption for loading and unloading(13)F3=∑k∈P(s)Nstw-stystk+∑k∈P(s)∑g∈G(k,t)Nsttgystk,whereThen, the total time consumption forNst can be formulated as
(14)Fst=F1+F2+F3.The objective function of wagon flow organization can be stated as(15)min⁡⁡∑s∈S∑t∈Q(s)Fst,
where  S is set of origin stations and s∈S. Q(s) is set of destinations of the wagon flows that originate from s. And Q(s)={t:Nst≠0}. t∈Q(s).


#### 3.2.5. Constraints

The wagon flow organization problem has two constrains.(
1) Loading and Unloading Capacity Constraint for Nonstop Train.The loading and unloading capacity must reach the minimum size of loading or unloading a whole nonstop train. If there are several wagon flows consolidated together to be shipped to a technical station in the itinerary by nonstop shipping scheme, the sum of their loading or unloading capacity must reach the average marshaling number of wagons in a nonstop shipping scheme
(16)m-stI(∑t∈Q(s)ystk)-∑t∈Q(s)mstystk≤0∀s∈S,k∈P(s),
wheremst is the smaller one between maximum loading capacity of station s and maximum unloading capacity of station t at certain period;m-st is the average marshaling number of wagons in nonstop shipping scheme;I(x) is a step function, I(x)={1ifx>00otherwise.(
2) Each Wagon Flow Is Served by One and Only One Train Service.
(17)
I
s
t
x
s
t
+
∑
k
∈
V
(
s
)
x
s
t
k
+
∑
k
∈
P
(
s
)
y
s
t
k
=
1
∀
s
∈
S
,
t
∈
Q
(
s
)
,
where Ist={1ifmst≥m-st0otherwise∀s∈S,t∈Q(s), which ensures that only the wagon flow that reaches the average marshaling number of wagon in nonstop shipping scheme can be shipped by it.


#### 3.2.6. Constraint Linearization

Constraint (16) can be linearized by introducing a variable ysk(18)ysk=I(∑t∈Q(s)ystk)={1∑t∈Q(s)ystk>00otherwisemmmmmmmmmmm∀s∈S,k∈P(s).Then constraint (16) can be formulated as follows:
(19)m-stysk-∑t∈Q(s)mstystk≤0∀s∈S,k∈P(s).Constraint (20) is also introduced to ensure that ysk equals 1 if and only if ystk=1(20)Mysk≥∑t∈Q(s)ystk∀s∈S,k∈P(s),
where M is a large positive constant.


#### 3.2.7. Mathematical Model

The wagon flow organization model can be stated as(M-II)(21)min⁡∑s∈S∑t∈Q(s)Fsts.t.Istxst+∑k∈V(s)xstk+∑k∈P(s)ystk=1∀s∈S,t∈Q(s)s.t.m-stysk-∑t∈Q(s)mstystk≤0∀s∈S,k∈P(s)s.t.Mysk≥∑t∈Q(s)ystk∀s∈S,k∈P(s)s.t.xst,xstk∈{0,1}∀s∈S,t∈Q(s),k∈V(s)s.t.ysk,ystk∈{0,1}∀s∈S,t∈Q(s),k∈P(s).


## 3.1. Model of Optimal Railway Freight Center Station Location Problem

As the location of railway freight center stations is similar to the hub location problem, customer specifies the ordinary railway freight station, while service point specifies the candidate railway freight center station.


### 3.1.1. Decision Variables

The objective of this model is to find the optimal location of service points and the assignment between customer and service point. The location decision and assignment are treated as decision variables. Those arexij equals 1 if customer i is assigned to service point j. Otherwise, it equals 0.yj equals 1 if service point is located at candidate service point j. Otherwise, it equals 0.


### 3.1.2. Objective Function

Customers hope that the service points are located close to themselves, so as to send the goods to the station quickly and cheaply, while the planners want to maximize the railway coverage with limited center stations, so as to reduce the total investment. Both the transport cost from costumers to service points and construction costs of stations are considered by the following objective function:(1)min⁡⁡Z=μ1∑i∈I∑j∈Jλdijqi′xij+μ2∑j∈JCjyj,
whereλ is unit cost to transport the goods from customer to service point;μ1 and μ2 are weight of transport cost and construction cost in objective function. The values are defined in advance. And μ1+μ2=1, μ1≥0, μ2≥0;dij is distance between customer i and candidate service point j;qi′ is transport demand of customer i;Cj is fixed cost to construct a service point at candidate service point j;I is set of customers, i∈I;J is set of candidate service points, j∈J.


### 3.1.3. Constraints

The coverage constraint of a service point is considered. The whole distance which is greater than the preestablished coverage distance at a service point should not exceed a previously chosen value. This constraint is related to the maximum covering location problem (MCLP), whereas the investment of service point may change the transport demand. The model also takes this situation into account.(i)
Each customer must be assigned to one service point(2)∑j∈Jxij=1∀i∈I.
(ii)
Candidate service pointj cannot serve any customer, if j is not chosen as a service point
(3)xij≤yj∀i∈I,j∈J.
(iii)
The total number of chosen service points should be constrained(4)∑j∈Jyj≤p.
p is the maximum number of the stations.
(iv)
Sum of the distance which is greater than coverage distance DC at a service point should not exceedδ. Both DC and δ are prespecified
(5)∑i∈Ilijxij≤δ∀j∈J.
l
i
j is defined as follows:
(6)lij={dijdij>DC0otherwise.
(v)
The goods serviced by pointj cannot exceed its capacity Capj. 													(7)∑i∈Iqi′xij≤Capj∀j∈J.
q
i
′ is defined as follows:
(8)qi′={εqipointiisnotaservicepointqiotherwise∀i∈I,where ε is a discount coefficient of demand at point i if i is not a service point, which describes the change of transport demand. qi is the expected transport demand at point i.


### 3.1.4. Mathematical Model

The location model of railway freight center stations can be stated as(M-I)(9)min⁡Z=μ1∑i∈I∑j∈Jλdijqi′xij+μ2∑j∈JCjyjs.t.∑j∈Jxij=1∀i∈Is.t.xij≤yj∀i∈I,j∈Js.t.∑j∈Jyj≤ps.t.∑i∈Ilijxij≤δ∀j∈Js.t.∑i∈Iqi′xij≤Capj∀j∈Js.t.xij,yj∈{0,1}∀i∈I,j∈J.


## 3.1.1. Decision Variables

The objective of this model is to find the optimal location of service points and the assignment between customer and service point. The location decision and assignment are treated as decision variables. Those arexij equals 1 if customer i is assigned to service point j. Otherwise, it equals 0.yj equals 1 if service point is located at candidate service point j. Otherwise, it equals 0.


## 3.1.2. Objective Function

Customers hope that the service points are located close to themselves, so as to send the goods to the station quickly and cheaply, while the planners want to maximize the railway coverage with limited center stations, so as to reduce the total investment. Both the transport cost from costumers to service points and construction costs of stations are considered by the following objective function:(1)min⁡⁡Z=μ1∑i∈I∑j∈Jλdijqi′xij+μ2∑j∈JCjyj,
whereλ is unit cost to transport the goods from customer to service point;μ1 and μ2 are weight of transport cost and construction cost in objective function. The values are defined in advance. And μ1+μ2=1, μ1≥0, μ2≥0;dij is distance between customer i and candidate service point j;qi′ is transport demand of customer i;Cj is fixed cost to construct a service point at candidate service point j;I is set of customers, i∈I;J is set of candidate service points, j∈J.


## 3.1.3. Constraints

The coverage constraint of a service point is considered. The whole distance which is greater than the preestablished coverage distance at a service point should not exceed a previously chosen value. This constraint is related to the maximum covering location problem (MCLP), whereas the investment of service point may change the transport demand. The model also takes this situation into account.(i)
Each customer must be assigned to one service point(2)∑j∈Jxij=1∀i∈I.
(ii)
Candidate service pointj cannot serve any customer, if j is not chosen as a service point
(3)xij≤yj∀i∈I,j∈J.
(iii)
The total number of chosen service points should be constrained(4)∑j∈Jyj≤p.
p is the maximum number of the stations.
(iv)
Sum of the distance which is greater than coverage distance DC at a service point should not exceedδ. Both DC and δ are prespecified
(5)∑i∈Ilijxij≤δ∀j∈J.
l
i
j is defined as follows:
(6)lij={dijdij>DC0otherwise.
(v)
The goods serviced by pointj cannot exceed its capacity Capj. 													(7)∑i∈Iqi′xij≤Capj∀j∈J.
q
i
′ is defined as follows:
(8)qi′={εqipointiisnotaservicepointqiotherwise∀i∈I,where ε is a discount coefficient of demand at point i if i is not a service point, which describes the change of transport demand. qi is the expected transport demand at point i.


## 3.1.4. Mathematical Model

The location model of railway freight center stations can be stated as(M-I)(9)min⁡Z=μ1∑i∈I∑j∈Jλdijqi′xij+μ2∑j∈JCjyjs.t.∑j∈Jxij=1∀i∈Is.t.xij≤yj∀i∈I,j∈Js.t.∑j∈Jyj≤ps.t.∑i∈Ilijxij≤δ∀j∈Js.t.∑i∈Iqi′xij≤Capj∀j∈Js.t.xij,yj∈{0,1}∀i∈I,j∈J.


## 3.2. Model of Wagon Flow Organization Problem

The outputs of M-I include the location of railway freight center stations and the transport demand of each station. Based on the average loaded weight of a wagon, the transport demand can be turned into wagon flows. Those are the input data of wagon flow organization.


### 3.2.1. An Example of Wagon Flow Organization Problem

The wagon flow organization problem can be illustrated by a simple network (see Figure1), which has three technical stations, two freight center stations, and three shipments (N12, N14, and N15).Figure 1
A simple rail network.There are 12 combinations for routing all the shipments on potential train services (see Figure2). Arcs in the network specify the available train connections. The OD pair will be reclassified at the technical stations in the itinerary if it is not served by through train. If transport demand between two adjacent stations is positive, the train service must be provided. So shipment N12 has only one service strategy that is served by the train service 1→2.Twelve combinations for routing all the shipments on potential train services.
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
(i)
(j)
(k)
(l)Strategy 1 (see Figure 2(a)). The OD pairs N15, N14, and N12 are consolidated in train service 1→2 at the origin station 1. In this case, N15 is reclassified at technical stations 2, 3, and 4. N14 is reclassified at technical stations 2 and 3.Strategy 2 (see Figure 2(b)). There are two through train services 1→4 and 1→5. In this case, both N15 and N14 are directly shipped to the destination.Strategy 3 (see Figure 2(c)). There are two through train services 1→3 and 1→4. In this case, N14 is directly shipped to the destination. While N15 is directly shipped to technical station 3, and N15 still needs to be reclassified at technical stations 3 and 4.Strategy 4 (see Figure 2(d)). Strategy 3 and Strategy 4 share the common train service set, but the train services are different. N15 is directly shipped to technical station 4 and reclassified there. N14 is directly shipped to technical station 3 and reclassified there.Strategy 5 (see Figure 2(e)). There is one through train service 1→4. In this case, N15 is directly shipped to the technical station 4 and reclassified there. N14 and N12 are consolidated in train service 1→2. N14 is reclassified at technical stations 2 and 3.Strategy 6 (see Figure 2(f)). Strategy 5 and Strategy 6 share the common train service set, but the train services are different. N14 is directly shipped to technical station 4 by train service 1→4. N15 and N12 are consolidated in train service 1→2. N15 is reclassified at technical stations 2, 3, and 4.Strategy 7 (see Figure 2(g)). There is one through train service 1→3. In this case, N15 is directly shipped to the technical station 3 and reclassified at technical stations 3 and 4. N14 and N12 are consolidated in train service 1→2. N14 is reclassified at technical stations 2 and 3.Strategy 8 (see Figure 2(h)). Strategy 7 and Strategy 8 share the common train service set, but the train services are different. N14 is directly shipped to technical station 3 by train service 1→3 and reclassified there. N15 and N12 are consolidated in train service 1→2. N15 is reclassified at technical stations 2, 3, and 4.Strategy 9 (see Figure 2(i)). There is one through train service 1→3. In this case, N15 and N14 are directly consolidated together and shipped to the technical station 3 by train service 1→3. N14 is reclassified at technical station 3. N15 is reclassified at technical station 3 and 4.Strategy 10 (see Figure 2(j)). There is one through train service 1→4. In this case, N15 and N14 are consolidated together and directly shipped to the technical station 4 by train service 1→4. N15 is reclassified there.Strategy 11 (see Figure 2(k)). There are two through train services 1→3 and 1→5. In this case, N15 is directly shipped to destination by train service 1→5. N14 is directly shipped to the technical station 3 by train service 1→3 and reclassified there.Strategy 12 (see Figure 2(l)). There is one through train service 1→5. In this case, N15 is directly shipped to destination by train service 1→5. N14 and N12 are consolidated in train service 1→2. N14 is reclassified at technical station 2 and 3.


### 3.2.2. Analysis of Wagon Flow Organization Problem

Among the twelve strategies above, there are two extreme ones, that is, Strategy 1 and Strategy 2.(1)
In Strategy 1, train connection services are only provided between adjacent stations. In this way, the number of through train services reaches minimum; that is, the total waiting time at the dispatching station reaches minimum. However, each shipment will be reclassified at each technical station on its itinerary. Thus the reclassification fee and time consumption will increase significantly. The increasing number of reclassified wagons may cause congestion and time delay at some technical stations.(2)
In Strategy 2, transport demands are all delivered to their destinations without reclassification, which will reduce the reclassification fee and time consumption. However, this strategy may be unworkable. Such strategy requires enough classification tracks to store outbound trains, and single OD pair must be large enough to dispatch a through train service. Waiting time at the dispatching station will increase if the OD pair is small.The solution of wagon flow organization problem is a tradeoff between Strategy 1 and Strategy 2, aiming at minimizing the total shipping and handling cost for all shipments. The time consumption is used to weigh the cost.


### 3.2.3. Decision Variables

From the example above, it can be concluded that there are three train services for a wagon flow: the first one is shipped to the destination by nonstop shipping scheme at the origin station, the second one is served by nonthrough shipment, and the third one is shipped to a technical station in the itinerary by nonstop shipping scheme consolidated with other OD pairs. These variables are(10)xst={1Nstisshippedbynonstopscheme0otherwisexstk={1Nstisshippedtofirsttechnicalstationkintheitineraryandreclassifiedthere0otherwisemmmmmmmnmmmmmnmmm∀k∈V(s)ystk={1Nstisshippedtotcehnicalstationkandreclassifiedthere0otherwisemmmmmmmmmmmmmmmnmmm∀k∈P(s),
whereNst is number of wagon flow from origin s to destination t;s is origin station;t is destination station;V(s) is set of technical stations next to the origin station s, k∈V(s);P(s) is set of technical stations in the itinerary of wagon flow Nst; exclude the technical station next to the origin station, P(s)∩V(s)=∅, k∈P(s).


### 3.2.4. Objective Function

The optimal goal is to obtain the wagon flow organization plan with minimum time consumption, which means the level of service. The travel time is not considered for it is identical. In order to simplify the problem, the following assumptions are proposed: (1) the supply of empty wagons is not considered, (2) the storage capacity of the origin and destination station are not considered, (3) the shipping routes for all the service are given, and (4) the train formation plans (TFP) of the technical stations in the itinerary are known. This information can be obtained previously.Time consumption of three train services forNst is as follows.(i)
For the nonstop service, the service time equals the loading and unloading times As there is no intermediate service such as reclassification(11)F1=Nstwstxst,wherewst is loading and unloading time at the origin and destination for a wagon served by nonstop shipping scheme.(ii)
Nonthrough shipment cost, which includes the reclassified time in the itinerary and loading and unloading time(12)F2=∑k∈V(s)Nstw--stxstk+∑k∈V(s)∑g∈G(k,t)Nsttgxstk,wherew--st is loading and unloading time at the origin and destination for a wagon shipped by non-through shipment;tg is reclassified time at technical station g;G(k,t) is set of technical stations in the itinerary after Nst has been reclassified at technical station k, and g∈G(k,t).(iii)
The wagon flow is shipped to a technical station in the itinerary by nonstop shipping scheme. This item includes the reclassified time consumption in the itinerary and time consumption for loading and unloading(13)F3=∑k∈P(s)Nstw-stystk+∑k∈P(s)∑g∈G(k,t)Nsttgystk,whereThen, the total time consumption forNst can be formulated as
(14)Fst=F1+F2+F3.The objective function of wagon flow organization can be stated as(15)min⁡⁡∑s∈S∑t∈Q(s)Fst,
where  S is set of origin stations and s∈S. Q(s) is set of destinations of the wagon flows that originate from s. And Q(s)={t:Nst≠0}. t∈Q(s).


### 3.2.5. Constraints

The wagon flow organization problem has two constrains.(
1) Loading and Unloading Capacity Constraint for Nonstop Train.The loading and unloading capacity must reach the minimum size of loading or unloading a whole nonstop train. If there are several wagon flows consolidated together to be shipped to a technical station in the itinerary by nonstop shipping scheme, the sum of their loading or unloading capacity must reach the average marshaling number of wagons in a nonstop shipping scheme
(16)m-stI(∑t∈Q(s)ystk)-∑t∈Q(s)mstystk≤0∀s∈S,k∈P(s),
wheremst is the smaller one between maximum loading capacity of station s and maximum unloading capacity of station t at certain period;m-st is the average marshaling number of wagons in nonstop shipping scheme;I(x) is a step function, I(x)={1ifx>00otherwise.(
2) Each Wagon Flow Is Served by One and Only One Train Service.
(17)
I
s
t
x
s
t
+
∑
k
∈
V
(
s
)
x
s
t
k
+
∑
k
∈
P
(
s
)
y
s
t
k
=
1
∀
s
∈
S
,
t
∈
Q
(
s
)
,
where Ist={1ifmst≥m-st0otherwise∀s∈S,t∈Q(s), which ensures that only the wagon flow that reaches the average marshaling number of wagon in nonstop shipping scheme can be shipped by it.


### 3.2.6. Constraint Linearization

Constraint (16) can be linearized by introducing a variable ysk(18)ysk=I(∑t∈Q(s)ystk)={1∑t∈Q(s)ystk>00otherwisemmmmmmmmmmm∀s∈S,k∈P(s).Then constraint (16) can be formulated as follows:
(19)m-stysk-∑t∈Q(s)mstystk≤0∀s∈S,k∈P(s).Constraint (20) is also introduced to ensure that ysk equals 1 if and only if ystk=1(20)Mysk≥∑t∈Q(s)ystk∀s∈S,k∈P(s),
where M is a large positive constant.


### 3.2.7. Mathematical Model

The wagon flow organization model can be stated as(M-II)(21)min⁡∑s∈S∑t∈Q(s)Fsts.t.Istxst+∑k∈V(s)xstk+∑k∈P(s)ystk=1∀s∈S,t∈Q(s)s.t.m-stysk-∑t∈Q(s)mstystk≤0∀s∈S,k∈P(s)s.t.Mysk≥∑t∈Q(s)ystk∀s∈S,k∈P(s)s.t.xst,xstk∈{0,1}∀s∈S,t∈Q(s),k∈V(s)s.t.ysk,ystk∈{0,1}∀s∈S,t∈Q(s),k∈P(s).


## 3.2.1. An Example of Wagon Flow Organization Problem

The wagon flow organization problem can be illustrated by a simple network (see Figure1), which has three technical stations, two freight center stations, and three shipments (N12, N14, and N15).Figure 1
A simple rail network.There are 12 combinations for routing all the shipments on potential train services (see Figure2). Arcs in the network specify the available train connections. The OD pair will be reclassified at the technical stations in the itinerary if it is not served by through train. If transport demand between two adjacent stations is positive, the train service must be provided. So shipment N12 has only one service strategy that is served by the train service 1→2.Twelve combinations for routing all the shipments on potential train services.
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
(i)
(j)
(k)
(l)Strategy 1 (see Figure 2(a)). The OD pairs N15, N14, and N12 are consolidated in train service 1→2 at the origin station 1. In this case, N15 is reclassified at technical stations 2, 3, and 4. N14 is reclassified at technical stations 2 and 3.Strategy 2 (see Figure 2(b)). There are two through train services 1→4 and 1→5. In this case, both N15 and N14 are directly shipped to the destination.Strategy 3 (see Figure 2(c)). There are two through train services 1→3 and 1→4. In this case, N14 is directly shipped to the destination. While N15 is directly shipped to technical station 3, and N15 still needs to be reclassified at technical stations 3 and 4.Strategy 4 (see Figure 2(d)). Strategy 3 and Strategy 4 share the common train service set, but the train services are different. N15 is directly shipped to technical station 4 and reclassified there. N14 is directly shipped to technical station 3 and reclassified there.Strategy 5 (see Figure 2(e)). There is one through train service 1→4. In this case, N15 is directly shipped to the technical station 4 and reclassified there. N14 and N12 are consolidated in train service 1→2. N14 is reclassified at technical stations 2 and 3.Strategy 6 (see Figure 2(f)). Strategy 5 and Strategy 6 share the common train service set, but the train services are different. N14 is directly shipped to technical station 4 by train service 1→4. N15 and N12 are consolidated in train service 1→2. N15 is reclassified at technical stations 2, 3, and 4.Strategy 7 (see Figure 2(g)). There is one through train service 1→3. In this case, N15 is directly shipped to the technical station 3 and reclassified at technical stations 3 and 4. N14 and N12 are consolidated in train service 1→2. N14 is reclassified at technical stations 2 and 3.Strategy 8 (see Figure 2(h)). Strategy 7 and Strategy 8 share the common train service set, but the train services are different. N14 is directly shipped to technical station 3 by train service 1→3 and reclassified there. N15 and N12 are consolidated in train service 1→2. N15 is reclassified at technical stations 2, 3, and 4.Strategy 9 (see Figure 2(i)). There is one through train service 1→3. In this case, N15 and N14 are directly consolidated together and shipped to the technical station 3 by train service 1→3. N14 is reclassified at technical station 3. N15 is reclassified at technical station 3 and 4.Strategy 10 (see Figure 2(j)). There is one through train service 1→4. In this case, N15 and N14 are consolidated together and directly shipped to the technical station 4 by train service 1→4. N15 is reclassified there.Strategy 11 (see Figure 2(k)). There are two through train services 1→3 and 1→5. In this case, N15 is directly shipped to destination by train service 1→5. N14 is directly shipped to the technical station 3 by train service 1→3 and reclassified there.Strategy 12 (see Figure 2(l)). There is one through train service 1→5. In this case, N15 is directly shipped to destination by train service 1→5. N14 and N12 are consolidated in train service 1→2. N14 is reclassified at technical station 2 and 3.


## 3.2.2. Analysis of Wagon Flow Organization Problem

Among the twelve strategies above, there are two extreme ones, that is, Strategy 1 and Strategy 2.(1)
In Strategy 1, train connection services are only provided between adjacent stations. In this way, the number of through train services reaches minimum; that is, the total waiting time at the dispatching station reaches minimum. However, each shipment will be reclassified at each technical station on its itinerary. Thus the reclassification fee and time consumption will increase significantly. The increasing number of reclassified wagons may cause congestion and time delay at some technical stations.(2)
In Strategy 2, transport demands are all delivered to their destinations without reclassification, which will reduce the reclassification fee and time consumption. However, this strategy may be unworkable. Such strategy requires enough classification tracks to store outbound trains, and single OD pair must be large enough to dispatch a through train service. Waiting time at the dispatching station will increase if the OD pair is small.The solution of wagon flow organization problem is a tradeoff between Strategy 1 and Strategy 2, aiming at minimizing the total shipping and handling cost for all shipments. The time consumption is used to weigh the cost.


## 3.2.3. Decision Variables

From the example above, it can be concluded that there are three train services for a wagon flow: the first one is shipped to the destination by nonstop shipping scheme at the origin station, the second one is served by nonthrough shipment, and the third one is shipped to a technical station in the itinerary by nonstop shipping scheme consolidated with other OD pairs. These variables are(10)xst={1Nstisshippedbynonstopscheme0otherwisexstk={1Nstisshippedtofirsttechnicalstationkintheitineraryandreclassifiedthere0otherwisemmmmmmmnmmmmmnmmm∀k∈V(s)ystk={1Nstisshippedtotcehnicalstationkandreclassifiedthere0otherwisemmmmmmmmmmmmmmmnmmm∀k∈P(s),
whereNst is number of wagon flow from origin s to destination t;s is origin station;t is destination station;V(s) is set of technical stations next to the origin station s, k∈V(s);P(s) is set of technical stations in the itinerary of wagon flow Nst; exclude the technical station next to the origin station, P(s)∩V(s)=∅, k∈P(s).


## 3.2.4. Objective Function

The optimal goal is to obtain the wagon flow organization plan with minimum time consumption, which means the level of service. The travel time is not considered for it is identical. In order to simplify the problem, the following assumptions are proposed: (1) the supply of empty wagons is not considered, (2) the storage capacity of the origin and destination station are not considered, (3) the shipping routes for all the service are given, and (4) the train formation plans (TFP) of the technical stations in the itinerary are known. This information can be obtained previously.Time consumption of three train services forNst is as follows.(i)
For the nonstop service, the service time equals the loading and unloading times As there is no intermediate service such as reclassification(11)F1=Nstwstxst,wherewst is loading and unloading time at the origin and destination for a wagon served by nonstop shipping scheme.(ii)
Nonthrough shipment cost, which includes the reclassified time in the itinerary and loading and unloading time(12)F2=∑k∈V(s)Nstw--stxstk+∑k∈V(s)∑g∈G(k,t)Nsttgxstk,wherew--st is loading and unloading time at the origin and destination for a wagon shipped by non-through shipment;tg is reclassified time at technical station g;G(k,t) is set of technical stations in the itinerary after Nst has been reclassified at technical station k, and g∈G(k,t).(iii)
The wagon flow is shipped to a technical station in the itinerary by nonstop shipping scheme. This item includes the reclassified time consumption in the itinerary and time consumption for loading and unloading(13)F3=∑k∈P(s)Nstw-stystk+∑k∈P(s)∑g∈G(k,t)Nsttgystk,whereThen, the total time consumption forNst can be formulated as
(14)Fst=F1+F2+F3.The objective function of wagon flow organization can be stated as(15)min⁡⁡∑s∈S∑t∈Q(s)Fst,
where  S is set of origin stations and s∈S. Q(s) is set of destinations of the wagon flows that originate from s. And Q(s)={t:Nst≠0}. t∈Q(s).


## 3.2.5. Constraints

The wagon flow organization problem has two constrains.(
1) Loading and Unloading Capacity Constraint for Nonstop Train.The loading and unloading capacity must reach the minimum size of loading or unloading a whole nonstop train. If there are several wagon flows consolidated together to be shipped to a technical station in the itinerary by nonstop shipping scheme, the sum of their loading or unloading capacity must reach the average marshaling number of wagons in a nonstop shipping scheme
(16)m-stI(∑t∈Q(s)ystk)-∑t∈Q(s)mstystk≤0∀s∈S,k∈P(s),
wheremst is the smaller one between maximum loading capacity of station s and maximum unloading capacity of station t at certain period;m-st is the average marshaling number of wagons in nonstop shipping scheme;I(x) is a step function, I(x)={1ifx>00otherwise.(
2) Each Wagon Flow Is Served by One and Only One Train Service.
(17)
I
s
t
x
s
t
+
∑
k
∈
V
(
s
)
x
s
t
k
+
∑
k
∈
P
(
s
)
y
s
t
k
=
1
∀
s
∈
S
,
t
∈
Q
(
s
)
,
where Ist={1ifmst≥m-st0otherwise∀s∈S,t∈Q(s), which ensures that only the wagon flow that reaches the average marshaling number of wagon in nonstop shipping scheme can be shipped by it.


## 3.2.6. Constraint Linearization

Constraint (16) can be linearized by introducing a variable ysk(18)ysk=I(∑t∈Q(s)ystk)={1∑t∈Q(s)ystk>00otherwisemmmmmmmmmmm∀s∈S,k∈P(s).Then constraint (16) can be formulated as follows:
(19)m-stysk-∑t∈Q(s)mstystk≤0∀s∈S,k∈P(s).Constraint (20) is also introduced to ensure that ysk equals 1 if and only if ystk=1(20)Mysk≥∑t∈Q(s)ystk∀s∈S,k∈P(s),
where M is a large positive constant.


## 3.2.7. Mathematical Model

The wagon flow organization model can be stated as(M-II)(21)min⁡∑s∈S∑t∈Q(s)Fsts.t.Istxst+∑k∈V(s)xstk+∑k∈P(s)ystk=1∀s∈S,t∈Q(s)s.t.m-stysk-∑t∈Q(s)mstystk≤0∀s∈S,k∈P(s)s.t.Mysk≥∑t∈Q(s)ystk∀s∈S,k∈P(s)s.t.xst,xstk∈{0,1}∀s∈S,t∈Q(s),k∈V(s)s.t.ysk,ystk∈{0,1}∀s∈S,t∈Q(s),k∈P(s).


## 4. Solution Algorithm

Tabu search (TS) [30, 31] uses the tabu list to record the information of local optimal solutions, which can help to enlarge the search space and avoid the locally optimal solutions. The adaptive clonal selection algorithm (ACSA) [32–34] is shown to be an evolutionary strategy capable of high convergence rate and diversified. In this section, TS and ACSA are combined into a new heuristics method, called T-ACSA, to solve the proposed models.


### 4.1. The Detail Techniques of T-ACSA

#### 4.1.1. Affinity Measure

Affinity measure of the algorithm is the objective of the model; the smaller the better. In order to extend the space of solution, the algorithm accepts solutions which fail to satisfy the constraints. However, penalty coefficient will be added to the affinity measure.


#### 4.1.2. The Design of Antibody

The antibody of model M-I is designed as follows.The length of antibody is equal to the amount of customers inI. The codes in antibody are in J, and the amount of codes is p. To better understand the antibody, a simple example consisting of seven customers and four service points is proposed. p equals three (see Figure 3). Service point 2 is not included in the antibody, which means that it is not chosen as a service point.Figure 3
The design of antibody for M-I.The antibody of model M-II is designed as follows.According to the optimal result of model M-I, the antibody of each service point can be designed. A piece of antibody is proposed (see Figure4).Figure 4
One piece of antibody for M-II.The number ofystk in each piece is decided by the amount of technical stations in P(s) in the itinerary s→t, which can be obtained when the route is set. The sum value of each piece is 1. The pieces are separated between each other. And their amount equals the number of wagon flows.


#### 4.1.3. Neighborhood Search Operation

The neighborhood search operation of model M-I is as follows: choose two positions in the antibody randomly and exchange the values. The neighborhood search operation of the antibody in Figure3 is shown in Figure 5.Figure 5
Neighborhood search operation of model M-I.The neighborhood search operation of model M-II is as follows: choose a piece in antibody randomly. Find a code whose value is 0  and assign it to 1. Change the other codes into 0. The other pieces of antibody remain the same. The neighborhood search operation of the antibody in Figure4 is shown in Figure 6.Figure 6
Neighborhood search operation of model M-II.


#### 4.1.4. Mutation Operation

Mutation operation of model M-I is shown in Figure7: the maximum number of codes is four. If the amount of the chosen candidate points reaches the maximum, choose a code e in the antibody randomly. Change both e and the codes whose value is the same as e (see Figure 7(a)). Else choose a code e and change its value randomly (see Figure 7(b)).The mutation operation of model M-I.
(a)
(b)The mutation operation of model M-II is the same as the neighborhood search operation.


#### 4.1.5. TS Operation

The tabu strategy is first in first out. The length of tabu list isH; the search times are R. The number of the candidate solutions is U. TS operation procedure is as follows.Step 1.
Choose an antibody and set it as the current solution. Initialize the tabu list and the best solution.Step 2.
Use the neighborhood search operation to update the current solution and generateU candidate solutions. Then sort them by the affinity measure values. Choose the best solution.Step 3.
Judge whether the best solution is tabued. If it is tabued, go to Step2. Else judge whether the length of tabu list reaches H. If not, add the best solution into the tabu list and tabu it. Else remove the tabu information of the first solution in the tabu list and add the best solution into the tabu list. Go to Step 4.Step 4.
Judge whether the current best solution is the best one in the history. If not, go to Step5. Else set it as the best one in the history. Go to Step 5.Step 5.
Judge whether the search times reachR. If not, go to Step 2. Else update the antibody with the best solution in the history.


### 4.2. The Process of T-ACSA

Based on the aforementioned detailed analysis, T-ACSA approach is designed as follows.Step 1.
Initialize the population of antibody. GenerateN antibodies and constitute the species group P.Step 2.
Count the affinities and sort the antibodies according to their affinities in an ascending order.Step 3.
Clone each antibody inP then get a new species group C. The number of clone is ni=[wmax⁡(1-((i-1)/N))] and ni≥wmin⁡, where i is the order of the antibody after sorting. wmax⁡ is the maximum clone number and wmin⁡ is the minimum clone number.Step 4.
Apply the TS operation to update each antibody inC and get the new species group C′.Step 5.
Use mutation operation to update each antibody inC′. And get the new species group C′′. The probability of mutation is inversely proportional to the evolution generation φl=[α(1-(l/L))]. Where α is the coefficient, l is the current generation and L is the maximum generation.Step 6.
Choose the firstdl antibodies in C′′ and replace the worst dl antibodies in P by them; dl=[(f--fmin⁡)(D/f¯)]. Where D is the coefficient, f- is the average value of affinities in C′′ and fmin⁡ is the minimum value of affinities in C′′.Step 7.
If current status does not meet the terminal condition (the maximum searching times), go to Step2. Otherwise, go to Step 8.Step 8.
Output the best solution, that is, the optimal location of service points or the wagon flow organization.


## 4.1. The Detail Techniques of T-ACSA

### 4.1.1. Affinity Measure

Affinity measure of the algorithm is the objective of the model; the smaller the better. In order to extend the space of solution, the algorithm accepts solutions which fail to satisfy the constraints. However, penalty coefficient will be added to the affinity measure.


### 4.1.2. The Design of Antibody

The antibody of model M-I is designed as follows.The length of antibody is equal to the amount of customers inI. The codes in antibody are in J, and the amount of codes is p. To better understand the antibody, a simple example consisting of seven customers and four service points is proposed. p equals three (see Figure 3). Service point 2 is not included in the antibody, which means that it is not chosen as a service point.Figure 3
The design of antibody for M-I.The antibody of model M-II is designed as follows.According to the optimal result of model M-I, the antibody of each service point can be designed. A piece of antibody is proposed (see Figure4).Figure 4
One piece of antibody for M-II.The number ofystk in each piece is decided by the amount of technical stations in P(s) in the itinerary s→t, which can be obtained when the route is set. The sum value of each piece is 1. The pieces are separated between each other. And their amount equals the number of wagon flows.


### 4.1.3. Neighborhood Search Operation

The neighborhood search operation of model M-I is as follows: choose two positions in the antibody randomly and exchange the values. The neighborhood search operation of the antibody in Figure3 is shown in Figure 5.Figure 5
Neighborhood search operation of model M-I.The neighborhood search operation of model M-II is as follows: choose a piece in antibody randomly. Find a code whose value is 0  and assign it to 1. Change the other codes into 0. The other pieces of antibody remain the same. The neighborhood search operation of the antibody in Figure4 is shown in Figure 6.Figure 6
Neighborhood search operation of model M-II.


### 4.1.4. Mutation Operation

Mutation operation of model M-I is shown in Figure7: the maximum number of codes is four. If the amount of the chosen candidate points reaches the maximum, choose a code e in the antibody randomly. Change both e and the codes whose value is the same as e (see Figure 7(a)). Else choose a code e and change its value randomly (see Figure 7(b)).The mutation operation of model M-I.
(a)
(b)The mutation operation of model M-II is the same as the neighborhood search operation.


### 4.1.5. TS Operation

The tabu strategy is first in first out. The length of tabu list isH; the search times are R. The number of the candidate solutions is U. TS operation procedure is as follows.Step 1.
Choose an antibody and set it as the current solution. Initialize the tabu list and the best solution.Step 2.
Use the neighborhood search operation to update the current solution and generateU candidate solutions. Then sort them by the affinity measure values. Choose the best solution.Step 3.
Judge whether the best solution is tabued. If it is tabued, go to Step2. Else judge whether the length of tabu list reaches H. If not, add the best solution into the tabu list and tabu it. Else remove the tabu information of the first solution in the tabu list and add the best solution into the tabu list. Go to Step 4.Step 4.
Judge whether the current best solution is the best one in the history. If not, go to Step5. Else set it as the best one in the history. Go to Step 5.Step 5.
Judge whether the search times reachR. If not, go to Step 2. Else update the antibody with the best solution in the history.


## 4.1.1. Affinity Measure

Affinity measure of the algorithm is the objective of the model; the smaller the better. In order to extend the space of solution, the algorithm accepts solutions which fail to satisfy the constraints. However, penalty coefficient will be added to the affinity measure.


## 4.1.2. The Design of Antibody

The antibody of model M-I is designed as follows.The length of antibody is equal to the amount of customers inI. The codes in antibody are in J, and the amount of codes is p. To better understand the antibody, a simple example consisting of seven customers and four service points is proposed. p equals three (see Figure 3). Service point 2 is not included in the antibody, which means that it is not chosen as a service point.Figure 3
The design of antibody for M-I.The antibody of model M-II is designed as follows.According to the optimal result of model M-I, the antibody of each service point can be designed. A piece of antibody is proposed (see Figure4).Figure 4
One piece of antibody for M-II.The number ofystk in each piece is decided by the amount of technical stations in P(s) in the itinerary s→t, which can be obtained when the route is set. The sum value of each piece is 1. The pieces are separated between each other. And their amount equals the number of wagon flows.


## 4.1.3. Neighborhood Search Operation

The neighborhood search operation of model M-I is as follows: choose two positions in the antibody randomly and exchange the values. The neighborhood search operation of the antibody in Figure3 is shown in Figure 5.Figure 5
Neighborhood search operation of model M-I.The neighborhood search operation of model M-II is as follows: choose a piece in antibody randomly. Find a code whose value is 0  and assign it to 1. Change the other codes into 0. The other pieces of antibody remain the same. The neighborhood search operation of the antibody in Figure4 is shown in Figure 6.Figure 6
Neighborhood search operation of model M-II.


## 4.1.4. Mutation Operation

Mutation operation of model M-I is shown in Figure7: the maximum number of codes is four. If the amount of the chosen candidate points reaches the maximum, choose a code e in the antibody randomly. Change both e and the codes whose value is the same as e (see Figure 7(a)). Else choose a code e and change its value randomly (see Figure 7(b)).The mutation operation of model M-I.
(a)
(b)The mutation operation of model M-II is the same as the neighborhood search operation.


## 4.1.5. TS Operation

The tabu strategy is first in first out. The length of tabu list isH; the search times are R. The number of the candidate solutions is U. TS operation procedure is as follows.Step 1.
Choose an antibody and set it as the current solution. Initialize the tabu list and the best solution.Step 2.
Use the neighborhood search operation to update the current solution and generateU candidate solutions. Then sort them by the affinity measure values. Choose the best solution.Step 3.
Judge whether the best solution is tabued. If it is tabued, go to Step2. Else judge whether the length of tabu list reaches H. If not, add the best solution into the tabu list and tabu it. Else remove the tabu information of the first solution in the tabu list and add the best solution into the tabu list. Go to Step 4.Step 4.
Judge whether the current best solution is the best one in the history. If not, go to Step5. Else set it as the best one in the history. Go to Step 5.Step 5.
Judge whether the search times reachR. If not, go to Step 2. Else update the antibody with the best solution in the history.


## 4.2. The Process of T-ACSA

Based on the aforementioned detailed analysis, T-ACSA approach is designed as follows.Step 1.
Initialize the population of antibody. GenerateN antibodies and constitute the species group P.Step 2.
Count the affinities and sort the antibodies according to their affinities in an ascending order.Step 3.
Clone each antibody inP then get a new species group C. The number of clone is ni=[wmax⁡(1-((i-1)/N))] and ni≥wmin⁡, where i is the order of the antibody after sorting. wmax⁡ is the maximum clone number and wmin⁡ is the minimum clone number.Step 4.
Apply the TS operation to update each antibody inC and get the new species group C′.Step 5.
Use mutation operation to update each antibody inC′. And get the new species group C′′. The probability of mutation is inversely proportional to the evolution generation φl=[α(1-(l/L))]. Where α is the coefficient, l is the current generation and L is the maximum generation.Step 6.
Choose the firstdl antibodies in C′′ and replace the worst dl antibodies in P by them; dl=[(f--fmin⁡)(D/f¯)]. Where D is the coefficient, f- is the average value of affinities in C′′ and fmin⁡ is the minimum value of affinities in C′′.Step 7.
If current status does not meet the terminal condition (the maximum searching times), go to Step2. Otherwise, go to Step 8.Step 8.
Output the best solution, that is, the optimal location of service points or the wagon flow organization.


## 5. Numerical Experiments

The models and algorithm are tested by a railway network with 48 railway freight stations. Station 1 to station 16 is candidate freight center stations. The parameters of this physical network are provided in Tables7 and 8, which include the distance information and the transport demand of each customer. Table 1 shows the investment to locate a center station at the candidate sites. Parameters of M-I model and T-ACSA are shown in Table 2.Table 1
The investment for each candidate station.
Candidate point
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
C
j
150
300
170
160
210
170
200
230
180
210
150
60
110
250
180
50
Unit: ten thousand CNY.Table 2
The parameters of model M-I and T-ACSA.
Parameters of model M-I
μ
1
μ
2
DC
p
δ
Cap
j
λ
ε
Value
0.3
0.7
182 (km)
8
200 (km)
1752000 (t)
0.67 (CNY/t)
0.8
Parameters of T-ACSA
w
max
⁡
w
min
⁡
L
D
α
N
H
R
U
Value
8
2
100
18
0.6~0.8
20~30
4~6
10~15
8We test the algorithm ten times by different combination of parameters. The best affinity value of objective function is 176776352.8 CNY. And the final location plan is 3, 5, 6, 13, and 14. The assignments for the forty-eight freight stations are 3, 3, 3, 3, 5, 6, 5, 5, 14, 6, 13, 13, 13, 14, 14, 13, 3, 3, 3, 3, 3, 5, 5, 5, 6, 5, 6, 5, 6, 6, 6, 5, 5, 5, 6, 13, 6, 14, 14, 6, 13, 14, 13, 13, 13, 14, 14, and 13. The iterative process of the algorithm is shown in Figure8.Figure 8
The iterative process of T-ACSA for M-I.For the second step, we can plan wagon flow organization with outputs of model M-I, such as the location of freight center stations, the transport volume between each center station. Assuming the average loaded weight of a rail wagon is 50 t, and the stations operate for 365 days per year. Then the total transport demand can converted into daily transport volume. The average volumes of railway freight center station 3, 5, 6, 13, and 14 are 95, 97, 84, 87, and 75, respectively. The railway network and the location of railway freight center stations are shown in Figure9. The reclassified time at technical stations is listed in Table 3. The wagon flows between railway freight center stations are shown in Table 4. And the parameters of model M-II and T-ACSA are shown in Table 5.Table 3
The reclassified time consumption at a technical station.
Technical station
t
A
t
B
t
C
t
D
t
E
t
F
t
G
t
g
4
6
4.5
5
4
5
4.5
Unit: car hour.Table 4
The wagon flows between railway freight center stations.
Origin
Destination
Sum of wagon flows
3
5
6
13
14
3
—
24
8
42
21
95
5
14
—
20
41
22
97
6
23
7
—
14
40
84
13
19
28
15
—
25
87
14
28
7
6
34
—
75Table 5
The parameters of model M-II and T-ACSA.
Parameters of model M-II
w
s
t
w
-
s
t
w
-
-
s
t
m
-
s
t
m
s
t
Value
20 (car hour)
15 (car hour)
12 (car hour)
40
40
Parameters of T-ACSA
w
′
max
⁡
w
′
min
⁡
L
′
D
′
α
′
N
′
H
′
R
′
U
′
Value
6
2
50
10
0.6~0.8
16~20
4
5~15
6Figure 9
The railway network and the location of candidate freight center stations.We test the algorithm five times by different combination of parameters. The final result of wagon flow organization is shown in Table6. The iterative process of the algorithm is shown in Figure 10.Table 6
The optimal result of wagon flow organization.
Origin
Destination
3
5
6
13
14
3
—
y
3,5
B
=
1
x
3,6
A
=
1
x
3,13
=
1
y
3,14
B
=
1
5
x
5,3
B
=
1
—
x
5,6
B
=
1
x
5,13
=
1
x
5,14
C
=
1
6
x
6,3
A
=
1
x
6,5
B
=
1
—
x
6,13
B
=
1
x
6,14
=
1
13
y
13,3
B
=
1
y
13,5
D
=
1
y
13,6
B
=
1
—
y
13,14
D
=
1
14
y
14,3
B
=
1
x
14,5
E
=
1
y
14,6
B
=
1
x
14,13
E
=
1
—Table 7
The distance between customer and service point.
Customer
The candidate service point
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
1
0
119
133
196
278
292
348
403
536
415
561
521
602
704
762
741
2
119
0
60
156
159
182
229
284
417
329
443
422
489
585
643
630
3
133
60
0
96
170
218
237
290
421
378
457
461
513
588
646
657
4
196
156
96
0
223
292
277
324
445
461
494
531
564
605
662
708
5
278
159
170
223
0
89
70
125
258
260
288
311
345
426
484
489
6
292
182
218
292
89
0
113
153
273
174
278
243
310
435
492
449
7
348
229
237
277
70
113
0
55
188
257
221
277
287
356
414
431
8
403
284
290
324
125
153
55
0
133
267
170
261
245
301
359
388
9
536
417
421
445
258
273
188
133
0
331
77
269
179
168
226
300
10
415
329
378
461
260
174
257
267
331
0
293
128
263
457
506
370
11
561
443
457
494
288
278
221
170
77
293
0
209
102
165
219
228
12
521
422
461
531
311
243
277
261
269
128
209
0
147
358
401
242
13
602
489
513
564
345
310
287
245
179
263
102
147
0
216
255
144
14
704
585
588
605
426
435
356
301
168
457
165
358
216
0
58
254
15
762
643
646
662
484
492
414
359
226
506
219
401
255
58
0
262
16
741
630
657
708
489
449
431
388
300
370
228
242
144
254
262
0
17
54
173
183
233
332
344
402
457
590
458
614
569
654
758
816
792
18
71
48
77
163
207
225
277
332
465
361
491
461
534
633
691
675
19
252
222
162
66
273
349
319
360
472
521
528
584
605
626
680
748
20
219
184
124
28
243
316
294
338
456
486
507
553
581
613
669
724
21
172
124
64
32
202
266
260
310
435
433
480
507
546
598
655
690
22
158
104
44
52
190
250
252
302
430
415
472
492
535
594
651
679
23
172
53
80
165
106
138
176
231
364
299
391
381
440
532
590
582
24
240
121
135
197
38
97
108
163
296
270
325
334
378
464
522
522
25
279
161
180
240
24
65
74
127
259
237
282
291
334
427
485
477
26
306
187
196
244
28
93
42
97
230
257
261
296
321
398
456
465
27
327
209
225
277
56
78
36
82
212
228
235
261
290
379
437
434
28
352
234
229
252
86
156
50
74
195
307
242
324
320
359
417
462
29
284
170
199
267
59
30
92
138
265
203
278
264
319
430
488
461
30
311
211
255
335
139
50
156
187
293
126
285
211
300
448
504
434
31
337
223
252
317
97
48
87
114
226
172
230
214
266
387
444
407
32
405
287
285
307
131
180
68
40
139
306
191
300
275
304
361
415
33
413
298
286
292
154
218
105
90
161
354
224
350
316
314
370
452
34
451
332
337
367
173
195
103
48
85
285
128
256
214
253
311
353
35
354
249
287
360
147
69
141
159
252
117
239
173
252
403
458
387
36
506
387
392
417
228
245
158
103
30
313
89
261
188
198
256
317
37
388
296
341
421
216
131
212
225
298
45
268
136
253
433
484
372
38
584
465
469
490
306
319
236
181
48
363
79
287
174
120
178
278
39
556
437
444
470
279
287
209
154
27
331
56
259
157
149
207
274
40
452
360
405
483
273
192
257
258
304
46
259
82
220
419
467
324
41
487
391
432
507
290
215
264
256
284
88
231
40
182
387
432
282
42
649
530
533
552
371
382
301
246
113
412
119
322
189
55
113
258
43
572
457
478
524
309
282
247
203
135
260
60
159
45
200
246
184
44
546
437
468
527
305
255
256
226
202
190
135
78
73
280
323
196
45
650
536
558
605
389
358
328
283
196
311
122
191
50
191
221
106
46
707
588
598
624
430
427
361
307
179
425
148
316
170
62
85
192
47
711
594
607
639
438
426
370
318
197
408
151
294
147
102
117
152
48
698
584
604
647
434
407
371
324
223
359
156
237
100
176
194
82
Unit: km.Table 8
The transport demand of each customer.
Customer
1
2
3
4
5
6
7
8
9
10
Demand
219000
109500
197100
262800
219000
153300
131400
262800
438000
175200
Customer
11
12
13
14
15
16
17
18
19
20
Demand
306600
219000
109500
131400
175200
262800
438000
175200
219000
175200
Customer
21
22
23
24
25
26
27
28
29
30
Demand
328500
219000
219000
175200
131400
153300
394200
109500
219000
109500
Customer
31
32
33
34
35
36
37
38
39
40
Demand
175200
262800
262800
153300
109500
240900
284700
438000
197100
131400
Customer
41
42
43
44
45
46
47
48
—
—
Demand
175200
350400
219000
131400
175200
175200
21900
131400
—
—
Unit: ton per a year.Figure 10
The iterative process of T-ACSA for M-II.To verify the algorithm, we solved the models M-I and M-II by ILOG Cplex at the same time. The final results of the present algorithms and Cplex are the same. The run time of models M-I and M-II is less than 2 s and 1 s.


## 6. Conclusion

A two-stage programming is proposed to describe and solve the location of railway freight center stations and wagon flow organization problem. The first stage determines the optimal location with the objective to minimize the total cost of service and investment. The second stage optimizes the wagon flow organization among different stations. Different from the research in literature, the first model considered the coverage distance constraint and the change of transport demand. The second model analyzed the cost of different schemes. A heuristic algorithm that combines TS with ACSA is designed. The numerical example of a network with 48 stations demonstrates that the method is workable.While the microoperations in wagon flow organization, like wagon flow route decision and the supply of empty wagons, have not been considered in the scheduling process, these aspects can be considered in the future research.



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*Source: 246791-2014-05-11.xml*