Bierlaire, Michel

Robenek, Tomá

Ropke, Stefan

Umang, Nitish

Infoscience - École polytechnique fédérale de Lausanne

Integrated berth allocation and yard assignment problemusing column generationTomáš RobenekTransp-OR,École Polytechnique Fédérale de LausanneDTU Transport,Technical University of DenmarkMay 2, 2012Introduction Problem Definition Branch and Price Results ConclusionAgenda1 Introduction2 Problem Definition3 Branch and Price4 Results5 ConclusionTomáš Robenek (EPFL, DTU) Port Optimization May 2, 2012 2 / 42Introduction Problem Definition Branch and Price Results Conclusion1 IntroductionBerth Allocation ProblemYard Assignment ProblemMotivation2 Problem Definition3 Branch and Price4 Results5 ConclusionTomáš Robenek (EPFL, DTU) Port Optimization May 2, 2012 3 / 42Introduction Problem Definition Branch and Price Results ConclusionBerth Allocation ProblemBerth Allocation ProblemFigure: Lacon ltd.’s plan for extension of the Riga’s port, LatviaTomáš Robenek (EPFL, DTU) Port Optimization May 2, 2012 4 / 42Introduction Problem Definition Branch and Price Results ConclusionYard Assignment ProblemYard Assignment ProblemFigure: Port of Weipa, Queensland, AustraliaTomáš Robenek (EPFL, DTU) Port Optimization May 2, 2012 5 / 42Introduction Problem Definition Branch and Price Results ConclusionMotivationMotivationFigure: Vessels queueing at Newcastle port, Australia (queue hits 60 vessels asmax)Tomáš Robenek (EPFL, DTU) Port Optimization May 2, 2012 6 / 42Introduction Problem Definition Branch and Price Results Conclusion1 Introduction2 Problem DefinitionInputVesselPortGeneral DataOutput3 Branch and Price4 Results5 ConclusionTomáš Robenek (EPFL, DTU) Port Optimization May 2, 2012 7 / 42Introduction Problem Definition Branch and Price Results ConclusionInputInput – VesselInformationNumber of VesselsArrival TimeLengthDraft (omitted)CargoQuantityCargo TypeTomáš Robenek (EPFL, DTU) Port Optimization May 2, 2012 8 / 42Introduction Problem Definition Branch and Price Results ConclusionInputInput – Draft OmittedTomáš Robenek (EPFL, DTU) Port Optimization May 2, 2012 9 / 42Introduction Problem Definition Branch and Price Results ConclusionInputInput – PortInformationNumber of SectionsLengthDraft (omitted)CoordinatesResourcesNumber of CargoLocationsCoordinatesNeighbouringLocationsTomáš Robenek (EPFL, DTU) Port Optimization May 2, 2012 10 / 42Introduction Problem Definition Branch and Price Results ConclusionInputInput – General DataInformationTime HorizonNumber of Cargo TypesIncompatible Cargo TypesDistancesTransfer RateCrane Handling RateBulk Ports (No Containers)Tomáš Robenek (EPFL, DTU) Port Optimization May 2, 2012 11 / 42Introduction Problem Definition Branch and Price Results ConclusionOutputOutputMinimizeHandling Time + Delay = Service TimeParalell HandlingTomáš Robenek (EPFL, DTU) Port Optimization May 2, 2012 12 / 42Introduction Problem Definition Branch and Price Results Conclusion1 Introduction2 Problem Definition3 Branch and PriceFrameworkInitial SolutionMaster ProblemSub-ProblemBranch and Bound4 Results5 ConclusionTomáš Robenek (EPFL, DTU) Port Optimization May 2, 2012 13 / 42Introduction Problem Definition Branch and Price Results ConclusionFrameworkFrameworkStartSolve Master ProblemInitial ColumnsPricing SolverDual VariablesNew Columns with Negative Reduced CostInteger Solution?No Columnswith NegativeReduced CostBranch & BoundNoOutput SolutionyesEndInitial SolutionColumn Generation – LowerBoundBranch and Bound – OptimalInteger SolutionTomáš Robenek (EPFL, DTU) Port Optimization May 2, 2012 14 / 42Introduction Problem Definition Branch and Price Results ConclusionInitial SolutionInitial SolutionStartSolve Master ProblemInitial ColumnsPricing SolverDual VariablesNew Columns with Negative Reduced CostInteger Solution?No Columnswith NegativeReduced CostBranch & BoundNoOutput SolutionyesEndStartIs Pool Empty?Add Current Column And Delete it From Initial Listyesj = 0j < columns.sizej++Is Current Column Compatible With Pool & column.vessel = i?nonoi = 0i < number of vesselsi++doneyesEnddoneTomáš Robenek (EPFL, DTU) Port Optimization May 2, 2012 15 / 42Introduction Problem Definition Branch and Price Results ConclusionMaster ProblemMaster Problem – ParametersStartSolve Master ProblemInitial ColumnsPricing SolverDual VariablesNew Columns with Negative Reduced CostInteger Solution?No Columnswith NegativeReduced CostBranch & BoundNoOutput SolutionyesEndParametersΩ – set of all feasible assignmentsΩ1 ⊂ Ω – current pool of columnsca – cost of assignment a ∈ Ω (Ω1)N = {1..n} – set of vesselsK = {1..m} – set of sectionsW = {1..w} – set of cargo typesT = {1..h} – set of time stepsL = {1..q} – set of locationsctw – number of vessels carrying cargo type wTomáš Robenek (EPFL, DTU) Port Optimization May 2, 2012 16 / 42Introduction Problem Definition Branch and Price Results ConclusionMaster ProblemMaster Problem – ParametersDecision Variblesλa ∈ (0, 1) – 1 if assignment a is selected, 0 otherwiseµlw ∈ (0, 1) – 1 if location l is storing cargo of type w , 0 otherwiserelaxedIdeaobtain berth and yard schedulesectionused by vessel at timelocationstored cargoused by vessel at timeTomáš Robenek (EPFL, DTU) Port Optimization May 2, 2012 17 / 42Introduction Problem Definition Branch and Price Results ConclusionMaster ProblemMaster Problem – ParametersParametersAia ={1 if vessel i is assigned in assignment a,0 otherwise.Bkta ={1 if section k is occupied at time t in assignment a,0 otherwise.C lwa ={1 if cargo w is stored at location l in assignment a,0 otherwise.Dlta ={1 if cargo location l is handling assignment a at time t,0 otherwise.Tomáš Robenek (EPFL, DTU) Port Optimization May 2, 2012 18 / 42Introduction Problem Definition Branch and Price Results ConclusionMaster ProblemMaster Problem – Objective Functionminimize∑a∈Ωca · λa (1)Tomáš Robenek (EPFL, DTU) Port Optimization May 2, 2012 19 / 42Introduction Problem Definition Branch and Price Results ConclusionMaster ProblemMaster Problem – ConstraintsAll Vessels Served∑a∈Ω1Aia · λa = 1, ∀i ∈ N, (2)Tomáš Robenek (EPFL, DTU) Port Optimization May 2, 2012 20 / 42Introduction Problem Definition Branch and Price Results ConclusionMaster ProblemMaster Problem – ConstraintsSection Occupation∑a∈Ω1Bkta · λa ≤ 1, ∀k ∈ K ,∀t ∈ T , (3)Figure: Illustration of Philadelphia ExperimentTomáš Robenek (EPFL, DTU) Port Optimization May 2, 2012 21 / 42Introduction Problem Definition Branch and Price Results ConclusionMaster ProblemMaster Problem – ConstraintsLocation Occupation∑a∈Ω1Dlta · λa ≤ 1, ∀l ∈ L, ∀t ∈ T , (4)Tomáš Robenek (EPFL, DTU) Port Optimization May 2, 2012 22 / 42Introduction Problem Definition Branch and Price Results ConclusionMaster ProblemMaster Problem – ConstraintsOne Cargo per Location∑a∈Ω1C lwa · λa − ctw · µlw ≤ 0, ∀l ∈ L,∀w ∈W , (5)∑w∈Wµlw ≤ 1, ∀l ∈ L, (6)Tomáš Robenek (EPFL, DTU) Port Optimization May 2, 2012 23 / 42Introduction Problem Definition Branch and Price Results ConclusionMaster ProblemMaster Problem – ConstraintsCompatible Neighboursµlw + µlw ≤ 1,∀l ∈ L, ∀l ∈ L,∀w ∈W ,∀w ∈W , (7)Tomáš Robenek (EPFL, DTU) Port Optimization May 2, 2012 24 / 42Introduction Problem Definition Branch and Price Results ConclusionSub-ProblemSub-Problem – ParametersStartSolve Master ProblemInitial ColumnsPricing SolverDual VariablesNew Columns with Negative Reduced CostInteger Solution?No Columnswith NegativeReduced CostBranch & BoundNoOutput SolutionyesEndIdearun for each vessel separatelyget n columns (one per vessel)SetsK = {1..m} – set of sectionsW = {1..w} – set of cargo typesT = {1..h} – set of time stepsL = {1..q} – set of locationsDual Variablesα, βkt , γlt , δlwTomáš Robenek (EPFL, DTU) Port Optimization May 2, 2012 25 / 42Introduction Problem Definition Branch and Price Results ConclusionSub-ProblemSub-Problem – Objective Functionminimize (c + s − a)− (α +∑k∈K ∑t∈T βkt · betakt+∑l∈L∑t∈T γlt · gammalt +∑l∈L∑w∈W δlw · deltalw )(8)Parametersa – arrival timeDecision Variablesc ≥ 0 – handling times ≥ 0 – start time of servicerelated to duals:betakt ∈ (0, 1) – 1 if vessel occupies section k at time t, 0 otherwisegammalt ∈ (0, 1) – 1 if vessel uses location l at time t, 0 otherwisedeltalw ∈ (0, 1) – 1 if cargo type w is stored at location l , 0 otherwiseTomáš Robenek (EPFL, DTU) Port Optimization May 2, 2012 26 / 42Introduction Problem Definition Branch and Price Results ConclusionSub-ProblemSub-Problem – Constraintss − a ≥ 0, (9)c ≥ htk · fractionjk −M · (1− ssj) , ∀k, j ∈ K , (10)Parametersfractionjk – fraction of cargo handled at section k, if the startingsection of the vessel is section jM – large enough number (set to 1 000 000, could be the largestquantity multiplied by the longest service time)Decision Variableshtk ≥ 0 – handling time of section kssj ∈ (0, 1) – 1 if section j is the starting section of the vesselTomáš Robenek (EPFL, DTU) Port Optimization May 2, 2012 27 / 42Introduction Problem Definition Branch and Price Results ConclusionSub-ProblemSub-Problem – Constraints∑j∈Kssj = 1, (11)∑j∈Kssj · scj + length ≤ ql , (12)Parametersscj – starting coordinate of section jlength – length of the vesselql – quay lengthTomáš Robenek (EPFL, DTU) Port Optimization May 2, 2012 28 / 42Introduction Problem Definition Branch and Price Results ConclusionSub-ProblemSub-Problem – Constraints∑l∈Lsplitl ≤ Z , (13)ParametersZ – maximum number of locations used by vesselDecision Variablessplitl ∈ (0, 1) – 1 if vessel uses location lTomáš Robenek (EPFL, DTU) Port Optimization May 2, 2012 29 / 42Introduction Problem Definition Branch and Price Results ConclusionSub-ProblemSub-Problem – Constraintssplitl ≤ deltalw , ∀l ∈ L, (14)∑l∈Lcsl = quantity , (15)csl ≤ splitl · quantity , ∀l ∈ L, (16)splitl ≤ csl , ∀l ∈ L, (17)Parametersw – cargo type carried on the vesselDecision Variablescsl ≥ 0 – quantity of cargo stored at location lTomáš Robenek (EPFL, DTU) Port Optimization May 2, 2012 30 / 42Introduction Problem Definition Branch and Price Results ConclusionSub-ProblemSub-Problem – Constraintstdk =∑l∈Ldkl · csl /quantity , ∀k ∈ K , (18)htk = F/cranesk + Vw · tdk , ∀k ∈ K , (19)Parametersdkl – distance between section k and location lcranesk – number of cranes in section kF – crane handling rate, Vw – cargo transfer rateDecision Variablestdk ≥ 0 – total average distance for section kTomáš Robenek (EPFL, DTU) Port Optimization May 2, 2012 31 / 42Introduction Problem Definition Branch and Price Results ConclusionSub-ProblemSub-Problem – Constraints∑t∈Ttimet = c, (20)t + M · (1− timet) ≥ s + 1, ∀t ∈ T , (21)t ≤ s + c + M · (1− timet) , ∀t ∈ T , (22)ParametersM in 21 – minimum value is s + 1M in 22 – minimum value is T − s + cDecision Variablestimet ∈ (0, 1) – 1 if the vessel is at time t served, 0 otherwiseTomáš Robenek (EPFL, DTU) Port Optimization May 2, 2012 32 / 42Introduction Problem Definition Branch and Price Results ConclusionSub-ProblemSub-Problem – Constraintsbetakt ≥xk + timet − 1, ∀k ∈ K ,∀t ∈ T , (23)betakt ≤xk , ∀k ∈ K ,∀t ∈ T , (24)betakt ≤timet , ∀k ∈ K ,∀t ∈ T , (25)gammalt ≥splitl + timet − 1, ∀l ∈ L, ∀t ∈ T , (26)gammalt ≤splitl , ∀l ∈ L, ∀t ∈ T , (27)gammalt ≤timet , ∀l ∈ L, ∀t ∈ T , (28)Tomáš Robenek (EPFL, DTU) Port Optimization May 2, 2012 33 / 42Introduction Problem Definition Branch and Price Results ConclusionBranch and BoundBranch and BoundStartSolve Master ProblemInitial ColumnsPricing SolverDual VariablesNew Columns with Negative Reduced CostInteger Solution?No Columnswith NegativeReduced CostBranch & BoundNoOutput SolutionyesEndλ = 0.734X = 0.333λ = 0X = 1λ = 0X = 0λ = 1X = 0λ = 1X = 1Tomáš Robenek (EPFL, DTU) Port Optimization May 2, 2012 34 / 42Introduction Problem Definition Branch and Price Results Conclusion1 Introduction2 Problem Definition3 Branch and Price4 ResultsStatusTests5 ConclusionTomáš Robenek (EPFL, DTU) Port Optimization May 2, 2012 35 / 42Introduction Problem Definition Branch and Price Results ConclusionStatusStatusMaster problem:able to solve small instancesfinishedSub - problem:validated with Opl and the minimum handling time of generatedassignmentsrunning time < 3 sec.Column Generation:without sub-problem, just reduced cost, able to solve small instancesstrategies:one column per turn – not able to solve in 3 hours (and still working)all negative columns per turn – solved below 20 iterationssub-problemBranch and Bound:to be implementedprobably existing function in CPLEXTomáš Robenek (EPFL, DTU) Port Optimization May 2, 2012 36 / 42Introduction Problem Definition Branch and Price Results ConclusionTestsTest InstanceCRANE OPERATING AREASection 1 Section 102 9Section 3 Section 84 7Section 5 Section 6250 50 200 100 200 200 100 200 50 250Location 11515100 Location 2 Location 3 Location 4 Location 5Location 10Location 9Location 8Location 7Location 6151600Connected to Section 1Connected to Section 10Tomáš Robenek (EPFL, DTU) Port Optimization May 2, 2012 37 / 42Introduction Problem Definition Branch and Price Results ConclusionTestsInitial Berth Plan291 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120SECTIONS1 VESSEL 1 VESSEL 2 VESSEL 34 VESSEL 9567810TIME HORIZONVESSEL 5VESSEL 6 VESSEL 7 VESSEL 8 VESSEL 103VESSEL 4Tomáš Robenek (EPFL, DTU) Port Optimization May 2, 2012 38 / 42Introduction Problem Definition Branch and Price Results ConclusionTestsDebuggingTomáš Robenek (EPFL, DTU) Port Optimization May 2, 2012 39 / 42Introduction Problem Definition Branch and Price Results Conclusion1 Introduction2 Problem Definition3 Branch and Price4 Results5 ConclusionTomáš Robenek (EPFL, DTU) Port Optimization May 2, 2012 40 / 42Thank you for your attention.Formulationmin z (1)s.t. mi − Ai ≥ 0 (2)∑kMsik = 1 (3)∑kM(sikbk) + Li ≤ L (4)∑pM(δi`ksi`) = xik ∀kM (5)σikt ≥ xik + θit − 1 ∀kM,∀tH (6)σikt ≤ xik ∀kM,∀tH (7)σikt ≤ θit ∀kM,∀tH (8)(dk −Di)xik ≥ 0 ∀kM (9)ci ≥ hwikρi`kQi −B(1− si`) ∀lM,∀kM,∀wWi (10)hwik = αwik + βwik ∀wWi, ∀kM (11)αwik = T/nwik ∀wWi, ∀kM (12)βwik = Vwrik ∀wWi,∀kM (13)∑lLφil ≤ 2 (14)rik =∑lL(rlkλil)/Qi ∀kM (15)∑wWpilw ≤ 1 ∀lL (16)φil ≤ pilw ∀wWi,∀lL (17)ωilt ≥ φil + θit − 1 ∀lL,∀tH (18)ωilt ≤ φil ∀lL,∀tH (19)ωilt ≤ θit ∀lL,∀tH (20)∑tHθit = ci (21)t+B(1− θit) ≥ mi + 1 ∀tH (22)t ≤ mi + ci + +B(1− θit) ∀tH (23)Qi =∑lLλil (24)λil ≤ φilQi ∀wWi,∀lL (25)φil ≤ Bλil ∀lL (26)λil ≤∑wWi∑tH(Rwωilt +B(1− pilw)) ∀lL (27)(28)4

Integrated berth allocation and yard assignment problem using column generation

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Integrated berth allocation and yard assignment problem using column generation

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