Research Article A Pragmatic Optimization Method for Motor ...downloads.hindawi.com/journals/ddns/2016/4540503.pdf · Research Article A Pragmatic Optimization Method for Motor Train

Research ArticleA Pragmatic Optimization Method for Motor Train SetAssignment and Maintenance Scheduling Problem

Jian Li1 Boliang Lin1 Zhongkai Wang2 Lei Chen1 and Jiaxi Wang1

1School of Traffic and Transportation Beijing Jiaotong University Beijing 100044 China2Institute of Computing Technologies China Academy of Railway Sciences Beijing 100081 China

Correspondence should be addressed to Boliang Lin bllinbjtueducn

Received 8 October 2015 Revised 10 February 2016 Accepted 14 February 2016

Academic Editor Alicia Cordero

Copyright copy 2016 Jian Li et al This is an open access article distributed under the Creative Commons Attribution License whichpermits unrestricted use distribution and reproduction in any medium provided the original work is properly cited

With the rapid development of high-speed railway inChina the problemofmotor train set assignment andmaintenance schedulingis becoming more and more important for transportation organizationThis paper focuses on considering the special maintenanceitems of motor train set and mainly meets the two maintenance cycle limits on aspects of mileage and time for each item Andthen a 0-1 integer programming model for motor train set assignment and maintenance scheduling is proposed which aims atmaximizing the accumulated mileage before each maintenance and minimizing the number of motor train sets Restrictions of themodel include the matching relation between motor train sets and routes as well as that between motor train sets and maintenanceitems andmaintenance capacity of motor train set depot A heuristic solution strategy based on particle swarm optimization is alsoproposed to solve the model In the end a case study is designed based on the background of Beijing south depot in China andthe result indicates that the model and algorithm proposed in this paper could solve the problem of motor train set assignment andmaintenance scheduling effectively

1 Introduction

In recent years the high-speed railway in China has beendeveloping quickly and the number of motor train setsand trains running on the high-speed railway have beenincreasing year by year at the same time By the end of year2014 the total operation mileage of China railway is 112000kilometers of which the operationmileage of high-speed rail-way is 16000 kilometers and there are 1411 motor train setsrunning for the high-speed railway passenger transportationTherefore in order to improve the service for high-speedrailway transportation organization further the problem ofhow to enhance the motor train set management includingmotor train set assignment and maintenance scheduling isbecoming a focus of attention for the railway transportationdepartment and it is difficult to be solved effectively Themotor train set has a few special characteristics comparedwith the general rolling stock which is reflected as followsFirstly the procurement cost of a motor train set is veryexpensive for example a normative ldquoCRHrdquo motor train set

is worth 31000000 dollars approximately which is muchmore expensive than the general rolling stock so it is of greatsignificance to minimize the number of motor train setsSecondly the inspection and repair system of motor trainset in China is extremely complex there are more than tentypes of ldquoCRHrdquo motor train sets and each type of motortrain set includes a lot of different maintenance items Whatis more the maintenance cycle and maintenance place ofeach item range a lot as well which together make it difficultto schedule the maintenance work Thirdly the mileage ofthe route ranges a lot some are long route while the othersare short route and the motor train set assignment for aroute relies on the type of motor train set Besides there isa strong relevance between motor train set assignment andmaintenance scheduling so we should take the two sides intoconsideration at the same time while scheduling the motortrain set assignment for a route and maintenance work Inpractice the motor train set assignment and maintenancescheduling is made by hand according to the experience ofdispatcher in the motor train set depot generally and it is

Hindawi Publishing CorporationDiscrete Dynamics in Nature and SocietyVolume 2016 Article ID 4540503 13 pageshttpdxdoiorg10115520164540503

2 Discrete Dynamics in Nature and Society

nearly impossible to get an optimization schedule Thereforeit is needed to take the train set assignment and maintenancescheduling into consideration simultaneously while makingup the motor train set circulation plan which is helpful toimprove the quality of the plan In this way it not onlyincreases themotor train set operation efficiency and reducesthe number of motor train set as well as the procurement costbut also reduces the maintenance times and the maintenancecost during a fixed period

In this paper we research the problem of motor train setassignment and maintenance scheduling based on the prac-tical problem existing in the motor train set depot in Chinaand propose a pragmatic method for optimizing the motortrain set assignment and maintenance scheduling whichcould produce an optimal result and provide a reference formotor train set assignment and maintenance scheduling inthe motor train set depot

The organization of this paper is as follows Section 1 is theintroduction Section 2 is about the related literature reviewIn Section 3 we introduce the problem about motor trainset assignment and maintenance scheduling In Section 4we analyze the optimization objective and constraints and a0-1 integer programming mathematical model is proposedIn Section 5 we discuss the solution algorithm for theoptimization model In Section 6 a practical case study isdesigned Section 7 is the conclusion and prospects of thispaper

2 Related Literatures

With the development of high-speed railway around theworld many experts and scholars have researched the motortrain set assignment and maintenance scheduling problemfrom different aspects and proposed different models andsolution methods for the motor train set circulation For theliteratures Kroon [1ndash4] researched the rolling stock assign-ment and connection problem extensively and profoundlyand these researchesmainly aimed atminimizing the numberof motor train sets and enhancing the robustness of themotor train set circulation plan which was based on thegiven train arcs and time table and considered the constraintsincluding the type of motor train set and the maintenancecondition And then a few diverse integer programmingmodels were proposed according to the different concreteissues and some algorithms such as branch and boundmethod were designed to solve the optimization modelsCordeau et al [5] proposed amulticommodity network flow-based model for assigning locomotives and cars to trainsin the context of passenger transportation and adopted abranch-and-bound method to solve the problem Lingayaet al [6] researched locomotives and cars assignment to aset of scheduled trains for the passenger railways and theydescribed a modeling and solution methodology for a carassignment problem Noori and Ghannadpour [7] studiedthe locomotive assignment problem which is modeled usingvehicle routing and scheduling problem in their research anda two-phase approach based on a hybrid genetic algorithmis used to solve the problem Maroti [8 9] focused on theregular preventivemaintenance of train unites atNSReizigers

and presented two integer programming models for solvingthe maintenance routing problem one is the interchangemodel and the other is the transition model which couldsolve the maintenance issue in the forthcoming one to threedays for the train unites Giacco et al [10] focused on themotor train set connection problem on a transportationservice network and then aimed at minimizing the numberof motor train sets and constructed a mixed integer linearprogramming model for optimizing the short-period main-tenance scheduling of motor train set and the model tookthe transportation tasks running without taking passengerswith short-period maintenance items into considerationMany experts and scholars have researched the same problemin other relevant industries such as the fleet assignmentproblem (FAP) ElMoudani andMora-Camino [11] proposeda dynamic approach for the problems of assigning planesto fights and of feet maintenance operations scheduling intheir research Sherali et al [12] presented a method tointegrate the FAP with schedule design aircraft maintenancerouting and crew scheduling and presented a randomizedsearch procedures Deris et al [13] researched the problem ofship maintenance scheduling and modelled it as a constraintsatisfaction problem (CSP) in their paper and a geneticalgorithm (GA) was adopted to solve the problem Go et al[14] researched the problem of operation and maintenancescheduling for a containership and developed amixed integerprogramming model for the problem based on which aheuristic algorithm was presented

There are also many Chinese experts and scholars whohave researched the relevant problem according to the actualsituation in China Nie et al [15] and Zhao and Tomii[16] comparatively early researched the operation problemof motor train set with mainly considering the followinginfluences empty motor train set dispatching diverse typesof motor train set multibase of motor train set and soon which laid the theoretical foundation for optimizingthe motor train set circulation Huang [17] mainly aimedat the routine maintenance issue and attempted to optimizethe motor train set operation and maintenance plan Inthis research the inspection and repair system the supplysystem of spare parts and the maintenance managementinformation system were considered Wang et al [18] ana-lyzed three statuses of motor train set including undertakingroute being in maintenance and waiting for maintenanceand then a connection network composed of undertakingroute and conducting maintenance is designed based onwhich the author proposed an optimization model aimingat maximizing the accumulated mileage before conductingthe corresponding maintenance Zhang et al [19] consideredthe constraints from two aspects particularly that is mainte-nance time period and maintenance mileage cycle of motortrain set and presented an optimization method to solve theproblem Wang et al [20] researched the integrated opti-mization method for operation and maintenance planningof motor train set which aimed at reducing the number ofmotor train sets and the maintenance cost and designed amax-min ant colony algorithm to solve the problem Li [21]presented a decomposition strategy to break down the motortrain set operation planning problem into three subproblems

Discrete Dynamics in Nature and Society 3

from diverse dimensionality which includes the time thespace and the process The motor train set plan was dividedinto the routine operation maintenance planning problemand the advanced maintenance planning problem and thenthe author proposed an integer programming model foroptimizing the former problem Wang et al [22] focusedon the multidepot vehicle routing problem and proposed anovel fitness-scaling adaptive genetic algorithm with localsearch to solve the problemwhichwas superior to someotheralgorithm such as the standard genetic algorithm Zhang etal [23] analyzed and summarized the application of particleswarm optimization in different areas which is helpful for thesolution strategy designing in this paper

From the literatures mentioned above we could get aconclusion that the related issues about motor train set cir-culation problem are researched extensively and profoundlyand a few research achievements are got such as optimizationmodels and solution algorithms However the researches andthe achievements are focused on some specific issues on thewhole What is more the condition of high-speed railway inchina is not the samewith that abroad and the correspondingresearch achievements abroad could not be used to serve thetransportation organization of high-speed railway in chinadirectly and those in chinamostly stay at a level of theoreticalresearch

3 Basic Problem Description

The problem discussed in this paper is the motor train setassignment and maintenance scheduling and it refers to afew key elements including motor train sets routes andmaintenance items Therefore the main tasks of motor trainset assignment and maintenance scheduling are to assigna well-conditioned motor train set to each route every dayand to arrange the maintenance work of motor train set ofwhich the accumulated mileage or time of the correspondingmaintenance item after the latest maintenance is to meet themaintenance cycle In this paper we define the route as theordered trainsrsquo circulation which are undertaken by the samemotor train set from the departure to the arrival at the motortrain set depot That is to say the departure depot and thearrival depot are the same depot which is the attachmentdepot of the motor train set In China there is a certain timereserved for infrastructure inspection of high-speed railwaywhich is generally 4 to 6 hours and the train is forbiddenduring this period So the motor train set regularly stays atthe attachment depot or the other depot at night Thereforethe departure time and the arrival time of a route may notbe in the same day so according to the departure time andthe arrival time of each route we divide the route into one-day route and multiday route in this paper In our opinion ifthe departure time and the arrival time of a route are on thesame day (0000sim2400) we call the route a one-day routeAnd if the arrival time of a route is on the next day relativeto the departure time we call the route a two-day routeand the others can be called multiday route by that analogyThe definition of route could be helpful for the optimizationmodel construction in the later research and it is much moretallied with the actual condition in the motor train set depot

Station A

Station C

Station B(the depot)

Day 1 Day 2 Day 3

T1 T2

T3T4

Figure 1 A diagrammatic sketch of route

According to the definition of route we give a diagram-matic sketch of route which is shown in Figure 1 In thediagrammatic sketch there are three stations named stationA station B and station C respectively and the motor trainset depot is at the same place with station B Station A sandstation C are both the adjacent stations of station B andthe two stations are the turn-back stations for the motortrain set while station B is the departure station Motortrain sets depart from station B with undertaking the trainsfrom station B to station A and from station B to station Crespectively and come back to station B with undertakingthe corresponding trains and they may go to the motor trainset depot for a fixed maintenance or staying So the routecomposed of train T1 and train T2 is one-day route and theroute composed of train T3 and train T4 is two-day route

It is relatively complex for the inspection and repair sys-tem ofmotor train set in China and themaintenance contentsare divided into five grades The grade one maintenanceis generally called the routine maintenance the grade twomaintenance is generally called the special maintenance andthe grades three to five maintenance are generally called theadvanced maintenance uniformly While the accumulatedmileage is to meet 4000 kilometers or the accumulatedtime is to meet 48 hours after the last routine maintenancethe motor train set has to go back to the depot for theroutine maintenance again Compared with the other twokinds of maintenance the maintenance period of the routinemaintenance is much shorter and it occurs generally at nightso it is of high maintenance frequency On the contrary theadvanced maintenance has a longer maintenance cycle andthe time spent on the maintenance is comparatively longer aswell At present the total accumulatedmileage of most motortrain set is not to meet the advanced maintenance periodParticularly the special maintenance has a few maintenanceitems (or being called maintenance packet) such as I2maintenance M1 maintenance flaw detection of hollow axleand traction engine greasing Besides eachmaintenance itemof the special maintenance ranges a lot in the aspects ofmaintenance cycle (including mileage and time cycle) andmaintenance timeThus the problem of special maintenancescheduling is much more complex than any other and itsquality has a deep effect on the operation efficiency and themaintenance cost of motor train set Because of this we focuson taking the special maintenance into consideration in thispaper and research the optimization method for train set


EMU1

EMU2

EMU3

Route 1

Route 1

Item A

EMU4Standby

Day 1 Day 2 Day 3

Route 1

Route 2

Route 2

Item B

Route 2

Route 2

Figure 2 An example of motor train set assignment and mainte-nance schedule

assignment and maintenance scheduling In Figure 2 we givean example of motor train set assignment and maintenanceschedule with considering the special maintenance

For the motor train set assignment and maintenanceschedule shown in Figure 2 the motor train set EMU1undertakes route 1 on the first day and conducts maintenanceitemA on the second day and undertakes route 1 again on thethird day The motor train set EMU2 undertakes route 2 onthe first two days and conducts maintenance item B on thethird day The motor train set EMU3 undertakes the route2 during the period The motor train set EMU4 is being instandby mode on the first day and undertakes the route 1 androute 2 respectively on the second day and the third day

4 Mathematical Optimization Model

41 Basic Assumptions for Modelling

Assumption 1 We set the ldquodayrdquo as the smallest unit of timein the scheduling process If a motor train set is arrangedto conduct more than one maintenance item on a certainday we select the longest maintenance time of the arrangedmaintenance items as the maintenance time

Assumption 2 We assume that the maintenance time of asixteen-marshalling motor train set is equal to that of aneight-marshalling motor train set with the same type forany maintenance items In practice this can be achieved bysending two maintenance groups for the sixteen-marshallingmotor train set while one group for eight-marshalling motortrain set

Assumption 3 Neglect the substitution among different typesof motor train sets In other words a certain type of motortrain set can only undertake the corresponding route

Assumption 4 Various information including initial statemaintenance records routes and maintenance items arealready known

42 Parameters and Variables Definition

(1) Sets Notation It is defined that119863 = 119905 | 119905 = 1 2 119873

119863 is

the set of dates for scheduling and119873

119863is the number of days

119905 is the index of date when 119905 = 0 and it represents the daybefore the planning cycle 119864 = 119890 | 119890 = 1 2 119873

119864 is the set

of motor train sets119873119864is the number of motor train sets and

119890 is the index of motor train set 119875 = 119901 | 119901 = 1 2 119873

119875

is the set of maintenance items 119873119875is the total number of

maintenance item and 119901 is the index of maintenance itemEach maintenance item has a restriction of maintenancemileage cycle 119878

119901(unit kilometer) andmaintenance time cycle

119879

119901(unit day) 120583

119901(unit day) is the length of maintenance

time for maintenance item 119901 119877 = 119903 | 119903 = 1 2 119873

119877 is

the set of routes where 119873

119877and 119903 are the number and index

of the route respectively Each route 119903 has two attributesincluding running mileage 119904

119903(unit kilometer) and running

time 120576

119903(unit day) We define 119864(119905) as the set of motor train

sets which are available to start undertaking a route on the119905th day and more specifically they are the remaining onesexcept the motor train sets in operation or maintenance Themotor train set of 119864(119905) can be used for starting to undertakea route or conducting a maintenance item in advance or evenbeing in standby state119864(119905) is generated during the schedulingprocess according to the initial state of the motor train set onthat very day

(2) Parameters Notation It is defined that 120575119890119903is the matching

parameter for motor train set 119890 and route 119903 If the motortrain set 119890 matches well with the route 119903 the value of 120575119890

119903is

1 otherwise the value of 120575119890119903is 0 120591119890119901is the matching parameter

for motor train set 119890 and maintenance item 119901 If the motortrain set 119890 matches well with the maintenance item 119901 thevalue of 120591

119890

119901is 1 otherwise the value of 120591

119890

119901is 0 119862

119901is the

capability of maintenance item 119901 for the motor train setdepot namely the number of the motor train sets that canbe arranged to conduct the maintenance item 119901 on the sameday 120582 is the percentage that the accumulated mileage or timecould exceed the maintenance cycle limit and it is set to 10generally in practice

(3) Variables Notation It is defined that 119909119890119903(119905) is the decision

variable which indicates whether the motor train set 119890 startsto undertake the route 119903 on the 119905th day or not If the answeris yes then let 119909119890

119903(119905) = 1 otherwise let 119909119890

119903(119905) = 0 119910119890

119901(119905) be

the decision variable which indicates whether themotor trainset 119890 starts to conduct the maintenance item 119901 on the 119905th dayor not If the answer is yes then let 119910119890

119901(119905) = 1 otherwise

let 119910119890119901(119905) = 0 120593119890

119903(119905) be the assistant decision variable which

indicates whether the motor train set 119890 undertakes the route119903 on the 119905th day or not If the answer is yes then let 120593119890

119903(119905) =

1 otherwise let 120593

119890

119903(119905) = 0 120601119890

119901(119905) be the assistant variable

which indicates whether the motor train set 119890 conducts themaintenance item 119901 on the 119905th day or not If the answer is yesthen let 120601119890

119901(119905) = 1 otherwise let 120601119890

119901(119905) = 0 120572119890(119905) be the state

decision variable which indicates whether the motor train set119890 is in operation on the 119905th day or not If the answer is yesthen let 120572119890(119905) = 1 otherwise let 120572119890(119905) = 0 120573119890(119905) be the statedecision variable which indicates whether the motor train set119890 is in maintenance on the 119905th day or not If the answer is yesthen let 120573119890(119905) = 1 otherwise let 120573119890(119905) = 0 120574119890(119905) be the statedecision variable which indicates whether the motor train set


119890 is in standby state on the 119905th day or not If the answer is yesthen let 120574119890(119905) = 1 otherwise let 120574119890(119905) = 0

The variable 119897

119890

119901(119905) (unit kilometer) is the accumulated

mileage of the motor train set 119890 until the 119905th day after thelatest maintenance of item 119901 The variable 119891

119890

119901(119905) (unit day)

is the accumulated time of motor train set 119890 until the 119905th dayafter the latest maintenance of item 119901

43 OptimizationObjective Themain optimization objectiveof this paper is to make the accumulated mileage producedafter the latest maintenance for a certain maintenance itemof a motor train set be close to the maximal maintenancemileage cycle as much as possible which could be helpful todecreasemaintenance frequency during the scheduling cyclewhich could reduce themaintenance cost and to improve theoperation efficiency So in this paper we aim at maximizingthe accumulated mileage which is described as the followingformula

max 119885

1= sum

119890isin119864

sum

119905isin119863

sum

119901isin119875

119897

119890

119901(119905 minus 1) 119910

119890

119901(119905) (1)

Because of the high procurement cost of motor train setwe aim at minimizing the number of motor train sets inoperation in order to reduce the procurement cost which isdescribed as the following formula

min 119885

2= sum

119890isin119864

119868 (120579

119890) (2)

In formula (2) 120579119890 is the times motor train set 119890 under-took a certain route during the scheduling cycle which isdescribed as formula (3) 119868(119909) is a common step functionwhich is given as formula (4)

120579

119890= sum

119903isin119877

sum

119905isin119863

119909

119890

119903(119905) 119890 isin 119864 (3)

119868 (119909) =

1 119909 gt 0

0 119909 le 0

(4)

In this paper we transform the problem of multiobjectiveprogramming to a single-objective programming problemto simplify the process of solution On the premise ofno delayed maintenance the maximization of accumulatedmileage before eachmaintenance is equal to theminimizationof 119863-value between the maximal mileage limit of a certainmaintenance item and the actual accumulated mileage Itcan be described as formula (5) and ldquo119905 minus 1rdquo in this formularepresents the day before the 119905th day

min 119885

3= sum

119890isin119864

sum

119905isin119863

sum

119901isin119875

(119878

119901 (1 + 120582) minus 119897

119890

119901(119905 minus 1)) 119910

119890

119901(119905) (5)

On the basis of objectives described above a conversioncoefficient 120596 is set for operating mileage and the number ofmotor train sets in operation Thus the number of motortrain sets in operation can be transformed to the equivalent

mileage and formulas (3) and (5) can be integrated to thefollowing formula

min 119885

= sum

119890isin119864

sum

119905isin119863

sum

119901isin119875

(119878

119901 (1 + 120582) minus 119897

119890

119901(119905 minus 1)) 119910

119890

119901(119905)

+ 120596sum

119890isin119864

119868 (120579

119890)

(6)

The value of formula (5) increases with the days of thescheduling cycle whereas function (3) has everything to dowith the number of motor train sets It is reasonable tointroduce scheduling days in 120596 to keep the weight of thesetwo formulas unaffected Therefore the value of 120596 can becalculated by formula (7) 119897 is the average daily operatingmileage of the motor train set

120596 = 119897 times 119873

119863 (7)

44 Constraints Analysis As the research contents in thispaper are based on the practical problem existing in themotor train set depot in China we determine the constraintsaccording to the actual condition of the motor train set depotin China and analyze influence factors as comprehensive aspossible

Only when the type and personnel quota of the motortrain set 119890 match well with the route 119903 the motor train set 119890could have the opportunity to undertake the route 119903 In otherwords the motor train set and the route have to match witheach other which can be described as the following formula

119909

119890

119903(119905) le 120575

119890

119903119890 isin 119864 (119905) 119903 isin 119877 119905 isin 119863 (8)

Similarly only when the motor train set 119890 matches withthe maintenance item 119901 the motor train set 119890 could havethe opportunity to conduct the maintenance item 119901 In otherwords the motor train set and the maintenance item have tosatisfy the matching relation which can be described as thefollowing formula

119910

119890

119901(119905) le 120591

119890

119901119890 isin 119864 (119905) 119901 isin 119875 119905 isin 119863 (9)

For any motor train set 119890 in the set 119864(119905) it can start toundertake just one route on the 119905th day at most which can bedescribed as the following formula

sum

119903isin119877

119909

119890

119903(119905) le 1 119890 isin 119864 (119905) 119905 isin 119863 (10)

For any route 119903 in the set 119877 one and just one motor trainset should be assigned to it on the 119905th day which can bedescribed as

sum

119890isin119864(119905)

119909

119890

119903(119905) = 1 119903 isin 119877 119905 isin 119863 (11)

The accumulated mileage 119897

119890

119901(119905) should not exceed the

maximal mileage limit of the corresponding maintenanceitem which can be described as

119897

119890

119901(119905) le (1 + 120582) 119878119901

119890 isin 119864 119901 isin 119875 119905 isin 119863 (12)


The accumulated time 119891

119890

119901(119905) after the latest maintenance

of each motor train set for a certain maintenance item shouldnot exceed the maximal time cycle limit which can bedescribed as

119891

119890

119901(119905) le (1 + 120582) 119879119901

(13)

The maintenance capacity limit of the motor train setdepot should be taken into account in the maintenancearrangement of the motor train set to avoid centralizedinspection and repair Therefore the following formula isobtained

sum

119890isin119864

120601

119890

119901(119905) le 119862

119901119901 isin 119875 119905 isin 119863 (14)

Each motor train set has its unique state which is beingin operation or in maintenance or in standby state and it canbe described as

120572

119890(119905) + 120573

119890(119905) + 120574

119890(119905) = 1 119890 isin 119864 119905 isin 119863 (15)

All the decision variables should satisfy the 0-1 integralconstraint which can be described as

119909

119890

119903(119905) 119910

119890

119901(119905) 120593

119890

119903(119905) 120601

119890

119901(119905) 120572

119890(119905) 120573

119890(119905) 120574

119890(119905)

isin 0 1

(16)

The accumulated mileage 119897119890119901(119905) and the accumulated time

119891

119890

119901(119905) can be calculated by formulas (17) and (18) respectively

If the motor train set 119890 is not arranged to conduct themaintenance item 119901 on the 119905th day then the accumulatedmileage 119897

119890

119901(119905) should be the sum of accumulated mileage on

the day before 119897119890119901(119905minus1) and the operating mileage of the route

which is undertaken by the motor train set 119890 on that very dayAt the same time the accumulated time119891119890

119901(119905) should increase

by one day on the previous basis On the contrary if themotortrain set 119890 is arranged to conduct the maintenance item 119901 onthe 119905th day then the corresponding accumulatedmileage 119897119890

119901(119905)

and the accumulated time 119891119890119901(119905) should return to zero

119897

119890

119901(119905) =

0 if 120601119890119901(119905) = 1

119897

119890

119901(119905 minus 1) + 120591

119890

119901sum

119903isin119877

119909

119890

119903(119905) 119904119903

else

119890 isin 119864 119901 isin 119875 119905 isin 119863

(17)

119891

119890

119901(119905) =

0 if 120601119890119901(119905) = 1

119891

119890

119901(119905 minus 1) + 1 times 120591

119890

119901 else

119890 isin 119864 119901 isin 119875 119905 isin 119863

(18)

Decision variables 119909

119890

119903(119905) 119910119890119901(119905) 120593119890119903(119905) 120601119890119901(119905) 120572119890(119905) 120573119890(119905)

and 120574

119890(119905) are complementary to each other If the motor train

set 119890 starts to undertake the route 119903 then it has to do so

throughout the operating time cycle 119905

119903 Otherwise the value

of 120593119890119903(119905) should be zero The relation can be described as

119905+120576119903minus1

sum

1199051015840=119905

120593

119890

119903(119905

1015840) = 120576

119903 if 119909

119890

119903(119905) = 1

120593

119890

119903(119905) = 0 else

119890 isin 119864 (119905) 119903 isin 119877 119905 isin 119863

(19)

Similarly if the motor train set 119890 starts to conduct themaintenance item119901 on the 119905th day then it should conduct thesamemaintenance item throughout themaintenance time 119905

119901

Otherwise the value of 120601119890119901(119905) should be zeroThe relation can

be described as

119905+120583119901minus1

sum

1199051015840=119905

120601

119890

119901(119905

1015840) = 120583

119901 if 119910119890

119901(119905) = 1

120601

119890

119901(119905) = 0 else

119890 isin 119864 (119905) 119901 isin 119875 119905 isin 119863

(20)

The decision variables 120572

119890(119905) and 120573

119890(119905) which decide

whether the motor train set 119890 is in maintenance or inoperation on the 119905th day are influenced by the value ofassistant variables 120593

119890

119903(119905) and 120601

119890

119901(119905) respectively The relation

can be described as

120572

119890(119905) = sum

119903isin119877

120593

119890

119903(119905) 119890 isin 119864 119905 isin 119863 (21)

120573

119890(119905) =

1 ifsum119901isin119875

120601

119890

119901(119905) ge 1

0 else119890 isin 119864 119905 isin 119863 (22)

Formula (21) indicates that if the motor train set isarranged to undertake a route on the 119905th day it will be inoperation state Formula (22) indicates that if the motor trainset 119890 is arranged to conduct at least one maintenance item onthe 119905th day it will be in maintenance state

45 Model Construction On the basis of abovementionedanalysis a 0-1 integer programming model for motor trainset assignment and maintenance collaboration scheduling isproposed in this paper which is shown as follows

(M) min 119885

= sum

119890isin119864

sum

119905isin119863

sum

119901isin119875

(119878

119901 (1 + 120582) minus 119897

119890

119901(119905 minus 1)) 119910

119890

119901(119905)

+ 120596sum

119890isin119864

119868 (120579

119890)

st (8) sim (16)

(23)

In this model the relevant intermediate variables canbe calculated by the corresponding formula on the basis ofoptimization objective and constraints above When 119905 = 0the relevant variables represent the initial state of the motortrain set In other words it indicates which state the motor


train set was in such as in operation or in maintenanceor in standby state Furthermore it also shows the progressof undertaking a route or conducting a maintenance itemThe model proposed in this paper is a nonlinear 0-1 integerprogramming model and cannot be directly solved by Lingoor any other business software Thus we design a heuristicsolution strategy to address this problem in this paper

5 Solution Strategy

As the motor train set assignment and maintenance schedul-ing is an extremely complex work and in order to providereference to the dispatchers for their work a fast solutionmethod should be proposed for this problem Particle swarmoptimization has the advantage of fast convergence speed andhigh accuracy solution and it is easy to be applied in mostareas so we use the PSO in this study and design a solutionstrategy for the optimization model based on analysis andpreprocess

51 Application Principles of PSO According to the funda-mental principles of particle swarm optimization and thepractical problem of motor train set assignment and main-tenance scheduling the characteristic of the optimizationmodel we set that each particle represents a motor trainset assignment schedule and the corresponding maintenanceschedule is produced by the motor train set schedule Sowe can conclude that the dimension of each particle is 119869 =

119873

119864times 119873

119863times 119873

119877 and according to the definition of decision

variable 119909

119890

119903(119905) of the model the dimension 119895 for a particle

denotes that themotor train set 119890 starts to undertake the route119903 on the 119905th day or not On this base we let parameter 119873

119872

represent the number of particle swarms and 119898 is the indexof each particle The motor train set assignment schedulewhich is represented by particle 119898 is expressed as 119883

119898=

(119909

1198981 119909

1198982 119909

119898119869) and each particle119898 has a fitness function

expressed as 119865(119909) During the process of iterative computa-tion each particle has a velocity vector expressed as 119881

119898=

(V1198981

V1198982

V119898119869

) andhas a historical optimalmotor train setassignment schedule expressed as 119875

119898= (119901

1198981 119901

1198982 119901

119898119869)

Besides there is a global optimal motor train set assignmentschedule expressed as 119875

119892= (119901

1198921 119901

1198922 119901

119892119869) for the whole

particle swarm The velocity update in the dimension 119895 ofparticle119898 is computed according to

V(119896+1)119898119895

= 120596

(119896)

119898V(119896)119898119895

+ 119888

1119903

1(119901

(119896)

119898119895minus 119909

(119896)

119898119895)

+ 119888

2119903

2(119901

(119896)

119892119895minus 119909

(119896)

119898119895)

(24)

The velocity of a particle is limited in the section[Vmin Vmax] and if the computation result is out of the rangewe set the boundary value of the velocity section as actualvelocity In formula (24) 119896 is the index of iteration timesduring the iteration process The parameters 119903

1and 119903

2are

the random number in the section [0 1] which could avoidfalling into the local optimum and help to search the globaloptimal solutionThe parameters 119888

1and 119888

2are called learning

factor On one hand the iterative result could inherit theadvantage of the historical optimal value of a particle through

the parameter 1198881 and on the other hand it could inherit the

advantage of the global optimal value of the particle swarmthrough the parameter 119888

2119901(119896)119898119895

and119901

(119896)

119892119895 respectively represent

the historical optimal value of a particle and the globaloptimal value of the particle swarm 120596(119896)

119898is called the inertia

weight the value of which is generated by a function with thelinear decreasing of the iteration times The computationalformula is shown as

120596

(119896)

119898= 120596max minus

120596max minus 120596min119896max

times 119896 (25)

In formula (26) 120596max represents the maximal inertiaweight and 120596min represents the minimal inertia weight andthe values are determinedmainly by referring to the empiricalresults gotten in our computational experiments as well asthe existing relevant research results The parameter 119896maxrepresents the maximal iteration times the value of whichalso relies on the empirical results

In this paper the decision variable 119909

119890

119903(119905) is a 0-1 integer

variable so the value of each particle119909(119896+1)119898119895

is also a 0-1 integerTherefore we adopt ambiguity function and randommethodto update the value of 119909(119896+1)

119898119895 and the computational formula

is shown as

119909

(119896+1)

119898119895=

1 120588 lt Sigmoid (V(119896+1)119898119895

)

0 others(26)

In formula (27) 120588 is a random number in the section[0 1] and the Sigmoid function is a common ambiguityfunction which is shown as

Sigmoid (V(119896)119898119895

) =

1

1 + exp (minusV(119896)119898119895

)

(27)

52 Constraints Preprocess In order to reduce the complexityand be easy to realize the optimization solution for themathematical model we propose to preprocess some of theconstraints in this paper

Firstly we remove the route constraint (11) of the modeland set a corresponding penalty coefficient 119876

1 If there is a

route that is not assigned with a motor train set on a certainday we add a value of 119876

1to the fitness function 119865(119909) For a

calculated motor train set assignment schedule we assumethat the number of routes which are not assigned with motortrain sets is119873

1 then the total penalty value is119873

1119876

1

In the same way we remove the maintenance capacityconstraint (14) of the model and set a corresponding penaltycoefficient 119876

2 If the number of motor train sets arranged

to conduct maintenance item 119901 exceeds the maintenancecapacity of the depot for the maintenance item 119901 we add avalue of 119876

2to the fitness function 119865(119909) for each exceeding

motor train set According to themotor train setmaintenanceschedule we assume that the extra number of motor trainsets being to conduct maintenance item 119901 is119873

119901(119905) on the 119905th

day then the total penalty value for exceeding maintenancecapacity is sum

119905isin119863sum

119901isin119875119873

119901(119905)119876

2


Based on the preprocess for some complex constraintswe take the optimization objective value of the model as theprimary component of the fitness function of a particle andadd the total penalty value to the fitness function So wecan get the expression for the fitness function shown as theformula (28) We aim at minimizing the function and take itas the decision reference for iteration process

min 119865 (119909)

= sum

119890isin119864

sum

119905isin119863

sum

119901isin119875

(119878

119901 (1 + 120582) minus 119897

119890

119901(119905 minus 1)) 119910

119890

119901(119905)

+ 120596sum

119890isin119864

119868 (120579

119890) + 119873

1119876

1+ sum

119905isin119863

sum

119901isin119875

119873

119901 (119905) 1198762

(28)

53 The Key Solution Steps On the bases of applicationprinciples analysis of PSO and constraints preprocess wedesign the detailed solution process which has four key stepsas follows

(1) Generate the Available Motor Train Set 119864(119905) For anyparticle 119898 on the 119905th day during the scheduling cycle wegenerate the motor train set 119864 according to the attachmentmotor train set of the depot at first And then we select theunavailablemotor train set 119890 in the set119864which is to undertakea route or to conduct a maintenance item on the 119905th day andremove it from the set 119864 If the assistant decision variable120572

119890(119905) = 1 or 120573

119890(119905) = 1 it indicates that the motor train set

119890 is in a state of being in operation or in maintenance on the119905th day so it is an unavailable motor train set for a routeand it should be removed from the set 119864 Then circulate themotor train set in the set 119864 and until the motor train set 119864 istraversed completely So far the available motor train set 119864(119905)on the 119905th day is generated successfully

(2) Arrange Motor Train set to Conduct Maintenance For anymotor train set 119890 in the available motor train set 119864(119905) weselect the route 119903 from the route set 119877 which is not only fitfor the motor train set 119890 but also has the shortest mileage Ifthemotor train set 119890 is unable to undertake the route 119903 duringthe maintenance cycle of each maintenance item we removeit from the set 119864(119905) arrange it to conduct the correspondingmaintenance item 119901 and let 119910119890

119901(119905) = 1 At the same time

we update the accumulated mileage 119897

119890

119901(119905) and time 119891

119890

119901(119905) We

repeat this process until each motor train set in the set 119864(119905) istraversed completely

(3) Arrange Motor Train set to Undertake Route (a) We adoptthe random strategy to assign a motor train set to a routeduring the process of generating initial solution For a route 119903in the route set119877 we select amotor train set 119890 from the set119864(119905)randomly and determine whether it is able to undertake theroute 119903 or not by the motor train set type and maintenancecycle limit and so forth If the motor train set 119890 is able toundertake the route 119903 we let 119909119890

119903(119905) = 1 and remove it from the

set 119864(119905) and update the accumulated mileage 119897

119890

119901(119905) and time

119891

119890

119901(119905) On the contrary if the motor train set 119890 is unable to

undertake the route 119903 we continue to select another motor

train set from the set 119864(119905) randomly and repeat this processIf there is no more motor train set which is able to undertakethe route 119903 we skip this route and continue to focus on thenext route until all the routes in the route set 119877 are traversedcompletely

(b) We adopt the particle swarm optimization to assign amotor train set to a route during the process of optimizationiteration For a route 119903 in the route set 119877 we select a motortrain set 119890 from the set 119864(119905) randomly If the motor train set119890 is fit for the route 119903 then we determine whether it is toundertake the route 119903 or not according to formula (26) Ifthe motor train set is assigned to undertake the route 119903 welet 119909119890119903(119905) = 1 remove it from the set 119864(119905) and update the

accumulated mileage 119897

119890

119901(119905) and time 119891

119890

119901(119905) If the motor train

set 119890 is not fit for the route 119903 or it is not assigned to undertakethe route according to the iteration update formula (26) weselect another motor train set from the set 119864(119905) randomly andrepeat this process If there is not any motor train set whichis assigned to undertake the route 119903 by the particle swarmoptimization we continue to adopt the random strategy toassign a motor train set to the route 119903

(4) Arrange the Remaining Motor Train sets to Conduct aMaintenance Item in Advance or to Be in Standby Mode Afterthe process of steps (2) and (3) the motor train sets in theset 119864(119905) are remaining motor train sets and we propose toarrange the remaining motor train sets in the set 119864(119905) toconduct a maintenance item or to be in standby mode byrandom strategy which indicates the maintenance of motortrain set in advance If a motor train set is arranged toconduct maintenance we propose to arrange it to conductthe maintenance item 119901 of which the accumulated mileageor time is mostly close to the maintenance cycle limit and let119910

119890

119901(119905) = 1 and update the accumulated mileage 119897119890

119901(119905) and time

119891

119890

119901(119905) at the same time If the motor train set is arranged to

be in standby mode only the accumulated time 119891119890119901(119905) is to be

updatedTo summarize the four key steps above we give a flow

chart for the solution process of motor train set assignmentand maintenance scheduling based on the particle swarmoptimization The flow chart is shown as Figure 3

6 Case Study

In this paper we design a case study on the background ofBeijing south motor train set depot in China and apply themathematical model and solution strategy to the practicalproblem of motor train set assignment and maintenancescheduling According to partial actual data that we can getfrom the depot we set that the scheduling cycle is fromOctober 30 2014 to November 5 2014 which is called weeklyschedule

We select 22 motor train sets with two types ofCRH380BL and CRH380CL in this case study and select 10routes according to the two types of motor train set Therouteswhich are suitable for themotor train set of CRH380BLtype are No R1simNo R6 and the other routes which aresuitable for the motor train set of CRH380CL type are No


Start

End

Initialize the basic data of motor train set and the calculatingparameter

Reach the maximal iteration times

Generate the initial solution and calculate the fitness function F(x) of each particle

Assign the initial solution to the historical optical value of each particle and select the global optical solution of the particle swarm

Update the value of inertia weight and velocity according to the calculation formulas and the iteration times

Update the solution by particle swarm optimization or random strategy and calculate the fitness function F(x) of each particle

Update the historical optical solution of each particle and the global optical solution of the particle swarm

No

Yes

Output the motor trainset assignment and maintenance schedule

Generate initial

solution

Iterativeoptimizationcalculation

Figure 3 The flow chart for solution process

R7simNo R10 Routes R1 R5 R7 and R10 are two-day routesand the others are one-day routes The basic information ofthese routes is shown in Table 1 which includes route namesuitable type and route mileage and route time

The basic information of the selected motor train sets isshown in Table 2 which includes train set name train settype accumulated mileage initial task and task process Theaccumulated mileage is the total mileage from the beginningof being in operation to the day before the scheduling cycle fora motor train set The initial task is the state of a motor trainset that it might undertake a route conduct a maintenanceitem or be in standby state We assume that the motor trainsets which did not undertake a route were all in standby stateThe task process is calculated by the number of days alreadyspent for this task For the two-day route it has two states oftask process includingDay 1 andDay 2 In this case study theday before the scheduling cycle is October 29 2014 If a motortrain set was in standby state we set its task process as 0

As the maintenance items of each type of motor train setare of great difference we select a few typical maintenanceitems for motor train set of CRH380BL and CRH380CLThebasic information of the selectedmaintenance items is shownin Table 3 It includes item name train set type mileage cycletime cycle and maintenance time From Table 3 we can learnthat each type of motor train set has sevenmaintenance itemsand themaintenance times of these maintenance items are allone day

The maintenance record of motor train set lasts toOctober 29 2015 which includes train set name item namedate and mileage The date of maintenance is the day for thelatest maintenance and the mileage is the total accumulatedmileage from the beginning of being in operation to the latestmaintenance for the motor train set As there are a lot ofmaintenance records we just list the maintenance recordsof four motor train sets for example which are shown inTable 4According to the accumulatedmileage inTables 2 and4 we can get the value of accumulated mileage by calculatingtheD-value between themileage data of lastmaintenance andthe day before the scheduling cycle

Other calculation parameters in this case study are set asfollows the days of planning cycle 119873

119863= 7 the maintenance

capacity for each maintenance item 119862

119901= 3 the extended

percentage 120582 = 10 and the average daily mileage of motortrain sets 119897 = 2000 In order to make sure that each routecan be assigned with a motor train set and the maintenancecapacity is not exceeded every day the penalty coefficient isset to infinity In this paper we set 119876

1= 119876

2= 1000000

According to the test results and experience we set thepopulation of particle swarm119873

119872= 40 themaximal iteration

times 119896max = 1000 the velocity section [minus4 4] the inertiaweight section [04 09] and the learning factor 119888

1= 119888

2= 20

In this paper we complete the program development byC++ programming language based on the visual studio 2010programming platform After 32 minutes of optimization


Table 1 Basic information of routes

Route name Suitable type Routemileage (km) Route time (d)

R1 CRH380BL 4342 2R2 CRH380BL 2670 1R3 CRH380BL 2080 1R4 CRH380BL 3216 1R5 CRH380BL 4480 2R6 CRH380BL 2902 1R7 CRH380CL 4342 2R8 CRH380CL 1994 1R9 CRH380CL 1994 1R10 CRH380CL 3616 2

Table 2 Basic information of motor train sets

Train setname Train set type Accumulated

mileage (km) Initial task Task process(d)

EMU1 CRH380BL 1895686 R5 1EMU2 CRH380BL 1940792 R2 1EMU3 CRH380BL 1868027 R5 2EMU4 CRH380BL 1895082 Standby 0EMU5 CRH380BL 1894356 Standby 0EMU6 CRH380BL 1885061 R4 1EMU7 CRH380BL 1854992 R3 1EMU8 CRH380BL 1849781 Standby 0EMU9 CRH380BL 1262735 Standby 0EMU10 CRH380BL 1159710 R1 1EMU11 CRH380BL 1216010 Standby 0EMU12 CRH380BL 1171145 R6 1EMU13 CRH380BL 1100308 R1 2EMU14 CRH380CL 572107 R8 1EMU15 CRH380CL 964404 R10 1EMU16 CRH380CL 884219 R10 2EMU17 CRH380CL 911571 R7 1EMU18 CRH380CL 883828 Standby 0EMU19 CRH380CL 763740 R7 2EMU20 CRH380CL 793955 Standby 0EMU21 CRH380CL 784786 Standby 0EMU22 CRH380CL 830930 R9 1

calculation a quasi-optimal motor train set assignmentand maintenance schedule is generated which is shown inFigure 4

From the quasi-optimal motor train set assignment andmaintenance schedule shown in Figure 4 we can get a con-clusion that there are 17 motor train sets in operation duringthe scheduling cycle 10 of which are of type CRH380BL and7 motor train sets are of type CRH380CL For the schedulewe arrange maintenance for 13 times in total for the motortrain sets including a 1-time M1 maintenance a 1-time M3

maintenance a 1-time traction enginemaintenance and a 10-time I2 maintenance For example the motor train set EMU9conducts I2 maintenance on the 2nd day the accumulatedmileage of the motor train set up to this maintenance is21452 kilometers relative to the latest I2 maintenance and theaccumulated time is 9 days This maintenance arranged forthe accumulated mileage is about to meet the maintenancemileage cycle limit of I2maintenance item For anothermotortrain set EMU16 it is arranged to conduct I2 maintenanceon the 4th day the accumulated mileage of the motor trainset up to this maintenance is 16018 kilometers relative to thelatest I2 maintenance and the accumulated time is 11 daysThismaintenance arranged for the accumulated time is aboutto meet the maintenance time cycle limit of I2 maintenanceitem It can be seen that becausewe take themaintenance timecycle into consideration some motor train sets are arrangedto conduct the corresponding maintenance item because theaccumulated time is to meet the maximal time cycle limit ofa certain maintenance item which may lead to a huge wasteof mileage before the maintenance In practice as the motortrain set assignment and maintenance schedule is made byhand in general the quality of the schedule mostly dependson the experience of scheduler and it is impossible to getan optimization schedule What is more the motor train setsometimes may not be arranged to conduct the correspond-ing maintenance item while the accumulated mileage oraccumulated time exceeds the maximal maintenance periodlimit In this case study through the optimization calculationthe accumulated mileage of the motor train set is as closeto the maximal maintenance mileage cycle as possible andthe operation efficiency of the motor train set is improvedcommendably The premise of that is the accumulated timedoes not exceed the maintenance time period limit

7 Conclusion

With the rapid development of high-speed railway theproblem of how to strengthen the management of motortrain set operation and maintenance is more and moreprominent In this paper on basis of practical situation ofBeijing south motor train set depot in China we focus onthe special maintenance item of motor train set and researchthe optimization method for motor train set assignment andmaintenance scheduling After the analysis of the optimiza-tion objectives and constraints the relationship between therelated intermediate variables is determined and then wepropose a 0-1 integer programming model for optimizingthe motor train set assignment and maintenance schedulingAccording to the optimization objectives and constraintswe design a solution strategy for the mathematical modelbased on the fundamental principles of particle swarmoptimization In the end we design a case study based onthe basic data of Beijing south motor train set depot and theoptimization result indicates that we can get a satisfactoryoptimization result through the mathematical model andthe solution strategy proposed in this paper However thepractical situation is rather complex For example somespecialmaintenance items are arranged at night together withthe routine maintenance so there is no need to arrange a full


Table 3 Basic information of maintenance items

Item name Train set type Mileage cycle (km) Time cycle (d) Maintenance time (d)I2 CRH380BL 20000 20 1M1 CRH380BL 100000 90 1M2 CRH380BL 400000 360 1M3 CRH380BL 800000 720 1Traction engine CRH380BL 200000 180 1Gearbox CRH380BL 400000 360 1Hollow axle CRH380BL 100000 90 1I2 CRH380CL 20000 10 1M1 CRH380CL 100000 45 1M2 CRH380CL 400000 180 1M3 CRH380CL 800000 360 1Traction engine CRH380CL 200000 90 1Gearbox CRH380CL 400000 180 1Hollow axle CRH380CL 100000 45 1

EMU1

EMU2

EMU3

EMU4

Oct 30 Oct 31 Nov 1

EMU5

EMU6

EMU7

EMU8

EMU9

EMU10

EMU11

EMU12

EMU13

EMU14

EMU15

EMU16

EMU17

EMU18

EMU19

EMU20

Nov 2 Nov 3 Nov 4 Nov 5

Traction engine

Standby

Standby Standby

Standby

Standby Standby Standby Standby Standby Standby



Standby

Standby


EMU21

EMU22

Standby Standby Standby Standby Standby StandbyStandby

Standby Standby Standby Standby StandbyStandby

R1R1

R1

R1

R1

R1

I2

I2

I2

I2

I2

I2

I2

I2

I2

I2

R1

R1

R3

R3 R3

R3

R3R3

R3R2

R6

R6

R6 R6

R6

R6

R7

R10

R10

R10R10

R10 R10

R10

R10

R7

R7

R7

R7

R7

R9

M3

R9

R9R9

R9 R9

R9

R7

R7 R8R8R8

R8

R8

R8 R8

R6

R2

R2

R2

R2

R2

R2R4

M1

R4

R4

R4 R4

R4

R4

R5

R5

R5R5R5

R5

R5R5

Figure 4 A quasi-optimal motor train set assignment and maintenance schedule


Table 4 Maintenance records of motor train sets

Train setname Item name Date Mileage

(km)

CRH6202B I2 20141026 1891206CRH6202B M1 20140909 1851620CRH6202B M2 20140718 1743716CRH6202B M3 20130906 1318710

CRH6202B Tractionengine

20141024 1886726

CRH6202B Gearbox 20140718 1743716CRH6202B Hollow axle 20140912 1851620CRH6230B I2 20141026 1936310CRH6230B M1 20140913 1839962CRH6230B M2 20140723 1732774CRH6230B M3 20131219 1317565


20140724 1732774

CRH6230B Gearbox 20140724 1732774CRH6230B Hollow axle 20140913 1839962CRH6301C I2 20141023 566125CRH6301C M1 20140916 494039CRH6301C M2 20140806 399489CRH6301C M3 20140806 399489

CRH6301C Tractionengine

20140806 399489

CRH6301C Gearbox 20140801 399489CRH6301C Hollow axle 20140918 494039CRH6303C I2 20141020 943358CRH6303C M1 20140925 893546CRH6303C M2 20140809 800586CRH6303C M3 20140809 800586


20140807 800586

CRH6303C Gearbox 20140807 800586CRH6303C Hollow axle 20140925 893546

day for the maintenance and it could undertake a route inthe daytime What is more there are situations which occurnow and then For example a certain type of motor trainset replaces another type of motor train set to undertakethe route and motor train set depots invoke motor train setsagainst each other Thus it can be seen that the motor trainset assignment and maintenance scheduling is affected by alot of influence factors and some related problems should beresearched further

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgments

This research is supported in part by the National ScienceFoundation of China (51378056) and the China RailwayScientific Research Project (2015J006-B 2015X004-C) Theauthors are thankful to all the personnel who provided helpfor this research

References

[1] E Abbink B van den Berg L Kroon andM Salomon ldquoAlloca-tion of railway rolling stock for passenger trainsrdquoTransportationScience vol 38 no 1 pp 33ndash41 2004

[2] A Alfieri R Groot L Kroon and A Schrijver ldquoEfficientcirculation of railway rolling stockrdquo Transportation Science vol40 no 3 pp 378ndash391 2006

[3] P-J Fioole L Kroon G Maroti and A Schrijver ldquoA rollingstock circulation model for combining and splitting of passen-ger trainsrdquo European Journal of Operational Research vol 174no 2 pp 1281ndash1297 2006

[4] M Peeters and L Kroon ldquoCirculation of railway rollingstock a branch-and-price approachrdquoComputers andOperationsResearch vol 35 no 2 pp 538ndash556 2008

[5] J-F Cordeau F Soumis and J Desrosiers ldquoSimultaneousassignment of locomotives and cars to passenger trainsrdquo Oper-ations Research vol 49 no 4 pp 531ndash548 2001

[6] N Lingaya J-F Cordeau G Desaulniers J Desrosiers andF Soumis ldquoOperational car assignment at VIA Rail CanadardquoTransportation Research Part B Methodological vol 36 no 9pp 755ndash778 2002

[7] S Noori and S F Ghannadpour ldquoLocomotive assignment prob-lem with trains precedence using genetic algorithmrdquo Journal ofIndustrial Engineering International vol 8 no 1 2012

[8] G Maroti and L Kroon ldquoMaintenance routing for train unitsthe transition modelrdquo Transportation Science vol 39 no 4 pp518ndash525 2005

[9] G Maroti and L Kroon ldquoMaintenance routing for train unitsthe interchange modelrdquo Computers and Operations Researchvol 34 no 4 pp 1121ndash1140 2007

[10] G L Giacco A DrsquoAriano and D Pacciarelli ldquoRolling stockrostering optimization under maintenance constraintsrdquo Journalof Intelligent Transportation Systems vol 18 no 1 pp 95ndash1052014

[11] W El Moudani and F Mora-Camino ldquoA dynamic approach foraircraft assignment and maintenance scheduling by airlinesrdquoJournal of Air Transport Management vol 6 no 4 pp 233ndash2372000

[12] H D Sherali E K Bish and X Zhu ldquoAirline fleet assignmentconcepts models and algorithmsrdquo European Journal of Opera-tional Research vol 172 no 1 pp 1ndash30 2006

[13] S Deris S Omatu H Ohta L C S Kutar and P AbdSamat ldquoShip maintenance scheduling by genetic algorithm andconstraint-based reasoningrdquo European Journal of OperationalResearch vol 112 no 3 pp 489ndash502 1999

[14] H Go J-S Kim and D-H Lee ldquoOperation and preven-tive maintenance scheduling for containerships mathematicalmodel and solution algorithmrdquo European Journal of OperationalResearch vol 229 no 3 pp 626ndash636 2013

[15] L Nie P Zhao H Yang and A Hu ldquoStudy on motor trainsetoperation in high speed railwayrdquo Journal of the China RailwaySociety vol 23 no 3 pp 1ndash7 2001


[16] P Zhao and N Tomii ldquoTrain-set scheduling and an algorithmrdquoJournal of the China Railway Society vol 25 no 3 pp 1ndash7 2003

[17] F HuangOptimization Research onMaintenance andOperationof Electric Multiple Unite (EMU) in China Tongji UniversityShanghai China 2008

[18] Y Wang J Liu and J Miao ldquoColumn generation algorithmsbased optimization method for maintenance scheduling ofmultiple unitsrdquoChina Railway Science vol 31 no 2 pp 115ndash1202010

[19] C-C Zhang W Hua and J-H Chen ldquoResearch on EMUscheduling under constraint of kilometrage and time for sched-uled inspection andmaintenancerdquo Journal of the China RailwaySociety vol 32 no 3 pp 16ndash19 2010

[20] ZWang T ShiW Zhang andHWang ldquoModel and algorithmfor the integrative scheduling of EMU utilization plan andmaintenance planrdquo China Railway Science vol 33 no 3 pp102ndash108 2012

[21] H Li Theory and Method Studies on EMU Scheduling Problemfor High Speed Railway Beijing Jiaotong University BeijingChina 2013

[22] S Wang Z Lu L Wei G Ji and J Yang ldquoFitness-scaling adap-tive genetic algorithm with local search for solving the MultipleDepot Vehicle Routing Problemrdquo Simulation Transactions ofthe Society for Modeling and Simulation International vol 91no 10 pp 1ndash16 2015

[23] Y Zhang S Wang and G Ji ldquoA comprehensive survey onparticle swarm optimization algorithm and its applicationsrdquoMathematical Problems in Engineering vol 2015 Article ID931256 38 pages 2015

Submit your manuscripts athttpwwwhindawicom

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MathematicsJournal of


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nearly impossible to get an optimization schedule Thereforeit is needed to take the train set assignment and maintenancescheduling into consideration simultaneously while makingup the motor train set circulation plan which is helpful toimprove the quality of the plan In this way it not onlyincreases themotor train set operation efficiency and reducesthe number of motor train set as well as the procurement costbut also reduces the maintenance times and the maintenancecost during a fixed period

In this paper we research the problem of motor train setassignment and maintenance scheduling based on the prac-tical problem existing in the motor train set depot in Chinaand propose a pragmatic method for optimizing the motortrain set assignment and maintenance scheduling whichcould produce an optimal result and provide a reference formotor train set assignment and maintenance scheduling inthe motor train set depot

The organization of this paper is as follows Section 1 is theintroduction Section 2 is about the related literature reviewIn Section 3 we introduce the problem about motor trainset assignment and maintenance scheduling In Section 4we analyze the optimization objective and constraints and a0-1 integer programming mathematical model is proposedIn Section 5 we discuss the solution algorithm for theoptimization model In Section 6 a practical case study isdesigned Section 7 is the conclusion and prospects of thispaper

2 Related Literatures

With the development of high-speed railway around theworld many experts and scholars have researched the motortrain set assignment and maintenance scheduling problemfrom different aspects and proposed different models andsolution methods for the motor train set circulation For theliteratures Kroon [1ndash4] researched the rolling stock assign-ment and connection problem extensively and profoundlyand these researchesmainly aimed atminimizing the numberof motor train sets and enhancing the robustness of themotor train set circulation plan which was based on thegiven train arcs and time table and considered the constraintsincluding the type of motor train set and the maintenancecondition And then a few diverse integer programmingmodels were proposed according to the different concreteissues and some algorithms such as branch and boundmethod were designed to solve the optimization modelsCordeau et al [5] proposed amulticommodity network flow-based model for assigning locomotives and cars to trainsin the context of passenger transportation and adopted abranch-and-bound method to solve the problem Lingayaet al [6] researched locomotives and cars assignment to aset of scheduled trains for the passenger railways and theydescribed a modeling and solution methodology for a carassignment problem Noori and Ghannadpour [7] studiedthe locomotive assignment problem which is modeled usingvehicle routing and scheduling problem in their research anda two-phase approach based on a hybrid genetic algorithmis used to solve the problem Maroti [8 9] focused on theregular preventivemaintenance of train unites atNSReizigers

and presented two integer programming models for solvingthe maintenance routing problem one is the interchangemodel and the other is the transition model which couldsolve the maintenance issue in the forthcoming one to threedays for the train unites Giacco et al [10] focused on themotor train set connection problem on a transportationservice network and then aimed at minimizing the numberof motor train sets and constructed a mixed integer linearprogramming model for optimizing the short-period main-tenance scheduling of motor train set and the model tookthe transportation tasks running without taking passengerswith short-period maintenance items into considerationMany experts and scholars have researched the same problemin other relevant industries such as the fleet assignmentproblem (FAP) ElMoudani andMora-Camino [11] proposeda dynamic approach for the problems of assigning planesto fights and of feet maintenance operations scheduling intheir research Sherali et al [12] presented a method tointegrate the FAP with schedule design aircraft maintenancerouting and crew scheduling and presented a randomizedsearch procedures Deris et al [13] researched the problem ofship maintenance scheduling and modelled it as a constraintsatisfaction problem (CSP) in their paper and a geneticalgorithm (GA) was adopted to solve the problem Go et al[14] researched the problem of operation and maintenancescheduling for a containership and developed amixed integerprogramming model for the problem based on which aheuristic algorithm was presented

There are also many Chinese experts and scholars whohave researched the relevant problem according to the actualsituation in China Nie et al [15] and Zhao and Tomii[16] comparatively early researched the operation problemof motor train set with mainly considering the followinginfluences empty motor train set dispatching diverse typesof motor train set multibase of motor train set and soon which laid the theoretical foundation for optimizingthe motor train set circulation Huang [17] mainly aimedat the routine maintenance issue and attempted to optimizethe motor train set operation and maintenance plan Inthis research the inspection and repair system the supplysystem of spare parts and the maintenance managementinformation system were considered Wang et al [18] ana-lyzed three statuses of motor train set including undertakingroute being in maintenance and waiting for maintenanceand then a connection network composed of undertakingroute and conducting maintenance is designed based onwhich the author proposed an optimization model aimingat maximizing the accumulated mileage before conductingthe corresponding maintenance Zhang et al [19] consideredthe constraints from two aspects particularly that is mainte-nance time period and maintenance mileage cycle of motortrain set and presented an optimization method to solve theproblem Wang et al [20] researched the integrated opti-mization method for operation and maintenance planningof motor train set which aimed at reducing the number ofmotor train sets and the maintenance cost and designed amax-min ant colony algorithm to solve the problem Li [21]presented a decomposition strategy to break down the motortrain set operation planning problem into three subproblems






Station A

Station C


Day 1 Day 2 Day 3

T1 T2

T3T4





EMU1

EMU2

EMU3

Route 1

Route 1

Item A

EMU4Standby

Day 1 Day 2 Day 3

Route 1

Route 2

Route 2

Item B

Route 2

Route 2












119863 is




119864 is the set



119875




119879

119901(unit day) 120583



119877 is





time 120576





119903is



119890


119890

119901is 0 119862

119901is the




119903(119905) = 1 otherwise let 119909119890

119903(119905) = 0 119910119890

119901(119905) be


119901(119905) = 1 otherwise

let 119910119890119901(119905) = 0 120593119890



119903(119905) =


119890

119903(119905) = 0 120601119890



119901(119905) = 1 otherwise let 120601119890

119901(119905) = 0 120572119890(119905) be the state




The variable 119897

119890



119890

119901(119905) (unit day)



max 119885

1= sum

119890isin119864

sum

119905isin119863

sum

119901isin119875

119897

119890

119901(119905 minus 1) 119910

119890

119901(119905) (1)


min 119885

2= sum

119890isin119864

119868 (120579

119890) (2)


120579

119890= sum

119903isin119877

sum

119905isin119863

119909

119890

119903(119905) 119890 isin 119864 (3)

119868 (119909) =

1 119909 gt 0

0 119909 le 0

(4)


min 119885

3= sum

119890isin119864

sum

119905isin119863

sum

119901isin119875

(119878

119901 (1 + 120582) minus 119897

119890

119901(119905 minus 1)) 119910

119890

119901(119905) (5)



min 119885

= sum

119890isin119864

sum

119905isin119863

sum

119901isin119875

(119878

119901 (1 + 120582) minus 119897

119890

119901(119905 minus 1)) 119910

119890

119901(119905)

+ 120596sum

119890isin119864

119868 (120579

119890)

(6)


120596 = 119897 times 119873

119863 (7)



119909

119890

119903(119905) le 120575

119890

119903119890 isin 119864 (119905) 119903 isin 119877 119905 isin 119863 (8)


119910

119890

119901(119905) le 120591

119890

119901119890 isin 119864 (119905) 119901 isin 119875 119905 isin 119863 (9)


sum

119903isin119877

119909

119890

119903(119905) le 1 119890 isin 119864 (119905) 119905 isin 119863 (10)


sum

119890isin119864(119905)

119909

119890

119903(119905) = 1 119903 isin 119877 119905 isin 119863 (11)


119890



119897

119890

119901(119905) le (1 + 120582) 119878119901

119890 isin 119864 119901 isin 119875 119905 isin 119863 (12)



119890



119891

119890

119901(119905) le (1 + 120582) 119879119901

(13)


sum

119890isin119864

120601

119890

119901(119905) le 119862

119901119901 isin 119875 119905 isin 119863 (14)


120572

119890(119905) + 120573

119890(119905) + 120574

119890(119905) = 1 119890 isin 119864 119905 isin 119863 (15)


119909

119890

119903(119905) 119910

119890

119901(119905) 120593

119890

119903(119905) 120601

119890

119901(119905) 120572

119890(119905) 120573

119890(119905) 120574

119890(119905)

isin 0 1

(16)


119891

119890



119890






119901(119905)


119897

119890

119901(119905) =

0 if 120601119890119901(119905) = 1

119897

119890

119901(119905 minus 1) + 120591

119890

119901sum

119903isin119877

119909

119890

119903(119905) 119904119903

else

119890 isin 119864 119901 isin 119875 119905 isin 119863

(17)

119891

119890

119901(119905) =

0 if 120601119890119901(119905) = 1

119891

119890

119901(119905 minus 1) + 1 times 120591

119890

119901 else

119890 isin 119864 119901 isin 119875 119905 isin 119863

(18)


119890

119903(119905) 119910119890119901(119905) 120593119890119903(119905) 120601119890119901(119905) 120572119890(119905) 120573119890(119905)

and 120574






119905+120576119903minus1

sum

1199051015840=119905

120593

119890

119903(119905

1015840) = 120576

119903 if 119909

119890

119903(119905) = 1

120593

119890

119903(119905) = 0 else

119890 isin 119864 (119905) 119903 isin 119877 119905 isin 119863

(19)


119901


be described as

119905+120583119901minus1

sum

1199051015840=119905

120601

119890

119901(119905

1015840) = 120583

119901 if 119910119890

119901(119905) = 1

120601

119890

119901(119905) = 0 else

119890 isin 119864 (119905) 119901 isin 119875 119905 isin 119863

(20)


119890(119905) and 120573



119890

119903(119905) and 120601

119890


can be described as

120572

119890(119905) = sum

119903isin119877

120593

119890

119903(119905) 119890 isin 119864 119905 isin 119863 (21)

120573

119890(119905) =

1 ifsum119901isin119875

120601

119890

119901(119905) ge 1

0 else119890 isin 119864 119905 isin 119863 (22)



(M) min 119885

= sum

119890isin119864

sum

119905isin119863

sum

119901isin119875

(119878

119901 (1 + 120582) minus 119897

119890

119901(119905 minus 1)) 119910

119890

119901(119905)

+ 120596sum

119890isin119864

119868 (120579

119890)

st (8) sim (16)

(23)




5 Solution Strategy



119873

119864times 119873

119863times 119873


variable 119909

119890



119872


119898=

(119909

1198981 119909

1198982 119909



119898=

(V1198981

V1198982

V119898119869


119898= (119901

1198981 119901

1198982 119901

119898119869)


119892= (119901

1198921 119901

1198922 119901



V(119896+1)119898119895

= 120596

(119896)

119898V(119896)119898119895

+ 119888

1119903

1(119901

(119896)

119898119895minus 119909

(119896)

119898119895)

+ 119888

2119903

2(119901

(119896)

119892119895minus 119909

(119896)

119898119895)

(24)


1and 119903

2are


1and 119888





2119901(119896)119898119895

and119901

(119896)





120596

(119896)

119898= 120596max minus


times 119896 (25)



119890

119903(119905) is a 0-1 integer




is shown as

119909

(119896+1)

119898119895=

1 120588 lt Sigmoid (V(119896+1)119898119895

)

0 others(26)


Sigmoid (V(119896)119898119895

) =

1

1 + exp (minusV(119896)119898119895

)

(27)



1 If there is a





1119876

1






119901(119905) on the 119905th


119905isin119863sum

119901isin119875119873

119901(119905)119876

2



min 119865 (119909)

= sum

119890isin119864

sum

119905isin119863

sum

119901isin119875

(119878

119901 (1 + 120582) minus 119897

119890

119901(119905 minus 1)) 119910

119890

119901(119905)

+ 120596sum

119890isin119864

119868 (120579

119890) + 119873

1119876

1+ sum

119905isin119863

sum

119901isin119875

119873

119901 (119905) 1198762

(28)



119890(119905) = 1 or 120573




119901(119905) = 1 At the same time


119890

119901(119905) and time 119891

119890

119901(119905) We





119890

119901(119905) and time

119891

119890






119890

119901(119905) and time 119891

119890




119890


119901(119905) and time

119891

119890





6 Case Study




Start

End








No

Yes


Generate initial

solution












1= 119876

2= 1000000




1= 119888

2= 20













7 Conclusion





EMU1

EMU2

EMU3

EMU4

Oct 30 Oct 31 Nov 1

EMU5

EMU6

EMU7

EMU8

EMU9

EMU10

EMU11

EMU12

EMU13

EMU14

EMU15

EMU16

EMU17

EMU18

EMU19

EMU20


Traction engine

Standby

Standby Standby

Standby




Standby

Standby


EMU21

EMU22



R1R1

R1

R1

R1

R1

I2

I2

I2

I2

I2

I2

I2

I2

I2

I2

R1

R1

R3

R3 R3

R3

R3R3

R3R2

R6

R6

R6 R6

R6

R6

R7

R10

R10

R10R10

R10 R10

R10

R10

R7

R7

R7

R7

R7

R9

M3

R9

R9R9

R9 R9

R9

R7

R7 R8R8R8

R8

R8

R8 R8

R6

R2

R2

R2

R2

R2

R2R4

M1

R4

R4

R4 R4

R4

R4

R5

R5

R5R5R5

R5

R5R5





(km)



20141024 1886726



20140724 1732774



20140806 399489



20140807 800586





Acknowledgments


References
































Volume 2014




Journal of











Journal of


Function Spaces






Algebra














Station A

Station C


Day 1 Day 2 Day 3

T1 T2

T3T4





EMU1

EMU2

EMU3

Route 1

Route 1

Item A

EMU4Standby

Day 1 Day 2 Day 3

Route 1

Route 2

Route 2

Item B

Route 2

Route 2












119863 is




119864 is the set



119875




119879

119901(unit day) 120583



119877 is





time 120576





119903is



119890


119890

119901is 0 119862

119901is the




119903(119905) = 1 otherwise let 119909119890

119903(119905) = 0 119910119890

119901(119905) be


119901(119905) = 1 otherwise

let 119910119890119901(119905) = 0 120593119890



119903(119905) =


119890

119903(119905) = 0 120601119890



119901(119905) = 1 otherwise let 120601119890

119901(119905) = 0 120572119890(119905) be the state




The variable 119897

119890



119890

119901(119905) (unit day)



max 119885

1= sum

119890isin119864

sum

119905isin119863

sum

119901isin119875

119897

119890

119901(119905 minus 1) 119910

119890

119901(119905) (1)


min 119885

2= sum

119890isin119864

119868 (120579

119890) (2)


120579

119890= sum

119903isin119877

sum

119905isin119863

119909

119890

119903(119905) 119890 isin 119864 (3)

119868 (119909) =

1 119909 gt 0

0 119909 le 0

(4)


min 119885

3= sum

119890isin119864

sum

119905isin119863

sum

119901isin119875

(119878

119901 (1 + 120582) minus 119897

119890

119901(119905 minus 1)) 119910

119890

119901(119905) (5)



min 119885

= sum

119890isin119864

sum

119905isin119863

sum

119901isin119875

(119878

119901 (1 + 120582) minus 119897

119890

119901(119905 minus 1)) 119910

119890

119901(119905)

+ 120596sum

119890isin119864

119868 (120579

119890)

(6)


120596 = 119897 times 119873

119863 (7)



119909

119890

119903(119905) le 120575

119890

119903119890 isin 119864 (119905) 119903 isin 119877 119905 isin 119863 (8)


119910

119890

119901(119905) le 120591

119890

119901119890 isin 119864 (119905) 119901 isin 119875 119905 isin 119863 (9)


sum

119903isin119877

119909

119890

119903(119905) le 1 119890 isin 119864 (119905) 119905 isin 119863 (10)


sum

119890isin119864(119905)

119909

119890

119903(119905) = 1 119903 isin 119877 119905 isin 119863 (11)


119890



119897

119890

119901(119905) le (1 + 120582) 119878119901

119890 isin 119864 119901 isin 119875 119905 isin 119863 (12)



119890



119891

119890

119901(119905) le (1 + 120582) 119879119901

(13)


sum

119890isin119864

120601

119890

119901(119905) le 119862

119901119901 isin 119875 119905 isin 119863 (14)


120572

119890(119905) + 120573

119890(119905) + 120574

119890(119905) = 1 119890 isin 119864 119905 isin 119863 (15)


119909

119890

119903(119905) 119910

119890

119901(119905) 120593

119890

119903(119905) 120601

119890

119901(119905) 120572

119890(119905) 120573

119890(119905) 120574

119890(119905)

isin 0 1

(16)


119891

119890



119890






119901(119905)


119897

119890

119901(119905) =

0 if 120601119890119901(119905) = 1

119897

119890

119901(119905 minus 1) + 120591

119890

119901sum

119903isin119877

119909

119890

119903(119905) 119904119903

else

119890 isin 119864 119901 isin 119875 119905 isin 119863

(17)

119891

119890

119901(119905) =

0 if 120601119890119901(119905) = 1

119891

119890

119901(119905 minus 1) + 1 times 120591

119890

119901 else

119890 isin 119864 119901 isin 119875 119905 isin 119863

(18)


119890

119903(119905) 119910119890119901(119905) 120593119890119903(119905) 120601119890119901(119905) 120572119890(119905) 120573119890(119905)

and 120574






119905+120576119903minus1

sum

1199051015840=119905

120593

119890

119903(119905

1015840) = 120576

119903 if 119909

119890

119903(119905) = 1

120593

119890

119903(119905) = 0 else

119890 isin 119864 (119905) 119903 isin 119877 119905 isin 119863

(19)


119901


be described as

119905+120583119901minus1

sum

1199051015840=119905

120601

119890

119901(119905

1015840) = 120583

119901 if 119910119890

119901(119905) = 1

120601

119890

119901(119905) = 0 else

119890 isin 119864 (119905) 119901 isin 119875 119905 isin 119863

(20)


119890(119905) and 120573



119890

119903(119905) and 120601

119890


can be described as

120572

119890(119905) = sum

119903isin119877

120593

119890

119903(119905) 119890 isin 119864 119905 isin 119863 (21)

120573

119890(119905) =

1 ifsum119901isin119875

120601

119890

119901(119905) ge 1

0 else119890 isin 119864 119905 isin 119863 (22)



(M) min 119885

= sum

119890isin119864

sum

119905isin119863

sum

119901isin119875

(119878

119901 (1 + 120582) minus 119897

119890

119901(119905 minus 1)) 119910

119890

119901(119905)

+ 120596sum

119890isin119864

119868 (120579

119890)

st (8) sim (16)

(23)




5 Solution Strategy



119873

119864times 119873

119863times 119873


variable 119909

119890



119872


119898=

(119909

1198981 119909

1198982 119909



119898=

(V1198981

V1198982

V119898119869


119898= (119901

1198981 119901

1198982 119901

119898119869)


119892= (119901

1198921 119901

1198922 119901



V(119896+1)119898119895

= 120596

(119896)

119898V(119896)119898119895

+ 119888

1119903

1(119901

(119896)

119898119895minus 119909

(119896)

119898119895)

+ 119888

2119903

2(119901

(119896)

119892119895minus 119909

(119896)

119898119895)

(24)


1and 119903

2are


1and 119888





2119901(119896)119898119895

and119901

(119896)





120596

(119896)

119898= 120596max minus


times 119896 (25)



119890

119903(119905) is a 0-1 integer




is shown as

119909

(119896+1)

119898119895=

1 120588 lt Sigmoid (V(119896+1)119898119895

)

0 others(26)


Sigmoid (V(119896)119898119895

) =

1

1 + exp (minusV(119896)119898119895

)

(27)



1 If there is a





1119876

1






119901(119905) on the 119905th


119905isin119863sum

119901isin119875119873

119901(119905)119876

2



min 119865 (119909)

= sum

119890isin119864

sum

119905isin119863

sum

119901isin119875

(119878

119901 (1 + 120582) minus 119897

119890

119901(119905 minus 1)) 119910

119890

119901(119905)

+ 120596sum

119890isin119864

119868 (120579

119890) + 119873

1119876

1+ sum

119905isin119863

sum

119901isin119875

119873

119901 (119905) 1198762

(28)



119890(119905) = 1 or 120573




119901(119905) = 1 At the same time


119890

119901(119905) and time 119891

119890

119901(119905) We





119890

119901(119905) and time

119891

119890






119890

119901(119905) and time 119891

119890




119890


119901(119905) and time

119891

119890





6 Case Study




Start

End








No

Yes


Generate initial

solution












1= 119876

2= 1000000




1= 119888

2= 20













7 Conclusion





EMU1

EMU2

EMU3

EMU4

Oct 30 Oct 31 Nov 1

EMU5

EMU6

EMU7

EMU8

EMU9

EMU10

EMU11

EMU12

EMU13

EMU14

EMU15

EMU16

EMU17

EMU18

EMU19

EMU20


Traction engine

Standby

Standby Standby

Standby




Standby

Standby


EMU21

EMU22



R1R1

R1

R1

R1

R1

I2

I2

I2

I2

I2

I2

I2

I2

I2

I2

R1

R1

R3

R3 R3

R3

R3R3

R3R2

R6

R6

R6 R6

R6

R6

R7

R10

R10

R10R10

R10 R10

R10

R10

R7

R7

R7

R7

R7

R9

M3

R9

R9R9

R9 R9

R9

R7

R7 R8R8R8

R8

R8

R8 R8

R6

R2

R2

R2

R2

R2

R2R4

M1

R4

R4

R4 R4

R4

R4

R5

R5

R5R5R5

R5

R5R5





(km)



20141024 1886726



20140724 1732774



20140806 399489



20140807 800586





Acknowledgments


References
































Volume 2014




Journal of











Journal of


Function Spaces






Algebra










EMU1

EMU2

EMU3

Route 1

Route 1

Item A

EMU4Standby

Day 1 Day 2 Day 3

Route 1

Route 2

Route 2

Item B

Route 2

Route 2












119863 is




119864 is the set



119875




119879

119901(unit day) 120583



119877 is





time 120576





119903is



119890


119890

119901is 0 119862

119901is the




119903(119905) = 1 otherwise let 119909119890

119903(119905) = 0 119910119890

119901(119905) be


119901(119905) = 1 otherwise

let 119910119890119901(119905) = 0 120593119890



119903(119905) =


119890

119903(119905) = 0 120601119890



119901(119905) = 1 otherwise let 120601119890

119901(119905) = 0 120572119890(119905) be the state




The variable 119897

119890



119890

119901(119905) (unit day)



max 119885

1= sum

119890isin119864

sum

119905isin119863

sum

119901isin119875

119897

119890

119901(119905 minus 1) 119910

119890

119901(119905) (1)


min 119885

2= sum

119890isin119864

119868 (120579

119890) (2)


120579

119890= sum

119903isin119877

sum

119905isin119863

119909

119890

119903(119905) 119890 isin 119864 (3)

119868 (119909) =

1 119909 gt 0

0 119909 le 0

(4)


min 119885

3= sum

119890isin119864

sum

119905isin119863

sum

119901isin119875

(119878

119901 (1 + 120582) minus 119897

119890

119901(119905 minus 1)) 119910

119890

119901(119905) (5)



min 119885

= sum

119890isin119864

sum

119905isin119863

sum

119901isin119875

(119878

119901 (1 + 120582) minus 119897

119890

119901(119905 minus 1)) 119910

119890

119901(119905)

+ 120596sum

119890isin119864

119868 (120579

119890)

(6)


120596 = 119897 times 119873

119863 (7)



119909

119890

119903(119905) le 120575

119890

119903119890 isin 119864 (119905) 119903 isin 119877 119905 isin 119863 (8)


119910

119890

119901(119905) le 120591

119890

119901119890 isin 119864 (119905) 119901 isin 119875 119905 isin 119863 (9)


sum

119903isin119877

119909

119890

119903(119905) le 1 119890 isin 119864 (119905) 119905 isin 119863 (10)


sum

119890isin119864(119905)

119909

119890

119903(119905) = 1 119903 isin 119877 119905 isin 119863 (11)


119890



119897

119890

119901(119905) le (1 + 120582) 119878119901

119890 isin 119864 119901 isin 119875 119905 isin 119863 (12)



119890



119891

119890

119901(119905) le (1 + 120582) 119879119901

(13)


sum

119890isin119864

120601

119890

119901(119905) le 119862

119901119901 isin 119875 119905 isin 119863 (14)


120572

119890(119905) + 120573

119890(119905) + 120574

119890(119905) = 1 119890 isin 119864 119905 isin 119863 (15)


119909

119890

119903(119905) 119910

119890

119901(119905) 120593

119890

119903(119905) 120601

119890

119901(119905) 120572

119890(119905) 120573

119890(119905) 120574

119890(119905)

isin 0 1

(16)


119891

119890



119890






119901(119905)


119897

119890

119901(119905) =

0 if 120601119890119901(119905) = 1

119897

119890

119901(119905 minus 1) + 120591

119890

119901sum

119903isin119877

119909

119890

119903(119905) 119904119903

else

119890 isin 119864 119901 isin 119875 119905 isin 119863

(17)

119891

119890

119901(119905) =

0 if 120601119890119901(119905) = 1

119891

119890

119901(119905 minus 1) + 1 times 120591

119890

119901 else

119890 isin 119864 119901 isin 119875 119905 isin 119863

(18)


119890

119903(119905) 119910119890119901(119905) 120593119890119903(119905) 120601119890119901(119905) 120572119890(119905) 120573119890(119905)

and 120574






119905+120576119903minus1

sum

1199051015840=119905

120593

119890

119903(119905

1015840) = 120576

119903 if 119909

119890

119903(119905) = 1

120593

119890

119903(119905) = 0 else

119890 isin 119864 (119905) 119903 isin 119877 119905 isin 119863

(19)


119901


be described as

119905+120583119901minus1

sum

1199051015840=119905

120601

119890

119901(119905

1015840) = 120583

119901 if 119910119890

119901(119905) = 1

120601

119890

119901(119905) = 0 else

119890 isin 119864 (119905) 119901 isin 119875 119905 isin 119863

(20)


119890(119905) and 120573



119890

119903(119905) and 120601

119890


can be described as

120572

119890(119905) = sum

119903isin119877

120593

119890

119903(119905) 119890 isin 119864 119905 isin 119863 (21)

120573

119890(119905) =

1 ifsum119901isin119875

120601

119890

119901(119905) ge 1

0 else119890 isin 119864 119905 isin 119863 (22)



(M) min 119885

= sum

119890isin119864

sum

119905isin119863

sum

119901isin119875

(119878

119901 (1 + 120582) minus 119897

119890

119901(119905 minus 1)) 119910

119890

119901(119905)

+ 120596sum

119890isin119864

119868 (120579

119890)

st (8) sim (16)

(23)




5 Solution Strategy



119873

119864times 119873

119863times 119873


variable 119909

119890



119872


119898=

(119909

1198981 119909

1198982 119909



119898=

(V1198981

V1198982

V119898119869


119898= (119901

1198981 119901

1198982 119901

119898119869)


119892= (119901

1198921 119901

1198922 119901



V(119896+1)119898119895

= 120596

(119896)

119898V(119896)119898119895

+ 119888

1119903

1(119901

(119896)

119898119895minus 119909

(119896)

119898119895)

+ 119888

2119903

2(119901

(119896)

119892119895minus 119909

(119896)

119898119895)

(24)


1and 119903

2are


1and 119888





2119901(119896)119898119895

and119901

(119896)





120596

(119896)

119898= 120596max minus


times 119896 (25)



119890

119903(119905) is a 0-1 integer




is shown as

119909

(119896+1)

119898119895=

1 120588 lt Sigmoid (V(119896+1)119898119895

)

0 others(26)


Sigmoid (V(119896)119898119895

) =

1

1 + exp (minusV(119896)119898119895

)

(27)



1 If there is a





1119876

1






119901(119905) on the 119905th


119905isin119863sum

119901isin119875119873

119901(119905)119876

2



min 119865 (119909)

= sum

119890isin119864

sum

119905isin119863

sum

119901isin119875

(119878

119901 (1 + 120582) minus 119897

119890

119901(119905 minus 1)) 119910

119890

119901(119905)

+ 120596sum

119890isin119864

119868 (120579

119890) + 119873

1119876

1+ sum

119905isin119863

sum

119901isin119875

119873

119901 (119905) 1198762

(28)



119890(119905) = 1 or 120573




119901(119905) = 1 At the same time


119890

119901(119905) and time 119891

119890

119901(119905) We





119890

119901(119905) and time

119891

119890






119890

119901(119905) and time 119891

119890




119890


119901(119905) and time

119891

119890





6 Case Study




Start

End








No

Yes


Generate initial

solution












1= 119876

2= 1000000




1= 119888

2= 20













7 Conclusion





EMU1

EMU2

EMU3

EMU4

Oct 30 Oct 31 Nov 1

EMU5

EMU6

EMU7

EMU8

EMU9

EMU10

EMU11

EMU12

EMU13

EMU14

EMU15

EMU16

EMU17

EMU18

EMU19

EMU20


Traction engine

Standby

Standby Standby

Standby




Standby

Standby


EMU21

EMU22



R1R1

R1

R1

R1

R1

I2

I2

I2

I2

I2

I2

I2

I2

I2

I2

R1

R1

R3

R3 R3

R3

R3R3

R3R2

R6

R6

R6 R6

R6

R6

R7

R10

R10

R10R10

R10 R10

R10

R10

R7

R7

R7

R7

R7

R9

M3

R9

R9R9

R9 R9

R9

R7

R7 R8R8R8

R8

R8

R8 R8

R6

R2

R2

R2

R2

R2

R2R4

M1

R4

R4

R4 R4

R4

R4

R5

R5

R5R5R5

R5

R5R5





(km)



20141024 1886726



20140724 1732774



20140806 399489



20140807 800586





Acknowledgments


References
































Volume 2014




Journal of











Journal of


Function Spaces






Algebra











The variable 119897

119890



119890

119901(119905) (unit day)



max 119885

1= sum

119890isin119864

sum

119905isin119863

sum

119901isin119875

119897

119890

119901(119905 minus 1) 119910

119890

119901(119905) (1)


min 119885

2= sum

119890isin119864

119868 (120579

119890) (2)


120579

119890= sum

119903isin119877

sum

119905isin119863

119909

119890

119903(119905) 119890 isin 119864 (3)

119868 (119909) =

1 119909 gt 0

0 119909 le 0

(4)


min 119885

3= sum

119890isin119864

sum

119905isin119863

sum

119901isin119875

(119878

119901 (1 + 120582) minus 119897

119890

119901(119905 minus 1)) 119910

119890

119901(119905) (5)



min 119885

= sum

119890isin119864

sum

119905isin119863

sum

119901isin119875

(119878

119901 (1 + 120582) minus 119897

119890

119901(119905 minus 1)) 119910

119890

119901(119905)

+ 120596sum

119890isin119864

119868 (120579

119890)

(6)


120596 = 119897 times 119873

119863 (7)



119909

119890

119903(119905) le 120575

119890

119903119890 isin 119864 (119905) 119903 isin 119877 119905 isin 119863 (8)


119910

119890

119901(119905) le 120591

119890

119901119890 isin 119864 (119905) 119901 isin 119875 119905 isin 119863 (9)


sum

119903isin119877

119909

119890

119903(119905) le 1 119890 isin 119864 (119905) 119905 isin 119863 (10)


sum

119890isin119864(119905)

119909

119890

119903(119905) = 1 119903 isin 119877 119905 isin 119863 (11)


119890



119897

119890

119901(119905) le (1 + 120582) 119878119901

119890 isin 119864 119901 isin 119875 119905 isin 119863 (12)



119890



119891

119890

119901(119905) le (1 + 120582) 119879119901

(13)


sum

119890isin119864

120601

119890

119901(119905) le 119862

119901119901 isin 119875 119905 isin 119863 (14)


120572

119890(119905) + 120573

119890(119905) + 120574

119890(119905) = 1 119890 isin 119864 119905 isin 119863 (15)


119909

119890

119903(119905) 119910

119890

119901(119905) 120593

119890

119903(119905) 120601

119890

119901(119905) 120572

119890(119905) 120573

119890(119905) 120574

119890(119905)

isin 0 1

(16)


119891

119890



119890






119901(119905)


119897

119890

119901(119905) =

0 if 120601119890119901(119905) = 1

119897

119890

119901(119905 minus 1) + 120591

119890

119901sum

119903isin119877

119909

119890

119903(119905) 119904119903

else

119890 isin 119864 119901 isin 119875 119905 isin 119863

(17)

119891

119890

119901(119905) =

0 if 120601119890119901(119905) = 1

119891

119890

119901(119905 minus 1) + 1 times 120591

119890

119901 else

119890 isin 119864 119901 isin 119875 119905 isin 119863

(18)


119890

119903(119905) 119910119890119901(119905) 120593119890119903(119905) 120601119890119901(119905) 120572119890(119905) 120573119890(119905)

and 120574






119905+120576119903minus1

sum

1199051015840=119905

120593

119890

119903(119905

1015840) = 120576

119903 if 119909

119890

119903(119905) = 1

120593

119890

119903(119905) = 0 else

119890 isin 119864 (119905) 119903 isin 119877 119905 isin 119863

(19)


119901


be described as

119905+120583119901minus1

sum

1199051015840=119905

120601

119890

119901(119905

1015840) = 120583

119901 if 119910119890

119901(119905) = 1

120601

119890

119901(119905) = 0 else

119890 isin 119864 (119905) 119901 isin 119875 119905 isin 119863

(20)


119890(119905) and 120573



119890

119903(119905) and 120601

119890


can be described as

120572

119890(119905) = sum

119903isin119877

120593

119890

119903(119905) 119890 isin 119864 119905 isin 119863 (21)

120573

119890(119905) =

1 ifsum119901isin119875

120601

119890

119901(119905) ge 1

0 else119890 isin 119864 119905 isin 119863 (22)



(M) min 119885

= sum

119890isin119864

sum

119905isin119863

sum

119901isin119875

(119878

119901 (1 + 120582) minus 119897

119890

119901(119905 minus 1)) 119910

119890

119901(119905)

+ 120596sum

119890isin119864

119868 (120579

119890)

st (8) sim (16)

(23)




5 Solution Strategy



119873

119864times 119873

119863times 119873


variable 119909

119890



119872


119898=

(119909

1198981 119909

1198982 119909



119898=

(V1198981

V1198982

V119898119869


119898= (119901

1198981 119901

1198982 119901

119898119869)


119892= (119901

1198921 119901

1198922 119901



V(119896+1)119898119895

= 120596

(119896)

119898V(119896)119898119895

+ 119888

1119903

1(119901

(119896)

119898119895minus 119909

(119896)

119898119895)

+ 119888

2119903

2(119901

(119896)

119892119895minus 119909

(119896)

119898119895)

(24)


1and 119903

2are


1and 119888





2119901(119896)119898119895

and119901

(119896)





120596

(119896)

119898= 120596max minus


times 119896 (25)



119890

119903(119905) is a 0-1 integer




is shown as

119909

(119896+1)

119898119895=

1 120588 lt Sigmoid (V(119896+1)119898119895

)

0 others(26)


Sigmoid (V(119896)119898119895

) =

1

1 + exp (minusV(119896)119898119895

)

(27)



1 If there is a





1119876

1






119901(119905) on the 119905th


119905isin119863sum

119901isin119875119873

119901(119905)119876

2



min 119865 (119909)

= sum

119890isin119864

sum

119905isin119863

sum

119901isin119875

(119878

119901 (1 + 120582) minus 119897

119890

119901(119905 minus 1)) 119910

119890

119901(119905)

+ 120596sum

119890isin119864

119868 (120579

119890) + 119873

1119876

1+ sum

119905isin119863

sum

119901isin119875

119873

119901 (119905) 1198762

(28)



119890(119905) = 1 or 120573




119901(119905) = 1 At the same time


119890

119901(119905) and time 119891

119890

119901(119905) We





119890

119901(119905) and time

119891

119890






119890

119901(119905) and time 119891

119890




119890


119901(119905) and time

119891

119890





6 Case Study




Start

End








No

Yes


Generate initial

solution












1= 119876

2= 1000000




1= 119888

2= 20













7 Conclusion





EMU1

EMU2

EMU3

EMU4

Oct 30 Oct 31 Nov 1

EMU5

EMU6

EMU7

EMU8

EMU9

EMU10

EMU11

EMU12

EMU13

EMU14

EMU15

EMU16

EMU17

EMU18

EMU19

EMU20


Traction engine

Standby

Standby Standby

Standby




Standby

Standby


EMU21

EMU22



R1R1

R1

R1

R1

R1

I2

I2

I2

I2

I2

I2

I2

I2

I2

I2

R1

R1

R3

R3 R3

R3

R3R3

R3R2

R6

R6

R6 R6

R6

R6

R7

R10

R10

R10R10

R10 R10

R10

R10

R7

R7

R7

R7

R7

R9

M3

R9

R9R9

R9 R9

R9

R7

R7 R8R8R8

R8

R8

R8 R8

R6

R2

R2

R2

R2

R2

R2R4

M1

R4

R4

R4 R4

R4

R4

R5

R5

R5R5R5

R5

R5R5





(km)



20141024 1886726



20140724 1732774



20140806 399489



20140807 800586





Acknowledgments


References
































Volume 2014




Journal of











Journal of


Function Spaces






Algebra











119890



119891

119890

119901(119905) le (1 + 120582) 119879119901

(13)


sum

119890isin119864

120601

119890

119901(119905) le 119862

119901119901 isin 119875 119905 isin 119863 (14)


120572

119890(119905) + 120573

119890(119905) + 120574

119890(119905) = 1 119890 isin 119864 119905 isin 119863 (15)


119909

119890

119903(119905) 119910

119890

119901(119905) 120593

119890

119903(119905) 120601

119890

119901(119905) 120572

119890(119905) 120573

119890(119905) 120574

119890(119905)

isin 0 1

(16)


119891

119890



119890






119901(119905)


119897

119890

119901(119905) =

0 if 120601119890119901(119905) = 1

119897

119890

119901(119905 minus 1) + 120591

119890

119901sum

119903isin119877

119909

119890

119903(119905) 119904119903

else

119890 isin 119864 119901 isin 119875 119905 isin 119863

(17)

119891

119890

119901(119905) =

0 if 120601119890119901(119905) = 1

119891

119890

119901(119905 minus 1) + 1 times 120591

119890

119901 else

119890 isin 119864 119901 isin 119875 119905 isin 119863

(18)


119890

119903(119905) 119910119890119901(119905) 120593119890119903(119905) 120601119890119901(119905) 120572119890(119905) 120573119890(119905)

and 120574






119905+120576119903minus1

sum

1199051015840=119905

120593

119890

119903(119905

1015840) = 120576

119903 if 119909

119890

119903(119905) = 1

120593

119890

119903(119905) = 0 else

119890 isin 119864 (119905) 119903 isin 119877 119905 isin 119863

(19)


119901


be described as

119905+120583119901minus1

sum

1199051015840=119905

120601

119890

119901(119905

1015840) = 120583

119901 if 119910119890

119901(119905) = 1

120601

119890

119901(119905) = 0 else

119890 isin 119864 (119905) 119901 isin 119875 119905 isin 119863

(20)


119890(119905) and 120573



119890

119903(119905) and 120601

119890


can be described as

120572

119890(119905) = sum

119903isin119877

120593

119890

119903(119905) 119890 isin 119864 119905 isin 119863 (21)

120573

119890(119905) =

1 ifsum119901isin119875

120601

119890

119901(119905) ge 1

0 else119890 isin 119864 119905 isin 119863 (22)



(M) min 119885

= sum

119890isin119864

sum

119905isin119863

sum

119901isin119875

(119878

119901 (1 + 120582) minus 119897

119890

119901(119905 minus 1)) 119910

119890

119901(119905)

+ 120596sum

119890isin119864

119868 (120579

119890)

st (8) sim (16)

(23)




5 Solution Strategy



119873

119864times 119873

119863times 119873


variable 119909

119890



119872


119898=

(119909

1198981 119909

1198982 119909



119898=

(V1198981

V1198982

V119898119869


119898= (119901

1198981 119901

1198982 119901

119898119869)


119892= (119901

1198921 119901

1198922 119901



V(119896+1)119898119895

= 120596

(119896)

119898V(119896)119898119895

+ 119888

1119903

1(119901

(119896)

119898119895minus 119909

(119896)

119898119895)

+ 119888

2119903

2(119901

(119896)

119892119895minus 119909

(119896)

119898119895)

(24)


1and 119903

2are


1and 119888





2119901(119896)119898119895

and119901

(119896)





120596

(119896)

119898= 120596max minus


times 119896 (25)



119890

119903(119905) is a 0-1 integer




is shown as

119909

(119896+1)

119898119895=

1 120588 lt Sigmoid (V(119896+1)119898119895

)

0 others(26)


Sigmoid (V(119896)119898119895

) =

1

1 + exp (minusV(119896)119898119895

)

(27)



1 If there is a





1119876

1






119901(119905) on the 119905th


119905isin119863sum

119901isin119875119873

119901(119905)119876

2



min 119865 (119909)

= sum

119890isin119864

sum

119905isin119863

sum

119901isin119875

(119878

119901 (1 + 120582) minus 119897

119890

119901(119905 minus 1)) 119910

119890

119901(119905)

+ 120596sum

119890isin119864

119868 (120579

119890) + 119873

1119876

1+ sum

119905isin119863

sum

119901isin119875

119873

119901 (119905) 1198762

(28)



119890(119905) = 1 or 120573




119901(119905) = 1 At the same time


119890

119901(119905) and time 119891

119890

119901(119905) We





119890

119901(119905) and time

119891

119890






119890

119901(119905) and time 119891

119890




119890


119901(119905) and time

119891

119890





6 Case Study




Start

End








No

Yes


Generate initial

solution












1= 119876

2= 1000000




1= 119888

2= 20













7 Conclusion





EMU1

EMU2

EMU3

EMU4

Oct 30 Oct 31 Nov 1

EMU5

EMU6

EMU7

EMU8

EMU9

EMU10

EMU11

EMU12

EMU13

EMU14

EMU15

EMU16

EMU17

EMU18

EMU19

EMU20


Traction engine

Standby

Standby Standby

Standby




Standby

Standby


EMU21

EMU22



R1R1

R1

R1

R1

R1

I2

I2

I2

I2

I2

I2

I2

I2

I2

I2

R1

R1

R3

R3 R3

R3

R3R3

R3R2

R6

R6

R6 R6

R6

R6

R7

R10

R10

R10R10

R10 R10

R10

R10

R7

R7

R7

R7

R7

R9

M3

R9

R9R9

R9 R9

R9

R7

R7 R8R8R8

R8

R8

R8 R8

R6

R2

R2

R2

R2

R2

R2R4

M1

R4

R4

R4 R4

R4

R4

R5

R5

R5R5R5

R5

R5R5





(km)



20141024 1886726



20140724 1732774



20140806 399489



20140807 800586





Acknowledgments


References
































Volume 2014




Journal of











Journal of


Function Spaces






Algebra











5 Solution Strategy



119873

119864times 119873

119863times 119873


variable 119909

119890



119872


119898=

(119909

1198981 119909

1198982 119909



119898=

(V1198981

V1198982

V119898119869


119898= (119901

1198981 119901

1198982 119901

119898119869)


119892= (119901

1198921 119901

1198922 119901



V(119896+1)119898119895

= 120596

(119896)

119898V(119896)119898119895

+ 119888

1119903

1(119901

(119896)

119898119895minus 119909

(119896)

119898119895)

+ 119888

2119903

2(119901

(119896)

119892119895minus 119909

(119896)

119898119895)

(24)


1and 119903

2are


1and 119888





2119901(119896)119898119895

and119901

(119896)





120596

(119896)

119898= 120596max minus


times 119896 (25)



119890

119903(119905) is a 0-1 integer




is shown as

119909

(119896+1)

119898119895=

1 120588 lt Sigmoid (V(119896+1)119898119895

)

0 others(26)


Sigmoid (V(119896)119898119895

) =

1

1 + exp (minusV(119896)119898119895

)

(27)



1 If there is a





1119876

1






119901(119905) on the 119905th


119905isin119863sum

119901isin119875119873

119901(119905)119876

2



min 119865 (119909)

= sum

119890isin119864

sum

119905isin119863

sum

119901isin119875

(119878

119901 (1 + 120582) minus 119897

119890

119901(119905 minus 1)) 119910

119890

119901(119905)

+ 120596sum

119890isin119864

119868 (120579

119890) + 119873

1119876

1+ sum

119905isin119863

sum

119901isin119875

119873

119901 (119905) 1198762

(28)



119890(119905) = 1 or 120573




119901(119905) = 1 At the same time


119890

119901(119905) and time 119891

119890

119901(119905) We





119890

119901(119905) and time

119891

119890






119890

119901(119905) and time 119891

119890




119890


119901(119905) and time

119891

119890





6 Case Study




Start

End








No

Yes


Generate initial

solution












1= 119876

2= 1000000




1= 119888

2= 20













7 Conclusion





EMU1

EMU2

EMU3

EMU4

Oct 30 Oct 31 Nov 1

EMU5

EMU6

EMU7

EMU8

EMU9

EMU10

EMU11

EMU12

EMU13

EMU14

EMU15

EMU16

EMU17

EMU18

EMU19

EMU20


Traction engine

Standby

Standby Standby

Standby




Standby

Standby


EMU21

EMU22



R1R1

R1

R1

R1

R1

I2

I2

I2

I2

I2

I2

I2

I2

I2

I2

R1

R1

R3

R3 R3

R3

R3R3

R3R2

R6

R6

R6 R6

R6

R6

R7

R10

R10

R10R10

R10 R10

R10

R10

R7

R7

R7

R7

R7

R9

M3

R9

R9R9

R9 R9

R9

R7

R7 R8R8R8

R8

R8

R8 R8

R6

R2

R2

R2

R2

R2

R2R4

M1

R4

R4

R4 R4

R4

R4

R5

R5

R5R5R5

R5

R5R5





(km)



20141024 1886726



20140724 1732774



20140806 399489



20140807 800586





Acknowledgments


References
































Volume 2014




Journal of











Journal of


Function Spaces






Algebra











min 119865 (119909)

= sum

119890isin119864

sum

119905isin119863

sum

119901isin119875

(119878

119901 (1 + 120582) minus 119897

119890

119901(119905 minus 1)) 119910

119890

119901(119905)

+ 120596sum

119890isin119864

119868 (120579

119890) + 119873

1119876

1+ sum

119905isin119863

sum

119901isin119875

119873

119901 (119905) 1198762

(28)



119890(119905) = 1 or 120573




119901(119905) = 1 At the same time


119890

119901(119905) and time 119891

119890

119901(119905) We





119890

119901(119905) and time

119891

119890






119890

119901(119905) and time 119891

119890




119890


119901(119905) and time

119891

119890





6 Case Study




Start

End








No

Yes


Generate initial

solution












1= 119876

2= 1000000




1= 119888

2= 20













7 Conclusion





EMU1

EMU2

EMU3

EMU4

Oct 30 Oct 31 Nov 1

EMU5

EMU6

EMU7

EMU8

EMU9

EMU10

EMU11

EMU12

EMU13

EMU14

EMU15

EMU16

EMU17

EMU18

EMU19

EMU20


Traction engine

Standby

Standby Standby

Standby




Standby

Standby


EMU21

EMU22



R1R1

R1

R1

R1

R1

I2

I2

I2

I2

I2

I2

I2

I2

I2

I2

R1

R1

R3

R3 R3

R3

R3R3

R3R2

R6

R6

R6 R6

R6

R6

R7

R10

R10

R10R10

R10 R10

R10

R10

R7

R7

R7

R7

R7

R9

M3

R9

R9R9

R9 R9

R9

R7

R7 R8R8R8

R8

R8

R8 R8

R6

R2

R2

R2

R2

R2

R2R4

M1

R4

R4

R4 R4

R4

R4

R5

R5

R5R5R5

R5

R5R5





(km)



20141024 1886726



20140724 1732774



20140806 399489



20140807 800586





Acknowledgments


References
































Volume 2014




Journal of











Journal of


Function Spaces






Algebra










Start

End








No

Yes


Generate initial

solution












1= 119876

2= 1000000




1= 119888

2= 20













7 Conclusion





EMU1

EMU2

EMU3

EMU4

Oct 30 Oct 31 Nov 1

EMU5

EMU6

EMU7

EMU8

EMU9

EMU10

EMU11

EMU12

EMU13

EMU14

EMU15

EMU16

EMU17

EMU18

EMU19

EMU20


Traction engine

Standby

Standby Standby

Standby




Standby

Standby


EMU21

EMU22



R1R1

R1

R1

R1

R1

I2

I2

I2

I2

I2

I2

I2

I2

I2

I2

R1

R1

R3

R3 R3

R3

R3R3

R3R2

R6

R6

R6 R6

R6

R6

R7

R10

R10

R10R10

R10 R10

R10

R10

R7

R7

R7

R7

R7

R9

M3

R9

R9R9

R9 R9

R9

R7

R7 R8R8R8

R8

R8

R8 R8

R6

R2

R2

R2

R2

R2

R2R4

M1

R4

R4

R4 R4

R4

R4

R5

R5

R5R5R5

R5

R5R5





(km)



20141024 1886726



20140724 1732774



20140806 399489



20140807 800586





Acknowledgments


References
































Volume 2014




Journal of











Journal of


Function Spaces






Algebra




















7 Conclusion





EMU1

EMU2

EMU3

EMU4

Oct 30 Oct 31 Nov 1

EMU5

EMU6

EMU7

EMU8

EMU9

EMU10

EMU11

EMU12

EMU13

EMU14

EMU15

EMU16

EMU17

EMU18

EMU19

EMU20


Traction engine

Standby

Standby Standby

Standby




Standby

Standby


EMU21

EMU22



R1R1

R1

R1

R1

R1

I2

I2

I2

I2

I2

I2

I2

I2

I2

I2

R1

R1

R3

R3 R3

R3

R3R3

R3R2

R6

R6

R6 R6

R6

R6

R7

R10

R10

R10R10

R10 R10

R10

R10

R7

R7

R7

R7

R7

R9

M3

R9

R9R9

R9 R9

R9

R7

R7 R8R8R8

R8

R8

R8 R8

R6

R2

R2

R2

R2

R2

R2R4

M1

R4

R4

R4 R4

R4

R4

R5

R5

R5R5R5

R5

R5R5





(km)



20141024 1886726



20140724 1732774



20140806 399489



20140807 800586





Acknowledgments


References
































Volume 2014




Journal of











Journal of


Function Spaces






Algebra












EMU1

EMU2

EMU3

EMU4

Oct 30 Oct 31 Nov 1

EMU5

EMU6

EMU7

EMU8

EMU9

EMU10

EMU11

EMU12

EMU13

EMU14

EMU15

EMU16

EMU17

EMU18

EMU19

EMU20


Traction engine

Standby

Standby Standby

Standby




Standby

Standby


EMU21

EMU22



R1R1

R1

R1

R1

R1

I2

I2

I2

I2

I2

I2

I2

I2

I2

I2

R1

R1

R3

R3 R3

R3

R3R3

R3R2

R6

R6

R6 R6

R6

R6

R7

R10

R10

R10R10

R10 R10

R10

R10

R7

R7

R7

R7

R7

R9

M3

R9

R9R9

R9 R9

R9

R7

R7 R8R8R8

R8

R8

R8 R8

R6

R2

R2

R2

R2

R2

R2R4

M1

R4

R4

R4 R4

R4

R4

R5

R5

R5R5R5

R5

R5R5





(km)



20141024 1886726



20140724 1732774



20140806 399489



20140807 800586





Acknowledgments


References
































Volume 2014




Journal of











Journal of


Function Spaces






Algebra












(km)



20141024 1886726



20140724 1732774



20140806 399489



20140807 800586





Acknowledgments


References
































Volume 2014




Journal of











Journal of


Function Spaces






Algebra

























Volume 2014




Journal of











Journal of


Function Spaces






Algebra
















Volume 2014




Journal of











Journal of


Function Spaces






Algebra









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Research Article A Pragmatic Optimization Method for Motor ...downloads.hindawi.com/journals/ddns/2016/4540503.pdf · Research Article A Pragmatic Optimization Method for Motor Train