research article an improved discrete pso for...

Research ArticleAn Improved Discrete PSO for Tugboat Assignment Problemunder a Hybrid Scheduling Rule in Container Terminal

Su Wang1 Min Zhu1 Ikou Kaku2 Guoyue Chen2 and Mingya Wang1

1Computer Center East China Normal University North Zhongshan Road Shanghai 200062 China2Faculty of Systems Science and Technology Akita Prefectural University 84-4 Tsuchiya-Ebinokuti Yulihonjo 015-0055 Japan

Correspondence should be addressed to Su Wang swangccecnueducn

Received 31 July 2014 Accepted 4 December 2014 Published 28 December 2014

Academic Editor Valder Steffen Jr

Copyright copy 2014 Su Wang et alThis is an open access article distributed under the Creative Commons Attribution License whichpermits unrestricted use distribution and reproduction in any medium provided the original work is properly cited

In container terminal tugboat plays vital role in safety of ship docking Tugboat assignment problem under a hybrid scheduling rule(TAP-HSR) is to determine the assignment between multiple tugboats and ships and the scheduling sequence of ships to minimizethe turnaround time of ships A mixed-integer programming model and the scheduling method are described for TAP-HSRproblemThen an improved discrete PSO (IDPSO) algorithm for TAP-HSR problem is proposed to minimize the turnaround timeof ships In particular some new redefined PSO operators and the discrete updating rules of position and velocity are developedThe experimental results show that the proposed IDPSO can get better solutions than GA and basic discrete PSO

1 Introduction

Nowadays more and more people have recognized theimportance of world trade via container terminals Shipdocking is one important operation having an influenceon the productivity of container terminal which includestwo types of operations One is the berth allocation oftendescribed as the problem in which the berthing time and theberthing position are specified to each ship [1] The other isthe tugboat assignment described as a decision problem tofind the best assignment of tugboats and scheduling sequenceof ships which minimizes the turnaround time of shipssubject to the assignment rules Tugboat is one kind of smallships with large engine used tomove ships in and out of berth[2] With the steadily growing size and rapidly increasingnumber of entry ships safe and efficient docking operationof ships is one of the most important issues in containerterminals [3] Because container terminal has finite tugboatswith different service abilities it is very important to developeffective tugboat assignment method to reduce turnaroundtime of ships and maintain high productivity of containerterminal

When a ship arrives at the container terminal the shipshould be assigned an available berth for unloading and

loading containers However ships cannot dock by them-selves to the berths because the port lane is narrow andthe port water is shallow The specified equipment calledtugboat is used to help ships dock safely Due to the strongstructure and large power tugboat can pull large-sized shipsIf a ship is so large that one tugboat cannot tug two or moretugboats can be used together to tow the ship It is consideredthat only with an available berth and proper tugboats aship can dock safely and prepare for unloading or loadingoperation Otherwise it has to wait at the anchorage outsidethe container terminal

Several studies have been conducted for equipmentassignment or scheduling problem in container terminalssuch as berth assignment [4 5] quay crane scheduling[6ndash8] yard crane scheduling [9 10] truck scheduling [11]and storage space assignment [12] However little researchwork on tugboat assignment problem can be found Mori[3] outlined tugboat development situation and tugboatbusiness in Asian countries especially in Japan Liu andWang introduced tugboat allocation simulation [13] and thenLiu studied the tugboat scheduling problem and employeda hybrid evolutionary algorithm to solve this problem [14]However it only focused on the tugboat scheduling ruleswithout mathematical model Furthermore the study did

Hindawi Publishing CorporationMathematical Problems in EngineeringVolume 2014 Article ID 714832 10 pageshttpdxdoiorg1011552014714832

2 Mathematical Problems in Engineering

not cover the optimization of the assignment plan betweentugboats and ships

Tugboat assignment is a typical kind of assignmentoptimization problem The existing approaches for solvingthe assignment problems can be divided into two categoriesThe traditional methods are mathematical programmingapproaches using Branch and Bound [15] and LagrangianRelaxation [16] to get the best solution efficiently but onlyfor small-sized problem Besides metaheuristic algorithmsinvolving genetic algorithms (GA) [17ndash19] ant colony opti-mization [20 21] simulated annealing (SA) [22] Tabu search(TS) [23 24] and variable neighborhood search (VNS) [25]are fast and efficient algorithms to get approximate solutionswith reasonable computational time for solving large scaleand more difficult combination optimization problem

Recently particle swarm optimization (PSO) is a novelevolutionary algorithm presented by Kennedy and Eberhart[26] Due to easy implementation and fast convergence PSOhas proved to be an effective and competitive algorithm fora wide variety of optimization problems However most ofthe published studies have concentrated on the continuousoptimization problems Recently some work has been doneto the discrete optimization problem In order to extend theapplication to discrete space Kennedy and Eberhart [27]proposed a discrete binary version of PSO where a particlemoved in a state space restricted to zero and one on eachdimension Yin et al [28] embedded a hill-climbing heuristicin PSO to assign application tasks to processors such thatthe resource demand of each task was satisfied and thesystem throughput was increased Lin et al [29] implementeda hybrid PSO (HPSO) to solve the biobjective personnelassignment problem The HPSO algorithm was combinedwith the random-key (RK) encoding scheme and individualenhancement (IE) scheme Similar hybrid PSO algorithmsare proposed by introducing operations of GAs into PSOsystems by Robinson et al [30] Lian et al [31] proposedsome new valid PSO operators for permutation job-shopscheduling to minimize makespan Prescilla and ImmanuelSelvakumar [32] applied the modified binary particle swarmoptimization algorithm to solve the real-time task assignmentin heterogeneous multiprocessor

To the best of our knowledge there is no publishedwork for dealing with tugboat assignment using PSO In thispaper we introduce the tugboat assignment problem under ahybrid scheduling rule (TAP-HSR) and give the mathemat-ical model Then an improved discrete PSO algorithm forTAP-HSR problem is proposed to minimize the turnaroundtime of ships In particular a newmethod for redefining PSOoperators is developed The experimental results show thatthe proposed method is effective and efficient compared toGA

The rest of this paper is organized as follows In thenext section tugboat assignment problem under a hybridscheduling rule (TAP-HSR) and the mathematical modelare described Section 3 proposes the improved discretePSO (IDPSO) for TAP-HSR problem Section 4 discussesthe experimental results Finally Section 5 provides someconclusions and the future work

2 Problem Formulation and Scheduling Rules

21 Problem Description The tugboat assignment problemunder a hybrid scheduling rule is an essential decision-making problem in container terminal The tugboat assign-ment operation has two procedures One is to find theassignment plan under the assignment rules The other isto generate the scheduling sequence of ships based on theassignment plan and some scheduling rules

Container terminal assigns tugboats to ships based onsome assignment rules Firstly each tugboat can serve atmostone ship and each ship can be assigned one or more shipsat any time Secondly the assigned number of tugboats isdetermined by ship length The ship with length equal to orless than 100 meters needs one tugboat at least Otherwisethe ship with the length more than 100 meters needs twotugboats at least Finally it is required to assign tugboat withproper horsepower (ps) Horsepower is a common unit tomeasure tugboat powerThe tugboat with higher horsepowerhas more service capacity to tug longer ship The assignmentrules between ships and tugboats are shown in Table 1

Assignment plan only describes the matching relation-ship of multiple tugboats and multiple ships It is requiredto use some scheduling rules to determine a schedulingsequence of ships according to the assignment plan Thispaper applies a hybrid scheduling rule combined with first-come-first-served (FCFS) rule and first-fit rule Based on thehybrid scheduling rule the ship with earlier arriving timeand the available assigned tugboats has the higher priority ofdocking

The goal of TAP-HSR is to minimize the turnaroundtime of ships under the premise that all ships finish dockingWith more and more ships arriving it is impossible tomake sure that ships can have available tugboats for dockingimmediately at arrival Some ships will wait outside theterminal at the anchorage until available tugboats can workfor them Good assignment can decrease the waiting time ofships so that the container terminal can serve more ships andget more profits

22 Model Formulation In this paper TAP-HSR is based onthe following assumptions

(1) The expected arriving time and the length of all shipsare known

(2) The docking operation time of ship depends on thelength of ship and is independent of tugboat

(3) Each ship has an available berth for unloading andloading containers

(4) Tugboat can work for only one ship at the same time(5) The preparation time of tugboat is zero which means

one tugboat can start to serve the next ship immedi-ately after it finishes the previous ship

(6) All tugboats are available when the first ship starts toberth

(7) The operation of tugboat requires an uninterruptedperiod of time

Mathematical Problems in Engineering 3

Table 1 Assignment rules between ships and tugboats

Length of ship (m) Required horsepower (ps) of tugboats Required number of tugboats Docking operation time (min)0ndash100 ⩾2600 ⩾1 40100ndash200 ⩾5200 ⩾2 48200ndash250 ⩾6400 ⩾2 60250ndash300 ⩾6800 ⩾2 75⩾300 ⩾8000 ⩾2 85

This paper describes TAP-HSR problem with the follow-ing notations

119898 the number of tugboats119899 the number of ships119869119895 the ship 119895 119895 isin 1 2 3 119899

ℎ119894 the horsepower of tugboat 119894119901119895 the state of tugboat 119894 (119901119895 = 0 denotes tugboat 119894 isavailable Otherwise 119901119895 = 1)119902119895 the state of ship 119895 (119902119895 = 1 denotes ship 119895 finisheddocking Otherwise 119902119895 = 0)119900119895 the scheduling order of ship 119895

119897119895 the length of ship 119895

ℎ119894 the horsepower of tugboat 119894 119894 isin 1 2 3 119898

119905119895 the operation time of ship 119895

119868119895 the arriving time of ship 119895

119878119895 the start docking time of ship 119895

119879119895 the completion of docking time of ship 119895

119867119895 the least required horsepower of ship 119895

119877119895 the least required tugboat number of ship 119895

119909119894119895 decision variable denotes tugboat 119894 assigned toship 119895

This paper gives the following definitions

Definition 1 (assignment matrix) 119883119898119899 is the decision matrixwhich describes the assignment between ships and tugboatsThe decision variable 119909119894119895 isin 0 1 denotes whether or nottugboat 119894 is assigned to ship 119895

119883119898119899 = (

11990911 11990912 sdot sdot sdot 1199091119899

11990921 11990922 sdot sdot sdot 1199092119899

sdot sdot sdot

1199091198981 1199091198982 sdot sdot sdot 119909119898119899

)

119909119894119895 = 1 Tugboat 119894 is assigned to ship 119895

0 Otherwise

(1)

Definition 2 (conflict ship) If tugboat 119894 is assigned to bothship 119895 and ship 119896 and ship 119895 is scheduled for docking beforeship 119896 ship 119895 is called the conflict ship of ship 119896 Let119862119896 denotethe conflict ship set of ship 119896

119862119896 = 119869119895 | 119909119894119895 sdot 119909119894119896 = 1 119900119895 lt 119900119896 (2)

119862119896 = represents that all of the assigned tugboats toship 119896 are available when ship 119896 arrives In that condition theship can dock immediately at arriving time Otherwise if theassigned tugboats are working for the conflict ships of ship 119896ship 119896 has to wait until all the conflict ships finish docking

Let 119879119878 denote the start docking time of the first ship andlet 119879119865 denote the completion docking time of the last shipThe objective function of TAP-HSR problem is to minimizethe turnaround time of ships described as

min (119879119865 minus 119879119878) (3)

where 119879119878 = 0 according to assumption (6) and 119879119865 is themaximum completion docking time of ships defined as

119879119865 =119899max119895=1

119879119895 (4)

The objective function is min119879119865 represented by thefollowing formula

min 119899max119895=1

119879119895 (5)

To find the maximum completion docking time of shipsit is required to calculate the completion docking time of eachship 119879119895 (119895 = 1 2 3 119899) represented as the start dockingtime of ship 119878119895 plus the operation time of ship 119905119895

119879119895 = 119878119895 + 119905119895 (6)

In TAP-HSR problem let 119869119895 denote ship 119895 According toDefinition 2 the maximum completion time of the conflictships of 119869119895 is max119869119896isin119862119895119879119896 The start docking time 119878119895 iscalculated by

119878119895 = max119868119895max119869119896isin119862119895

119879119896 (7)

Equation (7) states that if ship arrives after the timewhen all the conflict ships have done the docking operationthe start docking time is the arriving time Otherwise thestart docking time of this ship is the maximum completiondocking time of the conflict ships

The constraints in TAP-HSR problem are as follows119898

sum

119894=1

119909119894119895 ge 119877119895

119898

sum

119894=1

119909119894119895ℎ119894 ge 119867119895

(8)


While (119896 lt 119899)Set the state of each tugboat 119901119894 = 0For 119895 = 1 to 119899

If (119902119895 = 0 and the state of each assigned tugboat 119901119894 = 0)119900119895 = 119900 flagSet the state of each assigned tugboat 119901119894 = 1Calculate the conflict ship set 119888119895Calculate the start docking time 119904119895 according to (7)Calculate the completion docking time 119879119895 according to (6)Set the state of ship 119895 119902119895 = 1

119896 = 119896 + 1End if

Next 119895119900 flag = 119900 flag + 1

EndCalculate the turnaround time of ships according to (4)

Pseudocode 1

Constraints in (8) ensure that the total number andhorsepower of tugboats assigned to each ship shouldmeet theassignment rules to make sure that the tugboats have enoughhorsepower to tug the ship

23 The Scheduling Method The assignment matrix onlydescribes the assignment plan between ships and tugboatsIt is required to generate the scheduling sequence of shipsbased on the assignment matrix and scheduling rules Thescheduling sequence is generated as in the following steps

Step 1 Order the ships by the arriving time Initialize 119901119894 = 0119902119895 = 0 the number of finished ships 119896 = 0 and the initialscheduling order 119900 flag = 1

Step 2 Check the assignment matrix whether it is a feasiblesolution which satisfies constraints in (8) If it is a feasiblesolution go to Step 3

Step 3 Find the conflict ships of each ship and generatethe scheduling sequence based on first-fit rule The pseu-docode of generating scheduling sequence is described inPseudocode 1

Step 4 Output the scheduling sequence the start dockingtime and completion docking time of each ship and theturnaround time of ships

We give one example of scheduling ships based on assign-ment plan Table 2 shows the tugboat allocation Table 3shows the arriving ships

One assignment matrix between tugboats and ships is asfollows

119883119894 =(

0

1

0

0

0

1

1

0

0

0

1

1

1

0

0

0

1

0

0

1

)

1198691 1198692 1198693 1198694 1198695

(9)

Table 2 One example of tugboat allocation

Tugboat number Horsepower (ps) Number1 2600 12 3200 13 3400 14 4000 1

Table 3 One example of arriving ships

Ship number Length (m) Arriving time(min)

Operation time(min)

1 87 0 402 245 5 603 286 10 754 93 15 405 202 20 60

119883119894 satisfies constraints in (8) so that it is a feasible assign-ment According to the scheduling method the schedulingsequence 119874 = (1198691 1198693 1198694 1198692 1198695) and turnaround time of shipsis 135 minutes The scheduling Gant graph is as in Figure 1

3 Discrete PSO Algorithm forTAP-HSR Problem

TAP-HSR is in one kind of NP-hard problem and difficult forhuman schedulers The goal of our research is to achieve asgood as possible results in reasonable computational time

31 Standard PSO Algorithm Swarm optimization algorithmis a swarm intelligence algorithm proposed by Kennedy andEberhart [26] as a simulation of the social behavior of birdflocking PSO is composed of a number of individuals calledparticles Particles are initialized with random locations andvelocities Each particle tries to improve the location by


M2 40min

M2 M3 60min

M3 M4 75min

M1 40min

0 135

M1 M4 60min

(min)

1

2

3

4

5

Ship

num

ber

Figure 1 The scheduling Gant graph

following the best location of itself and the best particle ofthe whole swarm The position of 119894th particle is representedby 119872-dimensional vector denoted by 119883119894 = (1199091198941 1199091198942 119909119894119872)

(119894 = 1 2 119896) The velocity for the 119894th particle can bedescribed as 119881119894 = (V1198941 V1198942 V119894119872) (119894 = 1 2 119896)

A particlersquos movement is based on (10)

V119905+1119894 = 120596 times V119905119894 + 1198881 times rand1 ( )

times (119901119905119894 minus 119909119905119894) + 1198882 times rand2 ( ) times (119901

119905119892 minus 119909119905119894)

(10)

119909119905+1119894 = 119909

119905119894 + V119905+1119894 (11)

where120596 is called inertia weight [33] 1198881 and 1198882 are two constantnumbers which are often called social or cognitive confidencecoefficients the functions rand1( ) and rand2( ) can generatea random number in [0 1] 119909119905119894 is the position of 119894th particle atthe iteration 119905 V119905119894 is the velocity of 119894th particle at the iteration119905 119901119905119894 is the local best position that 119894th particle has reached atthe iteration 119905 and 119901

119905119892 is the global best position that all the

particles have reached at the iteration 119905

32 The Improved Discrete PSO Because of the continuousposition space of particles standard encoding method ofPSO cannot be directly adopted for discrete optimizationproblems It is a challenge to employ the algorithm in com-binatorial optimization problems especially for assignmentproblem In this paper we exploit a new redefining PSOoperator method and discrete updating rules of position andvelocity to get the solutions for TAP-HSR problem

321 Discrete Particle Representation The most importantproblem in applying PSO to TAP-HSR problem is to find asuitable particle representation method In assignment prob-lem each particle represents a candidate assignment matrixwith119898 times 119899 elements [34] Each element is an integer value in0 1 For example in (12) 119883119894 denotes one assignment thatassigns four tugboats to five ships Tugboat 1 is assigned toship 1 Tugboats 2 and 3 are assigned to ship 2 Tugboats 1and 2 are assigned to ship 3 Tugboat 4 is assigned to ship 4Tugboats 3 and 4 are assigned to ship 5

One example of particle represented by assignmentmatrix is as follows

119883119894 = (

1 0 1 0 0

0 1 1 0 0

0 1 0 0 1

0 0 0 1 1

) (12)

322 Operators Redefinition

Definition 3 The position subtraction operator 119883119894 ⊟ 119883119895 thesubtraction of two positions can be defined as XOR (exclusiveOR) operation of binary matrix

119883119894 ⊟ 119883119895 = (not119883119894 and 119883119895) or (119883119894 and not119883119895) (13)

Equation (14) gives one example of the subtraction ofpositions 119883119905119894 and 119883

119905+1119894

119883119905119894 = (

1 0 1 0 0

0 1 1 0 0

0 1 0 0 1

0 0 0 1 1

)

119883119905+1119894 = (

1 0 1 0 0

0 1 0 0 0

0 0 1 0 1

0 1 0 1 1

)

119883119905+1119894 ⊟ 119883

119905119894 = (

0 0 0 0 0

0 0 1 0 0

0 1 1 0 0

0 1 0 0 0

)

(14)

It can be seen that the subtraction of positions is to findthe different elements between two positions [34] Accordingto (11) the velocity is the subtraction of positions that can berepresented by a set of the numerical pairs (V119895)length denotesthe element number of V119895

V119895 = (119909 119910) | 119909 isin 1 2 119872 119910 isin 1 2 119873 (15)

In (12) V119905+1119894 = 119883119905+1119894 ⊟ 119883

119905119894 = (2 3) (3 2) (3 3) (4 2)

(V119905+1119894 )length = 4

Definition 4 The velocity addition operator V119894 ⊞ V119895 theaddition of velocities is considered as a union of the elementsin V119894 or in V119895

V119894 ⊞ V119895 = (119909 119910) | (119909 119910) isin V119894 or (119909 119910) isin V119895 (16)

Definition 5 Thevelocity subtraction operator V119894⊟V119895 the sub-traction of velocities is considered as the relative complementelements of V119895 with respect to V119894 V119894 ⊟ V119895 is denoted by the setof elements in V119894 but not in V119895

V119894 ⊟ V119895 = (119909 119910) | (119909 119910) isin V119894 (119909 119910) notin V119895 (17)


Table 4 Examples of 119888 ⊠ V2 and V1 ⊞ (119888 ⊠ V2)

119888 119886 119887 119888 ⊠ V2 V1 ⊞ (119888 ⊠ V2)

119888 isin (0 05]

1 3 1003817100381710038171003817V21003817100381710038171003817

3

1= (2 3) (3 2) (3 3) V1 ⊞

1003817100381710038171003817V21003817100381710038171003817

3

1= (1 5) (2 3) (3 2) (3 3)

2 3 1003817100381710038171003817V21003817100381710038171003817

3

2= (3 2) (3 3) V1 ⊞

1003817100381710038171003817V21003817100381710038171003817

3

2= (1 5) (2 3) (3 2) (3 3)

4 4 1003817100381710038171003817V21003817100381710038171003817

4

4= (4 2) V1 ⊞

1003817100381710038171003817V21003817100381710038171003817

4

4= (1 5) (3 2) (2 3) (4 2)

119888 isin (05 1)

1 3 ⊟1003817100381710038171003817V2

1003817100381710038171003817

3

1= (2 3) (3 2) (3 3) V1 ⊟

1003817100381710038171003817V21003817100381710038171003817

3

1= (1 5)

2 3 ⊟1003817100381710038171003817V2

1003817100381710038171003817

3

2= (3 2) (3 3) V1 ⊟

1003817100381710038171003817V21003817100381710038171003817

3

2= (1 5) (2 3)

4 4 ⊟1003817100381710038171003817V2

1003817100381710038171003817

4

4= (4 2) V1 ⊟

1003817100381710038171003817V21003817100381710038171003817

4

4= (1 5) (3 2) (2 3)

One example of the addition and subtraction of twovelocities is as follows

V119894 = (2 3) (3 2) (3 3) (4 2)

V119895 = (1 5) (3 2)

V119894 ⊞ V119895 = (2 3) (3 2) (3 3) (4 2) (1 5)

V119894 ⊟ V119895 = (2 3) (3 3) (4 2)

(18)

Definition 6 Themultiply operator of coefficient and velocity119888 ⊠ V let 119888 be coefficient and let V be velocity 119888 is a randomreal number in (0 1)

119888 ⊠ V =119888

1003817100381710038171003817V21003817100381710038171003817

119887

119886=

V119887119886 119888 isin (0 05]

⊟ V119887119886 119888 isin (05 1) (19)

where 119886 and 119887 are random integers and 119886 119887 isin [1 (V)length]V119887119886 represents a subelement string of V2 from the 119886th elementto the 119887th element in (20) If 119886 = 1 119887 = (V)length V

(V)length1 = V

Examples of V119887119886 are as follows

V (2 3) (3 2) (3 3) (4 2)

V31 |1(2 3) (3 2) (3 3) |

3(4 2)

(2 3) (3 2) (3 3)

V32 (2 3) |2(3 2) (3 3) |

3(4 2)

(3 2) (3 3)

V41 |1(2 3) (3 2) (3 3) (4 2) |

4

(2 3) (3 2) (3 3) (4 2)

(20)

Based on (19) V1 ⊞ (119888 ⊠ V2) can be represented by

V1 ⊞ (119888 ⊠ V2) = V1 ⊞

1003817100381710038171003817V21003817100381710038171003817

119887

119886119888 isin (0 05]

V1⊟1003817100381710038171003817V2

1003817100381710038171003817

119887

119886119888 isin (05 1)

(21)

For illustration V1 = (1 5) (3 2) (2 3) and V2 =

(2 3) (3 2) (3 3) (4 2) Table 4 shows some examples of119888 ⊠ V2 and V1 ⊞ (119888 ⊠ V2)

From Table 4 it is known that if 119888 isin (0 05] V1 ⊞ (119888 ⊠

V2) can be added to some new elements If 119888 isin (05 1) some

elements might be deleted from V2This method can keep thediversity of the swarm and exploit the global search ability ofPSO thoroughly

Definition 7 The multiply operator of inertia weight andvelocity 120596 ⊠ V let 120596 = 1 and let 120596 ⊠ V = V

Based on Definitions 3ndash7 the new discrete movements ofparticles are followed by the following equations

V119905+1119894 = V119905119894 ⊞ 119888110038171003817100381710038171003817119901119905119894 ⊟ 119909119905119894

10038171003817100381710038171003817

1198871

1198861⊞1198882

10038171003817100381710038171003817119901119905119892 ⊟ 119909119905119894

10038171003817100381710038171003817

1198872

1198862(22)

119909119905+1119894 = 119909

119905119894 ⊞ V119905+1119894 (23)

323The Procedure of IDPSO for TAP-HSR Thepseudocodeof the improved discrete PSO is listed in Pseudocode 2

4 Experimental Results

In this section we present the computational experimentsto evaluate the performance of the improved discrete PSOalgorithm in this paper The algorithms are coded in Javalanguage with the environment of Eclipse and simulatedon a PC with a 280GHz Intel CPU and 4G RAM In theexperiments three algorithms are implemented for TAP-HSRproblems including GA the basic discrete PSO (BDPSO)and the improved discrete PSO (IDPSO) in Section 3 Thebasic discrete PSO is implemented without the multipleoperator of coefficient and velocity 119888 ⊠ V The movements ofparticles are followed by the following equations

V119905+1119894 = V119905119894 ⊞ (119901119905119894 ⊟ 119909119905119894) ⊞ (119901

119905119892 ⊟ 119909119905119894)

119909119905+1119894 = 119909

119905119894 ⊞ V119905+1119894

(24)

In IDPSO there are two important parameters the maxiteration119873119862max and swarm size119870Through the experimentsit can be seen that when themax iteration is fixed the optimalsolution is improved with the growth of swarm size becausemore particles have stronger space search capabilities How-ever if swarm size is excessive the optimal solution qualitycan be decreased so that it is required to increase the maxiteration to get a better solution The reason is that too manyparticles cause the solution dispersion so that it is difficultto converge to an optimal solution On the other hand theprogram running timewill increase with the increasing of themax iteration and swarm size Therefore it is important toselect the proper max iteration and swarm size according tothe problem size shown in Table 5


Step 1 (Initializing)Initialize the parameters the max iteration 119873119862max the no-improvement iteration 119866119862max swarm size 119870each particlersquos location and velocity the local best location 119901119894 and the global best location 119901119892Step 2 (Updating and Fitness Calculation)Set the iteration number t = 1Repeat

For each particle generate random parameter 1198881 1198882 1198861 1198862 1198871 1198872Update the velocity of each particle according to (22)Update the position of each particle according to (23)Check the new position 119883

119905+1119894 is a feasible assignment according to (8)

If (119883119905+1119894 is not a feasible assignment) thenMutation 119883

119905+1119894 to a feasible assignment

End ifGenerate the scheduling sequenceCalculate the turnaround time as fitness of 119883119905+1119894Update 119901119894 119901119892Check the improvement of 119901119892If 119901119892 is improved then

119888 flag = 0Else

119888 flag = 119888 flag + 1End if119905 = 119905 + 1Untill 119905 gt 119873119862max or c flag = 119866119862max

Step 3 (Final stage)Output the best tugboat assignment the ship scheduling sequence the start docking time and thecompletion docking time of eachship the turnaround time of the best assignment

Pseudocode 2

Table 5 The max iteration 119873119862max and swarm size 119870

Problem size 119873119862max 119870

Small-sized problem 1000ndash2500 50ndash150Middle-sized problem 2500ndash3500 150ndash250

To compare the performance of IDPSO BDPSO andGA the experiments consisted of 12 small-sized problemsand 10 middle-sized problems that are conducted In bothIDPSO andBDPSOwe use a swarmof 100 particles for small-sized experiments The max iteration is 2000 and the no-improvement iteration is 1000 In middle-sized experimentwe use a swarm of 200 particles The max iteration is 3000and the no-improvement iteration is 1500 In GA populationsize is 100Themax iteration is 3000 and the no-improvementiteration is 1500 The probability of crossover is 08 Theprobability of mutation is 001 For the three algorithmstwo stopping rules are adopted One is max-iteration ruleand the other is no-improvement rule [34] The comparativeexperimental results obtained by the IDPSO BDPSO andGA on the different size problems are shown in Tables 6 and7 Since PSO and GA algorithms are stochastic algorithmseach separate run of the program could result in a differentresult We run each algorithm 20 times for each problemcaseThe values of119891119887 min represent theminimum turnaroundtime of 20 runs The values of 119891avg min are the average

turnaround time of the 20 runs The values of mean time arethe computational time of average computational time

Through this comparison it is observed that the qual-ity of solutions obtained by IDPSO is always better thanthat obtained using BDPSO and GA in terms of the bestturnaround time and the average turnaround time Further-more the improvement of solutions actually becomes greateras the size of the problems increases The results obtainedby GA are bit better than BDPSO in most of the instancesThe proposed IDPSO can obtain the optimal or near-optimalsolutions in all cases

As far as computational efficiency is concerned it is worthmentioning that GA requires less time when compared toBDPSO and IDPSO On the other hand IDPSO reports solu-tions for all problem sizes within reasonable time ThereforeIDPSO is recommended for use in TAP-HSR problem asit yields satisfactory results with reasonable computationaltime All these experiments showed that the improved dis-crete PSO in this paper is an effective mechanism for solvingthe tugboat assignment in container terminals

5 Conclusions

Although there is a huge amount of literature on discretePSO the PSO algorithm for tugboat assignment problemdoes not have rich literature In this paper we present amathematic model of tugboat assignment problem under ahybrid scheduling rule to minimize the turnaround time of


Table 6 The compared results for small-sized problem

Problem size Tugboats Algorithm 119891119887 min 119891avg min Mean time(ship times tugboat) 2600 3200 3400 4000 (min) (min) (s)

5 times 4 1 1 1 1IDPSO 135 135 2BDPSO 175 183 3GA 135 159 2












ships Moreover the improved discrete PSO is proposed toget solutions with good quality in a reasonable computationaltime Some new redefined PSO operators and discrete updat-ing rules of position and velocity are described in detail Theproposed IDPSO algorithm in this paper can be consideredas more effective approach for TAP-HSR problem comparedwith GA

The proposed IDPSO in this paper for small-sized andmiddle-sized TAP-HSR is very effective but for large-size it

is not approving It needs to be improved in the future worksuch as combining heuristic rule with IDPSO Furthermoreapplying IDPSO to other discrete combinatorial optimizationproblems is also possible in further research

Conflict of Interests

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


Table 7 The compared results for middle-sized problem












Acknowledgment

This work is supported by the National High TechnologyResearch and Development Program of China (863 Programno 2013AA01A211)

References

[1] E Nishimura A Imai and S Papadimitriou ldquoBerth allocationplanning in the public berth system by genetic algorithmsrdquoEuropean Journal of Operational Research vol 131 no 2 pp282ndash292 2001

[2] httpwwwsearchcomreferenceTUGb

[3] T Mori ldquoThe present situation and the issues on tugboat busi-ness in Japanrdquo httpwwwh2dionnejpsimt-morironbunhtml

[4] C Bierwirth and F Meisel ldquoA survey of berth allocationand quay crane scheduling problems in container terminalsrdquoEuropean Journal of Operational Research vol 202 no 3 pp615ndash627 2010

[5] M H Elwany I Ali and Y Abouelseoud ldquoA heuristics-based solution to the continuous berth allocation and craneassignment problemrdquo Alexandria Engineering Journal vol 52no 4 pp 671ndash677 2013

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

[7] Y-M Fu A Diabat and I-T Tsaim ldquoA multi-vessel quay craneassignment and scheduling problem formulation and heuristicsolution approachrdquoExpert Systemswith Applications vol 41 no15 pp 6959ndash6965 2014

[8] A Diabat and E Theodorou ldquoAn integrated quay craneassignment and scheduling problemrdquo Computers and IndustrialEngineering vol 73 pp 115ndash123 2014

[9] W C Ng ldquoCrane scheduling in container yards with inter-crane interferencerdquo European Journal of Operational Researchvol 164 no 1 pp 64ndash78 2005

[10] M E H Petering and K G Murty ldquoEffect of block lengthand yard crane deployment systems on overall performanceat a seaport container transshipment terminalrdquo Computers ampOperations Research vol 36 no 5 pp 1711ndash1725 2009


[11] V D Nguyen and K H Kim ldquoA dispatching method for auto-mated lifting vehicles in automated port container terminalsrdquoComputers amp Industrial Engineering vol 56 no 3 pp 1002ndash1020 2009

[12] M Bazzazi N Safaei and N Javadian ldquoA genetic algorithmto solve the storage space allocation problem in a containerterminalrdquoComputers amp Industrial Engineering vol 56 no 1 pp44ndash52 2009

[13] Z-X Liu and S-M Wang ldquoThe computer simulation study ofport tugboat operationrdquo Journal of System Simulation vol 16no 1 pp 45ndash48 2004

[14] Z-X Liu ldquoHybrid evolutionary strategy optimization forport tugboat operation schedulingrdquo in Proceedings of the 3rdInternational Symposium on Intelligent Information TechnologyApplication (IITA rsquo09) pp 511ndash515 November 2009

[15] W C Ng and K L Mak ldquoYard crane scheduling in portcontainer terminalsrdquo Applied Mathematical Modelling vol 29no 3 pp 263ndash276 2005

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

[17] F Li L D Xu C Jin and H Wang ldquoRandom assignmentmethod based on genetic algorithms and its application inresource allocationrdquo Expert Systems with Applications vol 39no 15 pp 12213ndash12219 2012

[18] R K Ahuja J B Orlin and A Tiwari ldquoA greedy geneticalgorithm for the quadratic assignment problemrdquo Computersand Operations Research vol 27 no 10 pp 917ndash934 2000

[19] U Tosun ldquoA new recombination operator for the genetic algo-rithm solution of the quadratic assignment problemrdquo ProcediaComputer Science vol 32 pp 29ndash36 2014

[20] N CDemirel andMDToksarı ldquoOptimization of the quadraticassignment problem using an ant colony algorithmrdquo AppliedMathematics and Computation vol 183 no 1 pp 427ndash4352006

[21] C Ozkale and A Fıglalı ldquoEvaluation of the multiobjectiveant colony algorithm performances on biobjective quadraticassignment problemsrdquo Applied Mathematical Modelling vol 37no 14-15 pp 7822ndash7838 2013

[22] Y Hamam andK S Hindi ldquoAssignment of programmodules toprocessors a simulated annealing approachrdquo European Journalof Operational Research vol 122 no 2 pp 509ndash513 2000

[23] M S Hussin and T Stutzle ldquoTabu search versus simulatedannealing as a function of the size of quadratic assignmentproblem instancesrdquo Computers amp Operations Research vol 43pp 286ndash291 2014

[24] J Skorin-kapov ldquoTabu search applied to quadratic assignmentproblemrdquo ORSA Journal on Computing vol 2 no 1 pp 33ndash401990

[25] Y Kuo ldquoOptimizing truck sequencing and truck dock assign-ment in a cross docking systemrdquo Expert Systems with Applica-tions vol 40 no 14 pp 5532ndash5541 2013

[26] J Kennedy and R Eberhart ldquoParticle swarm optimizationrdquoin Proceedings of the IEEE International Conference on NeuralNetworks pp 1942ndash1948 December 1995

[27] J Kennedy and R C Eberhart ldquoA discrete binary versionof the particle swarm algorithmrdquo in Proceedings of the IEEEInternational Conference on Systems Man and Cybernetics vol5 pp 4104ndash4108 Orlando Fla USA October 1997

[28] P-Y Yin S-S Yu P-PWang and Y-TWang ldquoA hybrid particleswarm optimization algorithm for optimal task assignment in

distributed systemsrdquo Computer Standards amp Interfaces vol 28no 4 pp 441ndash450 2006

[29] S-Y Lin S-J Horng T-W Kao et al ldquoAn efficient bi-objectivepersonnel assignment algorithm based on a hybrid particleswarm optimization modelrdquo Expert Systems with Applicationsvol 37 no 12 pp 7825ndash7830 2010

[30] J Robinson S Sinton and Y Rahmat-Samii ldquoParticle swarmgenetic algorithm and their hybrids optimization of a profiledcorrugated horn antennardquo in Proceedings of the IEEE Antennasand Propagation Society International Symposium and URSINational Radio Science Conference pp 168ndash175 June 2002

[31] Z Lian B Jiao and X Gu ldquoA similar particle swarm optimiza-tion algorithm for job-shop scheduling tominimizemakespanrdquoAppliedMathematics and Computation vol 183 no 2 pp 1008ndash1017 2006

[32] K Prescilla and A Immanuel Selvakumar ldquoModified binaryparticle swarm optimization algorithm application to real-timetask assignment in heterogeneousmultiprocessorrdquoMicroproces-sors and Microsystems vol 37 no 6-7 pp 583ndash589 2013

[33] Y Shi and R Eberhart ldquoModified particle swarm optimizerrdquoin Proceedings of the IEEE World Congress on ComputationalIntelligence Conference pp 69ndash73 May 1998

[34] Y Han J Tang I Kaku and LMu ldquoSolving uncapacitatedmul-tilevel lot-sizing problems using a particle swarm optimizationwith flexible inertial weightrdquo Computers and Mathematics withApplications vol 57 no 11-12 pp 1748ndash1755 2009

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


not cover the optimization of the assignment plan betweentugboats and ships

Tugboat assignment is a typical kind of assignmentoptimization problem The existing approaches for solvingthe assignment problems can be divided into two categoriesThe traditional methods are mathematical programmingapproaches using Branch and Bound [15] and LagrangianRelaxation [16] to get the best solution efficiently but onlyfor small-sized problem Besides metaheuristic algorithmsinvolving genetic algorithms (GA) [17ndash19] ant colony opti-mization [20 21] simulated annealing (SA) [22] Tabu search(TS) [23 24] and variable neighborhood search (VNS) [25]are fast and efficient algorithms to get approximate solutionswith reasonable computational time for solving large scaleand more difficult combination optimization problem

Recently particle swarm optimization (PSO) is a novelevolutionary algorithm presented by Kennedy and Eberhart[26] Due to easy implementation and fast convergence PSOhas proved to be an effective and competitive algorithm fora wide variety of optimization problems However most ofthe published studies have concentrated on the continuousoptimization problems Recently some work has been doneto the discrete optimization problem In order to extend theapplication to discrete space Kennedy and Eberhart [27]proposed a discrete binary version of PSO where a particlemoved in a state space restricted to zero and one on eachdimension Yin et al [28] embedded a hill-climbing heuristicin PSO to assign application tasks to processors such thatthe resource demand of each task was satisfied and thesystem throughput was increased Lin et al [29] implementeda hybrid PSO (HPSO) to solve the biobjective personnelassignment problem The HPSO algorithm was combinedwith the random-key (RK) encoding scheme and individualenhancement (IE) scheme Similar hybrid PSO algorithmsare proposed by introducing operations of GAs into PSOsystems by Robinson et al [30] Lian et al [31] proposedsome new valid PSO operators for permutation job-shopscheduling to minimize makespan Prescilla and ImmanuelSelvakumar [32] applied the modified binary particle swarmoptimization algorithm to solve the real-time task assignmentin heterogeneous multiprocessor

To the best of our knowledge there is no publishedwork for dealing with tugboat assignment using PSO In thispaper we introduce the tugboat assignment problem under ahybrid scheduling rule (TAP-HSR) and give the mathemat-ical model Then an improved discrete PSO algorithm forTAP-HSR problem is proposed to minimize the turnaroundtime of ships In particular a newmethod for redefining PSOoperators is developed The experimental results show thatthe proposed method is effective and efficient compared toGA

The rest of this paper is organized as follows In thenext section tugboat assignment problem under a hybridscheduling rule (TAP-HSR) and the mathematical modelare described Section 3 proposes the improved discretePSO (IDPSO) for TAP-HSR problem Section 4 discussesthe experimental results Finally Section 5 provides someconclusions and the future work

2 Problem Formulation and Scheduling Rules

21 Problem Description The tugboat assignment problemunder a hybrid scheduling rule is an essential decision-making problem in container terminal The tugboat assign-ment operation has two procedures One is to find theassignment plan under the assignment rules The other isto generate the scheduling sequence of ships based on theassignment plan and some scheduling rules

Container terminal assigns tugboats to ships based onsome assignment rules Firstly each tugboat can serve atmostone ship and each ship can be assigned one or more shipsat any time Secondly the assigned number of tugboats isdetermined by ship length The ship with length equal to orless than 100 meters needs one tugboat at least Otherwisethe ship with the length more than 100 meters needs twotugboats at least Finally it is required to assign tugboat withproper horsepower (ps) Horsepower is a common unit tomeasure tugboat powerThe tugboat with higher horsepowerhas more service capacity to tug longer ship The assignmentrules between ships and tugboats are shown in Table 1

Assignment plan only describes the matching relation-ship of multiple tugboats and multiple ships It is requiredto use some scheduling rules to determine a schedulingsequence of ships according to the assignment plan Thispaper applies a hybrid scheduling rule combined with first-come-first-served (FCFS) rule and first-fit rule Based on thehybrid scheduling rule the ship with earlier arriving timeand the available assigned tugboats has the higher priority ofdocking

The goal of TAP-HSR is to minimize the turnaroundtime of ships under the premise that all ships finish dockingWith more and more ships arriving it is impossible tomake sure that ships can have available tugboats for dockingimmediately at arrival Some ships will wait outside theterminal at the anchorage until available tugboats can workfor them Good assignment can decrease the waiting time ofships so that the container terminal can serve more ships andget more profits

22 Model Formulation In this paper TAP-HSR is based onthe following assumptions

(1) The expected arriving time and the length of all shipsare known

(2) The docking operation time of ship depends on thelength of ship and is independent of tugboat

(3) Each ship has an available berth for unloading andloading containers

(4) Tugboat can work for only one ship at the same time(5) The preparation time of tugboat is zero which means

one tugboat can start to serve the next ship immedi-ately after it finishes the previous ship

(6) All tugboats are available when the first ship starts toberth

(7) The operation of tugboat requires an uninterruptedperiod of time


















119883119898119899 = (

11990911 11990912 sdot sdot sdot 1199091119899

11990921 11990922 sdot sdot sdot 1199092119899

sdot sdot sdot

1199091198981 1199091198982 sdot sdot sdot 119909119898119899

)


0 Otherwise

(1)


119862119896 = 119869119895 | 119909119894119895 sdot 119909119894119896 = 1 119900119895 lt 119900119896 (2)



min (119879119865 minus 119879119878) (3)


119879119865 =119899max119895=1

119879119895 (4)


min 119899max119895=1

119879119895 (5)


119879119895 = 119878119895 + 119905119895 (6)


119878119895 = max119868119895max119869119896isin119862119895

119879119896 (7)



sum

119894=1

119909119894119895 ge 119877119895

119898

sum

119894=1

119909119894119895ℎ119894 ge 119867119895

(8)




119896 = 119896 + 1End if

Next 119895119900 flag = 119900 flag + 1


Pseudocode 1









119883119894 =(

0

1

0

0

0

1

1

0

0

0

1

1

1

0

0

0

1

0

0

1

)

1198691 1198692 1198693 1198694 1198695

(9)





Operation time(min)

1 87 0 402 245 5 603 286 10 754 93 15 405 202 20 60






M2 40min

M2 M3 60min

M3 M4 75min

M1 40min

0 135

M1 M4 60min

(min)

1

2

3

4

5

Ship

num

ber







119905119892 minus 119909119905119894)

(10)

119909119905+1119894 = 119909

119905119894 + V119905+1119894 (11)







119883119894 = (

1 0 1 0 0

0 1 1 0 0

0 1 0 0 1

0 0 0 1 1

) (12)



119883119894 ⊟ 119883119895 = (not119883119894 and 119883119895) or (119883119894 and not119883119895) (13)


119905+1119894

119883119905119894 = (

1 0 1 0 0

0 1 1 0 0

0 1 0 0 1

0 0 0 1 1

)

119883119905+1119894 = (

1 0 1 0 0

0 1 0 0 0

0 0 1 0 1

0 1 0 1 1

)

119883119905+1119894 ⊟ 119883

119905119894 = (

0 0 0 0 0

0 0 1 0 0

0 1 1 0 0

0 1 0 0 0

)

(14)


V119895 = (119909 119910) | 119909 isin 1 2 119872 119910 isin 1 2 119873 (15)

In (12) V119905+1119894 = 119883119905+1119894 ⊟ 119883

119905119894 = (2 3) (3 2) (3 3) (4 2)

(V119905+1119894 )length = 4


V119894 ⊞ V119895 = (119909 119910) | (119909 119910) isin V119894 or (119909 119910) isin V119895 (16)


V119894 ⊟ V119895 = (119909 119910) | (119909 119910) isin V119894 (119909 119910) notin V119895 (17)



119888 119886 119887 119888 ⊠ V2 V1 ⊞ (119888 ⊠ V2)

119888 isin (0 05]

1 3 1003817100381710038171003817V21003817100381710038171003817

3

1= (2 3) (3 2) (3 3) V1 ⊞

1003817100381710038171003817V21003817100381710038171003817

3

1= (1 5) (2 3) (3 2) (3 3)

2 3 1003817100381710038171003817V21003817100381710038171003817

3

2= (3 2) (3 3) V1 ⊞

1003817100381710038171003817V21003817100381710038171003817

3

2= (1 5) (2 3) (3 2) (3 3)

4 4 1003817100381710038171003817V21003817100381710038171003817

4

4= (4 2) V1 ⊞

1003817100381710038171003817V21003817100381710038171003817

4

4= (1 5) (3 2) (2 3) (4 2)

119888 isin (05 1)

1 3 ⊟1003817100381710038171003817V2

1003817100381710038171003817

3

1= (2 3) (3 2) (3 3) V1 ⊟

1003817100381710038171003817V21003817100381710038171003817

3

1= (1 5)

2 3 ⊟1003817100381710038171003817V2

1003817100381710038171003817

3

2= (3 2) (3 3) V1 ⊟

1003817100381710038171003817V21003817100381710038171003817

3

2= (1 5) (2 3)

4 4 ⊟1003817100381710038171003817V2

1003817100381710038171003817

4

4= (4 2) V1 ⊟

1003817100381710038171003817V21003817100381710038171003817

4

4= (1 5) (3 2) (2 3)


V119894 = (2 3) (3 2) (3 3) (4 2)

V119895 = (1 5) (3 2)

V119894 ⊞ V119895 = (2 3) (3 2) (3 3) (4 2) (1 5)

V119894 ⊟ V119895 = (2 3) (3 3) (4 2)

(18)


119888 ⊠ V =119888

1003817100381710038171003817V21003817100381710038171003817

119887

119886=

V119887119886 119888 isin (0 05]

⊟ V119887119886 119888 isin (05 1) (19)


(V)length1 = V


V (2 3) (3 2) (3 3) (4 2)

V31 |1(2 3) (3 2) (3 3) |

3(4 2)

(2 3) (3 2) (3 3)

V32 (2 3) |2(3 2) (3 3) |

3(4 2)

(3 2) (3 3)

V41 |1(2 3) (3 2) (3 3) (4 2) |

4

(2 3) (3 2) (3 3) (4 2)

(20)


V1 ⊞ (119888 ⊠ V2) = V1 ⊞

1003817100381710038171003817V21003817100381710038171003817

119887

119886119888 isin (0 05]

V1⊟1003817100381710038171003817V2

1003817100381710038171003817

119887

119886119888 isin (05 1)

(21)








V119905+1119894 = V119905119894 ⊞ 119888110038171003817100381710038171003817119901119905119894 ⊟ 119909119905119894

10038171003817100381710038171003817

1198871

1198861⊞1198882

10038171003817100381710038171003817119901119905119892 ⊟ 119909119905119894

10038171003817100381710038171003817

1198872

1198862(22)

119909119905+1119894 = 119909

119905119894 ⊞ V119905+1119894 (23)




V119905+1119894 = V119905119894 ⊞ (119901119905119894 ⊟ 119909119905119894) ⊞ (119901

119905119892 ⊟ 119909119905119894)

119909119905+1119894 = 119909

119905119894 ⊞ V119905+1119894

(24)









119888 flag = 0Else



Pseudocode 2








5 Conclusions



































Acknowledgment


References












































Volume 2014




Journal of











Journal of


Function Spaces






Algebra


























119883119898119899 = (

11990911 11990912 sdot sdot sdot 1199091119899

11990921 11990922 sdot sdot sdot 1199092119899

sdot sdot sdot

1199091198981 1199091198982 sdot sdot sdot 119909119898119899

)


0 Otherwise

(1)


119862119896 = 119869119895 | 119909119894119895 sdot 119909119894119896 = 1 119900119895 lt 119900119896 (2)



min (119879119865 minus 119879119878) (3)


119879119865 =119899max119895=1

119879119895 (4)


min 119899max119895=1

119879119895 (5)


119879119895 = 119878119895 + 119905119895 (6)


119878119895 = max119868119895max119869119896isin119862119895

119879119896 (7)



sum

119894=1

119909119894119895 ge 119877119895

119898

sum

119894=1

119909119894119895ℎ119894 ge 119867119895

(8)




119896 = 119896 + 1End if

Next 119895119900 flag = 119900 flag + 1


Pseudocode 1









119883119894 =(

0

1

0

0

0

1

1

0

0

0

1

1

1

0

0

0

1

0

0

1

)

1198691 1198692 1198693 1198694 1198695

(9)





Operation time(min)

1 87 0 402 245 5 603 286 10 754 93 15 405 202 20 60






M2 40min

M2 M3 60min

M3 M4 75min

M1 40min

0 135

M1 M4 60min

(min)

1

2

3

4

5

Ship

num

ber







119905119892 minus 119909119905119894)

(10)

119909119905+1119894 = 119909

119905119894 + V119905+1119894 (11)







119883119894 = (

1 0 1 0 0

0 1 1 0 0

0 1 0 0 1

0 0 0 1 1

) (12)



119883119894 ⊟ 119883119895 = (not119883119894 and 119883119895) or (119883119894 and not119883119895) (13)


119905+1119894

119883119905119894 = (

1 0 1 0 0

0 1 1 0 0

0 1 0 0 1

0 0 0 1 1

)

119883119905+1119894 = (

1 0 1 0 0

0 1 0 0 0

0 0 1 0 1

0 1 0 1 1

)

119883119905+1119894 ⊟ 119883

119905119894 = (

0 0 0 0 0

0 0 1 0 0

0 1 1 0 0

0 1 0 0 0

)

(14)


V119895 = (119909 119910) | 119909 isin 1 2 119872 119910 isin 1 2 119873 (15)

In (12) V119905+1119894 = 119883119905+1119894 ⊟ 119883

119905119894 = (2 3) (3 2) (3 3) (4 2)

(V119905+1119894 )length = 4


V119894 ⊞ V119895 = (119909 119910) | (119909 119910) isin V119894 or (119909 119910) isin V119895 (16)


V119894 ⊟ V119895 = (119909 119910) | (119909 119910) isin V119894 (119909 119910) notin V119895 (17)



119888 119886 119887 119888 ⊠ V2 V1 ⊞ (119888 ⊠ V2)

119888 isin (0 05]

1 3 1003817100381710038171003817V21003817100381710038171003817

3

1= (2 3) (3 2) (3 3) V1 ⊞

1003817100381710038171003817V21003817100381710038171003817

3

1= (1 5) (2 3) (3 2) (3 3)

2 3 1003817100381710038171003817V21003817100381710038171003817

3

2= (3 2) (3 3) V1 ⊞

1003817100381710038171003817V21003817100381710038171003817

3

2= (1 5) (2 3) (3 2) (3 3)

4 4 1003817100381710038171003817V21003817100381710038171003817

4

4= (4 2) V1 ⊞

1003817100381710038171003817V21003817100381710038171003817

4

4= (1 5) (3 2) (2 3) (4 2)

119888 isin (05 1)

1 3 ⊟1003817100381710038171003817V2

1003817100381710038171003817

3

1= (2 3) (3 2) (3 3) V1 ⊟

1003817100381710038171003817V21003817100381710038171003817

3

1= (1 5)

2 3 ⊟1003817100381710038171003817V2

1003817100381710038171003817

3

2= (3 2) (3 3) V1 ⊟

1003817100381710038171003817V21003817100381710038171003817

3

2= (1 5) (2 3)

4 4 ⊟1003817100381710038171003817V2

1003817100381710038171003817

4

4= (4 2) V1 ⊟

1003817100381710038171003817V21003817100381710038171003817

4

4= (1 5) (3 2) (2 3)


V119894 = (2 3) (3 2) (3 3) (4 2)

V119895 = (1 5) (3 2)

V119894 ⊞ V119895 = (2 3) (3 2) (3 3) (4 2) (1 5)

V119894 ⊟ V119895 = (2 3) (3 3) (4 2)

(18)


119888 ⊠ V =119888

1003817100381710038171003817V21003817100381710038171003817

119887

119886=

V119887119886 119888 isin (0 05]

⊟ V119887119886 119888 isin (05 1) (19)


(V)length1 = V


V (2 3) (3 2) (3 3) (4 2)

V31 |1(2 3) (3 2) (3 3) |

3(4 2)

(2 3) (3 2) (3 3)

V32 (2 3) |2(3 2) (3 3) |

3(4 2)

(3 2) (3 3)

V41 |1(2 3) (3 2) (3 3) (4 2) |

4

(2 3) (3 2) (3 3) (4 2)

(20)


V1 ⊞ (119888 ⊠ V2) = V1 ⊞

1003817100381710038171003817V21003817100381710038171003817

119887

119886119888 isin (0 05]

V1⊟1003817100381710038171003817V2

1003817100381710038171003817

119887

119886119888 isin (05 1)

(21)








V119905+1119894 = V119905119894 ⊞ 119888110038171003817100381710038171003817119901119905119894 ⊟ 119909119905119894

10038171003817100381710038171003817

1198871

1198861⊞1198882

10038171003817100381710038171003817119901119905119892 ⊟ 119909119905119894

10038171003817100381710038171003817

1198872

1198862(22)

119909119905+1119894 = 119909

119905119894 ⊞ V119905+1119894 (23)




V119905+1119894 = V119905119894 ⊞ (119901119905119894 ⊟ 119909119905119894) ⊞ (119901

119905119892 ⊟ 119909119905119894)

119909119905+1119894 = 119909

119905119894 ⊞ V119905+1119894

(24)









119888 flag = 0Else



Pseudocode 2








5 Conclusions



































Acknowledgment


References












































Volume 2014




Journal of











Journal of


Function Spaces






Algebra












119896 = 119896 + 1End if

Next 119895119900 flag = 119900 flag + 1


Pseudocode 1









119883119894 =(

0

1

0

0

0

1

1

0

0

0

1

1

1

0

0

0

1

0

0

1

)

1198691 1198692 1198693 1198694 1198695

(9)





Operation time(min)

1 87 0 402 245 5 603 286 10 754 93 15 405 202 20 60






M2 40min

M2 M3 60min

M3 M4 75min

M1 40min

0 135

M1 M4 60min

(min)

1

2

3

4

5

Ship

num

ber







119905119892 minus 119909119905119894)

(10)

119909119905+1119894 = 119909

119905119894 + V119905+1119894 (11)







119883119894 = (

1 0 1 0 0

0 1 1 0 0

0 1 0 0 1

0 0 0 1 1

) (12)



119883119894 ⊟ 119883119895 = (not119883119894 and 119883119895) or (119883119894 and not119883119895) (13)


119905+1119894

119883119905119894 = (

1 0 1 0 0

0 1 1 0 0

0 1 0 0 1

0 0 0 1 1

)

119883119905+1119894 = (

1 0 1 0 0

0 1 0 0 0

0 0 1 0 1

0 1 0 1 1

)

119883119905+1119894 ⊟ 119883

119905119894 = (

0 0 0 0 0

0 0 1 0 0

0 1 1 0 0

0 1 0 0 0

)

(14)


V119895 = (119909 119910) | 119909 isin 1 2 119872 119910 isin 1 2 119873 (15)

In (12) V119905+1119894 = 119883119905+1119894 ⊟ 119883

119905119894 = (2 3) (3 2) (3 3) (4 2)

(V119905+1119894 )length = 4


V119894 ⊞ V119895 = (119909 119910) | (119909 119910) isin V119894 or (119909 119910) isin V119895 (16)


V119894 ⊟ V119895 = (119909 119910) | (119909 119910) isin V119894 (119909 119910) notin V119895 (17)



119888 119886 119887 119888 ⊠ V2 V1 ⊞ (119888 ⊠ V2)

119888 isin (0 05]

1 3 1003817100381710038171003817V21003817100381710038171003817

3

1= (2 3) (3 2) (3 3) V1 ⊞

1003817100381710038171003817V21003817100381710038171003817

3

1= (1 5) (2 3) (3 2) (3 3)

2 3 1003817100381710038171003817V21003817100381710038171003817

3

2= (3 2) (3 3) V1 ⊞

1003817100381710038171003817V21003817100381710038171003817

3

2= (1 5) (2 3) (3 2) (3 3)

4 4 1003817100381710038171003817V21003817100381710038171003817

4

4= (4 2) V1 ⊞

1003817100381710038171003817V21003817100381710038171003817

4

4= (1 5) (3 2) (2 3) (4 2)

119888 isin (05 1)

1 3 ⊟1003817100381710038171003817V2

1003817100381710038171003817

3

1= (2 3) (3 2) (3 3) V1 ⊟

1003817100381710038171003817V21003817100381710038171003817

3

1= (1 5)

2 3 ⊟1003817100381710038171003817V2

1003817100381710038171003817

3

2= (3 2) (3 3) V1 ⊟

1003817100381710038171003817V21003817100381710038171003817

3

2= (1 5) (2 3)

4 4 ⊟1003817100381710038171003817V2

1003817100381710038171003817

4

4= (4 2) V1 ⊟

1003817100381710038171003817V21003817100381710038171003817

4

4= (1 5) (3 2) (2 3)


V119894 = (2 3) (3 2) (3 3) (4 2)

V119895 = (1 5) (3 2)

V119894 ⊞ V119895 = (2 3) (3 2) (3 3) (4 2) (1 5)

V119894 ⊟ V119895 = (2 3) (3 3) (4 2)

(18)


119888 ⊠ V =119888

1003817100381710038171003817V21003817100381710038171003817

119887

119886=

V119887119886 119888 isin (0 05]

⊟ V119887119886 119888 isin (05 1) (19)


(V)length1 = V


V (2 3) (3 2) (3 3) (4 2)

V31 |1(2 3) (3 2) (3 3) |

3(4 2)

(2 3) (3 2) (3 3)

V32 (2 3) |2(3 2) (3 3) |

3(4 2)

(3 2) (3 3)

V41 |1(2 3) (3 2) (3 3) (4 2) |

4

(2 3) (3 2) (3 3) (4 2)

(20)


V1 ⊞ (119888 ⊠ V2) = V1 ⊞

1003817100381710038171003817V21003817100381710038171003817

119887

119886119888 isin (0 05]

V1⊟1003817100381710038171003817V2

1003817100381710038171003817

119887

119886119888 isin (05 1)

(21)








V119905+1119894 = V119905119894 ⊞ 119888110038171003817100381710038171003817119901119905119894 ⊟ 119909119905119894

10038171003817100381710038171003817

1198871

1198861⊞1198882

10038171003817100381710038171003817119901119905119892 ⊟ 119909119905119894

10038171003817100381710038171003817

1198872

1198862(22)

119909119905+1119894 = 119909

119905119894 ⊞ V119905+1119894 (23)




V119905+1119894 = V119905119894 ⊞ (119901119905119894 ⊟ 119909119905119894) ⊞ (119901

119905119892 ⊟ 119909119905119894)

119909119905+1119894 = 119909

119905119894 ⊞ V119905+1119894

(24)









119888 flag = 0Else



Pseudocode 2








5 Conclusions



































Acknowledgment


References












































Volume 2014




Journal of











Journal of


Function Spaces






Algebra










M2 40min

M2 M3 60min

M3 M4 75min

M1 40min

0 135

M1 M4 60min

(min)

1

2

3

4

5

Ship

num

ber







119905119892 minus 119909119905119894)

(10)

119909119905+1119894 = 119909

119905119894 + V119905+1119894 (11)







119883119894 = (

1 0 1 0 0

0 1 1 0 0

0 1 0 0 1

0 0 0 1 1

) (12)



119883119894 ⊟ 119883119895 = (not119883119894 and 119883119895) or (119883119894 and not119883119895) (13)


119905+1119894

119883119905119894 = (

1 0 1 0 0

0 1 1 0 0

0 1 0 0 1

0 0 0 1 1

)

119883119905+1119894 = (

1 0 1 0 0

0 1 0 0 0

0 0 1 0 1

0 1 0 1 1

)

119883119905+1119894 ⊟ 119883

119905119894 = (

0 0 0 0 0

0 0 1 0 0

0 1 1 0 0

0 1 0 0 0

)

(14)


V119895 = (119909 119910) | 119909 isin 1 2 119872 119910 isin 1 2 119873 (15)

In (12) V119905+1119894 = 119883119905+1119894 ⊟ 119883

119905119894 = (2 3) (3 2) (3 3) (4 2)

(V119905+1119894 )length = 4


V119894 ⊞ V119895 = (119909 119910) | (119909 119910) isin V119894 or (119909 119910) isin V119895 (16)


V119894 ⊟ V119895 = (119909 119910) | (119909 119910) isin V119894 (119909 119910) notin V119895 (17)



119888 119886 119887 119888 ⊠ V2 V1 ⊞ (119888 ⊠ V2)

119888 isin (0 05]

1 3 1003817100381710038171003817V21003817100381710038171003817

3

1= (2 3) (3 2) (3 3) V1 ⊞

1003817100381710038171003817V21003817100381710038171003817

3

1= (1 5) (2 3) (3 2) (3 3)

2 3 1003817100381710038171003817V21003817100381710038171003817

3

2= (3 2) (3 3) V1 ⊞

1003817100381710038171003817V21003817100381710038171003817

3

2= (1 5) (2 3) (3 2) (3 3)

4 4 1003817100381710038171003817V21003817100381710038171003817

4

4= (4 2) V1 ⊞

1003817100381710038171003817V21003817100381710038171003817

4

4= (1 5) (3 2) (2 3) (4 2)

119888 isin (05 1)

1 3 ⊟1003817100381710038171003817V2

1003817100381710038171003817

3

1= (2 3) (3 2) (3 3) V1 ⊟

1003817100381710038171003817V21003817100381710038171003817

3

1= (1 5)

2 3 ⊟1003817100381710038171003817V2

1003817100381710038171003817

3

2= (3 2) (3 3) V1 ⊟

1003817100381710038171003817V21003817100381710038171003817

3

2= (1 5) (2 3)

4 4 ⊟1003817100381710038171003817V2

1003817100381710038171003817

4

4= (4 2) V1 ⊟

1003817100381710038171003817V21003817100381710038171003817

4

4= (1 5) (3 2) (2 3)


V119894 = (2 3) (3 2) (3 3) (4 2)

V119895 = (1 5) (3 2)

V119894 ⊞ V119895 = (2 3) (3 2) (3 3) (4 2) (1 5)

V119894 ⊟ V119895 = (2 3) (3 3) (4 2)

(18)


119888 ⊠ V =119888

1003817100381710038171003817V21003817100381710038171003817

119887

119886=

V119887119886 119888 isin (0 05]

⊟ V119887119886 119888 isin (05 1) (19)


(V)length1 = V


V (2 3) (3 2) (3 3) (4 2)

V31 |1(2 3) (3 2) (3 3) |

3(4 2)

(2 3) (3 2) (3 3)

V32 (2 3) |2(3 2) (3 3) |

3(4 2)

(3 2) (3 3)

V41 |1(2 3) (3 2) (3 3) (4 2) |

4

(2 3) (3 2) (3 3) (4 2)

(20)


V1 ⊞ (119888 ⊠ V2) = V1 ⊞

1003817100381710038171003817V21003817100381710038171003817

119887

119886119888 isin (0 05]

V1⊟1003817100381710038171003817V2

1003817100381710038171003817

119887

119886119888 isin (05 1)

(21)








V119905+1119894 = V119905119894 ⊞ 119888110038171003817100381710038171003817119901119905119894 ⊟ 119909119905119894

10038171003817100381710038171003817

1198871

1198861⊞1198882

10038171003817100381710038171003817119901119905119892 ⊟ 119909119905119894

10038171003817100381710038171003817

1198872

1198862(22)

119909119905+1119894 = 119909

119905119894 ⊞ V119905+1119894 (23)




V119905+1119894 = V119905119894 ⊞ (119901119905119894 ⊟ 119909119905119894) ⊞ (119901

119905119892 ⊟ 119909119905119894)

119909119905+1119894 = 119909

119905119894 ⊞ V119905+1119894

(24)









119888 flag = 0Else



Pseudocode 2








5 Conclusions



































Acknowledgment


References












































Volume 2014




Journal of











Journal of


Function Spaces






Algebra











119888 119886 119887 119888 ⊠ V2 V1 ⊞ (119888 ⊠ V2)

119888 isin (0 05]

1 3 1003817100381710038171003817V21003817100381710038171003817

3

1= (2 3) (3 2) (3 3) V1 ⊞

1003817100381710038171003817V21003817100381710038171003817

3

1= (1 5) (2 3) (3 2) (3 3)

2 3 1003817100381710038171003817V21003817100381710038171003817

3

2= (3 2) (3 3) V1 ⊞

1003817100381710038171003817V21003817100381710038171003817

3

2= (1 5) (2 3) (3 2) (3 3)

4 4 1003817100381710038171003817V21003817100381710038171003817

4

4= (4 2) V1 ⊞

1003817100381710038171003817V21003817100381710038171003817

4

4= (1 5) (3 2) (2 3) (4 2)

119888 isin (05 1)

1 3 ⊟1003817100381710038171003817V2

1003817100381710038171003817

3

1= (2 3) (3 2) (3 3) V1 ⊟

1003817100381710038171003817V21003817100381710038171003817

3

1= (1 5)

2 3 ⊟1003817100381710038171003817V2

1003817100381710038171003817

3

2= (3 2) (3 3) V1 ⊟

1003817100381710038171003817V21003817100381710038171003817

3

2= (1 5) (2 3)

4 4 ⊟1003817100381710038171003817V2

1003817100381710038171003817

4

4= (4 2) V1 ⊟

1003817100381710038171003817V21003817100381710038171003817

4

4= (1 5) (3 2) (2 3)


V119894 = (2 3) (3 2) (3 3) (4 2)

V119895 = (1 5) (3 2)

V119894 ⊞ V119895 = (2 3) (3 2) (3 3) (4 2) (1 5)

V119894 ⊟ V119895 = (2 3) (3 3) (4 2)

(18)


119888 ⊠ V =119888

1003817100381710038171003817V21003817100381710038171003817

119887

119886=

V119887119886 119888 isin (0 05]

⊟ V119887119886 119888 isin (05 1) (19)


(V)length1 = V


V (2 3) (3 2) (3 3) (4 2)

V31 |1(2 3) (3 2) (3 3) |

3(4 2)

(2 3) (3 2) (3 3)

V32 (2 3) |2(3 2) (3 3) |

3(4 2)

(3 2) (3 3)

V41 |1(2 3) (3 2) (3 3) (4 2) |

4

(2 3) (3 2) (3 3) (4 2)

(20)


V1 ⊞ (119888 ⊠ V2) = V1 ⊞

1003817100381710038171003817V21003817100381710038171003817

119887

119886119888 isin (0 05]

V1⊟1003817100381710038171003817V2

1003817100381710038171003817

119887

119886119888 isin (05 1)

(21)








V119905+1119894 = V119905119894 ⊞ 119888110038171003817100381710038171003817119901119905119894 ⊟ 119909119905119894

10038171003817100381710038171003817

1198871

1198861⊞1198882

10038171003817100381710038171003817119901119905119892 ⊟ 119909119905119894

10038171003817100381710038171003817

1198872

1198862(22)

119909119905+1119894 = 119909

119905119894 ⊞ V119905+1119894 (23)




V119905+1119894 = V119905119894 ⊞ (119901119905119894 ⊟ 119909119905119894) ⊞ (119901

119905119892 ⊟ 119909119905119894)

119909119905+1119894 = 119909

119905119894 ⊞ V119905+1119894

(24)









119888 flag = 0Else



Pseudocode 2








5 Conclusions



































Acknowledgment


References












































Volume 2014




Journal of











Journal of


Function Spaces






Algebra
















119888 flag = 0Else



Pseudocode 2








5 Conclusions



































Acknowledgment


References












































Volume 2014




Journal of











Journal of


Function Spaces






Algebra










































Acknowledgment


References












































Volume 2014




Journal of











Journal of


Function Spaces






Algebra










































Volume 2014




Journal of











Journal of


Function Spaces






Algebra









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