Research Article A Transient Queuing Model for Analyzing ...downloads.hindawi.com/journals/mpe/2014/914706.pdf · a general condition that the service time follows the Erlang distribution

Research ArticleA Transient Queuing Model for Analyzing and Optimizing GateCongestion of Railway Container Terminals

Ming Zeng Wenming Cheng and Peng Guo

School of Mechanical Engineering Southwest Jiaotong University Chengdu 610031 China

Correspondence should be addressed to Ming Zeng zengming1209gmailcom

Received 3 April 2014 Revised 12 June 2014 Accepted 12 June 2014 Published 23 July 2014

Academic Editor Michael Lutjen

Copyright copy 2014 Ming Zeng et al This is an open access article distributed under the Creative Commons Attribution Licensewhich permits unrestricted use distribution and reproduction in any medium provided the original work is properly cited

As the significant connection between the external and internal of the railway container terminal the operation performance ofthe gate system plays an important role in the entire system So the gate congestion will bring many losses to the railway containerterminal even the entire railway container freight system In this paper the queue length and the average waiting time of the railwaycontainer terminal gate system as well as the optimal number of service channels during the different time period are investigatedAn119872119864119896119899 transient queuingmodel is developed based on the distribution of the arrival time interval and the service time besidesthe transient solutions are acquired by the equally likely combinations (ELC) heuristic method Then the model is integrated intoan optimization framework to obtain the optimal operation schemes Finally some computational experiments are conducted formodel validation sensitivity testing and systemoptimization Experimental results indicate that themodel can provide the accuratereflection to the operation situation of the railway container terminal gate system and the approach can yield the optimal numberof service channels within the reasonable computation time

1 Introduction

Nowadays the container transportation has been fast devel-oped throughout the world It becomes the best mode oftransportation for the commercial intercourse of variouscountries and the containerization ratio of the general cargosin some countries has exceeded 80 The statistics of theUnited Nations Conference on Trade and Development(UNCTAD) present that Germany is the first country ofEuropean Union containerized exports and imports whichhas 304 million TEUS in export and 284 million TEUSin import in 2010 Meanwhile China ranks as the first ofall the exporters of containerized cargo in 2010 which has313 million TEUS and it also has 120 million TEUS importvolume of containerized cargo in 2010 which is ranked thesecond The first country is the United States which has 176million TEUS [1] The railway container transportation wasfirstly carried out in the United States in 1853 and Chinabegan this practice in the 1950s In the recent years therailway container transportation volume of China is on arapid growth Compared with 2000 the dispatch volume of

railway container cargos increased by 601 in 2005 andthe average growth rate per annum of railway containertransportation volume in China is 116 in 2000ndash2004 In2006 the volume has a 167 increase to the year 2005 that is6891million tons Furthermore the container transportationvolume is as much as 471 TEU which is equivalent to 9065million tons in 2012 [2]

According to the huge demand China increases theinvestment to the relevant facilities and the infrastructureconstruction and has established a railway container trans-portation support system which takes 18 railway containerterminals as the hub Railway container terminals are alwaysestablished in the main economic centers and the impor-tant ports which are the collecting and distributing centreof the railway container transportation Railway containerterminal has many functions as marshalling declassificationhandling warehousing and some other logistics serviceNonetheless with the constant development of the regionaleconomy and the rapid growth of transportation volumethere still are many challenges to the railway container termi-nals Generally the maximum backlog time of the combined

Hindawi Publishing CorporationMathematical Problems in EngineeringVolume 2014 Article ID 914706 13 pageshttpdxdoiorg1011552014914706

2 Mathematical Problems in Engineering

type marshalling cargos in the container terminals may beup to 15ndash20 days Hence the improving of the operationefficiency in railway container terminal under the existingconditions is a serious problem at present As one of the mostimportant procedures in the railway container terminal oper-ation system the through capacity and service level of thegate system have significant impacts on the efficiency of therailway container terminal even the whole railway containerfreight system Currently the operation management of therailway container terminal gate system relies on the empiricalestimation and the real time information feedback in themost cases However these methods are not accurate andefficient in reality so they are not able to solve this problemfundamentally

Recently the researchers pay more attention to thestudy of container terminals Meanwhile their researchareas mainly focus on the operation management and theequipment scheduling Mattfeld and Kopfer [3] developed anintegral decision model for the operations of the terminalKim et al [4] proposed a dynamic programming model toevaluate the delay time of outside trucks in port containerterminals and discussed the different situations respectivelyLee et al [5] presented a mixed integer programming modeltogether with a genetic algorithm to solve the quay cranescheduling problem and then extended the problem withthe noninterference constraints [6] Similar problemwas alsostudied by using different constrains and algorithms [7ndash9]In addition the relevant layout planning [10 11] the berthallocation [12] and the programming problems [13ndash18] arealso researched comprehensively

Similarly the researches of the railway container termi-nals and the intermodal terminals are conducted in thesefields and the loadingdischarging equipment schedulingproblems have become the hot issues [19ndash21] Guo et al [22]presented a mixed integer programming model with a dis-crete artificial bee colony algorithm to solve the gantry cranescheduling problem in railway container terminals Dorndorfand Schneider [23] studied the triple cross-over stackingcranes scheduling problem in order to increase the productiv-ity and reduce the delays Besides the operation researchesfor example the loading planning [24 25] the shunting of railcontainer wagons [26] the container stacking [27] and thelayout planning like the crane areas determination [28 29]are some other research interests in this field The strategicplanning and the scenario generating for the railway con-tainer terminals are usually implemented by the simulationmethods [30ndash32] Guo et al [33] developed a discrete eventsimulation model for the container handling process of therailway container logistic center to make the evaluation andimprovementHowever there exist somedisadvantages in thesimulation methods for example the quantitative solutionscannot be provided and they are unable to be nested withthe optimizationmodelsTherefore Edmond andMaggs [34]have pointed out the importance of queuing theory in therelated decision making of container terminals many yearsbefore Recently some literature began to use the queuingmethods to analyze the problems For instance Canonacoet al [35] proposed a queuing network model to solve themachine operations problem of the container terminal But

the researches of queuing models for the railway containerterminal gate system are still relatively small because therailway container terminal gate system cannot be describedcompletely and accurately by the traditional queuing modelswhich are also incapable of analyzing the instantaneousfeatures of this system so there are some limitations in thisfield

The queuing models can be classified into stationaryqueuing models and transient queuing models Because itis difficult to describe and calculate all the system states ofthe transient queuing models the majority of the queuingmodels which are applied in various important research areasonly take the steady solution into consideration [36 37]However the theory and application research of transientqueuing models has gone through some development in thepast ten years Ausin et al [38] used the Bayesian analysismethod to solve the optimized number of servers in theGIMc queuing system on the basis of the minimum totalcosts Besides they utilized the Bayesian inference to predictthe transient features of the GIG1 system [39] Czachorskiet al [40] studied the transient features of the GG1 queuingsystem from the diffusion approximation model Parlar andSharafali [41] created a time-related queuing model to opti-mize the airport security checkpoints and briefly discusseda general condition that the service time follows the Erlangdistribution

In this paper a transient queuing model and an opti-mization model are established to analyze and optimize thegate congestion according to the arrival time interval ofthe external container trucks and the distribution regularityof the service time for the railway container terminal gatesystem Then the equally likely combinations heuristic solu-tion and the optimization solution methods are adopted tosolve these models After that the system simulation andsensitivity analysis are conducted to verify the rationalityand validity of our approach Finally the optimal num-bers of the service channels in different time periods aredetermined by the optimization experiment The remainderof this paper is structured as follows Section 2 providessome background information to the case studyThe queuingmodel and optimization model together with the solutionmethods are presented in Section 3 The model validationsensitivity analysis and the optimization experiments aregiven in Section 4 Finally Section 5 concludes the paper withan outline of future study

2 Chengdu Railway Container Terminal

Chengdu railway container terminal is located in the north-east of Chengdu where it is near the Chengxiang railwaystation which connects with many key rail routes of ChinaChengdu railway container terminal is the largest railwaycontainer terminal of Asia which is also the containerlogistics hinge of southwest China Therefore it acts as animportant part to the entire container transportation net-work The terminal was built in 2010 which is 84 kilometerslong and 850 meters wide The annual cargo throughput is 1millionTEU in recent period and the forward cargo handlingcapacity will be 25 million TEU

Mathematical Problems in Engineering 3

Figure 1 Chengdu railway container terminal gate system

Service

Move tocontainer yard

Assigned

Service channel

Container truck

station

route

Figure 2 Schematic representation of the railway container termi-nal gate system

Railway container terminal gate system is the access ofthe container trucks to go through There are some servicechannels which have the technical facilities to conduct thenecessary processes in the railway container terminal gatesystem A picture of Chengdu railway container terminal gatesystem can be seen in Figure 1Themajor works at the servicechannels are container number identification informationverification inspection and position assignmentThe railwaycontainer terminal gate system is schematically depicted inFigure 2 However because of the limitations of the facilitiesand the service conditions the container trucks have to lineup to complete the processes before they enter into or get outof the terminalTherefore the whole processes can be seen asa queuing process in which the container trucks are the inputflows and the channels are the service stations

3 Railway Container Terminal GateSystem Modelling

31 119872119864119896119899 Queuing Model All the data with respect tothe time intervals of two container trucks that arrive insuccession and the service time of the service channels werecollected to ascertain the best fitting distributions 560 arrivaltime intervals of container trucks and 350 service times ofthe service channels were field-measured in Chengdu railwaycontainer terminal gate system for the observation period oftwo weeks and then probability information of the collected

0 5 10 15 20 25 30 35 400

002

004

006

008

01

012

014

Arrival time interval (min)

Prob

abili

ty

The collected dataExponential distribution

Figure 3 The distribution of arrival time interval

0 5 10 15 20 25 30 35 400

001

002

003

004

005

006

007

008

009

01

Service time (min)

Prob

abili

ty

The collected dataErlang distribution

Figure 4 The distribution of service time

data along with the relevant distribution curves was acquiredas follows

The best fitting distribution of the arrival time intervals asdisplayed in Figure 3 is the exponential distribution 119891(119909) =0212119890

minus0212119909 119909 ge 0 and the average arrival time intervalis 4714 minutes The fit of arrival time interval distributionto the collected data is calculated by 119877-square which equals09542 and RMSE (root mean square error) which is 0012The distribution of the service time as shown in Figure 4 isbest fitted by the Erlang distribution 119891(119909) = 00625 sdot 119909 sdot

119890minus025119909 119909 ge 0 with 119896 equal to 2 and the mean servicerate is 025 The 119877-square value for fitting the service timedistribution to the collected data is 09078 and the RMSE is0096 Thus the appropriate queuing model for the railway


container terminal gate system is exponential arrival intervalswith Erlang service times andmultiple service stations whichis represented as119872119864119896=2119899 The solution for this model canbe divided into two types the exact solution method andthe approximate solution method However the former hasa large number of possible states and the state transitionsare complicated the calculation is quite difficult So here weutilize the approximate solutionmethod to solve the119872119864119896119899queuing model Equally likely combinations (ELC) heuristicsolution technique has been put forward by Escobar et al[42] the basic assumption of which is the combinations of thestages unfinished in the servers are equally likelyThe detaileddescription of this heuristic solution method is as follows

311 System States Description For the sake of simplifyingthe Erlang distribution and then forming an efficient solutionthe system states are represented by three elements and(119904 V 119898) 119904 means the stages have to be finished by thecontainer trucks in the system at this moment V is the totalnumber of container trucks in the system at present119898 refersto the mode number The reason for introducing the thirdelement is to distinguish the different modes with the same119904 and V For example the state (7 4) as shown in Figure 5eithermeans four trucks in the systemwith two trucks havingone stage unfinished one having two stages unfinished andthe other one just arriving (ie Mode 1) or represents onetruck having one stage left over and the other three all havingtwo stages (ie Mode 2)

312 State Transition Probability Before doing the descrip-tion of the state transition probability firstly let 119875119904V representthe probability of state (119904 V) because the relevant calculationwill be conducted so the mode number is discarded forconciseness Compared with the original ELC solution whichconsiders new trucks will arrive at each time step in thispaper we take the average arrival time interval into accountin order to reduce the state space dimension and calculationdifficulty

Suppose there are V container trucks and 119904 stages have tobe finished at current time 119905 for the next time step (ie 119905 + 1)the possible states and state transition are as below

Let 119905119886 denote the average time interval between two trucksarriving at the gate system and let 119905119873 represent the119873th truckhad arrived For the next time step before the (119873+1)th truckarrived there are two possible transitions that may happenwhen a truck finishes one stage

(a) The stage is the last one to finish for a truck and thetruck leaves the gate system

Under this situation the state is transferred from (119904 V) to(119904minus1 Vminus1) As mentioned in Section 311 there are differentmodes for the same state and different combinations for thesame modeTherefore the combinations for one given mode119868 can be calculated as follows

119862119894 =120596

12059611205962 sdot sdot sdot 120596119901 (1)

where 120596 is the minV 119899 which means the active servicechannels 119901 is the number of combinations of stages in

Stage 1 Stage 2 Stage 3








Mode 1 Mode 2

Figure 5 An example of different modes and stages for the samestate

the service channels 120596119895 is the number of channels that havethe same number of unfinished stages 119895 = 1 2 119901

Accordingly the total number of combinations for aspecific state is

119862total = sum119894isin119872

119862119894 (2)

where119872 is a set of all the mode numbers for this stateThen the probability of this situation (ie the last stage is

finished and the truck leaves the gate system) is calculated by

120590119904V =1

120596119862totalsum

119894isin119872

1198641119894119862119894 (3)

where 119862119894119862total means the probability of mode 119894 1198641119894 is thenumber of service channels with one stage unfinished inmode 119894 so 1198641119894120596 means the probability of service channelswith only one stage left in mode 119894

Suppose the service rate is 120582(119905) the transition probabilityfrom state (119904 V) to (119904 minus 1 V minus 1) can thus be given by

119875(119904V)rarr (119904minus1Vminus1) = 120590119904V times 120582 (119905) times 119896 times V (4)

(b) The truck is still in the gate system to finish the otherstages

Because the probability that the last stage is finished andthe truck leaves the gate system is 120590(119904V) the probability thatone more stage is completed but there is no truck leaving thegate system can be simply represented by

120575119904V = 1 minus 120590119904V (5)

Consequently the transition probability from state (119904 V)to (119904 minus 1 V) is calculated as

119875(119904V)rarr (119904minus1V) = 120575119904V times 120582 (119905) times 119896 times V (6)

On the other hand when the next time step (ie 119905 + 1)exactly equals 119905119873+119905119886 whichmeans the (119873+1)th truck has justarrived the only possible state transition is from (119904 V) to (119904 +119896 V + 1) 119896 is the total number of stages for a container truckto finish in the gate system Hence the transition probabilityis 119875(119904V)rarr (119904+119896V+1) = 1

The entire state transitionmentioned above for119872119864119896=2119899queuing model can be explained by the state transitiondiagram of Figure 6


3210Trucks

0 2 4 6

1 3 5

2 4

3

n

n n + 1

n + 1

n + 2

n + 3n + 1

2n

2n minus 1 2n + 1

2n + 2

(s ) rarr (s + k + 1)

(s ) rarr (s minus 1 minus 1)

(s ) rarr (s minus 1 )

middot middot middot













Figure 6 State transition diagram of the119872119864119896=2119899 queuing model

For Figure 6 the numbers in the first row denote the totalnumber of trucks in the system (ie V) and the numbers inthe circles represent the unfinished stages in the system (ie119904) a state (119904 V) can thus be expressed The solid arrows thatpoint to the right indicate the transitions that one truck justarrives The slanted arrows represent the transitions that onetruck finishes the last stage and leaves the system And thedownward dotted arrows are used to denote the transitionsthat one more stage is finished but the truck is still in thesystem

313 State to State Transition Probabilities Calculation Onthe basis of what is discussed in Section 312 the stateto state transition probabilities at each time step 119905119894 thencan be calculated and updated accordingly by the followingequations For the time steps 119905119894 = 119905 119905 + 119905119886 minus 1 thatis the time period when the 119873th truck had arrived andone time step before the (119873 + 1)th truck arrives Let 119875119904V(119905119894)represent the probability of state (119904 V) at current time 119905 andlet 1198751015840119904V(119905119894) represent the variation of 119875119904V(119905119894) at each time step

so obviously the state probabilities of the next time step canbe given by

119875119904V (119905119894 + 1) = 119875119904V (119905119894) + 1198751015840

119904V (119905119894) (7)

The calculation processes of the variation of probabilitiesare as below

(a)When the number of trucks is less than the number ofservice channels in the system V lt 119899 we have the following

For the state (0 0)

1198751015840

00(119905119894) = 120582 (119905119894) times 119896 times 11987511 (119905119894) (8)

For V = 1 2 119899 minus 1

1198751015840

V119896V (119905119894) = minus120582 (119905119894) sdot 119896 sdot V sdot 119875V119896V (119905119894) + 120590V119896+1V+1

sdot 120582 (119905119894) sdot 119896 sdot (V + 1) sdot 119875V119896+1V+1 (119905119894) (9)

For V = 1 2 119899 minus 1 119886 = 1 2 V sdot (119896 minus 1)

1198751015840

V119896minus119886V (119905119894) = minus120582 (119905119894) sdot 119896 sdot V sdot 119875V119896minus119886V (119905119894) + 120575V119896minus119886+1V

sdot 120582 (119905119894) sdot 119896 sdot V sdot 119875V119896minus119886+1V (119905119894) + 120590V119896minus119886+1V+1

sdot 120582 (119905119894) sdot 119896 sdot (V + 1) sdot 119875V119896minus119886+1V+1 (119905119894)

(10)

(b) When the number of trucks equals or is greater thanthe number of service channels in the system V ge 119899 we havethe following

For V = 119899 119899 + 1 119873

1198751015840

V119896V (119905119894) = minus120582 (119905119894) sdot 119896 sdot 119899 sdot 119875V119896V (119905119894) + 120590(119899minus1)119896+1119899

sdot 120582 (119905119894) sdot 119896 sdot 119899 sdot 119875V119896+1V+1 (119905119894) (11)

For V = 119899 119899 + 1 119873 119886 = 1 2 (119899 minus 1) sdot (119896 minus 1)

1198751015840

V119896minus119886V (119905119894) = minus120582 (119905119894) sdot 119896 sdot 119899 sdot 119875V119896minus119886V (119905119894) + 120575119899119896minus119886+1119899

sdot 120582 (119905119894) sdot 119896 sdot 119899 sdot 119875V119896minus119886+1V (119905119894) + 120590(119899minus1)119896minus119886+1119899

sdot 120582 (119905119894) sdot 119896 sdot 119899 sdot 119875V119896minus119886+1V+1 (119905119894)

(12)

For V = 119899 119899 + 1 119873 119887 = (119899 minus 1) sdot (119896 minus 1) + 1 (119899 minus 1) sdot (119896 minus1) + 2 119899(119896 minus 1)

1198751015840


sdot 120582 (119905119894) sdot 119896 sdot 119899 sdot 119875V119896minus119887+1V (119905119894) (13)

Then for the time point 119905119894 = 119905 + 119905119886 the (119873 + 1)thtruck arrives in the system Asmentioned in Section 312 theprobability of state transition from (119904 V) to (119904 + 119896 V + 1)is equal to 1 Therefore the state probabilities under thissituation can be calculated as below

For V = 0 1 119873 119873 + 1 119886 = 0 1 (119899 minus 1) sdot (119896 minus1) 119899 sdot (119896 minus 1) if the states (V119896 minus 119886 minus 119896 V minus 1) exist then

119875V119896minus119886V (119905119894) = 119875V119896minus119886minus119896Vminus1 (119905119894) (14)


The probabilities of the states which are not included in(8)ndash(14) are all equal to 0 The processes mentioned aboveare repeated until the time step reaches the end of analysistime period Then all the transient state probabilities can beacquired using the decoding procedure in Algorithm 1

314 The Calculation of Gate Congestion Indicators Aftercalculating the entire probabilities for each time step in theanalysis timeperiod some significant performance indicatorssuch as the probability of V container trucks in the system thequeue length and the average waiting time for the V containertrucks at any time point 119905 all could be calculated accordinglyMoreover these results can be obtained promptly withoutdoing the unnecessary or repetitive calculation as the sim-ulation models The value of the performance indicators isprovided by the following equations separately

(a) The probability of V container trucks in the system attime 119905

119875V (119905) =

119896V

sum

119904=V119875119904V (119905) 0 le V le 119899

119896V

sum

119904=119899+119896sdot(Vminus119899)119875119904V (119905) V gt 119899

(15)

(b) The queue length at time 119905

119871V (119905) = 119897 timessumV(119875V (119905) times V) 0 le V le 119873 (16)

where 119897 is the average length of the container trucks(c) The average waiting time for the V container trucks in

the system at time 119905

119882V (119905) = 120583sumV(119875V (119905) times V) (17)

32 Optimization Model As mentioned before this119872119864119896119899transient queuing model is easy to be integrated into anoptimization framework in order to get some useful resultsto support the related decision making So in this section asimple and valid optimizationmodel for the railway containerterminal gate system is proposed

The objective of the optimization is to minimize the totalcost of the railway container terminal gate system in a certainanalysis time period 119879 The total cost here consists of twoparts the operation expenses of the supply side and thewaiting cost of the demand side Therefore in other wordsthe goal of this optimization problem is to find the optimalnumber of service channels which strike a balance betweenthe supply and demand Then the objective function andthe constraints can be determined accordingly As shown infunction (18) the operating cost of the service channels isexpressed by the multiplication of operation cost per hourof a service channel the number of service channels andthe certain analysis time period 119879 The waiting cost of thecontainer trucks in the queue is calculated by multiplying thehourly waiting value of each container truck by the numberof container trucks waiting in the queue by the analysis timeperiod 119879 There are two constraints represented in functions

(19) and (20) separately All the above mentioned expressionsare presented as below

Let 119862119905 denotes the total cost of the gate system during thetime period 119905 Then

min 119862119905 = (119862119899 times 119899119905 + 119862119908 times 119877119905) times 119879 (18)

st120583 times 119877119905

119899119905

le 120576 (19)

119899min le 119899119905 le 119899max (20)

where 119862119899 is the operation cost per hour of a service channel119862119908 is the waiting cost per hour of a container truck 119899119905 is thenumber of service channels opened in the time period 119905 119899minand 119899max are the minimum and maximum number of servicechannels available in the system119877119905 is the number of containertrucks waiting in the system in the time period 119905 119879 is thetime span of time period 119905 120583 is the average service time ofthe service channel 120576 is the threshold of the average waitingtime (120583times119877119905)119899119905 le 120576 ensures the average waiting time of eachcontainer truck does not exceed the threshold 119899min le 119899119905 le119899max indicates that the number of service channels opened inthe time period 119905 cannot go beyond the available range

For the solution of this optimization problem firstintegrate the transient queuing model into the optimizationframework and then conduct an optimization calculation ineach time period 119905 That is to say the optimal number of theservice channels that satisfied the objective function and con-strains is searched at each round of the transient calculationprocedure in Section 313 This computation process will becarried out in MATLAB R2008a and the detailed results anddiscussions are listed in Section 43

4 Computational Experiments

41 Model Validation For the purpose of verifying thetransient queuing model and the corresponding solutionmethod a simulation method is adopted simultaneously toacquire the estimators of the railway container terminal gatesystem eM Plant is a professional object-oriented simulationsoftware formanagement industrial engineering and systemengineering It is able to analyze the operation conditionsof the railway container terminal gate system by runningthe visual simulation models One simulation model of therailway container terminal gate system is shown in Figure 7

In the simulation model the Container trucks arrivalis the producer of container trucks which is set at theexponential arrival time interval to represent the truck arrivalsituationsTheQueuing place and waiting signs represent thecontainer trucks in the queueThe Service station denotes theservice channels in the railway container terminal gate systemwhich is commanded to satisfy the Erlang service time TheContainer trucks move is the saver of the produced containertrucks and indicates the trucks move to the container yardEvent controller is used to determine the beginning andending time of the simulation

The experiments are conducted by using the 119872119864119896119899transient queuingmodel and the eM Plant simulationmodelrespectively Take 30 trucks per hour as a possible volume


Procedure the state to state transition probabilities transient calculationInitializationInput the initial value of 119896 119899 119905 max 119905119886 120582 Set 119905 = 0119873 = 0 119904 = 0 11987500(0) = 1 isend = 0Initialize the 119875119904V(119905119894) and 119875

1015840

119904V(119905119894)

while isend = 0 dofor 119905119894 = 119905 119905 + 119905119886 minus 1 do

if 119873 lt 119899 thencalculate 119875101584000(119905119894) by (8) and 11987500(119905119894 + 1) by (7)for V = 1 119873 do

calculate 1198751015840V119896V(119905119894) by (9) and 119875V119896V(119905119894 + 1) by (7)for 119886 = 1 V(119896 minus 1) do

calculate 1198751015840V119896minus119886V(119905119894) by (10) and 119875V119896minus119886V(119905119894 + 1) by (7)end for

end forelse if 119873ge 119899 then

calculate 119875101584000(119905119894) and 11987500(119905119894 + 1)for V = 1 119899 minus 1 do calculate the part of probabilities that V lt 119899

calculate 1198751015840V119896V(119905119894) 1198751015840

V119896minus119886V(119905119894) and 119875V119896V(119905119894 + 1) 119875V119896minus119886V(119905119894 + 1)end forfor V = 119899 119873 do calculate the part of probabilities that V ge 119899

calculate 1198751015840V119896V(119905119894) by (11) and 119875V119896V(119905119894 + 1) by (7)for 119886 = 1 (119899 minus 1)(119896 minus 1) do

calculate 1198751015840V119896minus119886V(119905119894) by (12) and 119875V119896minus119886V(119905119894 + 1) by (7)end forfor 119887 = (119899 minus 1)(119896 minus 1) + 1 119899(119896 minus 1) do


end forend if

end for119905119894 = 119905 + 119905119886 V = 119873 + 1calculate the probabilities when a new container truck arrives 119875V119896minus119886V(119905119894) by (14)V = V minus 1 119905 = 119905119894119873 = 119873 + 1if 119905 ge 119905 max thenisend = 1end if

end while

Algorithm 1 The state transition probabilities transient calculation procedure

Event controller

Container Queuing

Waiting 1

Waiting 2

Waiting 3

Waiting 4

Service station 1

Service station 2

Service station 3

Service station 4

placetrucks arrivalContainer

trucks move

Figure 7 Simulation model in eM Plant

of the container trucks for the experiment The service timefollows the 119864119896=2 distribution and the mean value is setas 8 minutes The analysis time period is instructed to be60 minutes Besides to relieve the impact of randomnessthe simulation model needs to be run several times to get

the mean value and standard variance for each case Mean-while in consideration of the running time the simulationmodel is run 20 times and all the results are shown in Table 1

It can be seen from the comparison experiment resultsthat the average number of trucks in the queue at the end of60 minutes calculated from the119872119864119896=2119899 transient queuingmodel is approximate to the simulation model This meansthe transient queuing model is correct and the equally likelycombinations (ELC) heuristic method is effective Thereforethe model together with the solution method can be appliedto analyze the queuing situation of the railway containerterminal gate system

42 Sensitivity Analysis The sensitivity tests of the119872119864119896119899transient queuing model are carried out in three ways whichare to analyze the changes of the results while the number ofservice channels the mean arrival volume of the containertrucks and the average service time are different From


Table 1 Results of the queuing method and the simulation method

Arrival volume(trucksh)

Number ofservice channels

Average number of trucks in the queue

Queuing model Simulation model Relative errorMean value Standard

variance

30

1 2300 2216 694 00382 1628 1473 575 01053 996 868 362 01474 473 441 228 0073

these tests the results of the model can be demonstratedand can provide some insights to the railway containergate system management The calculated values of the meanarrival volume of the container trucks and the average servicetime are determined in accordance with the distributions inSection 31 that is the mean arrival time interval equals 4714minutes (ie the mean arrival volume of the container trucksis 12 trucks per hour) and the average service rate is 025(ie the average service time is 4 minutes) Besides in orderto confirm the results and present the trend more obviouslyanother set of severe values are also adopted as a comparisonThe mean arrival volume of the container trucks and theaverage service time are set as 30 trucks per hour and 10minutes respectively

421 The Influence of Increasing the Number of ServiceChannels This sensitivity analysis test is on the purposeof observing the change of the average waiting time whenincreasing the number of service channels As mentionedabove the calculated mean arrival volume of the containertrucks is 12 trucks per hour and the average service time is 4minutes which followed the mean values of the distributionsin Section 31 Meanwhile the comparison analysis which hasthemean arrival volume of 30 trucks per hour and the averageservice time of 10minutes is also conducted According to thepractical condition of the current railway container terminalgate system the maximum number of service channels is 8Therefore let the number of service channels range from 1to 8 to see the changes of the average waiting time for thecontainer trucks at the end of 60 minutes The results of thecalculated value and the comparison analysis tests are shownin Figures 8 and 9 respectively

It can be observed from the results in Figure 8 thatthe average waiting time has a dramatic decrease while thenumber of service channels increases from 1 to 2 and thereis also an obvious decrease when another service channel isadded the average waiting time goes down to 23 minutesHowever there has hardly been any change when the numberof service channels increases further A similar situationcan be seen in Figure 9 There is 14892 minutesrsquo reductionof the average waiting time when the number of servicechannels increases from 1 to 2 and the time decreases 4876minutes while one more service channel is increased Thenthe declines slow down and the average waiting time has beenreduced to 1144minutes when 5 service channels are opened

1 2 3 4 5 6 7 82

25

3

35

4

45

5

55

6

Number of service channels

Aver

age w

aitin

g tim

e (m

in)

Figure 8 Results of the calculated value analysis test

Aver

age w

aitin

g tim

e (m

in)

0

50

100

150

200

250

1 2 3 4 5 6 7 8Number of service channels

Figure 9 Results of the comparison analysis test

After this the changes are not very obvious Consequentlythe results can be used to decide the number of servicechannels under a certain average waiting time limitation Forexample the most reasonable number of service channelsmust be 5 if the average waiting time needs to be around 10minutes in the second case


Table 2 Results of the calculated value analysis test

Mean arrival volume (trucksh) 10 12 15 20 30 60Average number of trucks 111 120 135 163 250 313Average waiting time (minutes) 223 240 27 327 50 625

Table 3 Results of the comparison analysis test


422 The Impact of Raising the Mean Arrival Volume of theContainer Trucks The objective of this sensitivity analysistest is to research the variation of the queue length and theaverage waiting timewhen raising themean arrival volume ofthe container trucksThe average service time is also set by thecalculated value (ie 4 minutes) and the comparison value(ie 10 minutes) but the mean arrival time interval in thistest is reduced from 6 to 1 minute in decrement of 1 minute(ie the mean arrival volumes of the container trucks are 1012 15 20 30 and 60 trucks per hour) by considering themeanarrival time interval of the exponential distribution (ie 4714minutes) Assume the number of service channels equals2 Then obtain the number of the trucks in the queue andthe average waiting time 60 minutes later The results of thecalculated value analysis tests and the comparison analysistests can be found in Tables 2 and 3

The results in Table 2 show a linear increasing trend alongwith the raise in the mean arrival volume of container trucksThe average waiting time is 223 minutes at the end of 60minutes when the mean arrival volume is 10 trucks per hourHowever there exists a severe congestion (the averagewaitingtime is as long as 625minutes) when themean arrival volumecomes to 60 trucks per hour under this condition The sametrend but with a more significant increase appears in thecomparison analysis test It can be seen from Table 3 that theaverage waiting time quadruples (from 1165 minutes to 4748minutes) while the mean arrival volume is increased from 10trucks per hour to 20 trucks per hour And more notably theaverage waiting time reaches up to 244minutes which cannotbe accepted by the people in practice Therefore the relevantauthorities can use the results to make some decisions tocontrol the average waiting time of the container trucks in therailway container terminal gate system For example in thesecond case if we want to maintain the average waiting timeless than 20minutes thenmore service channels are requiredor the service rate must be increased when the mean arrivalvolume is equal to or greater than 15 trucks per hour

423 The Effect of Altering the Average Service Time The lastsensitivity analysis test is aimed at investigating the influenceof varying the average service time To be more specific thistest is to seek the appropriate number of service channelswhen altering the average service time and also ensuring thatthe average waiting time remains at the same level The meanarrival volume of the container trucks is assigned as well by

2 3 4 5 6 7 8 9 10 110

1

2

3

4

5

6

Average service time (min)

Num

ber o

f ser

vice

chan

nels


the calculated value of 12 trucks per hour and the comparisonvalue of 30 trucks per hour respectively Considering thecalculated value (ie 4 minutes) and the comparison value(ie 10 minutes) of the average service time the range of theaverage service time in this test is set from 3 to 10 minutes inincrement of 1 minute Make the average waiting time of thecontainer trucks around 20 minutes and then determine thenumber of service channels needed at the end of 60 minutesThe computation results of the calculated value analysis testand the comparison analysis test are presented in Figures 10and 11 as below

A stepped growth trend can be seen from the results ofthe calculated value analysis test in Figure 10 There is onlyneed for 1 service channel to be openwhen the average servicetime is less than 5 minutes under the first condition Anotherservice channel is needed while the average service time ismore than 5 minutes but less than 8 minutes However thenumber of service channels shows a linear increasing whenthe average service time increases more that is with one-minute increase in average service time one more servicechannel is required The growth trend of the comparisonanalysis test in Figure 11 is corresponding to the calculatedvalue analysis test in Figure 10 there is also a rapid increaseafter the stepped growth with the average service time goingupOn thewhole the necessary number of service channels is


2 3 4 5 6 7 8 9 10 11Average service time (min)

Num

ber o

f ser

vice

chan

nels

0

1

2

3

4

5

6

7


threefoldwhen the average service time varies from3minutesto 9minutes in the second case It can be concluded fromall ofthese results that if the average waiting time of the containertrucks should be controlled in a certain rangewithout varyingthe mean arrival volume more service channels are neededwhen the average service time increases Meanwhile theresults are capable of helping the administrators to makesome relevant policies for these cases For example theappropriate number of service channels is equal to 3 whenthe mean arrival volume of the container trucks is 30 trucksper hour and the acceptable average waiting time is around20 minutes

43 Optimization Results and Discussion The model vali-dation and the analysis tests have verified the established119872119864119896119899 transient queuing model is reasonable and effectiveHence a real data experiment is put forward accordingly forthe solution of the optimization model Moreover throughthe discussion of the experiment results some features inpractical operations of the railway container terminal gatesystem are also revealed at the end

The aim of the optimization experiment here is to find theoptimal number of service channels for each time period andmake some conclusions from the results In this experimentthe hourly arrival volume of container trucks is the real datacollected from 400 am to 600 pm of Chengdu railwaycontainer terminal gate system for a certain day the detailscan be seen in Figure 12 Thus the analysis time period is14 hours and the time period 119905 is 1 hour (ie the systemperformance indicators such as the queue length the averagewaiting time and the optimal number of service channels arecalculated every one hour until all the results of 14 hours havebeen output) The average service time of Chengdu railwaycontainer terminal gate system is the mean value of Erlangdistribution in Section 31 For comparison the operationcost per hour of a service channel 119862119899 is set as two differentvalues one is $40 per hour and another is $80 per hourAssume the waiting cost per hour of a container truck 119862119908

1 2 3 4 5 6 7 8 9 10 11 12 13 145

10

15

20

25

30

35

Sequence number of time period

Arr

ival

vol

ume (

truc

ksh

)

Figure 12 Arrival volume during the analysis time period

is $15 per hour Meanwhile based on the practical situationof the Chengdu railway container terminal gate system theminimum and maximum number of service channels 119899minand 119899max are set as 1 and 8 In addition the acceptable averagewaiting time for each container truck is about 10 minutes inpracticeTherefore the threshold 120576 is set as 10minutes All theoptimization results for the duration of 14 hours are presentedin Figure 13

The sequence number of time period in Figure 13 is cor-responding to the above mentioned time period 119905 the initialnumber indicates the time period from 400 am to 500 amand the rest are expressed in the same way Graph (a) is theminimum total cost of Chengdu railway container terminalgate system in different time period which is calculated fromthe established optimization model The solid line and themarked dash line are generated from the two cases that theoperation cost per hour of a service channel is $80 and $40respectively Graph (b) is the optimal numbers of servicechannels during all the time periods under these two casesOn the whole the optimization results have a similar trendwith the arrival volume of the container trucks for the analysistime period which is the same as the intuition When thearrival volume of the container trucks is low the numberof service channels and the total cost are small With thegrowth of the arrival volume the number of service channelsand the total cost are increased It also can be concludedfrom this optimization experiment that when the operationcost of the service channel is relatively high compared to thewaiting cost of the container truck the average waiting timeof the container trucks should be sacrificed so as to obtainthe minimum total cost of the system As seen in Graph (b)the optimal number of service channels when the hourly costof operating a service channel is $80 is less than the numberof service channels in the other case (ie the operation costper hour of a service channel is $40) By comparison betweenthe two Graphs although opening more service channelscauses more expenses the total cost is still less when thelevel of operation cost of a service channel for an hour iscomparatively low


1 2 3 4 5 6 7 8 9 10 11 12 13 1450

100

150

200

250

300

350

400

450


Tota

l cos

t ($)

Cn =

Cn =

= $80= $40

(a) The minimum total cost

1 2 3 4 5 6 7 8 9 10 11 12 13 14Sequence number of time period

005

115

225

335

445

555

6

Num

ber o

f ser

vice

chan

nel

Cn

Cn

= $40= $80

(b) The optimal number of service channels

Figure 13 Optimization results for the analysis time period

Table 4 Optimization results of the railway container terminal gate system in different time sections

Time section Number of service channels Average number of trucks in the queue Average total cost ($)400ndash600 1 156 10341600ndash800 2 180 18693800ndash1200 3 329 289381200ndash1400 2 220 192991400ndash1800 4 489 39336

Beyond the analysis conclusions above the other usefulinformation also can be obtained from the further study ofthis optimization experiment Take the case that the opera-tion cost of a service channel is $80 per hour as an examplealthough the optimal number of service channels is alteredin time with the variation of the service demand at differenttime period a general pattern still can be summarized fora typical day of Chengdu railway container terminal gatesystem Then accordingly the analysis time period is dividedinto several sections on the basis of the same optimal numberof service channels For example the optimal numbers ofservice channels are all equal to 1 for the time periods 1 and2 (ie 400-500 and 500-600) therefore they are taken asone separate time section And the rest are treated in thesame way All the number of service channels the averagenumber of trucks in the queue and the average total cost ofdifferent time sections are provided in Table 4 As shown inthe table the number of service channels is the least in theearly morning of the railway container terminal gate systemand is gradually increasing until noon then after a littledecrease the number reaches the peak in the late afternoonThese results make sense to the relevant optimization of therailway container terminal gate system and give the decisionmakers some reference to grasp the pattern of the variation

and help in forming the most economic and reasonableoperation scheme

5 Conclusion

This paper contributes to analyzing the railway containerterminal gate system by the119872119864119896119899 transient queuingmodelwith the equally likely combinations (ELC) heuristic solutionThe comparison experiment was then conducted with theeM Plant simulation method The experiment results indi-cate that the119872119864119896119899model is capable of providing a precisereflection to the queuing process of the railway container ter-minal gate system and the relevant performance indicators ofgate congestion such as the number of container trucks in thesystem the queue length and the average waiting time canbe obtained by the ELC transient solution dynamically andeffectivelyMoreover a variety of sensitivity tests were carriedout for analysis It turned out that the model is efficient underdifferent conditions with low computational consumptionso it is able to be applied to assist decision making Inaddition because of the flexibility of the analyticmethod thismodel can be easily integrated in an optimization frameworkTherefore the optimization of the railway container terminalgate system was implemented ultimately All these results


can provide some support to the relevant authorities so themodel and method of this paper have the value of practicalapplication

Furthermore there are some related works that can bedone for the following research For example the applicabilityof the model and methods for the other railway containerterminals needs further observations In addition take therailway container terminal system as a whole combine thegate system optimization to the other parts of the entiresystem

Conflict of Interests

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

Acknowledgments

This research is partially supported by the National Nat-ural Science Foundation of China (no 51175442 and no51205328) the Fundamental Research Funds for the CentralUniversities (2010ZT03 and 2682014BR019) and the SpecialFund of the Chinese Scholarship Council of the Ministry ofEducation of China ([2012]3013)

References

[1] ldquoThe global trade statistics about the industry in world shippingcouncilrdquo httpwwwworldshippingorgabout-the-industryglobal-tradetrade-statistics1

[2] Railway Statistical Bulletin Ministry of Railway China 2014(Chinese) httpwwwxinhuanetcomzhengfuzf temp4htm

[3] D C Mattfeld and H Kopfer ldquoTerminal operations manage-ment in vehicle transshipmentrdquo Transportation Research A vol37 no 5 pp 435ndash452 2003

[4] K H Kim K M Lee and H Hwang ldquoSequencing deliveryand receiving operations for yard cranes in port containerterminalsrdquo International Journal of Production Economics vol84 no 3 pp 283ndash292 2003

[5] D Lee H Q Wang and L Miao ldquoQuay crane schedulingwith handling priority in port container terminalsrdquo EngineeringOptimization vol 40 no 2 pp 179ndash189 2008

[6] D Lee H Q Wang and L Miao ldquoQuay crane schedulingwith non-interference constraints in port container terminalsrdquoTransportation Research Part E Logistics and TransportationReview vol 44 no 1 pp 124ndash135 2008

[7] Y Zhu and A Lim ldquoCrane scheduling with non-crossingconstraintrdquo Journal of the Operational Research Society vol 57no 12 pp 1464ndash1471 2006

[8] E K Bish ldquoA multiple-crane-constrained scheduling problemin a container terminalrdquo European Journal of OperationalResearch vol 144 no 1 pp 83ndash107 2003

[9] K H Kim and Y Park ldquoA crane scheduling method for portcontainer terminalsrdquo European Journal of Operational Researchvol 156 no 3 pp 752ndash768 2004

[10] M E H Petering ldquoEffect of block width and storage yard layouton marine container terminal performancerdquo TransportationResearch E Logistics and Transportation Review vol 45 no 4pp 591ndash610 2009

[11] J Wiese L Suhl and N Kliewer ldquoMathematical models andsolution methods for optimal container terminal yard layoutsrdquoOR Spectrum vol 32 no 3 pp 427ndash452 2010

[12] A Imai E Nishimura and S Papadimitriou ldquoThe dynamicberth allocation problem for a container portrdquo TransportationResearch Part B Methodological vol 35 no 4 pp 401ndash417 2001

[13] M Kia E Shayan and F Ghotb ldquoInvestigation of port capacityunder a new approach by computer simulationrdquoComputers andIndustrial Engineering vol 42 no 2ndash4 pp 533ndash540 2002

[14] W Huang T Kuo and S Wu ldquoA comparison of analyticalmethods and simulation for container terminal planningrdquoJournal of the Chinese Institute of Industrial Engineers vol 24no 3 pp 200ndash209 2007

[15] B Ha E Park and C Lee ldquoA simulation model with a lowlevel of detail for container terminals and its applicationsrdquo inProceedings of the Winter Simulation Conference (WSC rsquo07) pp2003ndash2011 Piscataway NJ USA December 2007

[16] B K Lee S O Park and J H Seo ldquoA simulation study fordesigning a rail terminal in a container portrdquo in Proceedingsof the Winter Simulation Conference (WSC rsquo06) pp 1388ndash1397Monterey Calif USA December 2006

[17] S P Sgouridis D Makris and D C Angelides ldquoSimulationanalysis for midterm yard planning in container terminalrdquoJournal of Waterway Port Coastal and Ocean Engineering vol129 no 4 pp 178ndash187 2003

[18] P Legato and R MMazza ldquoBerth planning and resources opti-misation at a container terminal via discrete event simulationrdquoEuropean Journal of Operational Research vol 133 no 3 pp537ndash547 2001

[19] N Boysen F Jaehn and E Pesch ldquoScheduling freight trains inrail-rail transshipment yardsrdquo Transportation Science vol 45no 2 pp 199ndash211 2011

[20] N Boysen F Jaehn and E Pesch ldquoNew bounds and algorithmsfor the transshipment yard scheduling problemrdquo Journal ofScheduling vol 15 no 4 pp 499ndash511 2012

[21] G Froyland T Koch N Megow E Duane and H WrenldquoOptimizing the landside operation of a container terminalrdquoORSpectrum Quantitative Approaches in Management vol 30 no1 pp 53ndash75 2008

[22] P Guo W Cheng Z Zhang M Zhang and J Liang ldquoGantrycrane scheduling with interference constraints in railway con-tainer terminalsrdquo International Journal of Computational Intel-ligence Systems vol 6 no 2 pp 244ndash260 2013

[23] U Dorndorf and F Schneider ldquoScheduling automated triplecross-over stacking cranes in a container yardrdquo OR Spectrumvol 32 no 3 pp 617ndash632 2010

[24] F Bruns and S Knust ldquoOptimized load planning of trainsin intermodal transportationrdquo OR Spectrum QuantitativeApproaches in Management vol 34 no 3 pp 511ndash533 2012

[25] P Corry and E Kozan ldquoAn assignment model for dynamicload planning of intermodal trainsrdquo Computers and OperationsResearch vol 33 no 1 pp 1ndash17 2006

[26] I A Hansen ldquoAutomated shunting of rail container wagonsin ports and terminal areasrdquo Transportation Planning andTechnology vol 27 no 5 pp 385ndash401 2004

[27] R Dekker P Voogd and E van Asperen ldquoAdvanced methodsfor container stackingrdquoORSpectrum vol 28 no 4 pp 563ndash5862006

[28] N Boysen M Fliedner and M Kellner ldquoDetermining fixedcrane areas in rail-rail transshipment yardsrdquo TransportationResearch E Logistics and Transportation Review vol 46 no 6pp 1005ndash1016 2010


[29] N Boysen and M Fliedner ldquoDetermining crane areas inintermodal transshipment yards The yard partition problemrdquoEuropean Journal of Operational Research vol 204 no 2 pp336ndash342 2010

[30] M Gronalt T Benna and M Posset ldquoStrategic planning ofhinterland container terminals a simulation based procedurerdquoin Proceedings of Informatik Trifft Logistik Band 1 Beitrageder 37 Jahrestagung der Gesellschaft fur Informatik eV (GI)(INFORMATIK rsquo07) pp 425ndash428 Bremen Germany Septem-ber 2007

[31] S Hartmann ldquoA general framework for scheduling equipmentand manpower at container terminalsrdquo OR Spectrum vol 26no 1 pp 51ndash74 2004

[32] A E Rizzoli N Fornara and L M Gambardella ldquoA simulationtool for combined railroad transport in intermodal terminalsrdquoMathematics and Computers in Simulation vol 59 no 1ndash3 pp57ndash71 2002

[33] P Guo W Cheng and M Zhang ldquoSimulation on the operationinfluence by external factors for railway container logisticcenterrdquo inProceedings of the International Conference of LogisticsEngineering and Management Logistics for Sustained EconomicDevelopmentmdashInfrastructure Information Integration (ICLEMrsquo10) pp 2915ndash2922 Chengdu China October 2010

[34] E D Edmond and R P Maggs ldquoHow useful are queue modelsin port investment decision for container berthsrdquo The Journalof the Operational Research Society vol 29 no 8 pp 741ndash7501978

[35] P Canonaco P Legato R M Mazza et al ldquoA queuing networkmodel for themanagement of berth crane operationsrdquoComput-ers and Operations Research vol 35 no 8 pp 2432ndash2446 2008

[36] Z G Zhang H P Luh and C H Wang ldquoModeling security-check queuesrdquo Management Science vol 57 no 11 pp 1979ndash1995 2011

[37] S Kim ldquoThe toll plaza optimization problem design opera-tions and strategiesrdquo Transportation Research E Logistics andTransportation Review vol 45 no 1 pp 125ndash137 2009

[38] M C Ausin R E Lillo and M P Wiper ldquoBayesian control ofthe number of servers in a119866119868119872119888 queueing systemrdquo Journal ofStatistical Planning and Inference vol 137 no 10 pp 3043ndash30572007

[39] M C Ausin M P Wiper and R E Lillo ldquoBayesian predictionof the transient behaviour and busy period in short-and long-tailed GIG1 queueing systemsrdquo Computational Statistics ampData Analysis vol 52 no 3 pp 1615ndash1635 2008

[40] T Czachorski T Nycz and F Pekergin ldquoTransient states of pri-ority queuesmdasha diffusion approximation studyrdquo in Proceedingsof the 5th Advanced International Conference on Telecommuni-cations (AICT rsquo09) pp 44ndash51 Venice Italy May 2009

[41] M Parlar and M Sharafali ldquoDynamic allocation of airlinecheck-in counters a queueing optimization approachrdquoManage-ment Science vol 54 no 8 pp 1410ndash1424 2008

[42] M Escobar A ROdoni andE Roth ldquoApproximate solution formulti-server queueing systems with Erlangian service timesrdquoComputers and Operations Research vol 29 no 10 pp 1353ndash1374 2002

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of


Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of


Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of


Mathematical PhysicsAdvances in

Complex AnalysisJournal of


OptimizationJournal of


CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of


Operations ResearchAdvances in

Journal of


Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences


The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014


Algebra

Discrete Dynamics in Nature and Society



Decision SciencesAdvances in

Discrete MathematicsJournal of


Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of


type marshalling cargos in the container terminals may beup to 15ndash20 days Hence the improving of the operationefficiency in railway container terminal under the existingconditions is a serious problem at present As one of the mostimportant procedures in the railway container terminal oper-ation system the through capacity and service level of thegate system have significant impacts on the efficiency of therailway container terminal even the whole railway containerfreight system Currently the operation management of therailway container terminal gate system relies on the empiricalestimation and the real time information feedback in themost cases However these methods are not accurate andefficient in reality so they are not able to solve this problemfundamentally

Recently the researchers pay more attention to thestudy of container terminals Meanwhile their researchareas mainly focus on the operation management and theequipment scheduling Mattfeld and Kopfer [3] developed anintegral decision model for the operations of the terminalKim et al [4] proposed a dynamic programming model toevaluate the delay time of outside trucks in port containerterminals and discussed the different situations respectivelyLee et al [5] presented a mixed integer programming modeltogether with a genetic algorithm to solve the quay cranescheduling problem and then extended the problem withthe noninterference constraints [6] Similar problemwas alsostudied by using different constrains and algorithms [7ndash9]In addition the relevant layout planning [10 11] the berthallocation [12] and the programming problems [13ndash18] arealso researched comprehensively

Similarly the researches of the railway container termi-nals and the intermodal terminals are conducted in thesefields and the loadingdischarging equipment schedulingproblems have become the hot issues [19ndash21] Guo et al [22]presented a mixed integer programming model with a dis-crete artificial bee colony algorithm to solve the gantry cranescheduling problem in railway container terminals Dorndorfand Schneider [23] studied the triple cross-over stackingcranes scheduling problem in order to increase the productiv-ity and reduce the delays Besides the operation researchesfor example the loading planning [24 25] the shunting of railcontainer wagons [26] the container stacking [27] and thelayout planning like the crane areas determination [28 29]are some other research interests in this field The strategicplanning and the scenario generating for the railway con-tainer terminals are usually implemented by the simulationmethods [30ndash32] Guo et al [33] developed a discrete eventsimulation model for the container handling process of therailway container logistic center to make the evaluation andimprovementHowever there exist somedisadvantages in thesimulation methods for example the quantitative solutionscannot be provided and they are unable to be nested withthe optimizationmodelsTherefore Edmond andMaggs [34]have pointed out the importance of queuing theory in therelated decision making of container terminals many yearsbefore Recently some literature began to use the queuingmethods to analyze the problems For instance Canonacoet al [35] proposed a queuing network model to solve themachine operations problem of the container terminal But

the researches of queuing models for the railway containerterminal gate system are still relatively small because therailway container terminal gate system cannot be describedcompletely and accurately by the traditional queuing modelswhich are also incapable of analyzing the instantaneousfeatures of this system so there are some limitations in thisfield

The queuing models can be classified into stationaryqueuing models and transient queuing models Because itis difficult to describe and calculate all the system states ofthe transient queuing models the majority of the queuingmodels which are applied in various important research areasonly take the steady solution into consideration [36 37]However the theory and application research of transientqueuing models has gone through some development in thepast ten years Ausin et al [38] used the Bayesian analysismethod to solve the optimized number of servers in theGIMc queuing system on the basis of the minimum totalcosts Besides they utilized the Bayesian inference to predictthe transient features of the GIG1 system [39] Czachorskiet al [40] studied the transient features of the GG1 queuingsystem from the diffusion approximation model Parlar andSharafali [41] created a time-related queuing model to opti-mize the airport security checkpoints and briefly discusseda general condition that the service time follows the Erlangdistribution

In this paper a transient queuing model and an opti-mization model are established to analyze and optimize thegate congestion according to the arrival time interval ofthe external container trucks and the distribution regularityof the service time for the railway container terminal gatesystem Then the equally likely combinations heuristic solu-tion and the optimization solution methods are adopted tosolve these models After that the system simulation andsensitivity analysis are conducted to verify the rationalityand validity of our approach Finally the optimal num-bers of the service channels in different time periods aredetermined by the optimization experiment The remainderof this paper is structured as follows Section 2 providessome background information to the case studyThe queuingmodel and optimization model together with the solutionmethods are presented in Section 3 The model validationsensitivity analysis and the optimization experiments aregiven in Section 4 Finally Section 5 concludes the paper withan outline of future study

2 Chengdu Railway Container Terminal

Chengdu railway container terminal is located in the north-east of Chengdu where it is near the Chengxiang railwaystation which connects with many key rail routes of ChinaChengdu railway container terminal is the largest railwaycontainer terminal of Asia which is also the containerlogistics hinge of southwest China Therefore it acts as animportant part to the entire container transportation net-work The terminal was built in 2010 which is 84 kilometerslong and 850 meters wide The annual cargo throughput is 1millionTEU in recent period and the forward cargo handlingcapacity will be 25 million TEU



Service


Assigned

Service channel

Container truck

station

route





0 5 10 15 20 25 30 35 400

002

004

006

008

01

012

014


Prob

abili

ty



0 5 10 15 20 25 30 35 400

001

002

003

004

005

006

007

008

009

01

Service time (min)

Prob

abili

ty















119862119894 =120596

12059611205962 sdot sdot sdot 120596119901 (1)










Mode 1 Mode 2




119862total = sum119894isin119872

119862119894 (2)



120590119904V =1


119894isin119872

1198641119894119862119894 (3)






120575119904V = 1 minus 120590119904V (5)






3210Trucks

0 2 4 6

1 3 5

2 4

3

n

n n + 1

n + 1

n + 2

n + 3n + 1

2n

2n minus 1 2n + 1

2n + 2

(s ) rarr (s + k + 1)




















119875119904V (119905119894 + 1) = 119875119904V (119905119894) + 1198751015840

119904V (119905119894) (7)



For the state (0 0)

1198751015840

00(119905119894) = 120582 (119905119894) times 119896 times 11987511 (119905119894) (8)

For V = 1 2 119899 minus 1

1198751015840




1198751015840




(10)


For V = 119899 119899 + 1 119873

1198751015840




1198751015840




(12)


1198751015840










119875V (119905) =

119896V

sum

119904=V119875119904V (119905) 0 le V le 119899

119896V

sum

119904=119899+119896sdot(Vminus119899)119875119904V (119905) V gt 119899

(15)





119882V (119905) = 120583sumV(119875V (119905) times V) (17)






st120583 times 119877119905

119899119905

le 120576 (19)

119899min le 119899119905 le 119899max (20)









1015840

119904V(119905119894)







calculate 1198751015840V119896V(119905119894) 1198751015840





end forend if


end while


Event controller

Container Queuing

Waiting 1

Waiting 2

Waiting 3

Waiting 4

Service station 1

Service station 2

Service station 3

Service station 4


trucks move












variance

30

1 2300 2216 694 00382 1628 1473 575 01053 996 868 362 01474 473 441 228 0073




1 2 3 4 5 6 7 82

25

3

35

4

45

5

55

6


Aver

age w

aitin

g tim

e (m

in)


Aver

age w

aitin

g tim

e (m

in)

0

50

100

150

200

250












2 3 4 5 6 7 8 9 10 110

1

2

3

4

5

6


Num

ber o

f ser

vice

chan

nels






Num

ber o

f ser

vice

chan

nels

0

1

2

3

4

5

6

7





1 2 3 4 5 6 7 8 9 10 11 12 13 145

10

15

20

25

30

35


Arr

ival

vol

ume (

truc

ksh

)





1 2 3 4 5 6 7 8 9 10 11 12 13 1450

100

150

200

250

300

350

400

450


Tota

l cos

t ($)

Cn =

Cn =

= $80= $40



005

115

225

335

445

555

6

Num

ber o

f ser

vice

chan

nel

Cn

Cn

= $40= $80







5 Conclusion







Acknowledgments


References



















































Volume 2014




Journal of











Journal of


Function Spaces






Algebra











Service


Assigned

Service channel

Container truck

station

route





0 5 10 15 20 25 30 35 400

002

004

006

008

01

012

014


Prob

abili

ty



0 5 10 15 20 25 30 35 400

001

002

003

004

005

006

007

008

009

01

Service time (min)

Prob

abili

ty















119862119894 =120596

12059611205962 sdot sdot sdot 120596119901 (1)










Mode 1 Mode 2




119862total = sum119894isin119872

119862119894 (2)



120590119904V =1


119894isin119872

1198641119894119862119894 (3)






120575119904V = 1 minus 120590119904V (5)






3210Trucks

0 2 4 6

1 3 5

2 4

3

n

n n + 1

n + 1

n + 2

n + 3n + 1

2n

2n minus 1 2n + 1

2n + 2

(s ) rarr (s + k + 1)




















119875119904V (119905119894 + 1) = 119875119904V (119905119894) + 1198751015840

119904V (119905119894) (7)



For the state (0 0)

1198751015840

00(119905119894) = 120582 (119905119894) times 119896 times 11987511 (119905119894) (8)

For V = 1 2 119899 minus 1

1198751015840




1198751015840




(10)


For V = 119899 119899 + 1 119873

1198751015840




1198751015840




(12)


1198751015840










119875V (119905) =

119896V

sum

119904=V119875119904V (119905) 0 le V le 119899

119896V

sum

119904=119899+119896sdot(Vminus119899)119875119904V (119905) V gt 119899

(15)





119882V (119905) = 120583sumV(119875V (119905) times V) (17)






st120583 times 119877119905

119899119905

le 120576 (19)

119899min le 119899119905 le 119899max (20)









1015840

119904V(119905119894)







calculate 1198751015840V119896V(119905119894) 1198751015840





end forend if


end while


Event controller

Container Queuing

Waiting 1

Waiting 2

Waiting 3

Waiting 4

Service station 1

Service station 2

Service station 3

Service station 4


trucks move












variance

30

1 2300 2216 694 00382 1628 1473 575 01053 996 868 362 01474 473 441 228 0073




1 2 3 4 5 6 7 82

25

3

35

4

45

5

55

6


Aver

age w

aitin

g tim

e (m

in)


Aver

age w

aitin

g tim

e (m

in)

0

50

100

150

200

250












2 3 4 5 6 7 8 9 10 110

1

2

3

4

5

6


Num

ber o

f ser

vice

chan

nels






Num

ber o

f ser

vice

chan

nels

0

1

2

3

4

5

6

7





1 2 3 4 5 6 7 8 9 10 11 12 13 145

10

15

20

25

30

35


Arr

ival

vol

ume (

truc

ksh

)





1 2 3 4 5 6 7 8 9 10 11 12 13 1450

100

150

200

250

300

350

400

450


Tota

l cos

t ($)

Cn =

Cn =

= $80= $40



005

115

225

335

445

555

6

Num

ber o

f ser

vice

chan

nel

Cn

Cn

= $40= $80







5 Conclusion







Acknowledgments


References



















































Volume 2014




Journal of











Journal of


Function Spaces






Algebra

















119862119894 =120596

12059611205962 sdot sdot sdot 120596119901 (1)










Mode 1 Mode 2




119862total = sum119894isin119872

119862119894 (2)



120590119904V =1


119894isin119872

1198641119894119862119894 (3)






120575119904V = 1 minus 120590119904V (5)






3210Trucks

0 2 4 6

1 3 5

2 4

3

n

n n + 1

n + 1

n + 2

n + 3n + 1

2n

2n minus 1 2n + 1

2n + 2

(s ) rarr (s + k + 1)




















119875119904V (119905119894 + 1) = 119875119904V (119905119894) + 1198751015840

119904V (119905119894) (7)



For the state (0 0)

1198751015840

00(119905119894) = 120582 (119905119894) times 119896 times 11987511 (119905119894) (8)

For V = 1 2 119899 minus 1

1198751015840




1198751015840




(10)


For V = 119899 119899 + 1 119873

1198751015840




1198751015840




(12)


1198751015840










119875V (119905) =

119896V

sum

119904=V119875119904V (119905) 0 le V le 119899

119896V

sum

119904=119899+119896sdot(Vminus119899)119875119904V (119905) V gt 119899

(15)





119882V (119905) = 120583sumV(119875V (119905) times V) (17)






st120583 times 119877119905

119899119905

le 120576 (19)

119899min le 119899119905 le 119899max (20)









1015840

119904V(119905119894)







calculate 1198751015840V119896V(119905119894) 1198751015840





end forend if


end while


Event controller

Container Queuing

Waiting 1

Waiting 2

Waiting 3

Waiting 4

Service station 1

Service station 2

Service station 3

Service station 4


trucks move












variance

30

1 2300 2216 694 00382 1628 1473 575 01053 996 868 362 01474 473 441 228 0073




1 2 3 4 5 6 7 82

25

3

35

4

45

5

55

6


Aver

age w

aitin

g tim

e (m

in)


Aver

age w

aitin

g tim

e (m

in)

0

50

100

150

200

250












2 3 4 5 6 7 8 9 10 110

1

2

3

4

5

6


Num

ber o

f ser

vice

chan

nels






Num

ber o

f ser

vice

chan

nels

0

1

2

3

4

5

6

7





1 2 3 4 5 6 7 8 9 10 11 12 13 145

10

15

20

25

30

35


Arr

ival

vol

ume (

truc

ksh

)





1 2 3 4 5 6 7 8 9 10 11 12 13 1450

100

150

200

250

300

350

400

450


Tota

l cos

t ($)

Cn =

Cn =

= $80= $40



005

115

225

335

445

555

6

Num

ber o

f ser

vice

chan

nel

Cn

Cn

= $40= $80







5 Conclusion







Acknowledgments


References



















































Volume 2014




Journal of











Journal of


Function Spaces






Algebra










3210Trucks

0 2 4 6

1 3 5

2 4

3

n

n n + 1

n + 1

n + 2

n + 3n + 1

2n

2n minus 1 2n + 1

2n + 2

(s ) rarr (s + k + 1)




















119875119904V (119905119894 + 1) = 119875119904V (119905119894) + 1198751015840

119904V (119905119894) (7)



For the state (0 0)

1198751015840

00(119905119894) = 120582 (119905119894) times 119896 times 11987511 (119905119894) (8)

For V = 1 2 119899 minus 1

1198751015840




1198751015840




(10)


For V = 119899 119899 + 1 119873

1198751015840




1198751015840




(12)


1198751015840










119875V (119905) =

119896V

sum

119904=V119875119904V (119905) 0 le V le 119899

119896V

sum

119904=119899+119896sdot(Vminus119899)119875119904V (119905) V gt 119899

(15)





119882V (119905) = 120583sumV(119875V (119905) times V) (17)






st120583 times 119877119905

119899119905

le 120576 (19)

119899min le 119899119905 le 119899max (20)









1015840

119904V(119905119894)







calculate 1198751015840V119896V(119905119894) 1198751015840





end forend if


end while


Event controller

Container Queuing

Waiting 1

Waiting 2

Waiting 3

Waiting 4

Service station 1

Service station 2

Service station 3

Service station 4


trucks move












variance

30

1 2300 2216 694 00382 1628 1473 575 01053 996 868 362 01474 473 441 228 0073




1 2 3 4 5 6 7 82

25

3

35

4

45

5

55

6


Aver

age w

aitin

g tim

e (m

in)


Aver

age w

aitin

g tim

e (m

in)

0

50

100

150

200

250












2 3 4 5 6 7 8 9 10 110

1

2

3

4

5

6


Num

ber o

f ser

vice

chan

nels






Num

ber o

f ser

vice

chan

nels

0

1

2

3

4

5

6

7





1 2 3 4 5 6 7 8 9 10 11 12 13 145

10

15

20

25

30

35


Arr

ival

vol

ume (

truc

ksh

)





1 2 3 4 5 6 7 8 9 10 11 12 13 1450

100

150

200

250

300

350

400

450


Tota

l cos

t ($)

Cn =

Cn =

= $80= $40



005

115

225

335

445

555

6

Num

ber o

f ser

vice

chan

nel

Cn

Cn

= $40= $80







5 Conclusion







Acknowledgments


References



















































Volume 2014




Journal of











Journal of


Function Spaces






Algebra













119875V (119905) =

119896V

sum

119904=V119875119904V (119905) 0 le V le 119899

119896V

sum

119904=119899+119896sdot(Vminus119899)119875119904V (119905) V gt 119899

(15)





119882V (119905) = 120583sumV(119875V (119905) times V) (17)






st120583 times 119877119905

119899119905

le 120576 (19)

119899min le 119899119905 le 119899max (20)









1015840

119904V(119905119894)







calculate 1198751015840V119896V(119905119894) 1198751015840





end forend if


end while


Event controller

Container Queuing

Waiting 1

Waiting 2

Waiting 3

Waiting 4

Service station 1

Service station 2

Service station 3

Service station 4


trucks move












variance

30

1 2300 2216 694 00382 1628 1473 575 01053 996 868 362 01474 473 441 228 0073




1 2 3 4 5 6 7 82

25

3

35

4

45

5

55

6


Aver

age w

aitin

g tim

e (m

in)


Aver

age w

aitin

g tim

e (m

in)

0

50

100

150

200

250












2 3 4 5 6 7 8 9 10 110

1

2

3

4

5

6


Num

ber o

f ser

vice

chan

nels






Num

ber o

f ser

vice

chan

nels

0

1

2

3

4

5

6

7





1 2 3 4 5 6 7 8 9 10 11 12 13 145

10

15

20

25

30

35


Arr

ival

vol

ume (

truc

ksh

)





1 2 3 4 5 6 7 8 9 10 11 12 13 1450

100

150

200

250

300

350

400

450


Tota

l cos

t ($)

Cn =

Cn =

= $80= $40



005

115

225

335

445

555

6

Num

ber o

f ser

vice

chan

nel

Cn

Cn

= $40= $80







5 Conclusion







Acknowledgments


References



















































Volume 2014




Journal of











Journal of


Function Spaces






Algebra











1015840

119904V(119905119894)







calculate 1198751015840V119896V(119905119894) 1198751015840





end forend if


end while


Event controller

Container Queuing

Waiting 1

Waiting 2

Waiting 3

Waiting 4

Service station 1

Service station 2

Service station 3

Service station 4


trucks move












variance

30

1 2300 2216 694 00382 1628 1473 575 01053 996 868 362 01474 473 441 228 0073




1 2 3 4 5 6 7 82

25

3

35

4

45

5

55

6


Aver

age w

aitin

g tim

e (m

in)


Aver

age w

aitin

g tim

e (m

in)

0

50

100

150

200

250












2 3 4 5 6 7 8 9 10 110

1

2

3

4

5

6


Num

ber o

f ser

vice

chan

nels






Num

ber o

f ser

vice

chan

nels

0

1

2

3

4

5

6

7





1 2 3 4 5 6 7 8 9 10 11 12 13 145

10

15

20

25

30

35


Arr

ival

vol

ume (

truc

ksh

)





1 2 3 4 5 6 7 8 9 10 11 12 13 1450

100

150

200

250

300

350

400

450


Tota

l cos

t ($)

Cn =

Cn =

= $80= $40



005

115

225

335

445

555

6

Num

ber o

f ser

vice

chan

nel

Cn

Cn

= $40= $80







5 Conclusion







Acknowledgments


References



















































Volume 2014




Journal of











Journal of


Function Spaces






Algebra















variance

30

1 2300 2216 694 00382 1628 1473 575 01053 996 868 362 01474 473 441 228 0073




1 2 3 4 5 6 7 82

25

3

35

4

45

5

55

6


Aver

age w

aitin

g tim

e (m

in)


Aver

age w

aitin

g tim

e (m

in)

0

50

100

150

200

250












2 3 4 5 6 7 8 9 10 110

1

2

3

4

5

6


Num

ber o

f ser

vice

chan

nels






Num

ber o

f ser

vice

chan

nels

0

1

2

3

4

5

6

7





1 2 3 4 5 6 7 8 9 10 11 12 13 145

10

15

20

25

30

35


Arr

ival

vol

ume (

truc

ksh

)





1 2 3 4 5 6 7 8 9 10 11 12 13 1450

100

150

200

250

300

350

400

450


Tota

l cos

t ($)

Cn =

Cn =

= $80= $40



005

115

225

335

445

555

6

Num

ber o

f ser

vice

chan

nel

Cn

Cn

= $40= $80







5 Conclusion







Acknowledgments


References



















































Volume 2014




Journal of











Journal of


Function Spaces






Algebra

















2 3 4 5 6 7 8 9 10 110

1

2

3

4

5

6


Num

ber o

f ser

vice

chan

nels






Num

ber o

f ser

vice

chan

nels

0

1

2

3

4

5

6

7





1 2 3 4 5 6 7 8 9 10 11 12 13 145

10

15

20

25

30

35


Arr

ival

vol

ume (

truc

ksh

)





1 2 3 4 5 6 7 8 9 10 11 12 13 1450

100

150

200

250

300

350

400

450


Tota

l cos

t ($)

Cn =

Cn =

= $80= $40



005

115

225

335

445

555

6

Num

ber o

f ser

vice

chan

nel

Cn

Cn

= $40= $80







5 Conclusion







Acknowledgments


References



















































Volume 2014




Journal of











Journal of


Function Spaces






Algebra











Num

ber o

f ser

vice

chan

nels

0

1

2

3

4

5

6

7





1 2 3 4 5 6 7 8 9 10 11 12 13 145

10

15

20

25

30

35


Arr

ival

vol

ume (

truc

ksh

)





1 2 3 4 5 6 7 8 9 10 11 12 13 1450

100

150

200

250

300

350

400

450


Tota

l cos

t ($)

Cn =

Cn =

= $80= $40



005

115

225

335

445

555

6

Num

ber o

f ser

vice

chan

nel

Cn

Cn

= $40= $80







5 Conclusion







Acknowledgments


References



















































Volume 2014




Journal of











Journal of


Function Spaces






Algebra










1 2 3 4 5 6 7 8 9 10 11 12 13 1450

100

150

200

250

300

350

400

450


Tota

l cos

t ($)

Cn =

Cn =

= $80= $40



005

115

225

335

445

555

6

Num

ber o

f ser

vice

chan

nel

Cn

Cn

= $40= $80







5 Conclusion







Acknowledgments


References



















































Volume 2014




Journal of











Journal of


Function Spaces






Algebra














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









Documents

Research Article A Transient Queuing Model for Analyzing ...downloads.hindawi.com/journals/mpe/2014/914706.pdf · a general condition that the service time follows the Erlang distribution