research article a heuristic procedure for the outbound ...research article a heuristic procedure...

Research ArticleA Heuristic Procedure for the Outbound Container RelocationProblem during Export Loading Operations

Roberto Guerra-Olivares1 Rosa G Gonzaacutelez-Ramiacuterez2 and Neale R Smith1

1Escuela de Ingenierıa y Ciencias Tecnologico de Monterrey Avenida Eugenio Garza Sada 2501 64849 Monterrey NL Mexico2Escuela de Ingenierıa Industrial Pontificia Universidad Catolica de Valparaıso Avenida Brasil 2950Casilla 4059 2374631 Valparaıso Chile

Correspondence should be addressed to Neale R Smith nsmithitesmmx

Received 3 December 2014 Revised 22 February 2015 Accepted 22 February 2015

Academic Editor Seungik Baek

Copyright copy 2015 Roberto Guerra-Olivares et al This is an open access article distributed under the Creative CommonsAttribution License which permits unrestricted use distribution and reproduction in any medium provided the original work isproperly cited

During export ship loading operations it is often necessary to perform relocation movements with containers that interfere withaccess to the desired container in the ship loading sequence This paper presents a real-time heuristic procedure for the containerrelocation problem employing reachstacker vehicles as container handling equipment The proposed heuristic searches for goodrelocation coordinates within a set of nearby bays The heuristic has a parameter that determines how far from the original bay acontainer may be relocatedThe tradeoff between reducing relocation movements and limiting vehicle travel distances is examinedand the performance of the heuristic is compared with a common practice in the smaller container terminals in Chile and MexicoFinally a mathematical model for the container relocation problem is presented

1 Introduction and Literature Review

Maritime terminals are facilities with a constant need toplan space to receive ships in the quay and to store theircorresponding inbound and outbound containers in theport yard Outbound containers arrive to the port and aretemporally stored in the yard until their ship arrives andberths at the quay Efficient space planning yields shorter shipturnaround times in the quay avoiding delays in the shiproute

A port can be divided into three main areas gateyard and quay The gate is the interface of the port withexternal trucks which claim inbound containers and deliveroutbound containersThe yard is a container storage area andacts as a buffer to absorb the difference in arrival times ofthe external trucks and the ship The quay is the interface ofthe port with ships and can be organized in several berthingsites A container can be categorized as inbound if it arrivesto the port on a ship and is requested by an external truck oras outbound if it arrives to the port in an external truck anddeparts on a ship Containers can be classified into familiesof containers or segregations having similar attributes such

as length being inbound or outbound being full or emptyassociated ship port of destination and weight

The port yard is organized into a three-dimensionalsystem known as BarotiThe Baroti system defines the yard asbeing divided in blocks and each block is subdivided in baysThen each bay is organized in rows with each row having adefined maximum number of tiers or stack height Figure 1shows an illustration of this coordinate system

A common approach found in the literature to addressthe storage space allocation problem is by splitting it intotwo stages The first stage typically employs a mathematicalmodel to determine in what bays to store each containerusually based on containers having the same ship and portof destination The second stage is a heuristic procedure toassign a storage location to each individual container in thebay determined by the model in the previous stage The firststage of this approach is useful for the tactical level decisionswhile the second stage makes decisions at an operationallevel Tapia et al [1] propose a mathematical model to solvethe first stage of the container space allocation problem anddetermine a set of bays to store each container

Hindawi Publishing CorporationMathematical Problems in EngineeringVolume 2015 Article ID 201749 13 pageshttpdxdoiorg1011552015201749

2 Mathematical Problems in Engineering

Bays Rows

Tiers

Figure 1 Illustration of the Baroti system

The shipping company sends to the port a list with alloutbound containers that have booked a space in the shipThe list includes container attributes such as length port ofdestination and weightThis information is input for the firststage of the approach to address the problem By the time theoutbound containers start to arrive at the gate the port knowsthe total number of containers of each segregation that willarrive and inwhat bays they should be stored In practice hubports assign time windows to receive outbound containersso that the arrival sequence can be estimated but that is notthe case in other smaller ports The space assignment policydeals with an unknown container arrival sequence and seeksto assign good available Baroti locations in real time as thecontainers arrive at the port

The storage space allocation problem (SSAP) focuses onmaking decisions about the first Baroti coordinate assigned tooutbound containers when they arrive at the terminal Dur-ing the loading operation it is often the case that access to thenext required container is obstructed by another containeror by several containers These obstructing or interferingcontainers must be removed and relocated to gain accessto the next required container Thus a relevant decision isthe relocation of interfering containers during the loadingoperation onto the ship When a relocation movement isperformed a new Baroti coordinate is determined for theinterfering containerThis relocation problem is known in theliterature as the block relocation problem (BRP) A relatedbut different problem is the premarshalling problem whichinvolves moving containers to locations that allow them tobe loaded on a ship directly in the stowage sequence Thepremarshalling operation is usually performed before theloading operations are initiated not in real time as addressedherein The premarshalling literature will not be reviewedhere but the interested reader may wish to see the work byBortfeldt and Forster [2] as an example In the followingparagraphs a literature review is presented considering firstthe SSAP contributions and then those to the BRP

For an overview of port storage yard operations see thepaper by Carlo et al [3] who provide a literature overviewtrends and research directionsThey distinguish between thefollowing main decision problems that arise in the storageyard operations (1) yard design (2) storage space assignmentfor containers (3) material handling equipment dispatchingand routing to serve the storage and retrieval requests and(4) optimizing the relocation (or reshuffling) of containers

The storage space allocation problem is addressed inthe literature by several researchers For example K HKim and H B Kim [4] propose a method for minimizing

the number of relocation movements under the constraint ofsatisfying estimated space requirements An algorithm basedon Lagrangian relaxation is provided which achieves near-to-optimal solutions Considering the productivity of yardequipment Young Kim andHwan Kim [5] study the problemof routing a single straddle carrier to pick up outboundcontainers They formulate an integer programming modelwhich minimizes the total distance traveled by the straddlecarrier within the yard and determines both the bay visitingsequence and the number of containers that should be pickedup from each bay

Since for some ports the container weight informationavailable before container arrival is only an estimate Kanget al [6] present a method for deriving a strategy forstacking containers with uncertain weight informationTheyimplemented a simulated annealing algorithmwhich can finda good stacking strategy in a reasonable time Simulationexperiments show that their proposed strategy reduces thenumber of relocationmovements compared to the traditionalsame-weight-group stacking strategy

Lee et al [7] propose a simulated annealing algorithmto schedule two Rubber-Tired-Gantry cranes for loadingoutbound containers A strategy for reserving space foroutbound containers is analyzed by Woo and Kim [8] whosuggest a series of heuristic rules to reserve a collection ofadjacent stacks for each group of containers with the sameattributes

Park et al [9] develop an online search algorithm tooptimize the stacking policy in an automated terminal andintroduce a set of criteria that must be considered fordetermining a good stacking position for each incomingcontainer Their algorithm continuously tries and evaluatesvariants of the best-so-far policy during the real operation ofthe terminal Simulation experiments show that the proposedalgorithm is effective in reducing the quay crane delay

Chen and Lu [10] study the storage location assignmentproblem for outbound containers by splitting the probleminto two stages The first stage determines the yard baysand the number of locations in each yard bay which will beassigned to containers associated with different ships Thesecond stage determines the exact storage location for eachcontainer The first stage is formulated as a lineal integerprogrammingmodel thatminimizes the workload imbalancein each container block of the yard and the total distancetraveled by internal trucks The second stage is a heuristicprocedure which decides the exact location for each individ-ual container in the preassigned yard bays determined by thefirst stage upon its arrival at the terminal The objective is tominimize the total number of relocation movements duringthe loading operation The authors propose three stackingstrategies namely diagonal vertical and horizontal andrecommend the diagonal strategy because it generated theleast relocation movements under their testing conditions

Petering [11] presents several real-time container storagesystems and evaluates them using a discrete-event simulationmodelThe author analyzes realistic scenarios and simulationexperiments show that the systems can be applied inmaritimeterminals

Mathematical Problems in Engineering 3

More recently Wang et al [12] propose a two-stageoptimization model to solve the storage space allocationproblem for inbound containers in railway container termi-nal The objectives of the models are balancing the workloadof inbound containers blocks and reducing the overlappingamounts They solve the models employing a rolling horizonapproach The authors assume that the arrival and departuretimes of containers are known in advance and that RTGcranes are available to handle containers

The BRP is addressed by several researchers For exampleYang and Kim [13] propose a mathematical model whichminimizes relocation movements in block stacking systemsIt is important to point out that in [13] the generic term blockis used by the authors to refer indistinctly to a containera box or a pallet They analyze a static case that considersknown arrival and departure times of the storage demandunits and a dynamic case in which the arrival and departuretimes are not knownThe authors develop a genetic algorithmto find good solutions for the static case and suggest simpleheuristic rules to solve the dynamic case

The first linear integer programming model for the staticcontainer relocation problem was successfully formulatedby Wan et al [14] and is referred to as MRIP (minimizerelocation integer program) By design the MRIP retrievesall containers in a stack from a given initial configurationwith the minimum number of relocation movements Theauthors also extended some heuristic algorithms used inthe literature and in practice in order to consider furtherstack configurations resulting from storing or relocating theinterfering containers to various feasible slots Later Lixinet al [15] improve the static version of the MRIP modelformulated by Wan et al [14] and propose five heuristics tosolve it They develop a discrete-event simulation model totest the performance of the proposed heuristics The authorsinclude an analysis of the worst case performance of theheuristics The experimental results show that the improvedmodel can obtain optimal or feasible solutions faster than themodel formulated by Wan et al [14] The contribution of theheuristic proposed herein can be adjusted to find solutionswith different amounts of total distance traveled by internaltrucks This is done depending on the congestion level at theyard For a case in which there is a high level of congestionin the yard the heuristic can be adjusted to find solutionsin which the interfering container will be relocated onlyto nearby bays to reduce the distance traveled by internaltrucks On the other hand in a low congestion season theinterfering container can be relocated to nearby or moredistant positions which may result in the minimization ofadditional relocations

A branch and bound algorithm to minimize the numberof relocation movements during the pickup operation inblock-stacking warehouses is suggested by Kim and Hong[16] They also propose a decision rule based on probabilitytheory and compare the performance of the decision rulewith the performance of the branch and bound algorithmThe comparison shows that the total number of relocationmovements calculated by the decision rule exceeded thatfound by branch and bound by 73 on average and thecomputational time of the heuristic rule is within the level

that can be used in real time The authors consider someprecedence relationships among the pickups of the blocksin warehouses However it may not always be possible toapply this strategy in terminal ports due to the constraint ofmaintaining the shiprsquos physical balance in the water

Caserta et al [17] suggest a binary description encodingfor stacking areas where homogeneous blocks are stored onstacks It is important to point out that in [17 18] the genericterm ldquoblockrdquo is used by the authors to refer indistinctlyto a container a box or a pallet Fast access to data onthe current stacking area and an efficient transformationinto neighboring states is required for it They state a basefor developing metaheuristic search strategies by making arandomly (roulette-wheel) guided look ahead mechanism

Caserta et al [18] present a metaheuristic approach forthe block relocation problem (BRP)The authors indicate thatthis algorithmmaybe applied not only for stacking containersin port terminals but also for stacking boxes or pallets inwarehouses The objective is to find the block relocationpattern that minimizes the total number of movementsrequired to comply with a given retrieving sequence Theauthors consider that the blocks are handled by equipmentwith the ability to reach the position located at the top ofany stack which is consistent with RTG cranes In this paperwe propose an alternative method to solve this problememploying reachstacker vehicles

Borgman et al [19] analyze two concepts using a discrete-event simulation toolThe first concept is to assume containerdeparture times as known in order to limit the numberof relocation movements They stack containers leavingshortly on the top of the stacks The second concept isthe trade-off between stacking further away in the terminaland stacking close to the berth sites and accepting morerelocation movements They have employed data from realsituations to generate scenarios of container movements fora mechanized container terminal by means of a simulationmodelThedifference between the approach of Borgman et al[19] and the heuristic approach presented herein is that ourprocedure does not assume the container departure times asknown allowing more flexibility to unexpected changes inthe stowage sequence due to ship imbalance

More recently Gharehgozli et al [20] develop a decision-tree heuristic and a dynamic programming model focusingon minimizing the expected number of relocations whenarriving containers should be stacked in a block For small-scale problems with a small number of piles they comparethe decision-tree heuristic with the dynamic programmingmodel results and they use the dynamic programmingmodelto solve large-scale problems They contrast the performanceof a shared stacking policy with a dedicated stacking policyThe shared stacking policy allows containers ofmultiple shipsto be stacked on top of each other Their heuristic proceduresolves simplified instances with a small number of ships eachwith a single port of destination

Ries et al [21] address the storage space allocationproblem and the block relocation problem in two-stageframework in combination with a fuzzy logic rule-basedstrategy The framework proposed has the aim of providingreal-time decision support to deal with the uncertain arrival


sequence of containers to the yard that is operatedwith RTGsIn contrast with the paper of Ries et al [21] where the authorspropose a fuzzy logic rule we propose a heuristic approachto solve the problem Another difference is the employmentof reachstacker vehicles to handle containers in the heuristicpresented herein

In contrast to the previous literature which considersRubber-Tired-Gantry (RTG) cranes for container handling inthe port yard this paper provides a solution to the containerrelocation problem employing reachstacker vehicles openinga new line of research A reachstacker vehicle has a morelimited slot access compared to an RTG craneWhile an RTGcrane can access any slot located at the top of any stack of thebay the reachstacker vehicle can access only the top slot of thestack located at the end of the bay Reachstacker vehicles arethe principal container handling equipment in many LatinAmerican ports and in ports with low cargo volume Theprocedures available in the existing literature may not beapplicable in ports that employ only reachstacker vehicles

It is relevant for this paper to define two types of con-tainersThe desired container is the container that is requiredto be loaded to the ship according to a stowage planThe inter-fering container is the container that needs to be relocatedin order to gain access to the desired container A desiredcontainer may have one or more interfering containers

It is desirable that each outbound container remain storedin one single Baroti location during its stay in the yard Arelocation movement is performed when it is necessary tomove a container that is blocking access to another containerthat needs to be retrieved This study considers that all theoutbound containers have been assigned to a Baroti locationfor storage when they arrive to the yard according to aport operation policy When outbound containers are loadedonto the ship some relocation movements could be requiredin order to gain access to a container stacked below othercontainers This paper proposes a heuristic which suggestsBaroti locations for the interfering containers so that furtherrelocation movements are minimized

The main contribution of this paper is a real-time heuris-tic procedure that determines the new Baroti coordinatesfor relocated containers when relocation movements areperformed assuming that the containers are handled withreachstacker vehicles and the bays are accessed from onlyone side It is also important to mention that this procedureincorporates a parameter which defines the proximity of thenew Baroti location with respect to the original locationThis parameter allows the heuristic to be tuned to adapt thesolutions to varying congestion levels in the port yard Thetotal number of relocation movements resulting using thisstrategy is compared with the strategy used in real practicein some Latin American ports

2 Problem Description

Export containers arrive at the port several days before theestimated time of arrival of the ship and during that intervalof time they are stored in the yard Figure 2 shows a typicaltimeline for the container reception period Observe that theperiod ends 24 hours prior to the estimated time of arrival

Days0 121110987654321

ETA of the ship

Container reception period

Figure 2 Timeline for the container reception period and the ETAof the ship

(ETA) of the ship A container may be received in the yardafter the container reception period has ended but will incurin a penalty for the late arrival

The stowage plan is the sequence to load containersonto a ship and is defined in a hierarchical structure Forinstance for a given ship containers with the farthest portof destination are loaded before containers with the nearestport of destination Furthermore for each port of destinationheavier containers are loaded before lighter ones

The actual container loading sequence onto the ship maybe different from the stowage plan described above becauseit is mandatory that the distribution of the weight of the con-tainers on the ship remain uniformly distributed When thedistribution of the weight on the ship becomes unbalancedsome containers may need to be loaded as counterweight notnecessarily in the sequence determined in the stowage planSuch deviations from the original stowage plan may requirerelocation movements in order to gain access to requiredcontainers in the modified stowage sequence

During the container loading operation outbound con-tainers are retrieved from the yard by a reachstacker vehicleand loaded into an internal truck to be transported to thequay where the ship is berthed for the transfer of cargo Theoperation of a reachstacker vehicle requires a driver and atracker operatorThe driver operates the reachstacker vehicleand the tracker operator manipulates a container trackingdevice to indicate to the driver the location of the containerto be retrievedWhen relocationmovements are required thetracker operator introduces the new Baroti coordinate of theinterfering containers in the tracking device

Lack of information of potential good Baroti locationsfor relocating interfering containers may lead to an increasein the number of further relocation movements when aninterfering and relocated container obstructs the retrieval ofanother container in the newBaroti location In this paper wepropose a heuristic procedure that determines the new Baroticoordinates of relocated containers In order to implementthis heuristic the function of the container tracking devicewould be expanded to call the heuristic to suggest Baroticoordinates for interfering containers when they are relo-catedThe tracker operator could type the id of the interferingcontainer into the tracking device The required informationwould then be provided to the heuristic which would thensuggest a Baroti coordinate This coordinate would then betransferred back to the tracker operator through the tracking


Zone 3 Zone 2 Zone 1

Zone 4

Zone 5Quay

Gate

(a)


Zone 4

Zone 5Quay

Gate

(b)

Figure 3 Layout of the export yard section and illustration of two possible routes for internal trucks

device allowing the reachstacker driver to be informedwhereto place the interfering container

The strategy employed in practice in someports is to placetemporarily the interfering containers in the aisle near theretrieving bay (based on in-site interviews with yard man-agers at some container terminals inChile andMexico) Oncethe desired container is retrieved the interfering containersare returned to the original bay in the opposite sequenceas they were removed The heuristic proposed in this papersuggests Baroti coordinates in other bays for interferingcontainers instead of placing them in the aisle while thedesired container is retrieved In this way the number ofmovements per container is minimized

The advantage of storing the interfering containers in theaisle close to the original bay is that total distance traveledby internal trucks is minimized However the total numberof relocation movements may be greater because a total oftwo relocation movements are required for each interferingcontainer one for retrieving the interfering container fromthe bay to situate it in the aisle and another one to returnit to the original bay On the other hand if interferingcontainers are stored in distant bays the distance traveled byinternal trucks may increase but the relocation movementsmay decrease significantly as the container can be placed in abetter position and the movement in which the container isplaced in the aisle is eliminated

The port yard managers should define a strategy toemploy according to the storage capacity and congestionlevels of the port terminal In a high congestion level seasonit may be advantageous to employ a strategy which storesinterfering containers in nearby or adjacent bays Converselyin a season with a low or moderate congestion level inter-fering containers may be relocated into more distant bayspotentially reducing the number of relocationmovements Inorder to illustrate this tradeoff Figure 3 shows an example ofa container terminal export yard

In the example depicted in Figure 3 let us assume thatthe internal truck needs to retrieve a container from a bay in

Zone 1 An interfering container may be relocated to Zone 2(see Figure 3(a)) or to Zone 5 (see Figure 3(b)) after whichthe truck will return to Zone 1 for the desired containerto transport it to the quay Let us further assume that therelocation to Zone 2 results in fewer later relocations than therelocation to Zone 5 During a season with a high congestionlevel the relocation to Zone 2may be preferred since it resultsin shorter distances and lower travel times even thoughmorerelocations will be required In contrast during a seasonwith a low congestion level the relocation to Zone 5 maybe preferred since fewer relocations will be required and thelow congestion level implies that the longer travel distance toZone 5 will not result in much longer travel times

This paper proposes a real-time heuristic procedure todetermine new Baroti coordinates to interfering containerswhen relocation movements are performed during loadingoperation taking into consideration the congestion levelof the yard The criteria employed vary according to thecongestion level of the yard that should be indicated by theyard manager

3 Mathematical Model for the ContainerRelocation Problem

Amathematical model to determine a new Baroti coordinatefor an interfering container is presented in this section Thismodel receives as input the initial inventory of containersin the yard including their weights and positions and theweight of the interfering container

The assumptions of the model are the following

(1) The interfering container and the containers stored inthe yard have the same length either 20 or 40 ft

(2) This model is executed each time a relocation move-ment is performed to define the new Baroti coordi-nate to the interfering container

(3) This model avoids storing the interfering containerin an empty bay when it is possible to store it in


another bay obtaining the same number of relocationmovements

(4) Port yard uses a reachstacker vehicle to handle con-tainers

(5) Containers may only be placed at ground level orresting on top of other containers (ie containers arenot supported by a rack)

(6) Heavier containers are loaded to the ship beforelighter ones Hence the weight categories of thecontainers imply their loading sequence

(7) Containers are classified into 5 categories accordingto their weight Category number 1 is the lightest andnumber 5 the heaviest

(8) In each yard bay a container cannot be assignedto a given stack until the stack behind has beencompletely filled This assumption results in goodspace utilization in the bay because it avoids emptyslots which may be unreachable by the reachstackervehicle

(9) The yard bays are accessed by the reachstacker vehiclefrom one end

(10) To determine the total number of relocation move-ments it is assumed that after positioning the inter-fering container in a Baroti coordinate all containersstored in the yard are (virtually) loaded onto the ship

The following notation is defined

Parameters are as follows

119861 total number of bays available to allocate contain-ers119876 container storage capacity of each bay119882IC weight of the interfering container119864 number of empty slots available in all the bays afterrelocating the interfering container119868119894119895 weight of the container stored in position (119894 119895) as

the initial inventory this parameter is set to 0 whenno container is stored in the position (119894 119895)119871 max

(119894119895)119868119894119895

119872 a big constantCOMB set of ordered pairs of positions in a bay thisset is used to inspect if the arrangement will denote arelocation movement

COMB = (119886 119887) | 119886 in 1 sdot sdot sdot 119876 minus 1 119887 in 119886 + 1 119876

(1)

Variables are as follows

119910119894119895 1 if the interfering container is stored in the

position 119895 of the bay 119894 0 otherwise

119889(119886119887)

119894 1 if theweight of the container stored in position

119886 of the bay 119894 is greater than theweight of the containerstored in position 119887 of the same bay 0 otherwise

119863119894 total number of relocations incurred in bay 119894

119875119894119895 weight of the container stored in position 119895 of bay

119894 layout after positioning the interfering container119885119894119895 1 if the position 119895 of the bay 119894 is occupied by a

container from initial inventory 0 otherwise119867119894 1 if bay 119894 is not empty 0 otherwise

119865119896

119894119895 1 if 119896th container of weight 119872 is stored in position

119895 of bay 119894 0 otherwise

The mathematical formulation of the model is as follows

Min119861

sum

119894=1

119863119894+

119861

sum

119894=1

119867119894 (2)

st119861

sum

119894=1

119876

sum

119895=1

119910119894119895

= 1 (3)

119861

sum

119894=1

119876

sum

119895=1

119865119896

119894119895= 1 119896 isin 1 2 119864 (4)

119875119894119886

minus 119875119894119887

le 119872 lowast 119889(119886119887)

119894

119894 isin 1 2 119861 (119886 119887) isin COMB

(5)

119863119894= sum

(119886119887)isinCOMB119889(119886119887)

119894119894 isin 1 2 119861 (6)

119875119894119895

= 119910119894119895

lowast 119882IC + 119868119894119895

+ 119872 lowast

119864

sum

119896=1

119865119896

119894119895

119894 isin 1 2 119861 119895 isin 1 2 119876

(7)

119868119894119895

le 119871 lowast 119885119894119895

119894 isin 1 2 119861 119895 isin 1 2 119876

(8)

119910119894119895

+ 119885119894119895

+

119864

sum

119896=1

119865119896

119894119895le 1

119894 isin 1 2 119861 119895 isin 1 2 119876

(9)

119876

sum

119895=1

(119885119894119895

+ 119910119894119895) le 2 lowast 119867

119894119894 isin 1 2 119861 (10)

119910119894119895

isin 0 1 119894 isin 1 2 119861 119895 isin 1 2 119876

(11)

119889(119886119887)

119894isin 0 1 119894 isin 1 2 119861 (119886 119887) isin COMB

(12)

119863119894ge 0 119894 isin 1 2 119861 (13)

119875119894119895

ge 0 119894 isin 1 2 119861 119895 isin 1 2 119876 (14)

119885119894119895

isin 0 1 119894 isin 1 2 119861 119895 isin 1 2 119876

(15)


119867119894isin 0 1 119894 isin 1 2 119861 (16)

119865119896

119894119895isin 0 1

119894 isin 1 2 119861 119895 isin 1 2 119876

119896 isin 1 2 119864

(17)

The first term of objective function (2) minimizes the totalnumber of relocation movements in all bays of the yard Thesecond term of (2) minimizes the number of opened baysConstraint (3) ensures that the interfering container is storedin the yard Constraint (4) ensures that exactly 119864 containers(with weight 119872) are stored in the yard This constraintensures that the interfering container is not relocated toa position above an empty slot The employment of a bigweight for virtual containers does not impact the countof relocation movements in the first term of the objectivefunction Constraint (5) sets the variables 119889

(119886119887)

119894to the value

of 1 when a relocation movement is required and constraint(6) defines the variable 119863

119894as the total number of relocation

movements incurred in bay 119894 Constraint (7) defines the vari-ables 119875

119894119895as the final inventory after relocating the interfering

container Constraint (8) activates the binary variables 119885119894119895

when a container is stored in position (119894 119895) as initial inventoryObserve that 119871 is adjusted to take the value of the maximumweight of the containers stored as the initial inventory in theyard Constraint (9) states that in each position either theinterfering container a container from the initial inventoryor a (virtual) container of weight119872 can be stored Constraint(10) defines the variables 119867

119894 Finally constraints (11) to (17)

define the domain of the decision variables

4 Description of the Heuristic

A real-time heuristic is proposed to assign Baroti locations tointerfering containers assuming that reachstacker vehicles arethe only type of available container handling equipment Theproposed heuristic assumes that the current Baroti locationsof the outbound containers are known Since this procedureis applied in real time the heuristic is useful at the operationallevel of the decision making process

The proposed heuristic considers segregations of con-tainers with similar attributes such as the ship in whichthe containers will be loaded container weight and portof destination Containers of the same segregation can behandled the same way as any other container in the samesegregation allowing more flexibility in the stacking policiesin the yard than if each container were considered to beunique

The heuristic assumes that at the beginning of the proce-dure the current Baroti locations of the containers are knownand that the sequence to load containers onto the ship isdefined according to the container weight In general heaviercontainers are loaded on the ship before lighter ones Thecontainer ship should remain balanced and in practice thecontainer loading sequence may not be strictly followed due

Tier 4 4 8 12 16 20 24

Tier 3 3 7 11 15 19 23

Tier 2 2 6 10 14 18 22

Tier 1 1 5 9 13 17 21

Row 1 Row 2 Row 3 Row 4 Row 5 Row 6

Figure 4 Sequence to fill a yard bay using a reachstacker vehicleaccessing the bay from one end

to the need to balance the ship in the berth site The heuristicprocedure described in the remainder of this paper is referredto as the smart-relocation (S-R) heuristic since it providesBaroti coordinates for interfering containers when relocationmovements are executed

Figure 4 shows a sequence to fill a yard bay with contain-ers using a reachstacker vehicle as handling equipment andassumes that the reachstacker vehicle can access the bay onlyfrom the right side It is important to emphasize that this isnot the only feasible sequence to fill a bay with containers(eg if RTG cranes are available) but it is a representative ofthe practice in many Latin American ports The numbers inthe figure represent the sequence in which each slot is filledThis policy to fill yard bays implies that a container is notallocated in a position with sequence 119896 if the positions withsequences 1 2 119896 minus 1 have not been filled before

Some parameters should be defined as input informationto the procedure The parameters of the S-R heuristic are asfollows

IC interfering container that should be relocatedDC desired container that should be retrieved fromthe yard119861 set of available bays in the yard for container ICOB original bay in which the container IC is storedbefore relocation119877119887 range of bays relative toOB inwhich the container

IC can be relocated119879 maximum tier to stack containers in each row ofeach bay119876119894 container storage capacity of bay 119894

1198771 set of bays that are close enough to the bay OB

considering the range specified in 119877119887

1198772 subset of 119877

1including only the bays that are

neither empty nor full1198771015840 set of bays that are located beyond the range

specified in 119877119887

Inventory119894 number of containers stored in bay 119894

drect(119886 119887) rectilinear distance between bays 119886 and 119887119872 a large constant

The parameter 119877119887indicates the range of allowed bays to

inspect in search of a Baroti coordinate for the interfering


R1 = i | i in B lb le inum le ub i ne OB

Start

(4) Determine

(5) Compute

(7) (15)No (17) ComputeNo (18) Determine

(19) Update

(21) Compute

(22) Store the container IC

(16) Store the container ICin the empty

bay that is closest to the

bay OB

Yes

Yes

(10)

Yes

(12)No (13) Store the container IC in the empty bay that is closest to the bay OB

Yes

No

Stop

(3) Determine

(11) Store the container IC in bneg

Crneg ne M |EB| gt 0

(14) Store the container IC in the bay bpos

(9) Update the values of Crneg and Crpos as follows

in the bay be

be = i

R998400= (R

998400FB998400)

(20) Compute for all i in R998400Cri = 10 lowast Inventoryi + drect(i OB)

Cre = min Cri | i in R998400| i in R998400 Cri = CreCrneg = min Cri | i in R2 difi le 0 bneg = i | i in R2 Cri = Crneg

Crpos = min Cri | i in R2 difi gt 0 bpos = i | i in R2 Cri = Crpos

|R2| gt 0 |EB| gt 0

R2 = R1(

R998400 = B(R1cup OB)

(1) Set Crneg = M Crpos = M

(6) Determine li for all i in R2 where li is theweight of the last container allocated in bay i

EB = i | i in R1 Inventoryi = 0FB = i | i in R1 Inventoryi = Qi

ub = + RbOBnum(2) Set lb = minus RbOBnum

(8) Compute Cri for all i in R2 as followsdifi = li minus ICweight

Cri = difi lowast (minus1000) + 10 lowast Inventoryi + drect(i OB) difi le 0

Cri = difi lowast (1000) + 10 lowast Inventoryi + drect(i OB) difi gt 0

EB cup FB)

FB998400 = i | i in R998400 Inventoryi = Qi

Figure 5 Flow diagram of the S-R heuristic

container relative to the bayOB For instance if an interferingcontainer is stacked in bay number 3 (OB = bay 3) and 119877

119887

is defined as 1 then the range of bays is 3 plusmn 1 that is 1198771

=

bay 2 bay 4 Bay number 3 is not a candidate because it isthe original bay (Recall that this would involve two relocationmovements one to remove the container and another toreturn the container to the original bay) As 119877

119887increases

the heuristic procedure is able to inspect more distant baysbut the travel time of the interfering container to the newcoordinate increases The flow diagram of the S-R heuristicshown in Figure 5 indicates that it is preferable to store theinterfering container in a bay of the set 119877

2which includes

the nearby bays according to the parameter 119877119887 When it is

not possible to store the interfering container in a bay of theset 1198772 the heuristic suggests storing it in the nearest empty

bay In the case when it is not possible to store the interferingcontainer neither in 119877

2nor in an empty bay the S-R heuristic

suggests storing it in a bay even though it may not be closeto the bay OB The S-R heuristic is executed each time arelocation movement is performed

A numerical example is introduced to illustrate theoperation of the S-R heuristic Consider a port yard with

3 bays each bay with 2 rows Assume that the weight of thedesired container (DC) is 3 and the maximum tier height ineach stack is 2The 119877

119887is specified in this case as 1 Remember

that this heuristic procedure is used to make decisions atan operational level and must be run each time that arelocation movement is performed to determine the newBaroti coordinate of the interfering container Consider theyard layout of Figure 6 The numbers in the figure representthe weight of the containers The desired container is locatedin bay 2 row 2 and tier 1 The container in bay 2 row 2 andtier 2 is interfering and it needs to be relocated

The capacity of each bay is 4 containers The rectilineardistance between two bays is calculated as the difference intheir bay numbers As the container IC is located in baynumber 2 that bay represents the parameter OB The valuesof 119897119887and 119906119887in step 2 are 119897

119887= 2minus1 = 1 and 119906

119887= 2+1 = 3 Now

the set of bays 1198771can be determined as 119877

1= bay 1 bay 3

Since all bays in 1198771are not empty or full EB = FB = Oslash

and 1198772

= 1198771 The weight of the interfering container is 2 and

the heuristic assumes that the bays can be accessed only fromright side Since bay number 1 has only one container storedand its weight is 2 119897

1= 2 Bay number 3 has two containers


Tier 2

Tier 1 2

Row 1 Row 2

Tier 2 3 2

Tier 1 3 3

Row 1 Row 2

Bay 3Bay 1 Bay 2

Tier 2 2

Tier 1 1

Row 1 Row 2

DC

IC

Figure 6 Initial configuration for numerical example 1

Tier 2 2

Tier 1 2

Row 1 Row 2

Tier 2 3

Tier 1 3 3

Row 1 Row 2

Bay 3

Tier 2 2

Tier 1 1

Row 1 Row 2

Bay 1 Bay 2

Figure 7 Final configuration for numerical example 1

Tier 2 5 4

Tier 1 5 5

Row 1 Row 2

Tier 2 1 1

Tier 1 1 1

Row 1 Row 2

Tier 2

Tier 1 3

Row 1 Row 2

DC

IC Bay nBay 1 Bay 2


stored but the container with weight 2 is stacked above thecontainer with weight 1 so the weight of the last containerallocated in this bay is 2 defining 119897

3= 2 The cardinality of

the set 1198772is greater than zero and step number 8 is executed

as follows dif1

= 2 minus 2 = 0 dif2

= 2 minus 2 = 0 Therectilinear distance between bay number 2 and bay number1 is 1 Similarly the rectilinear distance between bay number2 and bay number 3 is 1 Cr

1= (0)(minus1000) + (10)(1) + 1 = 11

and Cr2

= (0)(minus1000) + (10)(2) + 1 = 21 Step number9 determines that Crneg = min11 21 so Crneg is 11 and119887neg is bay 1 The container IC can be stacked in either row1 or row 2 of bay number 1 but row number 1 is preferredto avoid empty spaces according to the sequence shown inFigure 4 The final decision in step number 11 is to relocatethe interfering container IC to bay number 1 Figure 7 showsthe configuration after relocating the interfering container

Another relevant situation is shown in Figure 8 Observethat in this case the yard also has 3 bays but one bay is labeledldquobay 119899rdquo to indicate that it is relatively far from bay 1 and bay2 The capacities of bays are equal to 4 containers and 119877

119887is

defined as 1 The weight of desired container (DC) is 5 andthe weight of the interfering container (IC) is 4

In this case OB is bay 1 and the values of 119897119887and 119906

119887are

defined as 119897119887

= 1 minus 1 = 0 and 119906119887

= 1 + 1 = 2 1198771contains

the set of near bays to OB according to 119877119887 but in this case

there is no bay 0 and bay number 1 is the original bay (OB)so1198771

= bay 2 Since bay 2 is full of containers FB = bay 2

and 1198772

= (1198771

FB) = Oslash There are no empty bays in 1198771 so

EB = Oslash In this case the cardinality of the set1198772is not greater

than 0 and the cardinality of the set EB is 0 so step number17 is executed and 119877

1015840 is determined as 1198771015840

= bay 119899 As FB1015840in step 18 is Oslash 1198771015840 remains with no change after the update instep 19 Bay 119899 has only one container stored and the rectilineardistance between bay 119899 and bay 1 is 119899 minus 1 so Cr

119899= (10)(1) +

119899 minus 1 = 9 + 119899 Step number 21 defines Cr119890as the minimum

among the values of Cr calculated for all bays in 1198771015840 but in

this case 1198771015840 consists only in one bay so Cr

119890= 9 + 119899 and 119887

119890=

bay 119899 The final decision is to store the interfering containerin bay 119899 Figure 9 shows the configuration after relocating theinterfering container

5 Determination of the Number of RelocationMovements in the Worst Case Scenario

In order to calculate the total number of relocation move-ments it is assumed that all the containers required to bemoved to retrieve a given container are relocated in the sameconfiguration as they were previously stacked at the bay For


Tier 2 5

Tier 1 5 5

Row 1 Row 2

Tier 2 1 1

Tier 1 1 1

Row 1 Row 2

Tier 2 4

Tier 1 3

Row 1 Row 2

Bay nBay 1 Bay 2


A

B

C

D

(a)

A

B

C

(b)

A

B

(c)

Figure 10 Configuration of a bay of capacity 4 during a container retrieval process

instance consider a bay with a capacity of 4 containers asillustrated in Figure 10(a)

Suppose that the sequence to retrieve the containersaccording to the stowage plan is DCBA To retrieve con-tainer D three relocation movements (ABC) are requiredOnce container D is retrieved the configuration of the bay isas shown in Figure 10(b)

Now in order to retrieve container C two additionalrelocation movements are required The configuration at thismoment is as shown in Figure 10(c)

Container B is the next one to be picked up so oneadditional relocation movement should be made to reachit The last container to be retrieved does not require anyrelocation movements so the total number of relocationmovements required to retrieve all the containers of this baywith capacity of four is 3 + 2 + 1 = 6 This sequence resultsin the greatest number of relocation movements because theorder of retrieving the containers is exactly the opposite of theorder of the containers stacked in the bay

The policy of returning to the original bay all interferingcontainers is referred to as the worst case scenario in theremainder of this paper It is quite similar to the actualpractice in many smaller ports and thus provides a closeapproximation to the current practice

6 Numerical Results

A set of experiments employing the S-R heuristic was per-formed and the results were compared against the worst casescenario described in previous section All the experimentspresented in this section were performed on a personalcomputer with an i5 processor and 6GB RAM The S-Rheuristic is coded in C

The ratio of relocation movements to total movementsis used as a performance metric Total movements aredefined as relocation movements + effective movements

In contrast with relocation movements an effective move-ment is performedwhen a container of the requested segrega-tion can be reached directly by the yard equipmentThe ratioof relocation movements to total movements is computed asfollows

number of relocation movements

sdot (number of relocation movements

+ number of effective movements)minus1

(18)

To illustrate this ratio consider a bay filledwith 30 containersand assume that a total of 10 relocation movements arerequired to empty the bayThis yields a ratio of 10(10+30) =

14 = 25We test four values of 119877

119887= 1 2 3 4 The 119877

119887value used

in each experiment is indicated in the header of Table 1 asS-R(119877

119887) A tight case is represented when 119877

119887= 1 because

the S-R heuristic seeks Baroti coordinates for interferingcontainers only in the immediate adjacent bays to OB Amedium case is represented when119877

119887= 2 3 and the relaxed

case is represented when 119877119887

= 4 indicating that the S-Rheuristic is able to seek Baroti coordinates in all bays of theyard We define 8 different types of instances and generate20 different container arrival sequences for each instancetype The number of containers maximum tier and weightlevels studied are specified in Table 1 for each instance typeAll instance types consider a port yard with 5 bays and 6rows in each bayThe average ratios of relocation movementsto total movements are reported in the table The proposedheuristic is able to suggest a new Baroti coordinate for eachinterfering container in less than 1 second enabling the useof this heuristic in real port operations

When 119877119887increases the ratio of relocation to total move-

ments decreases since the space to search Baroti coordinates


Table 1 Ratio of relocation movements to total movements of the procedures studied

Instance type Number of containers Number of tiers S-R(1) S-R(2) S-R(3) S-R(4) Worst case scenario1 96 4 456 394 375 390 7672 96 4 456 420 399 379 7523 120 5 495 453 421 465 8114 120 5 554 497 478 465 8265 80 4 392 333 301 308 6336 80 4 430 368 343 336 6687 100 5 419 366 330 323 7548 100 5 488 423 384 378 776

3538414447

80 90 100 110 120 130 140Distance traveled by internal trucks

Instance type 1

35

40

45

50

80 85 90 95 100 105 110 115 120 125

Reha

ndle

Distance traveled by internal trucks

Instance type 2

3540455055


Instance type 3

354045505560


Instance type 4

25

30

35

40

50 55 60 65 70 75 80 85Distance traveled by internal trucks

Instance type 5

2530354045


Instance type 6

283134374043

70 75 80 85 90 95 100 105 110 115Distance traveled by internal trucks

Instance type 7

353841444750


Instance type 8

Rb = 1

Rb = 2Rb = 4

Rb = 3

Rb = 2 Rb = 3

Rb = 4

Rb = 1

Rb = 2Rb = 1

Rb = 3

Rb = 1

Rb = 2

Rb = 4

Rb = 3Rb = 4

Rb = 1

Rb = 2

Rb = 3 Rb = 4

Rb = 1

Rb = 2

Rb = 3 Rb = 4

Rb = 1

Rb = 2Rb = 3

Rb = 4

Rb = 1

Rb = 2Rb = 3Rb = 4

mov

emen

ts (

)

Reha

ndle

m

ovem

ents

()

Reha

ndle

m

ovem

ents

()

Reha

ndle

m

ovem

ents

()

Reha

ndle

m

ovem

ents

()

Reha

ndle

m

ovem

ents

()

Reha

ndle

m

ovem

ents

()

Reha

ndle

m

ovem

ents

()

Figure 11 Relation between the ratio of relocation movements and the distance traveled by internal trucks

includes more bays Table 2 shows the gap between the S-Rheuristic and the worst case scenario of each instance type

The maximum gap reported between the S-R heuristicand the worst case scenario is 572 in instances with 119877

119887=

4 which is expected because in this case the heuristic cansuggest Baroti coordinates in any bay of the yard Theminimum gap is 329 and is reported in instances with119877119887

= 1 Observe that in this case the bigger the gap the betterthe performance because the comparison is against the worstcase scenario

The determination of relocation movements performedin the worst case scenario considers as one single relocationmovement the retrieval of the interfering container from thebay to situate it in the aisle and the movement to return itto the original bay However the movement from the bay tothe aisle and the movement from the aisle back to the bay are

actually two relocation movements Because of this the ratioof relocation to total movements in worst case scenario inactual practice is greater than or equal to the ratio reported inTable 1 This means that the benefit of using the S-R heuristicis actually greater than implied by the values in Table 1

The parameter 119877119887can be adjusted to achieve a tradeoff

between a reduction in relocationmovements and an increasein the distance traveled by internal trucks The rectilineardistance traveled by internal trucks and the ratio of relocationmovements for the instance types studied are shown inFigure 11 In this figure the relation between the ratio ofrelocation movements and the distance traveled by internaltrucks is shown

The general observed behavior is that the distance trav-eled by internal trucks increases with an increase of 119877

119887 The

percentage of relocation movements tends to decrease with


Table 2 Gap between results of S-R heuristic and the worst casescenario

Instance type S-R(1) S-R(2) S-R(3) S-R(4)1 405 487 512 4912 394 442 470 4973 389 441 480 4264 329 398 421 4365 381 474 525 5136 356 449 486 4967 444 514 562 5728 371 455 506 513

an increase in 119877119887 This is observed for all tested instance

types for 119877119887equal to 1 2 and 3 However for some instance

types the percentage of relocationmovements is smaller with119877119887

= 3 than with 119877119887

= 4 suggesting that an intermediatevalue of 119877

119887may be optimal in some cases

7 Conclusions and Recommendations forFurther Research

Wepropose a heuristic procedurewhich suggests Baroti coor-dinates for interfering containers during the loading oper-ation when reachstacker vehicles are the available handlingequipment To evaluate the performance of the proposedheuristic we compute the number of relocation movementsassuming the worst case scenario The proposed heuristichas a parameter 119877

119887 which determines the proximity of

candidate bays for container relocationWhen the yard wantsto perform the minimum number of relocation movementsthe interfering containers are relocated in any bay of the yardregardless of its proximity to the original bay in which thecontainer is stored On the other hand when the terminalneeds to minimize the distance traveled by internal trucksthe interfering containers are stored in a set of bays adjacentto the original bay but the number of relocation movementstends to be greater

For further research we propose developing heuristicproceduresminimizing the total cost of the loading operationconsidering both the distance traveled by internal trucks andthe relocationmovements Alternatively the ship turnaroundtime could be considered as a performance metric providedthat it may account for both relocations and distance metricsexpressed as the required time to serve the ship Anotherapproach to address this situation is to consider the problemas a biobjective optimization problem in which an efficientfrontier is sought rather than a single solution

Conflict of Interests

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

References

[1] F Tapia R Covarrubias P Miranda and R G Gonzalez-Ramırez ldquoOn the storage space allocation problemrdquo in Proceed-ings of the 22nd International Conference on Production Research(ICPR rsquo13) Iguassu Falls Brazil July-August 2013

[2] A Bortfeldt and F Forster ldquoA tree search procedure forthe container pre-marshalling problemrdquo European Journal ofOperational Research vol 217 no 3 pp 531ndash540 2012

[3] H J Carlo I F A Vis andK J Roodbergen ldquoStorage yard oper-ations in container terminals literature overview trends andresearch directionsrdquo European Journal of Operational Researchvol 235 no 2 pp 412ndash430 2014

[4] K H Kim and H B Kim ldquoSegregating space allocationmodels for container inventories in port container terminalsrdquoInternational Journal of Production Economics vol 59 no 1 pp415ndash423 1999

[5] K Young Kim and K Hwan Kim ldquoRouting algorithm fora single straddle carrier to load export containers onto acontainershiprdquo International Journal of Production Economicsvol 59 no 1 pp 425ndash433 1999

[6] J Kang K R Ryu and K H Kim ldquoDeriving stacking strategiesfor export containers with uncertain weight informationrdquoJournal of Intelligent Manufacturing vol 17 no 4 pp 399ndash4102006

[7] D-H Lee Z Cao andQMeng ldquoScheduling of two-transtainersystems for loading outbound containers in port containerterminals with simulated annealing algorithmrdquo InternationalJournal of Production Economics vol 107 no 1 pp 115ndash124 2007

[8] Y J Woo and K H Kim ldquoEstimating the space requirement foroutbound container inventories in port container terminalsrdquoInternational Journal of Production Economics vol 133 no 1 pp293ndash301 2011

[9] T Park R Choe Y H Kim and K R Ryu ldquoDynamic adjust-ment of container stacking policy in an automated containerterminalrdquo International Journal of Production Economics vol133 no 1 pp 385ndash392 2011

[10] L Chen and Z Lu ldquoThe storage location assignment problemfor outbound containers in a maritime terminalrdquo InternationalJournal of Production Economics vol 135 no 1 pp 73ndash80 2012

[11] M E H Petering ldquoReal-time container storage location assign-ment at an RTG-based seaport container transshipment ter-minal problem description control system simulation modeland penalty scheme experimentationrdquo Flexible Services andManufacturing Journal 31 pages 2013

[12] L Wang X Zhu and Z Xie ldquoStorage space allocation ofinbound container in railway container terminalrdquoMathematicalProblems in Engineering vol 2014 Article ID 956536 10 pages2014

[13] J H Yang and K H Kim ldquoA grouped storage method forminimizing relocations in block stacking systemsrdquo Journal ofIntelligent Manufacturing vol 17 no 4 pp 453ndash463 2006

[14] Y-W Wan J Liu and P-C Tsai ldquoThe assignment of storagelocations to containers for a container stackrdquo Naval ResearchLogistics vol 56 no 8 pp 699ndash713 2009

[15] T Lixin W Jiang J Liu and Y Dong ldquoResearch into containerreshuffling and stacking problems in container terminal yardsrdquoIIE Transactions 2014

[16] K H Kim and G-P Hong ldquoA heuristic rule for relocatingblocksrdquo Computers amp Operations Research vol 33 no 4 pp940ndash954 2006


[17] M Caserta S Schwarze and S Voszlig ldquoA new binary descriptionof the blocks relocation problem and benefits in a look aheadheuristicrdquo in Evolutionary Computation in Combinatorial Opti-mization vol 5482 pp 37ndash48 Springer Berlin Germany 2009

[18] M Caserta S Voszlig and M Sniedovich ldquoApplying the corridormethod to a blocks relocation problemrdquo OR Spectrum vol 33no 4 pp 915ndash929 2011

[19] B Borgman E van Asperen and R Dekker ldquoOnline rules forcontainer stackingrdquo OR Spectrum vol 32 no 3 pp 687ndash7162010

[20] A H Gharehgozli Y Yu R De Koster and J T UddingldquoA decision-tree stacking heuristic minimising the expectednumber of reshuffles at a container terminalrdquo InternationalJournal of Production Research vol 52 no 9 pp 2592ndash26112014

[21] J Ries R G Gonzalez-Ramırez and P Miranda ldquoA fuzzylogic model for the container stacking problem at containerterminalsrdquo inComputational Logistics vol 8760 ofLectureNotesin Computer Science pp 93ndash111 Springer Berlin Germany2014

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


Bays Rows

Tiers

Figure 1 Illustration of the Baroti system

The shipping company sends to the port a list with alloutbound containers that have booked a space in the shipThe list includes container attributes such as length port ofdestination and weightThis information is input for the firststage of the approach to address the problem By the time theoutbound containers start to arrive at the gate the port knowsthe total number of containers of each segregation that willarrive and inwhat bays they should be stored In practice hubports assign time windows to receive outbound containersso that the arrival sequence can be estimated but that is notthe case in other smaller ports The space assignment policydeals with an unknown container arrival sequence and seeksto assign good available Baroti locations in real time as thecontainers arrive at the port

The storage space allocation problem (SSAP) focuses onmaking decisions about the first Baroti coordinate assigned tooutbound containers when they arrive at the terminal Dur-ing the loading operation it is often the case that access to thenext required container is obstructed by another containeror by several containers These obstructing or interferingcontainers must be removed and relocated to gain accessto the next required container Thus a relevant decision isthe relocation of interfering containers during the loadingoperation onto the ship When a relocation movement isperformed a new Baroti coordinate is determined for theinterfering containerThis relocation problem is known in theliterature as the block relocation problem (BRP) A relatedbut different problem is the premarshalling problem whichinvolves moving containers to locations that allow them tobe loaded on a ship directly in the stowage sequence Thepremarshalling operation is usually performed before theloading operations are initiated not in real time as addressedherein The premarshalling literature will not be reviewedhere but the interested reader may wish to see the work byBortfeldt and Forster [2] as an example In the followingparagraphs a literature review is presented considering firstthe SSAP contributions and then those to the BRP

For an overview of port storage yard operations see thepaper by Carlo et al [3] who provide a literature overviewtrends and research directionsThey distinguish between thefollowing main decision problems that arise in the storageyard operations (1) yard design (2) storage space assignmentfor containers (3) material handling equipment dispatchingand routing to serve the storage and retrieval requests and(4) optimizing the relocation (or reshuffling) of containers

The storage space allocation problem is addressed inthe literature by several researchers For example K HKim and H B Kim [4] propose a method for minimizing

the number of relocation movements under the constraint ofsatisfying estimated space requirements An algorithm basedon Lagrangian relaxation is provided which achieves near-to-optimal solutions Considering the productivity of yardequipment Young Kim andHwan Kim [5] study the problemof routing a single straddle carrier to pick up outboundcontainers They formulate an integer programming modelwhich minimizes the total distance traveled by the straddlecarrier within the yard and determines both the bay visitingsequence and the number of containers that should be pickedup from each bay

Since for some ports the container weight informationavailable before container arrival is only an estimate Kanget al [6] present a method for deriving a strategy forstacking containers with uncertain weight informationTheyimplemented a simulated annealing algorithmwhich can finda good stacking strategy in a reasonable time Simulationexperiments show that their proposed strategy reduces thenumber of relocationmovements compared to the traditionalsame-weight-group stacking strategy

Lee et al [7] propose a simulated annealing algorithmto schedule two Rubber-Tired-Gantry cranes for loadingoutbound containers A strategy for reserving space foroutbound containers is analyzed by Woo and Kim [8] whosuggest a series of heuristic rules to reserve a collection ofadjacent stacks for each group of containers with the sameattributes

Park et al [9] develop an online search algorithm tooptimize the stacking policy in an automated terminal andintroduce a set of criteria that must be considered fordetermining a good stacking position for each incomingcontainer Their algorithm continuously tries and evaluatesvariants of the best-so-far policy during the real operation ofthe terminal Simulation experiments show that the proposedalgorithm is effective in reducing the quay crane delay

Chen and Lu [10] study the storage location assignmentproblem for outbound containers by splitting the probleminto two stages The first stage determines the yard baysand the number of locations in each yard bay which will beassigned to containers associated with different ships Thesecond stage determines the exact storage location for eachcontainer The first stage is formulated as a lineal integerprogrammingmodel thatminimizes the workload imbalancein each container block of the yard and the total distancetraveled by internal trucks The second stage is a heuristicprocedure which decides the exact location for each individ-ual container in the preassigned yard bays determined by thefirst stage upon its arrival at the terminal The objective is tominimize the total number of relocation movements duringthe loading operation The authors propose three stackingstrategies namely diagonal vertical and horizontal andrecommend the diagonal strategy because it generated theleast relocation movements under their testing conditions

Petering [11] presents several real-time container storagesystems and evaluates them using a discrete-event simulationmodelThe author analyzes realistic scenarios and simulationexperiments show that the systems can be applied inmaritimeterminals




















Days0 121110987654321

ETA of the ship










Zone 4

Zone 5Quay

Gate

(a)


Zone 4

Zone 5Quay

Gate

(b)




























(119894119895)119868119894119895



(1)




119889(119886119887)







119865119896




Min119861

sum

119894=1

119863119894+

119861

sum

119894=1

119867119894 (2)

st119861

sum

119894=1

119876

sum

119895=1

119910119894119895

= 1 (3)

119861

sum

119894=1

119876

sum

119895=1

119865119896

119894119895= 1 119896 isin 1 2 119864 (4)

119875119894119886

minus 119875119894119887

le 119872 lowast 119889(119886119887)

119894

119894 isin 1 2 119861 (119886 119887) isin COMB

(5)

119863119894= sum

(119886119887)isinCOMB119889(119886119887)

119894119894 isin 1 2 119861 (6)

119875119894119895

= 119910119894119895

lowast 119882IC + 119868119894119895

+ 119872 lowast

119864

sum

119896=1

119865119896

119894119895

119894 isin 1 2 119861 119895 isin 1 2 119876

(7)

119868119894119895

le 119871 lowast 119885119894119895

119894 isin 1 2 119861 119895 isin 1 2 119876

(8)

119910119894119895

+ 119885119894119895

+

119864

sum

119896=1

119865119896

119894119895le 1

119894 isin 1 2 119861 119895 isin 1 2 119876

(9)

119876

sum

119895=1

(119885119894119895

+ 119910119894119895) le 2 lowast 119867

119894119894 isin 1 2 119861 (10)

119910119894119895

isin 0 1 119894 isin 1 2 119861 119895 isin 1 2 119876

(11)

119889(119886119887)


(12)

119863119894ge 0 119894 isin 1 2 119861 (13)

119875119894119895

ge 0 119894 isin 1 2 119861 119895 isin 1 2 119876 (14)

119885119894119895

isin 0 1 119894 isin 1 2 119861 119895 isin 1 2 119876

(15)


119867119894isin 0 1 119894 isin 1 2 119861 (16)

119865119896

119894119895isin 0 1

119894 isin 1 2 119861 119895 isin 1 2 119876

119896 isin 1 2 119864

(17)


(119886119887)

119894to the value













Tier 4 4 8 12 16 20 24

Tier 3 3 7 11 15 19 23

Tier 2 2 6 10 14 18 22

Tier 1 1 5 9 13 17 21










1198772 subset of 119877










Start

(4) Determine

(5) Compute


(19) Update

(21) Compute




bay OB

Yes

Yes

(10)

Yes


Yes

No

Stop

(3) Determine





in the bay be

be = i

R998400= (R

998400FB998400)




|R2| gt 0 |EB| gt 0

R2 = R1(

R998400 = B(R1cup OB)








EB cup FB)




119887


=


119887increases


2which includes











119887= 2minus1 = 1 and 119906

119887= 2+1 = 3 Now


1= bay 1 bay 3


and 1198772





Tier 2

Tier 1 2

Row 1 Row 2

Tier 2 3 2

Tier 1 3 3

Row 1 Row 2

Bay 3Bay 1 Bay 2

Tier 2 2

Tier 1 1

Row 1 Row 2

DC

IC


Tier 2 2

Tier 1 2

Row 1 Row 2

Tier 2 3

Tier 1 3 3

Row 1 Row 2

Bay 3

Tier 2 2

Tier 1 1

Row 1 Row 2

Bay 1 Bay 2


Tier 2 5 4

Tier 1 5 5

Row 1 Row 2

Tier 2 1 1

Tier 1 1 1

Row 1 Row 2

Tier 2

Tier 1 3

Row 1 Row 2

DC

IC Bay nBay 1 Bay 2





as follows dif1



1= (0)(minus1000) + (10)(1) + 1 = 11

and Cr2



119887is



119887are

defined as 119897119887

= 1 minus 1 = 0 and 119906119887

= 1 + 1 = 2 1198771contains




and 1198772

= (1198771




1015840 is determined as 1198771015840


119899= (10)(1) +




119890= 9 + 119899 and 119887

119890=





Tier 2 5

Tier 1 5 5

Row 1 Row 2

Tier 2 1 1

Tier 1 1 1

Row 1 Row 2

Tier 2 4

Tier 1 3

Row 1 Row 2

Bay nBay 1 Bay 2


A

B

C

D

(a)

A

B

C

(b)

A

B

(c)







6 Numerical Results







(18)



119887= 1 2 3 4 The 119877

119887value used



119887= 1 because










3538414447


Instance type 1

35

40

45

50

80 85 90 95 100 105 110 115 120 125

Reha

ndle


Instance type 2

3540455055


Instance type 3

354045505560


Instance type 4

25

30

35

40


Instance type 5

2530354045


Instance type 6

283134374043


Instance type 7

353841444750


Instance type 8

Rb = 1

Rb = 2Rb = 4

Rb = 3

Rb = 2 Rb = 3

Rb = 4

Rb = 1

Rb = 2Rb = 1

Rb = 3

Rb = 1

Rb = 2

Rb = 4

Rb = 3Rb = 4

Rb = 1

Rb = 2

Rb = 3 Rb = 4

Rb = 1

Rb = 2

Rb = 3 Rb = 4

Rb = 1

Rb = 2Rb = 3

Rb = 4

Rb = 1

Rb = 2Rb = 3Rb = 4

mov

emen

ts (

)

Reha

ndle

m

ovem

ents

()

Reha

ndle

m

ovem

ents

()

Reha

ndle

m

ovem

ents

()

Reha

ndle

m

ovem

ents

()

Reha

ndle

m

ovem

ents

()

Reha

ndle

m

ovem

ents

()

Reha

ndle

m

ovem

ents

()




119887=








119887 The








= 3 than with 119877119887










References






























Volume 2014




Journal of











Journal of


Function Spaces






Algebra




























Days0 121110987654321

ETA of the ship










Zone 4

Zone 5Quay

Gate

(a)


Zone 4

Zone 5Quay

Gate

(b)




























(119894119895)119868119894119895



(1)




119889(119886119887)







119865119896




Min119861

sum

119894=1

119863119894+

119861

sum

119894=1

119867119894 (2)

st119861

sum

119894=1

119876

sum

119895=1

119910119894119895

= 1 (3)

119861

sum

119894=1

119876

sum

119895=1

119865119896

119894119895= 1 119896 isin 1 2 119864 (4)

119875119894119886

minus 119875119894119887

le 119872 lowast 119889(119886119887)

119894

119894 isin 1 2 119861 (119886 119887) isin COMB

(5)

119863119894= sum

(119886119887)isinCOMB119889(119886119887)

119894119894 isin 1 2 119861 (6)

119875119894119895

= 119910119894119895

lowast 119882IC + 119868119894119895

+ 119872 lowast

119864

sum

119896=1

119865119896

119894119895

119894 isin 1 2 119861 119895 isin 1 2 119876

(7)

119868119894119895

le 119871 lowast 119885119894119895

119894 isin 1 2 119861 119895 isin 1 2 119876

(8)

119910119894119895

+ 119885119894119895

+

119864

sum

119896=1

119865119896

119894119895le 1

119894 isin 1 2 119861 119895 isin 1 2 119876

(9)

119876

sum

119895=1

(119885119894119895

+ 119910119894119895) le 2 lowast 119867

119894119894 isin 1 2 119861 (10)

119910119894119895

isin 0 1 119894 isin 1 2 119861 119895 isin 1 2 119876

(11)

119889(119886119887)


(12)

119863119894ge 0 119894 isin 1 2 119861 (13)

119875119894119895

ge 0 119894 isin 1 2 119861 119895 isin 1 2 119876 (14)

119885119894119895

isin 0 1 119894 isin 1 2 119861 119895 isin 1 2 119876

(15)


119867119894isin 0 1 119894 isin 1 2 119861 (16)

119865119896

119894119895isin 0 1

119894 isin 1 2 119861 119895 isin 1 2 119876

119896 isin 1 2 119864

(17)


(119886119887)

119894to the value













Tier 4 4 8 12 16 20 24

Tier 3 3 7 11 15 19 23

Tier 2 2 6 10 14 18 22

Tier 1 1 5 9 13 17 21










1198772 subset of 119877










Start

(4) Determine

(5) Compute


(19) Update

(21) Compute




bay OB

Yes

Yes

(10)

Yes


Yes

No

Stop

(3) Determine





in the bay be

be = i

R998400= (R

998400FB998400)




|R2| gt 0 |EB| gt 0

R2 = R1(

R998400 = B(R1cup OB)








EB cup FB)




119887


=


119887increases


2which includes











119887= 2minus1 = 1 and 119906

119887= 2+1 = 3 Now


1= bay 1 bay 3


and 1198772





Tier 2

Tier 1 2

Row 1 Row 2

Tier 2 3 2

Tier 1 3 3

Row 1 Row 2

Bay 3Bay 1 Bay 2

Tier 2 2

Tier 1 1

Row 1 Row 2

DC

IC


Tier 2 2

Tier 1 2

Row 1 Row 2

Tier 2 3

Tier 1 3 3

Row 1 Row 2

Bay 3

Tier 2 2

Tier 1 1

Row 1 Row 2

Bay 1 Bay 2


Tier 2 5 4

Tier 1 5 5

Row 1 Row 2

Tier 2 1 1

Tier 1 1 1

Row 1 Row 2

Tier 2

Tier 1 3

Row 1 Row 2

DC

IC Bay nBay 1 Bay 2





as follows dif1



1= (0)(minus1000) + (10)(1) + 1 = 11

and Cr2



119887is



119887are

defined as 119897119887

= 1 minus 1 = 0 and 119906119887

= 1 + 1 = 2 1198771contains




and 1198772

= (1198771




1015840 is determined as 1198771015840


119899= (10)(1) +




119890= 9 + 119899 and 119887

119890=





Tier 2 5

Tier 1 5 5

Row 1 Row 2

Tier 2 1 1

Tier 1 1 1

Row 1 Row 2

Tier 2 4

Tier 1 3

Row 1 Row 2

Bay nBay 1 Bay 2


A

B

C

D

(a)

A

B

C

(b)

A

B

(c)







6 Numerical Results







(18)



119887= 1 2 3 4 The 119877

119887value used



119887= 1 because










3538414447


Instance type 1

35

40

45

50

80 85 90 95 100 105 110 115 120 125

Reha

ndle


Instance type 2

3540455055


Instance type 3

354045505560


Instance type 4

25

30

35

40


Instance type 5

2530354045


Instance type 6

283134374043


Instance type 7

353841444750


Instance type 8

Rb = 1

Rb = 2Rb = 4

Rb = 3

Rb = 2 Rb = 3

Rb = 4

Rb = 1

Rb = 2Rb = 1

Rb = 3

Rb = 1

Rb = 2

Rb = 4

Rb = 3Rb = 4

Rb = 1

Rb = 2

Rb = 3 Rb = 4

Rb = 1

Rb = 2

Rb = 3 Rb = 4

Rb = 1

Rb = 2Rb = 3

Rb = 4

Rb = 1

Rb = 2Rb = 3Rb = 4

mov

emen

ts (

)

Reha

ndle

m

ovem

ents

()

Reha

ndle

m

ovem

ents

()

Reha

ndle

m

ovem

ents

()

Reha

ndle

m

ovem

ents

()

Reha

ndle

m

ovem

ents

()

Reha

ndle

m

ovem

ents

()

Reha

ndle

m

ovem

ents

()




119887=








119887 The








= 3 than with 119877119887










References






























Volume 2014




Journal of











Journal of


Function Spaces






Algebra











Zone 4

Zone 5Quay

Gate

(a)


Zone 4

Zone 5Quay

Gate

(b)




























(119894119895)119868119894119895



(1)




119889(119886119887)







119865119896




Min119861

sum

119894=1

119863119894+

119861

sum

119894=1

119867119894 (2)

st119861

sum

119894=1

119876

sum

119895=1

119910119894119895

= 1 (3)

119861

sum

119894=1

119876

sum

119895=1

119865119896

119894119895= 1 119896 isin 1 2 119864 (4)

119875119894119886

minus 119875119894119887

le 119872 lowast 119889(119886119887)

119894

119894 isin 1 2 119861 (119886 119887) isin COMB

(5)

119863119894= sum

(119886119887)isinCOMB119889(119886119887)

119894119894 isin 1 2 119861 (6)

119875119894119895

= 119910119894119895

lowast 119882IC + 119868119894119895

+ 119872 lowast

119864

sum

119896=1

119865119896

119894119895

119894 isin 1 2 119861 119895 isin 1 2 119876

(7)

119868119894119895

le 119871 lowast 119885119894119895

119894 isin 1 2 119861 119895 isin 1 2 119876

(8)

119910119894119895

+ 119885119894119895

+

119864

sum

119896=1

119865119896

119894119895le 1

119894 isin 1 2 119861 119895 isin 1 2 119876

(9)

119876

sum

119895=1

(119885119894119895

+ 119910119894119895) le 2 lowast 119867

119894119894 isin 1 2 119861 (10)

119910119894119895

isin 0 1 119894 isin 1 2 119861 119895 isin 1 2 119876

(11)

119889(119886119887)


(12)

119863119894ge 0 119894 isin 1 2 119861 (13)

119875119894119895

ge 0 119894 isin 1 2 119861 119895 isin 1 2 119876 (14)

119885119894119895

isin 0 1 119894 isin 1 2 119861 119895 isin 1 2 119876

(15)


119867119894isin 0 1 119894 isin 1 2 119861 (16)

119865119896

119894119895isin 0 1

119894 isin 1 2 119861 119895 isin 1 2 119876

119896 isin 1 2 119864

(17)


(119886119887)

119894to the value













Tier 4 4 8 12 16 20 24

Tier 3 3 7 11 15 19 23

Tier 2 2 6 10 14 18 22

Tier 1 1 5 9 13 17 21










1198772 subset of 119877










Start

(4) Determine

(5) Compute


(19) Update

(21) Compute




bay OB

Yes

Yes

(10)

Yes


Yes

No

Stop

(3) Determine





in the bay be

be = i

R998400= (R

998400FB998400)




|R2| gt 0 |EB| gt 0

R2 = R1(

R998400 = B(R1cup OB)








EB cup FB)




119887


=


119887increases


2which includes











119887= 2minus1 = 1 and 119906

119887= 2+1 = 3 Now


1= bay 1 bay 3


and 1198772





Tier 2

Tier 1 2

Row 1 Row 2

Tier 2 3 2

Tier 1 3 3

Row 1 Row 2

Bay 3Bay 1 Bay 2

Tier 2 2

Tier 1 1

Row 1 Row 2

DC

IC


Tier 2 2

Tier 1 2

Row 1 Row 2

Tier 2 3

Tier 1 3 3

Row 1 Row 2

Bay 3

Tier 2 2

Tier 1 1

Row 1 Row 2

Bay 1 Bay 2


Tier 2 5 4

Tier 1 5 5

Row 1 Row 2

Tier 2 1 1

Tier 1 1 1

Row 1 Row 2

Tier 2

Tier 1 3

Row 1 Row 2

DC

IC Bay nBay 1 Bay 2





as follows dif1



1= (0)(minus1000) + (10)(1) + 1 = 11

and Cr2



119887is



119887are

defined as 119897119887

= 1 minus 1 = 0 and 119906119887

= 1 + 1 = 2 1198771contains




and 1198772

= (1198771




1015840 is determined as 1198771015840


119899= (10)(1) +




119890= 9 + 119899 and 119887

119890=





Tier 2 5

Tier 1 5 5

Row 1 Row 2

Tier 2 1 1

Tier 1 1 1

Row 1 Row 2

Tier 2 4

Tier 1 3

Row 1 Row 2

Bay nBay 1 Bay 2


A

B

C

D

(a)

A

B

C

(b)

A

B

(c)







6 Numerical Results







(18)



119887= 1 2 3 4 The 119877

119887value used



119887= 1 because










3538414447


Instance type 1

35

40

45

50

80 85 90 95 100 105 110 115 120 125

Reha

ndle


Instance type 2

3540455055


Instance type 3

354045505560


Instance type 4

25

30

35

40


Instance type 5

2530354045


Instance type 6

283134374043


Instance type 7

353841444750


Instance type 8

Rb = 1

Rb = 2Rb = 4

Rb = 3

Rb = 2 Rb = 3

Rb = 4

Rb = 1

Rb = 2Rb = 1

Rb = 3

Rb = 1

Rb = 2

Rb = 4

Rb = 3Rb = 4

Rb = 1

Rb = 2

Rb = 3 Rb = 4

Rb = 1

Rb = 2

Rb = 3 Rb = 4

Rb = 1

Rb = 2Rb = 3

Rb = 4

Rb = 1

Rb = 2Rb = 3Rb = 4

mov

emen

ts (

)

Reha

ndle

m

ovem

ents

()

Reha

ndle

m

ovem

ents

()

Reha

ndle

m

ovem

ents

()

Reha

ndle

m

ovem

ents

()

Reha

ndle

m

ovem

ents

()

Reha

ndle

m

ovem

ents

()

Reha

ndle

m

ovem

ents

()




119887=








119887 The








= 3 than with 119877119887










References






























Volume 2014




Journal of











Journal of


Function Spaces






Algebra






















(119894119895)119868119894119895



(1)




119889(119886119887)







119865119896




Min119861

sum

119894=1

119863119894+

119861

sum

119894=1

119867119894 (2)

st119861

sum

119894=1

119876

sum

119895=1

119910119894119895

= 1 (3)

119861

sum

119894=1

119876

sum

119895=1

119865119896

119894119895= 1 119896 isin 1 2 119864 (4)

119875119894119886

minus 119875119894119887

le 119872 lowast 119889(119886119887)

119894

119894 isin 1 2 119861 (119886 119887) isin COMB

(5)

119863119894= sum

(119886119887)isinCOMB119889(119886119887)

119894119894 isin 1 2 119861 (6)

119875119894119895

= 119910119894119895

lowast 119882IC + 119868119894119895

+ 119872 lowast

119864

sum

119896=1

119865119896

119894119895

119894 isin 1 2 119861 119895 isin 1 2 119876

(7)

119868119894119895

le 119871 lowast 119885119894119895

119894 isin 1 2 119861 119895 isin 1 2 119876

(8)

119910119894119895

+ 119885119894119895

+

119864

sum

119896=1

119865119896

119894119895le 1

119894 isin 1 2 119861 119895 isin 1 2 119876

(9)

119876

sum

119895=1

(119885119894119895

+ 119910119894119895) le 2 lowast 119867

119894119894 isin 1 2 119861 (10)

119910119894119895

isin 0 1 119894 isin 1 2 119861 119895 isin 1 2 119876

(11)

119889(119886119887)


(12)

119863119894ge 0 119894 isin 1 2 119861 (13)

119875119894119895

ge 0 119894 isin 1 2 119861 119895 isin 1 2 119876 (14)

119885119894119895

isin 0 1 119894 isin 1 2 119861 119895 isin 1 2 119876

(15)


119867119894isin 0 1 119894 isin 1 2 119861 (16)

119865119896

119894119895isin 0 1

119894 isin 1 2 119861 119895 isin 1 2 119876

119896 isin 1 2 119864

(17)


(119886119887)

119894to the value













Tier 4 4 8 12 16 20 24

Tier 3 3 7 11 15 19 23

Tier 2 2 6 10 14 18 22

Tier 1 1 5 9 13 17 21










1198772 subset of 119877










Start

(4) Determine

(5) Compute


(19) Update

(21) Compute




bay OB

Yes

Yes

(10)

Yes


Yes

No

Stop

(3) Determine





in the bay be

be = i

R998400= (R

998400FB998400)




|R2| gt 0 |EB| gt 0

R2 = R1(

R998400 = B(R1cup OB)








EB cup FB)




119887


=


119887increases


2which includes











119887= 2minus1 = 1 and 119906

119887= 2+1 = 3 Now


1= bay 1 bay 3


and 1198772





Tier 2

Tier 1 2

Row 1 Row 2

Tier 2 3 2

Tier 1 3 3

Row 1 Row 2

Bay 3Bay 1 Bay 2

Tier 2 2

Tier 1 1

Row 1 Row 2

DC

IC


Tier 2 2

Tier 1 2

Row 1 Row 2

Tier 2 3

Tier 1 3 3

Row 1 Row 2

Bay 3

Tier 2 2

Tier 1 1

Row 1 Row 2

Bay 1 Bay 2


Tier 2 5 4

Tier 1 5 5

Row 1 Row 2

Tier 2 1 1

Tier 1 1 1

Row 1 Row 2

Tier 2

Tier 1 3

Row 1 Row 2

DC

IC Bay nBay 1 Bay 2





as follows dif1



1= (0)(minus1000) + (10)(1) + 1 = 11

and Cr2



119887is



119887are

defined as 119897119887

= 1 minus 1 = 0 and 119906119887

= 1 + 1 = 2 1198771contains




and 1198772

= (1198771




1015840 is determined as 1198771015840


119899= (10)(1) +




119890= 9 + 119899 and 119887

119890=





Tier 2 5

Tier 1 5 5

Row 1 Row 2

Tier 2 1 1

Tier 1 1 1

Row 1 Row 2

Tier 2 4

Tier 1 3

Row 1 Row 2

Bay nBay 1 Bay 2


A

B

C

D

(a)

A

B

C

(b)

A

B

(c)







6 Numerical Results







(18)



119887= 1 2 3 4 The 119877

119887value used



119887= 1 because










3538414447


Instance type 1

35

40

45

50

80 85 90 95 100 105 110 115 120 125

Reha

ndle


Instance type 2

3540455055


Instance type 3

354045505560


Instance type 4

25

30

35

40


Instance type 5

2530354045


Instance type 6

283134374043


Instance type 7

353841444750


Instance type 8

Rb = 1

Rb = 2Rb = 4

Rb = 3

Rb = 2 Rb = 3

Rb = 4

Rb = 1

Rb = 2Rb = 1

Rb = 3

Rb = 1

Rb = 2

Rb = 4

Rb = 3Rb = 4

Rb = 1

Rb = 2

Rb = 3 Rb = 4

Rb = 1

Rb = 2

Rb = 3 Rb = 4

Rb = 1

Rb = 2Rb = 3

Rb = 4

Rb = 1

Rb = 2Rb = 3Rb = 4

mov

emen

ts (

)

Reha

ndle

m

ovem

ents

()

Reha

ndle

m

ovem

ents

()

Reha

ndle

m

ovem

ents

()

Reha

ndle

m

ovem

ents

()

Reha

ndle

m

ovem

ents

()

Reha

ndle

m

ovem

ents

()

Reha

ndle

m

ovem

ents

()




119887=








119887 The








= 3 than with 119877119887










References






























Volume 2014




Journal of











Journal of


Function Spaces






Algebra










119867119894isin 0 1 119894 isin 1 2 119861 (16)

119865119896

119894119895isin 0 1

119894 isin 1 2 119861 119895 isin 1 2 119876

119896 isin 1 2 119864

(17)


(119886119887)

119894to the value













Tier 4 4 8 12 16 20 24

Tier 3 3 7 11 15 19 23

Tier 2 2 6 10 14 18 22

Tier 1 1 5 9 13 17 21










1198772 subset of 119877










Start

(4) Determine

(5) Compute


(19) Update

(21) Compute




bay OB

Yes

Yes

(10)

Yes


Yes

No

Stop

(3) Determine





in the bay be

be = i

R998400= (R

998400FB998400)




|R2| gt 0 |EB| gt 0

R2 = R1(

R998400 = B(R1cup OB)








EB cup FB)




119887


=


119887increases


2which includes











119887= 2minus1 = 1 and 119906

119887= 2+1 = 3 Now


1= bay 1 bay 3


and 1198772





Tier 2

Tier 1 2

Row 1 Row 2

Tier 2 3 2

Tier 1 3 3

Row 1 Row 2

Bay 3Bay 1 Bay 2

Tier 2 2

Tier 1 1

Row 1 Row 2

DC

IC


Tier 2 2

Tier 1 2

Row 1 Row 2

Tier 2 3

Tier 1 3 3

Row 1 Row 2

Bay 3

Tier 2 2

Tier 1 1

Row 1 Row 2

Bay 1 Bay 2


Tier 2 5 4

Tier 1 5 5

Row 1 Row 2

Tier 2 1 1

Tier 1 1 1

Row 1 Row 2

Tier 2

Tier 1 3

Row 1 Row 2

DC

IC Bay nBay 1 Bay 2





as follows dif1



1= (0)(minus1000) + (10)(1) + 1 = 11

and Cr2



119887is



119887are

defined as 119897119887

= 1 minus 1 = 0 and 119906119887

= 1 + 1 = 2 1198771contains




and 1198772

= (1198771




1015840 is determined as 1198771015840


119899= (10)(1) +




119890= 9 + 119899 and 119887

119890=





Tier 2 5

Tier 1 5 5

Row 1 Row 2

Tier 2 1 1

Tier 1 1 1

Row 1 Row 2

Tier 2 4

Tier 1 3

Row 1 Row 2

Bay nBay 1 Bay 2


A

B

C

D

(a)

A

B

C

(b)

A

B

(c)







6 Numerical Results







(18)



119887= 1 2 3 4 The 119877

119887value used



119887= 1 because










3538414447


Instance type 1

35

40

45

50

80 85 90 95 100 105 110 115 120 125

Reha

ndle


Instance type 2

3540455055


Instance type 3

354045505560


Instance type 4

25

30

35

40


Instance type 5

2530354045


Instance type 6

283134374043


Instance type 7

353841444750


Instance type 8

Rb = 1

Rb = 2Rb = 4

Rb = 3

Rb = 2 Rb = 3

Rb = 4

Rb = 1

Rb = 2Rb = 1

Rb = 3

Rb = 1

Rb = 2

Rb = 4

Rb = 3Rb = 4

Rb = 1

Rb = 2

Rb = 3 Rb = 4

Rb = 1

Rb = 2

Rb = 3 Rb = 4

Rb = 1

Rb = 2Rb = 3

Rb = 4

Rb = 1

Rb = 2Rb = 3Rb = 4

mov

emen

ts (

)

Reha

ndle

m

ovem

ents

()

Reha

ndle

m

ovem

ents

()

Reha

ndle

m

ovem

ents

()

Reha

ndle

m

ovem

ents

()

Reha

ndle

m

ovem

ents

()

Reha

ndle

m

ovem

ents

()

Reha

ndle

m

ovem

ents

()




119887=








119887 The








= 3 than with 119877119887










References






























Volume 2014




Journal of











Journal of


Function Spaces






Algebra











Start

(4) Determine

(5) Compute


(19) Update

(21) Compute




bay OB

Yes

Yes

(10)

Yes


Yes

No

Stop

(3) Determine





in the bay be

be = i

R998400= (R

998400FB998400)




|R2| gt 0 |EB| gt 0

R2 = R1(

R998400 = B(R1cup OB)








EB cup FB)




119887


=


119887increases


2which includes











119887= 2minus1 = 1 and 119906

119887= 2+1 = 3 Now


1= bay 1 bay 3


and 1198772





Tier 2

Tier 1 2

Row 1 Row 2

Tier 2 3 2

Tier 1 3 3

Row 1 Row 2

Bay 3Bay 1 Bay 2

Tier 2 2

Tier 1 1

Row 1 Row 2

DC

IC


Tier 2 2

Tier 1 2

Row 1 Row 2

Tier 2 3

Tier 1 3 3

Row 1 Row 2

Bay 3

Tier 2 2

Tier 1 1

Row 1 Row 2

Bay 1 Bay 2


Tier 2 5 4

Tier 1 5 5

Row 1 Row 2

Tier 2 1 1

Tier 1 1 1

Row 1 Row 2

Tier 2

Tier 1 3

Row 1 Row 2

DC

IC Bay nBay 1 Bay 2





as follows dif1



1= (0)(minus1000) + (10)(1) + 1 = 11

and Cr2



119887is



119887are

defined as 119897119887

= 1 minus 1 = 0 and 119906119887

= 1 + 1 = 2 1198771contains




and 1198772

= (1198771




1015840 is determined as 1198771015840


119899= (10)(1) +




119890= 9 + 119899 and 119887

119890=





Tier 2 5

Tier 1 5 5

Row 1 Row 2

Tier 2 1 1

Tier 1 1 1

Row 1 Row 2

Tier 2 4

Tier 1 3

Row 1 Row 2

Bay nBay 1 Bay 2


A

B

C

D

(a)

A

B

C

(b)

A

B

(c)







6 Numerical Results







(18)



119887= 1 2 3 4 The 119877

119887value used



119887= 1 because










3538414447


Instance type 1

35

40

45

50

80 85 90 95 100 105 110 115 120 125

Reha

ndle


Instance type 2

3540455055


Instance type 3

354045505560


Instance type 4

25

30

35

40


Instance type 5

2530354045


Instance type 6

283134374043


Instance type 7

353841444750


Instance type 8

Rb = 1

Rb = 2Rb = 4

Rb = 3

Rb = 2 Rb = 3

Rb = 4

Rb = 1

Rb = 2Rb = 1

Rb = 3

Rb = 1

Rb = 2

Rb = 4

Rb = 3Rb = 4

Rb = 1

Rb = 2

Rb = 3 Rb = 4

Rb = 1

Rb = 2

Rb = 3 Rb = 4

Rb = 1

Rb = 2Rb = 3

Rb = 4

Rb = 1

Rb = 2Rb = 3Rb = 4

mov

emen

ts (

)

Reha

ndle

m

ovem

ents

()

Reha

ndle

m

ovem

ents

()

Reha

ndle

m

ovem

ents

()

Reha

ndle

m

ovem

ents

()

Reha

ndle

m

ovem

ents

()

Reha

ndle

m

ovem

ents

()

Reha

ndle

m

ovem

ents

()




119887=








119887 The








= 3 than with 119877119887










References






























Volume 2014




Journal of











Journal of


Function Spaces






Algebra










Tier 2

Tier 1 2

Row 1 Row 2

Tier 2 3 2

Tier 1 3 3

Row 1 Row 2

Bay 3Bay 1 Bay 2

Tier 2 2

Tier 1 1

Row 1 Row 2

DC

IC


Tier 2 2

Tier 1 2

Row 1 Row 2

Tier 2 3

Tier 1 3 3

Row 1 Row 2

Bay 3

Tier 2 2

Tier 1 1

Row 1 Row 2

Bay 1 Bay 2


Tier 2 5 4

Tier 1 5 5

Row 1 Row 2

Tier 2 1 1

Tier 1 1 1

Row 1 Row 2

Tier 2

Tier 1 3

Row 1 Row 2

DC

IC Bay nBay 1 Bay 2





as follows dif1



1= (0)(minus1000) + (10)(1) + 1 = 11

and Cr2



119887is



119887are

defined as 119897119887

= 1 minus 1 = 0 and 119906119887

= 1 + 1 = 2 1198771contains




and 1198772

= (1198771




1015840 is determined as 1198771015840


119899= (10)(1) +




119890= 9 + 119899 and 119887

119890=





Tier 2 5

Tier 1 5 5

Row 1 Row 2

Tier 2 1 1

Tier 1 1 1

Row 1 Row 2

Tier 2 4

Tier 1 3

Row 1 Row 2

Bay nBay 1 Bay 2


A

B

C

D

(a)

A

B

C

(b)

A

B

(c)







6 Numerical Results







(18)



119887= 1 2 3 4 The 119877

119887value used



119887= 1 because










3538414447


Instance type 1

35

40

45

50

80 85 90 95 100 105 110 115 120 125

Reha

ndle


Instance type 2

3540455055


Instance type 3

354045505560


Instance type 4

25

30

35

40


Instance type 5

2530354045


Instance type 6

283134374043


Instance type 7

353841444750


Instance type 8

Rb = 1

Rb = 2Rb = 4

Rb = 3

Rb = 2 Rb = 3

Rb = 4

Rb = 1

Rb = 2Rb = 1

Rb = 3

Rb = 1

Rb = 2

Rb = 4

Rb = 3Rb = 4

Rb = 1

Rb = 2

Rb = 3 Rb = 4

Rb = 1

Rb = 2

Rb = 3 Rb = 4

Rb = 1

Rb = 2Rb = 3

Rb = 4

Rb = 1

Rb = 2Rb = 3Rb = 4

mov

emen

ts (

)

Reha

ndle

m

ovem

ents

()

Reha

ndle

m

ovem

ents

()

Reha

ndle

m

ovem

ents

()

Reha

ndle

m

ovem

ents

()

Reha

ndle

m

ovem

ents

()

Reha

ndle

m

ovem

ents

()

Reha

ndle

m

ovem

ents

()




119887=








119887 The








= 3 than with 119877119887










References






























Volume 2014




Journal of











Journal of


Function Spaces






Algebra










Tier 2 5

Tier 1 5 5

Row 1 Row 2

Tier 2 1 1

Tier 1 1 1

Row 1 Row 2

Tier 2 4

Tier 1 3

Row 1 Row 2

Bay nBay 1 Bay 2


A

B

C

D

(a)

A

B

C

(b)

A

B

(c)







6 Numerical Results







(18)



119887= 1 2 3 4 The 119877

119887value used



119887= 1 because










3538414447


Instance type 1

35

40

45

50

80 85 90 95 100 105 110 115 120 125

Reha

ndle


Instance type 2

3540455055


Instance type 3

354045505560


Instance type 4

25

30

35

40


Instance type 5

2530354045


Instance type 6

283134374043


Instance type 7

353841444750


Instance type 8

Rb = 1

Rb = 2Rb = 4

Rb = 3

Rb = 2 Rb = 3

Rb = 4

Rb = 1

Rb = 2Rb = 1

Rb = 3

Rb = 1

Rb = 2

Rb = 4

Rb = 3Rb = 4

Rb = 1

Rb = 2

Rb = 3 Rb = 4

Rb = 1

Rb = 2

Rb = 3 Rb = 4

Rb = 1

Rb = 2Rb = 3

Rb = 4

Rb = 1

Rb = 2Rb = 3Rb = 4

mov

emen

ts (

)

Reha

ndle

m

ovem

ents

()

Reha

ndle

m

ovem

ents

()

Reha

ndle

m

ovem

ents

()

Reha

ndle

m

ovem

ents

()

Reha

ndle

m

ovem

ents

()

Reha

ndle

m

ovem

ents

()

Reha

ndle

m

ovem

ents

()




119887=








119887 The








= 3 than with 119877119887










References






























Volume 2014




Journal of











Journal of


Function Spaces






Algebra












3538414447


Instance type 1

35

40

45

50

80 85 90 95 100 105 110 115 120 125

Reha

ndle


Instance type 2

3540455055


Instance type 3

354045505560


Instance type 4

25

30

35

40


Instance type 5

2530354045


Instance type 6

283134374043


Instance type 7

353841444750


Instance type 8

Rb = 1

Rb = 2Rb = 4

Rb = 3

Rb = 2 Rb = 3

Rb = 4

Rb = 1

Rb = 2Rb = 1

Rb = 3

Rb = 1

Rb = 2

Rb = 4

Rb = 3Rb = 4

Rb = 1

Rb = 2

Rb = 3 Rb = 4

Rb = 1

Rb = 2

Rb = 3 Rb = 4

Rb = 1

Rb = 2Rb = 3

Rb = 4

Rb = 1

Rb = 2Rb = 3Rb = 4

mov

emen

ts (

)

Reha

ndle

m

ovem

ents

()

Reha

ndle

m

ovem

ents

()

Reha

ndle

m

ovem

ents

()

Reha

ndle

m

ovem

ents

()

Reha

ndle

m

ovem

ents

()

Reha

ndle

m

ovem

ents

()

Reha

ndle

m

ovem

ents

()




119887=








119887 The








= 3 than with 119877119887










References






























Volume 2014




Journal of











Journal of


Function Spaces






Algebra















= 3 than with 119877119887










References






























Volume 2014




Journal of











Journal of


Function Spaces






Algebra






















Volume 2014




Journal of











Journal of


Function Spaces






Algebra









research article a heuristic procedure for the outbound ...research article a heuristic procedure...

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