Vehicle Routing Problems - Massimo Paolucci Home Page · 2 VRP – Massimo Paolucci 3 Vehicle Routing Problems VRPs include: Static problems: the service de mand is fixed and a priori

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Vehicle Routing ProblemsMassimo Paolucci([email protected])

010-353 2996

DIST – Università di Genova

2VRP – Massimo Paolucci

Vehicle Routing ProblemsVehicle Routing Problems (VRP) deal with the management of pickup and/or delivery activities

Critical problems, in particular in short distance transportation

Operational decisions: how the available fleet of vehicles (resources) can be efficiently used to satisfy a given service demand according to a set of operational requirements?

Vehicle Routing: define the routes and possibly the schedule for the available vehicles

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Vehicle Routing ProblemsVRPs include:

Static problems: the service demand is fixed and a priori knownDynamic problems: all or part of the service demand may become known when the vehicles have already started their service (the vehicle routes may be defined or changed on-line)

Main assumptions for the problems considered in the following:

The problems are staticThe available vehicles are homogeneous (every vehicle provide the same kind of service)


VRP – Characteristics and ComponentsFreight transportation provided by vehicles through a route network

Main components:Road networkCustomersVehiclesDepotsDriversOperational constraints:

globalfor single routes

Optimization objectives

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VRP – Characteristics and ComponentsThe road network:

A graph G=(V, A), G=(V, E) or G=(V, A∪ E)Directed, undirected or mixedSparse vs Dense

Sparse ⇒ |A|=O(|V|)Dense ⇒ |A|=O(|V|2)

Undirected graph: large scale road network (countries, regions)

Directed graph:small scale road network (cities)


The road network:Vertices

depots, customers, road intersectionsV={0, 1, …, n}

Arcsroadsdirected (i, j)∈A or undirected e∈Elength, or travel cost cij ∀ (i, j)∈Atravel time tij ∀ (i, j)∈A

VRP – Characteristics and Components

cijj0 i

4



The road network:The triangle inequality

ji

k

cij

cik ckj

cij ≤ cik + ckj ∀i, j, k


VRP – Characteristics and ComponentsCustomers:

The entities requiring service Associated with vertices or arcs Service demand ⇒ quantity of goods or materials to be delivered or picked upCharacteristics:

Service time windowsDesired service start time (with penalties)Service time (load/unload)Vehicle compatibilityPossibility of splitting service

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Vehicles:Fleet size (fixed or variable)Company or outsourced fleet (fixed vehicle cost)Depot of reference (single, multiple, possibility of change)Vehicle capacity (maximum load allowed; weight, volume)Freight compatibility (perishable goods, dangerous materials)Compatibility with streetsLoad/unload procedureCosts (associated with mileage, time, fuel, journey, load)


VRP – Characteristics and ComponentsDrivers

Employee workers or vehicle ownersUnion and contract conditionsWorking periods, shifts and breaksAvailability and possibility of overtime

DepotsSingle or multipleNumber and type of available vehiclesSet of a priori assigned customers

⇒ decomposable problems

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VRP – Characteristics and ComponentsOperational constraints

Relevant to:the nature of transportthe quality of servicethe driver working contract

Two classes of constraints:Local constraints (single route)Global constraints (the whole set of routes)


VRP – Characteristics and ComponentsOperational constraints

Local constraints (single route)vehicle capacitymaximum allowed route distance/durationtime constraints (arrival, departure, time windows) kind of service (pickup, delivery or both)precedence among customers:

pickup and deliverylinehaul/backhaul

Global constraints (whole set of routes)maximum number of vehiclesmaximum number of routes (for vehicle or depot)workload balancingworking periods and shifts (minimum time between routes)

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VRP – Characteristics and ComponentsObjectives

Minimize: the global transportation cost + drivers and vehicles fixed costsMinimize: the number of vehicles and/or driversBalancing of the routesMinimize: penalties for not/partially served customers

⇒ Conflicting objectives


VRP – Characteristics and ComponentsOther characteristics

Service split on several daysMore routes for vehicles in a dayMore requests for a customerDemand partially or not a priori known (dynamic, on-line problems)Stochastic and/or time dependent arc costs/travel times

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The General Vehicle Routing Problem Problem formulation

Given a generic graph G=(V, A∪E) determine a minimum cost set of M cycles (routes) that serve the required vertices U⊆V and the required edges R⊆A∪E satisfying a set of operational constraintsCost of a route = the sum of the cost of the edges belonging to the route

Applications: collection and delivery of goodswaste collectionstreet cleaningschool-bus routingdial-a-ride systemstransportation of people with handicaprouting of salespeople


The General Vehicle Routing ProblemTwo main classes of problems

Node Routing Problem (NRP)Customers/demand concentrated in sites associated with vertices

only required vertices U, R=∅frequently denoted as VRP or Vehicle Scheduling Problem

Arc Routing Problem (ARP)Customers/demand evenly distributed along the edges

only required edges R, U=∅

The problems without operational constraints:NRP reduces to Traveling Salesman Problem (TSP)ARP reduces to Rural Chinese Postman Problem (RCPP)ARP with R=A reduces to Chinese Postman Problem (CPP)

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The Node Routing Problems-

The Traveling Salesman Problem (TSP)The Capacitated Vehicle Routing Problem (CVRP)

Characteristics

Models

Algorithms (heuristics)


VRP – Node routing problemsIt concerns the distribution/collection of goodsMain components:

VehiclesDepotsDriversRoad Network

Solution:A set of routes performed by a fleet of vehicles such that:

each route starts and ends at vehicles’ depotsthe customers’ requirements are satisfiedthe operational constraints are fulfilledthe global transportation cost is minimized

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VRP – Node routing problemsRelationships among the basic VRP (NRP) problems

TSPTW

CVRP

TSP

MTSP

TSPB

DCVRP

VRPPD

VRPTW

VRPB

VRPPDTWVRPBTW

Backhauling Time windows

M vehcles

Vehicle capacity

Route length

Time windows

Backhauling

Mixed service


VRP – Node routing problemsTraveling Salesman Problem (TSP)Traveling Salesman Problem with Backhauls (TSPB)Traveling Salesman Problem with Time Windows (TSPTW)Multiple Traveling Salesman Problem (MTSP)Capacitated Vehicle Routing Problem (CVRP)Distance Constrained Vehicle Routing Problem (DCVRP)Vehicle Routing Problem with Backhauls (VRPB)Vehicle Routing Problem with Time Windows (VRPTW)Vehicle Routing Problem with Pickup and Delivery (VRPPD)

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VRP – Node routing problemsRoad graph

G=(V,A) (strongly) connected → G’=(V, A’) completeV ={0,1,...,n} , |V| = n+1, set of vertices

0 the depot

1,..., n customers’ locations (cities)

∀(i, j)∈A’ cij≥0 is a positive minimum cost (distance) of traveling from city i to city j


VRP – Node routing problemsRoad graph

if G’=(V, E’) undirected define∀i∈V δ(i)={(i, j): (i, j)∈E’} star of i

if G’ =(V, A) directed define∀i∈V δ+(i)={(i, j): (i, j)∈A’} forward star of i

∀i∈V δ-(i)={(j, i): (j, i)∈A’} backward star of i

∀S⊆V

δ(S)={(i, j): (i, j)∈A’ i∈S j∉S}∪{(i, j): (i, j)∈A’ i ∉S j∈S}

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VRP – Node routing problems

Triangle Inequality (TI)Given G=(V, A’) complete ∀(i, j, k)∈V such that i≠j ≠k

cij ≤ cik + ckj

TI is satisfied if G’ is derived from G by the computation of the shortest paths between each pair of vertices

Euclidean problemif the vertices are associated with points in a plane and the cost is given by the euclidean distance between two points (the cost matrix is symmetric)


VRP – Node routing problemsTSP (node routing problems without operational constraints)

A traveling salesman must visit his customers located in different cities and come back homeSingle vehicleRoad graph G=(V, A) → G’=(V, A’) complete

V = customers’ cities and TS home (depot); |V|=n∀(i, j)∈A’ cij≥0 the minimum cost (distance) of traveling from city

i to city j

Solution: a minimum cost route which starts and ends at depot and reaches each customer

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VRP – Node routing problemsTSP (node routing problems without operational constraints) (cont.)

Feasible solutions in G’: any hamiltonian circuit in G’Hamiltonian circuit: a closed path which visit each vertex exactly once.Solution: the minimum cost hamiltonian circuit in G’Two cases:

G’ directed: asymmetric TSP (ATSP): cij≠cji

G’ undirected: symmetric TSP (STSP): cij=cji

Complexity: Strongly NP-hard optimization problem


VRP – Node routing problemsMTSP (multiple traveling salesmen)

M vehiclesA single common depotEach vehicle (salesman) must visit at least one customer

Solution: M minimum cost routes (tours) which start and end at depot so that each customer is visited exactly once

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VRP – Node routing problemsCVRP (Capacitated VRP)

K identical vehicles

C vehicle capacity

A single common depot

∀i∈V\{0} customer a demand di≥0 is defined (d0=0) such that di≤C

∀S⊆V define

Each vehicle performs at most one route

K≥Kmin where Kmin is the minimum number of vehicles to serve all the customers

Kmin may be determined solving a Bin Packing Problem (BPP) (NP-hard problem but with fast approximation algorithms)

∑=∈Si

idSd )(


VRP – Node routing problemsCVRP (Capacitated CRP) (cont.)

∀S⊆V\{0} define r(S) the minimum number of vehicles to serve the customers in S

trivial bound

Solution: a set of exactly K routes (circuits) with minimum cost such that:

(a) each circuit visits the depot

(b) each customer is visited by exactly a single route

(c) the sum of the customer demands visited by a route does not exceed C

⎥⎥⎤

⎢⎢⎡=

CSdSr )()(

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VRP – Node routing problemsCVRP (Capacitated CRP) (cont.)

Simple variants:If K>Kmin some vehicle may not be used

find at most K routes

fixed costs for using vehicles

find the minimum number of routes

Different vehicle capacities Ck k=1,...,K

The routes must contains more than a single customer

Complexitythe CVRP is a Strongly NP-hard optimization problem

it generalizes the TSP


VRP – Node routing problemsDCVRP (Distance Constrained VRP)

A variant of CVRP:the capacity constraints is replaced by a maximum route length (time) constraint

∀(i, j)∈A’ tij≥0 the length (time) to travel from i to j

T = maximum route length (time)

Tk k=1,..,K if the vehicles are different

∀i∈V customer, a service time si may be defined explicitly or added to the travel times (t’ij= tij + si/2 + sj/2 )

The cost usually coincides with length (time)

Solution: the minimum total length (time) solution as for CVRP

DC-CVRP: both distance and capacity of vehicles are constrained

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VRP – Node routing problemsVRPTW (VRP with time windows)

A variant of CVRP:∀i∈V customer, a time window (TW) is defined as the time interval [ai, bi]

∀i∈V\{0} customer, a service time si is given

∀(i, j)∈A’ tij≥0 a travel time is given

the service for each customer must start within his TW

in case of early arrival the vehicle must wait time instant ai before starting the service

the routes starts at time 0

Travel times usually coincide with costs

TWs induce an implicit orientation (an asymmetric model can be used)


VRP – Node routing problemsVRPTW (VRP with time windows) (cont.)

A variant of CVRP:Solution: a set of exactly K routes (circuits) with minimum cost such that:



(c) the sum of the customer demands visited by a route does not exceed C(e) for each customer the service starts within the TW [ai, bi] and the vehicle stops si time instants

Complexity: Strongly NP-hard generalizes CVRP (ai=0 bi=∞)

TSPTW is the special case for K=1 and C≥d(V)

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VRP – Node routing problemsVRPB (VRP with backhauls)

An extension of CVRP:the set of customers is partitioned into Linehaul Customers (LC) and Backhaul Customers (BC)

V=L∪B |L|=n |B|=m

LC require a quantity of goods to be delivered

BC require a quantity of goods to be picked up

Precedence constraint among the LC and BC served by the same route:

all LC must be served before any BC


VRP – Node routing problemsVRPB (VRP with backhauls) (cont.)

An extension of CVRP:∀i∈B BC, di is the demand to be collected

∀i∈L LC, di is the demand to be delivered

In general, routes including only BC are not allowed

K≥max[KL, KB] where Kθ is the minimum number of vehicles to serve the customers of type θSolution: a set of exactly K routes (circuits) with minimum cost such that:



(c) the total demand of LC and BC visited by a route do not exceed separately the vehicle capacity C

(d) in each route all the LC must preceede the BC, if any

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VRP – Node routing problemsVRPB (VRP with backhauls) (cont.)

An extension of CVRP:Complexity: Strongly NP-hard generalizes CVRP (B=∅)

TSPB is the special case for K=1 and C≥max[d(L), d(B)]


VRP – Node routing problemsVRPPD (VRP with Pickup and Delivery)

A further extension of CVRP:each customer is associated with two quantities:

di demand of commodities to be delivered

pi demand of commodities to be picked up

the commodities are assumed homogeneous (sometimes only di= di - pi is specified)

for each customer is defined:Oi vertices that are origin of the delivery demand

Di vertices that are destination of the picked up demand

at each customer location the delivery is performed before the pickup

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VRP – Node routing problemsVRPPD (VRP with Pickup and Delivery) (cont.)

A further extension of CVRP:Solution: a set of exactly K routes (circuits) with minimum cost such that:



(c) the current load of a vehicle along the circuit is non negative never exceed the vehicle capacity C

(d) ∀ customer i, the customer in Oi (different from the depot) are served in the same circuit before i

(e) ∀ customer i, the customer in Di (different from the depot) are served in the same circuit after i


VRP – Node routing problemsVRPPD (VRP with Pickup and Delivery) (cont.)

A further extension of CVRP:If the origin and destination of demands are common (e.g., the depot) they may be not explicitly considered ⇒ VRP with simultaneous P & D (VRPSPD)

Complexity: Strongly NP-hard generalizes CVRP (Oi= Di={0}, pi=0 ∀i) TSPPD specializes the VRPPD for K=1

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The Traveling Salesman Problem - TSP VRP (NRP) basic and very important particular case:

single depotsingle vehicleunlimited vehicle capacity (the demand quantities are neglected)objective: find a mininum cost route to serve all the customers

The TSP solution is a minimum cost Hamiltonian circuitComplete graph with n vertices: the solution ≡ a permutation of {1,…, n}The number of possible solutions is (n-1)!


The Traveling Salesman Problem - TSP The TSP is an extensively studied NP-hard problem"[...] the TSP is one of the most intensely studied problems in computational mathematics and yet no effective solution method is known for the general case. Indeed, the resolution of the TSP would settle the P versus NP problem and fetch a $1,000,000 prize from the Clay Mathematics Institute (www.claymath.org/millennium/).[...] for over 50 years its study has led the way to improved solution methods in many areas of mathematical optimization." (www.tsp.gatech.edu//index.html)

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The Traveling Salesman Problem - TSPThe currently largest instance ever solved:the tour visiting all the 24.978 cities in Sweden (2004)

"The majority of the work was carried out on a cluster of 96 dual processor Intel Xeon 2.8 GHz workstations at Georgia Tech's School of Industrial and Systems Engineering"...

“The cumulative CPU time used in the five branch-and-cut runs and in the cutting-plane procedures for the five root LPs was approximately 84.8 CPU years on a single Intel Xeon 2.8 GHz processor. ”(www.tsp.gatech.edu//index.html)


The Traveling Salesman Problem - TSP Two cases depending on the structure of the cost matrix:

ATSP : Asymmetric TSPdirected graph G=(V,A)∀(i, j)∈A cij≠cji or cij=cji

TSP o STSP : Symmetric TSPundirected graph G=(V,E)∀ (i, j)∈E cij=cji

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The Traveling Salesman Problem - TSP The TSP is a Strongly NP-hard problem

How to solve it?Exact algorithms are based on Mixed Integer Programming (MIP) formulations and branch and cut approachesHeuristics approaches:

constructive heuristicsimprovement heuristicsmetaheuristics


The Traveling Salesman Problem - TSP Heuristics vs Exact algorithms

Heuristic algorithms should be used when:the problem dimensions exceed the maximum dimensions which can be dealt with exact approaches in acceptable computation timesthe problem, although of limited dimensions, must be solved in a very short computation timethe problem data are approximated (seeking for optimality is not worthy)there are problem variations for which exact approach cannot be easily extendedthe problem is dynamic

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The Traveling Salesman Problem - TSP The study of MIP formulations can provide insights for a better understanding of the problem properties and lead to the computations of Lower Bounds

Lower bounds provide an estimation of the quality of sub-optimal solution and are fundamental in branch & bound or branch & cut approaches

Z*TSP optimal solutionZLB lower boundZTSP a feasible sub-optimal solutionZLB≤ Z*TSP≤ ZTSP

LB

LBTSP

Z

ZZ −


The Traveling Salesman Problem - TSP Two MIP formulations

MIP formulation for the ATSPMIP formulation for the STSP

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ATSP – MIP formulation Directed and complete graph G’=(V,A’)

|V|=n|A’|=n⋅(n-1) cij=cji or cij≠cji

The solution is an hamiltonian circuit with minimum cost

VariablesBinary variables xij ∀(i, j)∈A’

⎩⎨⎧

=01

ijxif vertex j is visited immediately after vertex i

otherwise


ATSP – MIP formulation Objective function

Find the solution that minimizes Z

ijAjiij xcZ ∑

∈=

'),(

2 1

65

4

3

0

12c32c

34c

45c56c

60c

01c

ijAjiijATSP xcZ ∑

∈=

'),(

* min

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ATSP – MIP formulation Constraints - Degree constraints

In a hamiltonian circuit each vertex is visited only once:

Vix

Vjx

iVjij

jViij

∈∀=

∈∀=

∑

∑

∈

∈

1

1

}{\

}{\

i

)(iδ)(i−δ )(i+δ


ATSP – MIP formulationConstraints - STE

Degree constraints are not sufficient ...

Subtour Elimination Constraints (STE)Two classes of connectivity STE (cardinality of exponential order)A class of polynomial order cardinality STE

2 0

65

4

3

7

8 1

Subtours

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ATSP – MIP formulationConstraints – STE – Connectivity contraints

Connectivity constraints are defined for any proper subset of vertices with at least two vertices

2||,1 ≥⊂∀≥∑ ∑∈ ∉

SVSxSi Sj

ijS

2||,1|| ≥⊂∀−≤∑ ∑∈ ∈

SVSSxSi Sj

ij S

Note that the orientations must satisfy the degree contraints

1)( ≥Sδ


ATSP – MIP formulationConstraints – STE – Connectivity constraints

Drawback: the number of connectivity constraints grows as the number of subset of a given set ⇒ O(2|V|)

Example: A choice of S satisfying the connectivity constraints

2 0

65

4

37

8 1

S

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ATSP – MIP formulationConstraints – STE – Connectivity constraints

2 0

65

4

37

8 1

S

Example: A choice of S not satisfying the connectivity constraints

2 0

65

4

37

8 1

S

STE

Degree


ATSP – MIP formulationConstraints – STE – Polynomial cardinality constraints (Miller, Tucker, Zemlin, 1960)

a new set of real variables is needed: ui ∀i∈V\{0}, ui ≥ 0(n-1)⋅(n-2) ⇒ O(n2) constraints

How these constraints work?they eliminate any circuit not including vertex 0

jiVjinxnuu ijji ≠∈∀−≤⋅+− },0{\,1

2 0

65

4

3

78

1

Example

Not allowed circuit

Allowed circuit

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ATSP – MIP formulationConstraints – STE – Polynomial cardinality constraints

014849

89898989

7337

5775

4554

3443

≤⇒⋅≤⋅

≤⋅+−≤⋅+−≤⋅+−≤⋅+−

xuuxuuxuuxuu

5

4

3

7

Example: not allowed circuit

73x34x

57x45x

Always false !


ATSP – MIP formulationConstraints – STE – Polynomial cardinality constraints

Only hamiltonian circuits are finally allowedThe values assigned to ui denote the order with which the vertices are visited

3,0..33839

898989

62

2662

1661

8118

2882

==⇒≥−⇒⋅≤⋅+−

≤⋅+−≤⋅+−≤⋅+−

uugeuuuu

xuuxuuxuu2 0

68

1

Example: allowed circuit

28x

81x 16x

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ATSP – MIP formulationLower bound

Assignment Problem (AP)linear relaxation the STE constraints are eliminated

The solution is binaryThe bound is good for strongly asymmetric cost matrices (<1% from the optimal solution)

10

1

1

min

}{\

}{\

'),(

≤≤

∈=

∈=

∑

∑

∑

∈

∈

∈

ij

iVjij

jViij

ijAjiij

x

Vix

Vjx

xc


ATSP – MIP formulationAn example

a complete graph

a MIP formulation with a mathematical modelling language and a solver (Lingo – www.lindo.com)

2

1

34

5

6

0 702 454 842 2396 1196

702 0 324 1093 2136 764

454 324 0 1137 2180 798

842 1093 1137 0 1616 1857

2396 2136 2180 1616 0 2900

1196 764 798 1857 2900 0

A (symmetric) cost matrix

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AP relaxation (Z*AP= 5668)

2

1

3 4

5

6

S={1,3}

11 1615141236353432 ≥+++++++⇒≥∑ ∑∈ ∉

xxxxxxxxxSi Sj

ij



STE cut for S={1,3} (Z* = 5950)

2

1

3 4

5

6

S

11 1615141236353432 ≥+++++++⇒≥∑ ∑∈ ∉

xxxxxxxxxSi Sj

ij

112 =x

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STE cut for S={1,3} (Z* = 5950)

2

1

34

56

S’={4,5}

11 5653525146434241 ≥+++++++⇒≥∑ ∑∈ ∉

xxxxxxxxxSi Sj

ij



STE cut for S={1,3} and for S’={4,5} (Z* = 6610)

2

1

34

56

S’={4,5}

11 5653525146434241 ≥+++++++⇒≥∑ ∑∈ ∉

xxxxxxxxxSi Sj

ij

152 =x

Optimal hamiltonian circuit

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STSP – MIP formulation Undirected and complete graph G’=(V,E’)

|V|=n|E’|=n⋅(n-1) cij=cji ⇒ ce ∀e∈E’The solution is an hamiltonian circuit with minimum costThe traveling direction is not imposed

VariablesBinary variables xe ∀e∈E’ (one variable for edge)

⎩⎨⎧

=01

exif edge e is in the solution

otherwise


STSP – MIP formulationThe MIP formulation

'

2

min

)(

'

EeBxXx

Vix

xcZ

e

STE

iee

EeeeSTSP

∈∀∈∈

∈=

=

∑

∑

∈

∈

δDegree constraints

i

)(iδ

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STSP – MIP formulationThe MIP formulation

STEXx ∈

1||)(

−≤∑∈

SxSe

eσ

2)(

≥∑∈ Se

exδ

⎡ ⎤2/||||2, VSVS ≤≤⊂∀

S

2)( ≥Sδ

S


STSP – MIP formulationLower bound

A relaxationMin Hamiltonian Circuit

Min Spanning Tree (MST)

Spanning Treer-Spanning Tree

r

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A relaxationMin hamiltonian circuit is

an r-spanning treeMin Spanning Tree (MST)

Z*STSP=ZST+ca+cb

r-Spanning Treer

ca cb

Z*MST

Z*STSP≥ Z*MST+ca+cb=Zr-ST

ZST

rca cb



The Z*MST objective is a LB for Z*STSP

The Z*r-ST is a better LB for Z*STSP

The MIP formulation for the Min r-ST (r arbitrary vertex)

'

1||},{\1

2

2

min

)(

)('\

)(

'

EeBx

SrVSx

nx

x

xc

e

See

rEee

ree

Eeee

∈∀∈

≥⊂∀≥∑

−=∑

=∑

∑

∈

∈

∈

∈

δ

δ

δ

Constraints defining a ST for the sub-graph

induced in G’ by V\{r}

Constraint imposing that 2 edges are incident in r

Note that this constraint may be neglected

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The Min r-ST lower bond can be improved:

Initialize ZLB=-∞

For each i∈V

Find the MST Ti for Gi(V\{i}, Ei) the graph induced in G’ by vertices V\{i}

Find the Min r-ST where r=i adding to Ti the vertex i with 2 edges at minimum cost, and compute Z*r-ST

If ZLB< Z*r-ST then ZLB= Z*r-ST



A simple example

1

34

25

810

3

352

4

1

34

25

810

3

352

4

Z*MST=16

1

34

25

810

3

352

4

Z1-ST=26

36


STSP – MIP formulationA better lower bound: the Held and Karp Bound

The Min r-ST lower bound can be further improved (Lagrangean

relaxation of degree constraints):

Associate a multiplier πi with each vertex i

Replace the costs: c’e= ce+ πi + πj where e=(i, j)∈E’:

every tour increases its cost of

the cost of optimal r-tree changes of where δi is the degree of

vertex i in the r-tree

Then

∑∈Vi

iπ2

∑∈Vi

iiπδ

)(max

)(])2([min

][min2

**

*

*

π

ππδ

πδπ

πwwZ

wZZ

ZZ

STSP

ViiiSTrSTSP

ViiiSTr

ViiSTSP

=≥

=−+≥

+≥+

∑

∑∑

∈−

∈−

∈

is the best lower bound


STSP – MIP formulationA better lower bound: the Held and Karp Bound

The bounding procedure (subgradient optimization)

1. πi=0, i =1, …, n; Set k=0, w(π)=0, and t0 according to (a)

2. Determine the Min r-ST and set w(π)=max{w(π), Z*r-ST(π)}

3. Determine δ ki (the vertices degree at iteration k) for i=1, …, n

4. If δ ki = 2 for i=1, …, n then Z*STSP=w(π) and stop;

Else set k=k + 1;

5. if k>kmax then stop

Else set πi =πi+ tk(δ ki −2) for i=1, …, n and continue with step 2.

20)2(

))((2 ≤<

∑ −−=

∈

αδ

πα

Vi

ki

kwUBt (a)

37



A simple example

Z*1-ST=29=w(π)

δ 1=(2,2,2,2,2)

⇒Z*STSP=29

1

34

25

812

5

443

2

1

34

25

8πi=10+2=12

3+2=5

3+2-1=45-1=42+2-1=3

4-1-1=2

1

34

25

810

3

352

4

πi=0 UB=30

Z*1-ST=26= w(π)

δ 0=(2,2,1,1,4)

1411

)2630(20 ≈

++−=t

πi= πi+t0(δ 0i – 2)


CVRP – MIP formulationCVRP (Capacitated VRP) – Asymmetric network

K identical vehiclesC vehicle capacityA single common depot∀i∈V\{0} customer a demand di≥0 is defined (d0=0) such that di≤CEach vehicle performs at most one routeThe problem: find a set of exactly K routes (circuits) with minimum cost such that:

(a) each circuit visits the depot(b) each customer is visited by exactly a single route(c) the sum of the customer demands visited by a route does not exceed C

A vehicle flow formulationBinary variables xij ∀(i, j)∈A’

⎩⎨⎧

=01

ijxif (i, j) is in the solution

otherwise

38


CVRP – MIP formulationThe solution structure

K=3

101

3 7

4

9

0

2

12

8

11

56CdSd

S

Sii ≤=

=

∑∈ 1

1

1)(

}10,6,1{

CdSdS

Sii ≤=

=

∑∈ 2

2

2)(

}12,11,8,3,2{

CdSdS

Sii ≤=

=

∑∈ 3

3

3)(

}9,7,5,4{


CVRP – MIP formulationA MIP formulation

'),(

|},0{\)(

}0{\1

}0{\1

min

}0{\0

}0{\0

}{\

}{\

'),(

AjiBx

SVSSrx

Kx

Kx

Vix

Vjx

xc

ij

Si Sjij

Vjj

Vii

iVjij

jViij

Ajiijij

∈∀∈

∅≠⊆∀≥

=

=

∈∀=

∈∀=

∑ ∑

∑

∑

∑

∑

∑

∉ ∈

∈

∈

∈

∈

∈

out-degree constr.

in-degree constr.

out-degree of depot (K vehicle must be used)

in-degree of depot

Capacity Cut Constraints (CCC)

39



The Capacity Cut Constraints impose both the connectivity and the vehicle capacity requirementsr(S) is the minimum number of vehicle needed to serve the vertices in S, found as solution of a BP problem or simply fixed

⎥⎥⎤

⎢⎢⎡=

CSdSr )()(

V\S - S Cut

S

2)( =Sr

V\S

implied by degree constraintsimposed by CCC

∑ ∑∑ ∑∈ ∉∉ ∈

=⇒Si Sj

ijSi Sj

ij xx



The CCC avoid subtours not including the depot ... e.g.:

S1)( =SrV\S

S

S’

CSdSd ≤+ )'()(S

CSdSdSSSVS

>+≠⊆∀

)'()(',\'

2)'( =∪ SSr S’

40



Alternative: generalized subtour elimination constraints (GSEC)

The cardinality of CCC and GSEC is exponential

Alternative: constraints with polynomial cardinality (Christofides, Mingozzi, Toth, 1979)

∑ ∑∈ ∈

∅≠⊆∀−≤Si Sj

ij SVSSrSx },0{\)(||

Cdd

jiVjidCxCuu

ji

jijji

≤+

≠∈∀−≤⋅+−

such that

},0{\,

}0{\, VjiCud ii ∈∀≤≤ui = the vehicle load after visiting customer i



ui = the vehicle load after visiting customer i

if

otherwise

jji dCuu −≤−⇒0=ijx always true

jij duu +≥⇒1=ijx

41


CVRP – MIP formulationCVRP extensions

Tracing vehicles need O(n2K) binary variables

O(nK) binary variables

which can be eliminated noting that ∑∈

=}{\ iVjijkik xy

⎩⎨⎧

=01

ijkxif (i, j) is traversed by vehicle k

otherwise

⎩⎨⎧

=01

ikyif customer i is served by vehicle k

otherwise



KkVixx

Viy

xc

iVjjik

iVjijk

K

kik

Aji

K

kijkij

,...,1,

}0{\1

min

}{\}{\

1

'),( 1

=∈∀=

∈∀=

∑∑

∑

∑ ∑

∈∈

=

∈ =

each customer is served by exactly one vehicle

the same vehicle that enters a customer vertex also leaves it

42



KkViuBy

KkAjiBx

Cud

Cdd

KkVjidCxCuu

Ky

ikik

ijk

kiki

kji

jkijkkjkik

K

kk

,...,1},0{\

,...,1,'),(

such that

,...,1},0{\,

10

=∈∀ℜ∈∈

=∈∀∈

≤≤

≤+

=∈∀−≤+−

≤∑=

at most K vehicles are used (some vehicle may be not used)

STE and capacity constraints (different vehicles)

vehicles’ load after serving customer i



Distance constrainedtij travel time from i to j

si service time at customer i

T maximum total service time for the vehicles

Basic LBs for CVRP derive from the extension of LBs for ATSP and STSP

Exact algorithms (B&B, B&C) are able to solve small instances (n<50 sometimes n<100)

TysxtVi

ikiVi Vj

ijkij ≤+ ∑∑ ∑∈∈ ∈ }0{\

43


VRP – Heuristic approaches Heuristics in general allow finding sub-optimal, usually good, solutions in an acceptable computation time

Three main classes of heuristicsConstructive (the solution progressively build) Improvement (a given solution is progressively improved by a local search process)Metaheuristic (extend local search, consider a population o solution)

Summary:Constructive heuristics for the TSPLocal search approaches for TSPConstructive heuristics for CVRP


TSP – Constructive heuristics Constructive heuristics for the TSP:

Nearest Neighbour algorithmInsertion algorithmPatching/Merger algorithmMinimum spanning tree (based) algorithmChristofides’ algorithm

44


TSP – Constructive heuristics Performance guarantees of TSP heuristics

Bad news for the general case

TheoremLet A be a heuristic alg., A(I) the tour length produced by A and Opt(I) the optimal tour length for an instance I of the TSP. If given a constant 1≤r<∞, for all instances I

A(I) ≤ r⋅ Opt(I) then P=NP.

That means ... no hope of defining a heuristic algorithm with any fixed performance guarantee for the general TSP case

The theorem does not hold true for the STSP restricted to distance matrices satisfying the triangle inequality (metric TSP), a property mostly verified for real world logistic problems


TSP – Constructive heuristics Nearest Neighbour (NN)

Start from an arbitrary vertex v (e.g., the depot), set k=1 and U={v}the set of visited vertices

While k<nselect the next vertex w to visit such that

update U=U∪{w}, set the current vertex v=w and k=k+1

Greedy heuristic simple to be implementedBad performance: it selects short arcs only in the first iterations and final arcs are usually quite longAverage solution are 25% greater than the optimum O(n2) complexity: too slow for more than 10.000 verticesCould be a good starting solution for improvement heuristics

vhUh

vw cc∉

= min

45


TSP – Constructive heuristicsNearest Neighbour (NN)

An example: euclidean distance matrix

Bad news for NN (Theorem):For every r>0 there exists an n-vertices instance I, for arbitrarily large n, satisfying the triangle inequality such that

NN(I)≥ r⋅Opt(I)

depot: starting vertex

customer

Note that in this case in an optimal tour no crossing would appear ...


TSP – Constructive heuristics Insertion Algorithm

Base algorithmStart from a partial circuitInsert a new vertex in the circuit until all the vertices have been considered

Insertion rules (average error):– the nearest vertex (20%)– the farthest vertex (10%)– the cheapest insertion (17%)– random (11%)

46



An example (with the farthest vertex rule)Distance Matrix (symmetric)

A B C D EA 0 85 47 57 87B 85 0 43 52 38C 47 43 0 48 58D 57 52 48 0 32E 87 38 58 32 0 A

B

C

DE



First step: start from A (depot)

Distance Matrix (symmetric)A B C D E

A 0 85 47 57 87B 85 0 43 52 38C 47 43 0 48 58D 57 52 48 0 32E 87 38 58 32 0 A

B

C

DE

the farthest

length(T)=174

47



Second step: the farthest from A or E


A 0 85 47 57 87B 85 0 43 52 38C 47 43 0 48 58D 57 52 48 0 32E 87 38 58 32 0 A

B

C

DE

Tour(T, k): a procedure which selects the edge (i, j) in the tour T where to insert vertex k as

length(T)=210

( )ijkjikji

ccc −+).(

minarg



Third step: the farthest from A or E or B


A 0 85 47 57 87B 85 0 43 52 38C 47 43 0 48 58D 57 52 48 0 32E 87 38 58 32 0 A

B

C

DE

Tour(T, C): (A, B): ∆c = 47 + 43 - 85 = 5(B, E): ∆c = 43 + 58 - 38 = 63(E, A): ∆c = 58 + 47 - 87 = 18

length(T)=215

48



Fourth step: where to insert D?


A 0 85 47 57 87B 85 0 43 52 38C 47 43 0 48 58D 57 52 48 0 32E 87 38 58 32 0 A

B

C

DE

Tour(T, C): (A, C): ∆c = 57 + 48 - 47 = 58(C, B): ∆c = 48 + 52 - 43 = 57(B, E): ∆c = 52 + 32 - 38 = 46(E, A): ∆c = 32 + 57 - 87 = 2

length(T)=217


TSP – Constructive heuristics Patching/Merger algorithm

MergerStart from n single-vertex tourWhile number of tours >1

select a pair of tours (T, T’) to be merged such that is minimum

merge the selected toursIf T and T’ are single-vertex tour then Tour(T, k’) or Tour(T’, k)Else determine (i, j)∈T and (k, h)∈T’ such that is minimum

h

ki

j

}',:min{ TjTicij ∈∈

khijhjik cccc −−+

T T’

49


TSP – Constructive heuristics Patching/Merger algorithm

Patching (ATSP)Start from the subtours generated by the AP relaxation of the ATSP While number of tours >1

select a pair of tours (T, T’) to be merged such that is minimum

being (i, j)∈T and (h, k)∈T’merge the selected tours

h

ki

j

hkijhjik cccc −−+

T

T’


TSP – Constructive heuristics Minimum Spanning Tree (MST) based algorithm

FoundationsGiven a complete undirected graph G, a Hamiltonian Path (HP) is a tree in G with cost not less than the cost c(MST) of the Minimum Spanning Tree (MST)

The MST can be found in O(n2)

A Hamiltonian Circuit (HC) is a HP with an added edge ...

then the cost of the minimum cost HC is c(TSP) and c(MST)≤ c(TSP)

A circuit (possibly visiting some vertex more than once) can be generated from a MST by a procedure which doubles all the MST edges

50



FoundationsDepth first MST traversal procedure

Start from a leaf vertex (current vertex) repeatIf there is any not traversed MST edge from the current vertex follow that edge to a new current vertexotherwise go back to the vertex from which the current vertex was first reached traversing backward the edge previously followeduntil all the MST edges have been traversed twice

The procedure terminates in the starting vertexThe cost of the circuit is 2⋅c(MST) then

c(MST) ≤ c(TSP) ≤ 2⋅c(MST)



An example

A

B

DE

C

AB

D

E

C

G (edges have been neglected) MST

51



An example

A

B

DE

C

Doubled MST (DMST) circuit: B-E-D-A-D-C-D-E-B



FoundationsIf the triangle inequality holds the DMST circuit can be transformed in a HC of lesser cost, c(HC) ≤ 2⋅c(MST), by introducing shortcuts

Shortcut traversal: whenever the depth first traversal would take back to an already-visited vertex, skip ahead in the traversal and go directly to the next unvisited vertex (if all vertices have been visited come back to the start)

Algorithm (MSTS)Find a MST of GBuilt a HC by a Depth First Traversal with Shortcuts

The HC cost is lower and upper bounded c(MST) ≤ c(TSP) ≤ c(HC) ≤ 2⋅c(MST)

MSTS(I)≤2⋅Opt(I)

52



FoundationsAn example

A

B

DE

C

Traversal with shortcuts: B-E-D-A-C-B

Note that following a different depth first traversal B-E-D-C-D-A-D-E-B and introducing shortcuts the final HC would be longer


TSP – Constructive heuristics Christofides’ algorithm

Algorithm (CA)Find a MST for GTransform the MST in an Eulerian graph by introducing minimum cost edges between pairs of vertices with odd degree (minimum weight matching problem)Build a HC by an Eulerian Traversal with Shortcuts

Note that:The number of vertices with odd degree in a graph is always evenThe minimum weight matching problem can be solved in polynomial time (O(n3))It can be proved that

C(I)≤ 3/2 ⋅ Opt(I)

53



An example

A

B

D

E CF

GH

I

MST

A

B

D

E CF

GH

I

Eulerian graph by minimum weighted matching



An example

A

B

D

E CCF

GH

I

Eulerian traversal of the graph HC (shortcuts)

start A

B

D

E CCF

GH

I

start

Note that the solution generated by the MSTS algorithm is differentABCDEFGHI and longer

54


TSP – Improvement heuristics Improvement heuristics for the TSP:

Local search based on:2-OPT3-OPT

Lin-Kerninghan


TSP – Improvement heuristics Basic Local Search (LS) Algorithm

Initialization:generate a initial solution xSet the current solution and objective xc=x; Zc= Z(x)

RepeatSet xb=xc; Zb= Zc

For each candidate solution x∈N(xb) If Z(x)<Zc then xc=x; Zc= Z(x)

Until Zc< Zb

N(x) is the neighbourhood of solution xA neighbourhood of x is made of solutions that can be generated by modifying xIf xa∈N(xb) then xb∈N(xa)

55


TSP – Improvement heuristics 2-OPT and 3-OPT

The N(x) is generated by 2-exchange (3-exchange) moves 2 (3) egdes in the current solution are removed and substituted with 2 (3) other edges so that a different HC is produced

Good performance (small % deviation from optimal in the average)

If all the possible moves are explored it may be time consuming

Best Improvement (default) vs First Improvement strategy


TSP – Improvement heuristics 2-OPT

An example (ATSP)

Start HC

Arcs to be removed

Partial circuits

56



An example (ATSP)

New arcs

Flipped path

HC at step 1



An example (ATSP)

Further steps

Arcs to be removed

New arcsFlipped path

57



An example (ATSP)

Further steps

Flipped path

Arcs to be removed

New arcs



An example (STSP)

Edges to be removed

New edges

58


TSP – Improvement heuristicsLin-Kernighan (variable k-OPT)

A trade-off is neededk-exchange moves provide solutions said k-optimal:

A tour is said to be k-Optimal (or simply k-OPT) if it is impossible to obtain a shorter tour by replacing any k of its links by any other set of k links

Any k-OPT tour is also k’-OPT with k’≤kA tour is optimal if is n-OPTThe larger is k the more likely the solution is optimalIncreasing k the required computation time also rapidly increase (for naive implementation O(nk) and k is limited to 2 and 3 in practiceDrawback: the value of k must be a priori specified



Lin-KernighanOne of the best heuristic availableVariable k-OPTAverage running time O(n2.2) (original implementation)Not easy to implement ⇒ extensively studiedThe algorithm changes the value of k during its execution deciding at each iteration which value should be assigned to k

Basic LK heuristic (LKH)At each iteration step examine for ascending values of k if an interchange of k edges may result in a shorter tour:

Given that the exchange of r edges is being considered, a series of tests is performed to determine whether r+1 edge exchanges should be considered

The iterations continue until some stopping condition is satisfied

59



At each step LKH considers a growing set of potential exchanges(starting with r = 2)The exchanges are chosen so that a feasible tour may be formed at any stage of the processIf the exploration finds a new shorter tour then the current tour is replacedThere are several variations of the basic LKH

Let:X the set of r edges to be removed and Y the set of r edges to be added (initially empty)(vi, vi+1) be an edge, x an edge in X and y an edge in YG=c(xi)-c(yi) the gain of replacing xi with yi


Lin-Kernighan (variable k-OPT)An example of LKH iteration

TSP – Improvement heuristics

Initial tour T

v1

x1

v2

v3v1

x1

v2y1

x2

X={x1, x2}Y={y1}

G1=c(x1)- c(y1)>0

v4

A tour must always be closed with an edge (v4,v1)

Such edge is not chosen since G<0

60


Lin-Kernighan (variable k-OPT)An example of LKH iteration

TSP – Improvement heuristics

v3v1

x1

v2y1

x2

v4

y2x3

X={x1, x2, x3}Y={y1, y2}

G2=c(x2)- c(y2)>0

v3v1

x1

v2y1

x2

v4

y2x3

v5

v6 v5y3X={x1, x2, x3}

Y={y1, y2, y3}G3=c(x3)- c(y3)>0

A tour can be closed with an edge (v6,v1)

v6

The step executes a k=3 exchange


Heuristics for the CVRP Exact algorithms for VRP are based on models generalizing the TSP onesOnly quite small instances can be optimally solved (e.g., 100 customers, and 5 vehicles)Heuristic approaches are always used in the practice:

Constructive heuristicsTwo phase methods:

Cluster First – Route SecondRoute First – Cluster Second

Improvement heuristics

Metaheuristics: Actually the most powerful ones (but more time consuming and difficult to tune, extend, adapt) ... will be covered in a dedicated lecture

61


Heuristics for the CVRP Constructive heuristics

Sequential: the solution is built one tour at a timeParallel: more tours are built in parallel

Two main techniquesMerge existing route (savings criterion)Assign customer to routes (insertion cost criterion)

Clark and Wright Savings Algorithm Both sequential and parallel versionIt can be applied when the number of vehicles is a decision variableIt merges pairs of routes so that the end of one route continue with the start of the other in order to maximise the distance savings derived from the merges


Heuristics for the CVRP Clark and Wright Savings Algorithm (sequential version)

InitializationFor each vertex i=1,.., n generate a route (0, i, 0)

For each i compute the savings sij=ci0+c0j-cij for j=1,…, n, j≠i, and sort the savings in non increasing order

Define route (0, 1, 0) as the current route

Depot

62


Heuristics for the CVRP Clark and Wright Savings Algorithm

Iteration (sequential version: route extension)Let route (0, i,…., j, 0) be the generic current route

Determine the first saving ski or sje so that the route can be feasibly merged with another route including edge (arc) (k,0) or (0,e).

If a feasible merge exists then implement it and iterate with the current routeElse

If there are other routes not considered so far, then select a next route as current route and iterateElse stop (no more feasible merge is possible)



An example

i j

1) Current route (0, i, 0)

ijjiij cccs −+= 001) Best saving

ij

2) Current route (0, i, j, 0)

k

63



An example

i

j

2) Current route (0, i, j, 0)

k

jkkjjk cccs −+= 00

2) Best saving

i

j

k

3) Current route (0, i, j, k, 0) cannot be extended anymore in a feasible way.

The next current route is (0, h, 0)

h



An example

ghhggh cccs −+= 00

3) Best saving

i

j

k

3) Current route (0, h, 0)

h

g

i

j

kh

g

4) Current route (0, g, h, 0)

No more merge is feasibly possible.

A 3-routes solution is found !

64


Heuristics for the CVRP Clark and Wright Savings Algorithm (parallel version)

InitializationFor each vertex i=1,.., n generate a route (0, i, 0)

Compute the savings sij=ci0+c0j-cij for j=1,…, n, j≠i, and sort the savings in non increasing order

Iteration (parallel version: best feasible merge)If a positive saving does exist then

Starting from the largest saving sij determine if the two routes, one including edge (arc) (i, 0) and the other (0, j) can be feasibly merged;If so, merge the two routes by deleting (i, 0) and (0, j) and iterate with the next saving



An example: Symmetric and euclidean distancesVehicles’ capacity C=8Customer demands associated with vertices

Depot

E,2

A,3

C,4F,2

B,2

D,3 0417191814041F410611121138158E71610729210341D91112720369554C81113923607058B40811039570064A4158415458640Dep

FEDCBADepcij

65



An example:

Initial routes and lengths

• c(0-A-0) = 128

• c(0-B-0) = 116

• c(0-C-0) = 108

• c(0-D-0) = 82

• c(0-E-0) = 116

• c(0-F-0) = 82

Total cost = 632

Number of vehicles = 6

Depot

E,2

A,3

C,4F,2

B,2

D,3



An example:

0581141865F580380341E113802372D

402307623C183776052B6541223520A

FEDCBAsij

Depot

E,2

A,3

C,4F,2

B,2

D,3

sAB= cADep+cDepB-cAB

= 64 + 58 – 70 = 52

64

58

70

66



An example:

0581141865F580380341E113802372D

402307623C183776052B6541223520A

FEDCBAsij

Depot

E,2

A,3

C,4F,2

B,2

D,3No more merges are possible without violating the capacity constraint

3 vehicles

total cost=433



An example:

0581141865F580380341E113802372D

402307623C183776052B6541223520A

FEDCBAsij

Depot

E,2

A,3

C,4F,2

B,2

D,3 The heuristic is not able to find a 2-vehicle solution (total cost=459)

67


Heuristics for the CVRP Two phase methods: Cluster First – Route Second

Phase 1: ClusteringA clustering problem is solved to assign each customer to a single vehicle

Phase 2: RoutingFind the route for each vehicle (solving a TSP problem)



Methods:Elementary clustering methods

Sweep algorithmFisher-Jaikumar Generalized Assignment (GA) based algorithmLocation-based heuristic

Truncated Branch-and-Bound approachesLevels of the exploration tree ⇒ vehicle routesEach level contains a set of (partial) feasible routes generated by one or more criteria (e.g., savings)Branching ⇒ route selection

Petal Algorithms

68



Sweep AlgorithmPlanar VRPFeasible cluster initially obtained by rotating a ray centred at the depotA vehicle route is found by solving a TSP problem for each cluster

ray

Cluster 1 (customers assigned to vehicle 1)

Cluster 2

Cluster 3

Cluster 4

Cluster 5


Two phase methods: Cluster First – Route SecondSweep Algorithm

Planar VRPFeasible cluster initially obtained by rotating a ray centred at the depotA vehicle route is found by solving a TSP problem for each cluster

Heuristics for the CVRP

Route 1

Route 2

Route 3

Route 4

Route 5

69



Fisher-Jaikumar Generalized Assignment based algorithm

Step 1 (seed selection): select a seed jk in V for each cluster k=1,...,KStep 2 (allocation of customers to seed): compute cik the cost of allocating customer i to k as the cost of inserting i in the route 0-jk-0

Step 3 (generalized assignment): Solve a GA problem with costs cij, weights for the customers di, and vehicle capacity QStep 4 (TSP solution): solve a TSP for each cluster found

)(},min{ 000000 kjkjiikjkjkjkijiik ccccccccc +−++++=



Generalized Assignment Problem

GAP is a strong NP-hard for which both exact and heuristic approach exist

i

Customers Vehicles/Clusters

kxik

dicik

Q

Kknix

KkQxdb

nixa

xc

ik

n

iiki

K

kik

n

i

K

kikik

,...,1;,...,1}1,0{

,...,1)(

,...,11)(

min

1

1

1 1

==∀∈

∑ =∀≤⋅

=∀=∑

∑ ∑ ⋅

=

=

= =

(a) assignment constraints(b) capacity contraints

70


Two phase methods: Cluster First – Route SecondLocation based heuristic

Step 1 (concentrators selection): select a concentrator jk in V for each cluster k=1,...,K such that the total distance of allocating the n customers to the closest concentrator is minimized, and the total capacity of each cluster does not exceed QStep 2: Build the vehicle route for each cluster by inserting a customer at a time with the minimum (estimated) insertion cost


K=2


Two phase methods: Cluster First – Route SecondPetal heuristic

Order the vertices (v1, v2,..., vn), e.g., use sweep order.

For every vertex vi create clusters {vi}, {vi, vi+1}, {vi, vi+1, vi+2}, ..., as long as it is possible to feasibly serve the cluster by a single vehicle (Note that subscripts should be interpreted modulo n)


swee

p dire

ction

Cluster 1

Cluster 2

Cluster 3

Cluster 4

71



Petal heuristic (cont.)Compute for each cluster k the cost ck by solving (exactly or with a heuristic) the associated TSP problemDefine end solve a set partitioning problem (SPP) associating a column with each cluster and assigning ck cost to it

Rkx

Vixa

xc

k

Rkkik

Rkkk

∈∀∈

∈∀=∑

∑

∈

∈

}1,0{

1

min

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

==

111111

1111111

][ ikaA

If the incidence matrix A is an interval matrix the SPP is polynomially solvable

RouteServed

customers


Heuristics for the CVRP Two phase methods: Route First – Cluster SecondBeasley's algorithm

Phase 1: RoutingSolve a single TSP problem relaxing the capacity (duration) constraints

Phase 2: ClusteringCut the TSP solution into routes that satisfy the capacity (duration) constraints

Note: Phase 2 can be optimally solved in polynomial time using the approach of the Petal heuristicRFCS heuristic is a special case of the Petal one

72


Heuristics for the CVRP Two phase methods: Route First – Cluster Second

An example: Vehicles’ capacity C=8

Depot

E,2

A,3

C,4F,2

B,2

D,30417191814041F

410611121138158E

71610729210341D

91112720369554C

81113923607058B

40811039570064A

4158415458640Dep

FEDCBADepcij

The TSP solution Route length=343 (not feasible)




Depot

E,2

A,3

C,4F,2

B,2

D,3

0417191814041F

410611121138158E

71610729210341D

91112720369554C

81113923607058B

40811039570064A

4158415458640Dep

FEDCBADepcij

First set of clusters/routes Total length=484 (feasible)

starting vertex

73




Depot

E,2

A,3

C,4F,2

B,2

D,3

Second set of clusters/routes Total length=433 (feasible)

starting vertex ... and so on

0417191814041F

410611121138158E

71610729210341D

91112720369554C

81113923607058B

40811039570064A

4158415458640Dep

FEDCBADepcij


Heuristics for the CVRP Improvement heuristics

Based on LS: exploration of a neighbourhood N(x) of solutionsN(x) is built using “moves”

Possible moves:Insert a customer in a different position in the sequence of visitSwap the positions of a pair of customersk-Opt

Two classes of methods:Single route improvement

the assignment of customers to routes (vehicles) not change (analogous to TSP improvement heuristic)

Multi route improvementthe moves may also change the customer-route assignment

74



Multi route improvements:String crossing

Two chains of customers are exchanged by exchanging the endpoints of two edges in two different routes



Multi route improvements:String relocation

A chain of at most k customers is moved from one route to anotherHigher value of k may produce better results but the neighbourhood exploration time also increases

75



Multi route improvements:String exchange

Two chain of at most k customers are exchanged between two routesIf k = n then string exchange contains string cross as well (but this is a large

neighborhood which is time consuming to evaluate).The values for k are usually small {1, 2, 3}

String relocation is a special case if k1>0 and k2=0String crossing is a special case if k1=k2 and the strings are at the routes’ ends.


The Arc Routing Problems-

The Chinese Problem (CPP)

Characteristics

Models

Algorithms (heuristics)

76


ARP – Arc Routing ProblemsARPs concern the distribution/collection of goods or materials along the arcs (edges) of a road networkSame main components as NRP The density of customers along streets is sufficiently high to consider the associated arc (edge) the key network element to be served Solution:

A set of routes performed a fleet of vehicles such that:each route starts and ends at vehicles’ depotsthe requests for service associated with arcs or edges are satisfiedthe operational constraints are fulfilledthe global transportation cost is minimized


ARP – Arc Routing ProblemsOperational constraints of the same kind of the ones for NRP (e.g., number of available vehicles, vehicles’ capacity, route duration ...)ARP without operational constraints:

All arcs (edges) must be served: the Chinese Postman Problem (CPP)Only a subset of arcs (edges) must be served: the Rural Chinese Postman Problem (RCPP)

Other ARP variations:The Windy Chinese Postman ProblemThe Stacker Crane Problem

77


ARP – Arc Routing ProblemsThe CPP is characterized by:

A single vehicleAll the arcs (edges) must be servedThe service demand is neglected (as well as the vehicle capacity)For each arc (edge) the service cost and the traversing cost (without service) are known

Solution: a minimum cost route that traverses at least once every arc (edge)

An example

1starting vertex = depot 2

3

45

678

9

10

111213Does a route exist which traverses

each edge exactly once?edge traversal without service = deadheading


ARP – Arc Routing ProblemsThe Könisberg’s Bridges problem (Euler, 1707-1783)

Pregel river

Since the end of World War II (1946), Königsberg became a Russian city named Kaliningrad

The capital of Eastern Prussia, Königsberg

78



Is it possible to cross each bridge exactly once starting and ending at the same point? (Euler, 1736)

The answer is ...

A

B C

D

A

B

D

CPregel river

The graph model is a Multi-Graph



The answer is ... the same of the following decision problem:

Is the graph in the figure an eulerian graph ?i.e., does it contain a tour (closed walk) that transits on every edge exactly once?

Unfortunately the answer is no ! (the graph is not an eulerian one)

A

B

D

C

79


ARP – Arc Routing ProblemsEuler showed that for an eulerian cycle to exist, all the vertices must have even degree

Note that an eulerian (not closed) path can exist if the graph includes at most two vertices with odd degree

Every vertex has an odd degreeNote that

A

B

D

C

|δ(A)|=3

|δ(C)|=3

|δ(D)|=3

|δ(B)|=5

∑ ⋅=∈Vi

i E ||2|| δ


ARP – Arc Routing ProblemsAn eulerian tour in a graph is a tour that contains each edge (arc) exactly once

A postman tour in a graph is a tour that contains each edge (arc) at leastonce

If the (multi) graph G(V,A) is eulerian the optimal solution to the CPP is trivially

and the CPP problem reduces to find the eulerian tour in the graph

If G is not eulerian the CPP solution is a tour that transit on a subset of arcs (edges) more than once

Then the problem reduces to add to G copies of the arcs (edges) so that G becomes eulerian while the increment of the cost so generated

∑=∈Ee

eCPP cZ *

80


ARP – Arc Routing ProblemsHow to determine if a connected graph G is eulerian?

An undirected G(V, E) is eulerian ⇔ ∀i∈V |δ(i)| is even (G is even)A directed G(V, A) is eulerian ⇔ ∀i∈V |δ -(i)|=|δ +(i)| (G is symmetric)A mixed G(V, A∪E) is eulerian ⇔ G is even and satisfies the following balanced set condition (balanced)

For every subset of vertices S⊂V, the difference between the number of directed arcs that cross the cut (S, V\S) from S to V\S and the number of the directed arcs from V\S to S must be less than or equal to the number of undirected edges crossing (S, V\S)

5

2

4

3

1

S V\SEulerian

mixed graph 5

2

4

3

1Non-Eulerianmixed graph

Examples


CPP – The Chinese Postman ProblemFormulation of the Undirected PP (UPP)

Notation:G(V, E) undirected graphVO⊆V the subset of odd verticesAG=[aie i∈V, e∈E] the vertex-edge adjacent matrix (aie=1 if e∈δ(i), 0 otherwise)

Variables:xe = number of times edge e is added to G to transform it into an eulerian graph (xe=0 for every e∈E if G is eulerian)wi for i∈V a positive integer

81


CPP – The Chinese Postman ProblemFormulation of the Undirected PP (UPP)

ViwEex

Viwxa

Viwxa

xc

i

e

OiEe

eie

OiEe

eie

Eeee

∈∀≥∈∀≥

∉∀=∑

∈∀+=∑

∑

∈

∈

∈

integer 0integer 0

2

12

min

Every vertex must have an even degree

∑⇒∑ +=∈∈ Ee

eEe

eeUPP cxcZ )1( ** for eulerian graphs


CPP – The Chinese Postman ProblemSolution of the UPP

The UPP is solvable in polynomial timeIf G is eulerian determine the eulerian tour

Fleury AlgorithmEnd-Pairing Algorithm

If G is not eulerian:First add copies of the edges in order to transform it into an eulerian graph G’ with the minimum additional cost by solving a Matching Problem (polynomial)Then determine the eulerian tour in G’

82



The Fleury Algorithm1. Let T=E, W=V. Select a starting vertex v=i2. Select randomly an edge e∈T incident in v, e∈δ(v), e=(v, h), such

that e is not a bridge (if no other choice exists the graph is not eulerian)

3. Traverse e and remove it from T; if δ(v)∩T=∅ then W=W\{v}; set v=h

4. If T≠∅ and v≠i then continue with step 2

Definition:An edge is a bridge if the number of connected components in F(W,T\{e}) is increased by 1



The Fleury AlgorithmAn example: G eulerian

starting vertex

21

traversed edge 1

3

2

1

3

forbidden traversal

bridge edge

2

1

34

5 67

89

10F is disconnected

83



The End-Pairing Algorithm1. Trace a first simple T1 tour in G (that may not contain all

vertices). If all the edges are traversed then stop2. Consider any vertex v on T1 which is incident to an edge not on

the tour, and form a second tour T2 from v not overlapping T1

3. Merge T2 with T1: starting from v follow T1 and then continue with T2

4. If all the edges are traversed then stop, otherwise continue with step 2

Both Fleury and End-Pairing algorithms are easily generalized to directed eulerian graphs



The End-Pairing AlgorithmAn example

A BC

D

E

F

GH

A BC

D

E

F

GH

T1=A-B-F-A

A BC

D

E

F

GH

T2=A-C-D-A

A BC

D

E

F

GH

T1 merged with T2: T1=A-B-F-A-C-D-A

T2=D-E-G-DT1 merged with T2:

T1=A-B-F-A-C-D-E-G-D-A

84



Undirected non eulerian G(V, E): the Matching Problem1. Let E’=E. Identify the set VO of odd degree vertices and compute the

shortest paths spij between all pairs of vertices i, j∈VO i<j2. Build a complete undirected graph GO(VO,EO), EO={(i, j): i, j∈VO i<j} ,

and associate the costs spij to each edge in EO

3. Find a minimum cost matching M in GO where M⊂EO such that every v∈VO is incident to exactly an edge in M

4. For each edge (i, j)∈M add to E’ a copy of the edge included in the shortest path between i and j

5. Form the eulerian graph G’(V,E’)6. The optimal solution of the UPP corresponds to the eulerian tour in G’

The minimum matching problem can be solved with polynomial algorithms with complexity of O(|V|3) or less


Solution of the UPPUndirected non eulerian G(V, E): the Matching Problem

An example

CPP – The Chinese Postman Problem

AB

5

DE

FG

H

A B

C E

eulerian graph with doubled

edges

3 2 1

6

524

34

2

3 1

CA

B

5

DE

FG

H

3 2 1

524

34

2

3 1

C

6

22

6

73

3

4

2

AB

DE

FG

H

C

A-B-E-F-H-G-A-C-G-F-D-B-E-D-A

85


CPP – The Chinese Postman ProblemFormulation of the Directed PP (DPP)

Notation:G(V,A) directed graph∀i∈V in-degree and out-degree∀i∈V

Variables:xij = number of additional copies of arc (i, j) added to G to transform it into an eulerian graph (xij=0 ∀(i, j)∈A if G is eulerian)

|)(|)( iid −− = δ |)(|)( iid ++ = δ)()()( ididib +− −=


CPP – The Chinese Postman ProblemFormulation of the CPP Directed PP (DPP)

Ajix

Viibxx

xc

ij

ijji

ijij

Ajiijij

∈∀≥

∈∀=∑−∑

∑

−∈+∈

∈

),(integer 0

)(

min

)()(

),(

δδ Every vertex must be symmetric

∑⇒∑ +=∈∈ Aji

ijAji

ijijDPP cxcZ),(),(

** )1( for eulerian graphs

86


CPP – The Chinese Postman ProblemSolution of the DPP

The DPP is solvable in polynomial timeIf G is eulerian determine the eulerian tour

Fleury or End-Pairing Algorithms adapted to directed graphsvan Aardenne-Ehrenfest and De Bruijn Algorithm

If G is not eulerian:First add copies of the arcs in order to transform it into an eulerian graph G’ with the minimum additional cost by solving a Transportation Problem (polynomial)Then determine the eulerian tour in G’



The van Aardenne-Ehrenfest and De Bruijn Algorithm1. Build an spanning in-tree arborescence routed in an arbitrary vertex v2. Label all arcs in G as follows: order and arbitrary label the arcs

outgoing from v; order and label the arcs outgoing from any other vertex consecutively in an arbitrary way but selecting as the last arc an arc used in the arborescence

3. Starting from an arbitrary vertex traverse the lowest labelled arc outgoing from it and continue by leaving any entered vertex following the arc not yet traversed with the lowest label. When all arcs have been traversed the eulerian tour is found

Definitions:An in-tree is an oriented tree in which a single vertex is reachable from any other one.An spanning arborescence of a directed graph G (if it exists) is an arborescence with include every vertex of G

87



The van Aardenne-Ehrenfest and De Bruijn AlgorithmAn example

A BC

D

E

F

GH

Starting from vertex D

D-G-E-D-C-A-H-F-B-A-D

in-tree spanning arborescence

A BC

D

E

F

GH

labelling of arcs

A BC

D

E

F

GH

7

6

5

12

4

10

38 9



Directed non eulerian G(V, A): the Transportation Prob.1. Define the subset of V

S={i∈V: b(i)>0} (supply vertices)D={i∈V: b(i)<0} (demand vertices)

2. Compute the shortest paths spij between all pairs of vertices (i, j) such that i∈S, j∈D

3. Solve a transportation problem for the supply and demand vertices defined with the spij costs, finding the optimal flow yij for each pair (i, j)

4. For each yij>0 augment the original graph G with the copies of the arcs associated to the shortest path between vertex i and j, forming the eulerian graph G’(V,A’)

5. The optimal solution of the UPP corresponds to the eulerian tour in G’

The transportation problem can be solved with complexity of O(|V|3)or O(|A||V|2)

88



Directed non eulerian G(V, A): the Transportation Prob.An example

S={B,D}

D={A,C,F}

the transportation

problem

B C

D

E

A4

1

85

10

35

F2

6

bB=1

bA=-1

bF=-1

bC=-1

bD=2 bE=0

∑ =∈Vi

ib 0)(

B

D

A

C

F

bB=1

bD=2

bA=-1

bC=-1

bF=-1

spBA=17 spBC=8spBF=19 spDA=5spDC=18 spDF=7

yBA=0 yBC=1yBF=0 yDA=1 yDC=0 yDF=1



Directed non eulerian G(V, A): the Transportation Prob.An example

Starting from vertex B

B-C-D-A-F-B-C-E-D-A-F-D-A-B

B C

D

E

A

F

B C

D

E

A

F

B C

D

E

A 813

6

112 9

7

F

3

10

2114

5

89


CPP – The Chinese Postman ProblemThe Mixed PP (MPP)

If G(V, A∪E) mixed graph is eulerian (even and balanced) the postman tour is found as follows:

assign a direction to some edges in order to make G symmetric

complete the orientation of the remaining edges to transform Ginto an eulerian directed graph

determine the postman tour with one of the available algorithms

If G is even but not balanced the problem is solved by the Minieka’s Algorithm

Note that if G is even and symmetric is also balanced but symmetry is not a necessary condition to eulerian



The Ford-Fulkerson’s procedure for transforming an eulerian mixed graph G(V, A∪E) into a eulerian directed graph

Transform G into a symmetric graph1. replace each (i, j)∈E with two directed arcs (i, j) and (j, i)

transforming the graph into G(V, A’). Assign to each arc in A a lower bound of 1 and to each arc in A’\A a lower bound of 0. Assign to any arc in A’ an upper bound of 1.

2. Determine a feasible circulation for the network flow model so defined. Let xij the flow in arc (i, j)

3. Orient the edge (i, j)∈E as arc (i, j), if xij=1 and xji=0, remove the edge from E and include the arc in A.

90


CPP – The Chinese Postman ProblemThe Mixed PP (MPP): eulerian graphs


Transform G into a symmetric graph: an example

10

11

≤≤

≤≤

jk

ij

x

xsolution of the network

flow problem (circulation)

i

j

k

i

j

k


CPP – The Chinese Postman ProblemThe Mixed PP (MPP): eulerian graphs


Transform the mixed symmetric graph G into a directed symmetric one1. If E is empty then stop2. Le i a vertex with at least an incident edge (i, j)∈E. Set v=i and w=j3. Orient (v, w) from v to w. If w=i go to step 14. Set v=w and identify an edge (v, w) incident to v. Go to step 3

91


CPP – The Chinese Postman ProblemThe Mixed PP (MPP): eulerian

The Ford-Fulkerson’s procedure for transforming an eulerianmixed graph G(V,A∪E) into a eulerian directed graph

Transform the mixed symmetric graph G into a directed symmetric one: an example

v1

w1

v2

w2v3

w3

v4

w4



Even graphThe Minieka’s Algorithm transforms an even unbalanced mixed graph G(V, A∪E) into an even and balanced one (i.e., into an euleriandirected graph)The algorithm is based on the definition and solution of a minimum cost flow problem:

if no feasible solution exists then the graph cannot be balancedand no postman tour exists

For not even mixed graph the problem is NP-hardA possible heuristic is a two-stage procedure:

(1) optimally adding edges (matching) to make the graph even(2) to apply the Minieka’s algorithm

The procedure may not find the optimal solution even if each single stage is performed optimally.

92


RCPP – The Rural Chinese Postman ProblemThe Rural CPP

Given a general G(V, A∪E) only a subset R⊂A∪E of arcs and edges must be servedSolution: a minimum cost circuit that traverses each arc and edge in R at least onceNotation: let GR the graph induced in G by R

Complexity:G directed or undirected and GR connected

⇒ RCPP polynomially solvableG directed or undirected and GR not connected

⇒ RCPP is NP-hard


RCPP – The Rural Chinese Postman ProblemThe Rural CPP

Algorithm for G undirected and GR connectedDetermine the minimum cost paths in G between every pair of odd vertices in GRSolve a minimum cost matching for such verticesAugment GR by adding the edges corresponding to the minimum matchingSolve a UPP on GR

RG

93


RCPP – The Rural Chinese Postman Problem

The RCPP: G undirected and GR not connectedBalance-and-Connect Heuristic

Build and extend GR by solving a matching problem to obtain and even graphConnect the GR components by solving a MST problem on a complete graph T(Vc, Ec) such that

the vertices in Vc correspond to the components of GR

the cost associated with each edge (r, s)∈T is computed as the minimum cost path connecting any pair of vertices in G belonging to the components of GR the node r and s are associated with

Apply the algorithm for RPP with connected GR


RCPP – The Rural Chinese Postman Problem

The RCPP: directed GThe approaches for the undirected case can be extended:

The exact algorithm when GR is connected is extended by solving a transportation problem to make GR symmetricThe balance-and-connect heuristic is extended by solving a minimum spanning arborescence problem to connect the components of GR

An example (Directed RCPP)

2

2 2

2

2

1 1

3

5 5

94


RCPP – The Rural Chinese Postman ProblemThe RCPP: directed G

A “reversed” heuristic:Connect-and-Balance

Strongly connect the components of GR

Obtain an even graph

2

2 2

2

2

1 1

3

5 5 2 2

2

1 15 5

2

2 2

2

2 3

Balance&

Connect

Non-required arcs cost=12

Non-required arcs cost=5


RCPP – The Rural Chinese Postman ProblemOther PP

The Stacker Crane ProblemA RPP on a mixed graph where R=A (NP-hard – close to TSP)

The Windy CPPA UPP where the costs of edge traversing depend on the travel direction (polynomially solvable if the graph is eulerian, NP-hard otherwise)

95


CARP – The Capacitated ARPARP with operational constraints

Components:A connected road network G =(V, A)A cost (distance) cij is associated with each arc (i, j)∈AA demand dij is associated with each arc (i, j)∈R⊆AA fleet made of K vehicle with capacity CA depot associated with vertex 0

Solution: A set of K routes, all including vertex 0, covering all the arcs in Rand such that:

the total demand served by each vehicle does not exceed the capacity Ceach arc in R is served by a unique route (vehicle)

Objective:Minimize the total cost (distance) of the K routes


CARP – The Capacitated ARPExamples of CARP applications:

Street sweepingWinter grittingRefuse collectionElectric meter readingAirline scheduling

96


CARP – MIP formulationCARP – Symmetric network G=(V, E)

R⊆E R={(i, j)∈E: dij>0}

k=1,...,K identical vehicles with capacity C

Binary variables

Formulation (Golden & Wong)

traversed),( if 1 Ejixijk ∈=served ),( if 1 Rjilijk ∈=

KkRjilx

KkVixx

xc

ijkijk

Ejijjikijk

K

k Ejiijkij

,...,1),(

,...,10)(

min

),(:

1 ),(

=∀∈∀≥

=∀∑ ∈∀=−

∑ ∑

∈

= ∈

(Route continuity)

(Objective)

(Serviced edge must be traversed)


CARP – MIP formulationCARP – Symmetric network G=(V,E)

Formulation (cont.)

KkVS

uyuy

ux

ynSx

KkCdll

Rjill

kS

kS

kS

kS

kS

Si Sjijk

kS

Sjiijk

Rjiijjikijk

K

kjikijk

,...,1}0{\

}1,0{,1

1

1||

,...,1)(

),(1)(

2

,

),(

1

=∀⊆∀

⎪⎪⎭

⎪⎪⎬

⎫

∈≤+

−≥

+−≤

=∀≤+

∈∀=+

∑ ∑

∑

∑

∑

∈ ∉

∈

∈

=

(Capacity)

(Required edges must be served)

(Elimination of disconnected subtours)

97


CARP – MIP formulationDisconnected subtour elimination constraints

Examples

00 == kS

kS uy

Simpose that:

• vehicle k cannot traverse more than 2 edges of S

• S is connected to V\S by at least 1 edge

S

01 == kS

kS uy

impose that:

• vehicle k is allowed to traverse the 3 required edges of S

• S is connected to V\S by at least 1 edge


CARP – The Capacitated ARPCARP is NP-hard even for directed and undirected graphs

Heuristics for the CARPClasses of heuristics:

Constructive heuristicsImprovement heuristicsTwo-phase methods

Cluster First – Route SecondRoute First – Cluster Second

An undirected graph G=(V, E) is assumed in the following

98


Heuristics for the CARPConstructive heuristics: The Path Scanning Algorithm

1. Let RU⊆R the subset of service edges not yet served.Initialize RU=R, k=1, v=0

2. (Decision at vertex v) If an edge can be selected from RU that can be served by route k, then

serve the edge and remove it from RU

update the vehicle capacity and the vehicle position v with the reached vertex wgo to step 2

Otherwise reach the depot following the shortest path SP(v,0), and set k=k+1 and v=0

3. If k≤K and RU≠∅ then go to step 2Otherwise stop


Heuristics for the CARPConstructive heuristics: The Path Scanning Algorithm

The edge selected at each iteration (step 2) from RU must satisfy the vehicle capacity constraint and can be chosen with one among the following criteria:

the cost increment is minimizedthe residual capacity is maximizedthe distance from the depot is either maximized or minimizedthe distance from the depot is maximized if half of the vehicle capacity is available, otherwise is minimizedrandom

99


Heuristics for the CARPConstructive heuristics: The Augment-Merge Algorithm

Build a distinct tour for serving each single edge in RAugment:

starting from the longest tour, serving, e.g., edge e, extend the service to edges different from e but included in the tour as long as it is allowed by the vehicle capacity

Merge: compute the savings caused by any possible merge of pairs of routessort the savings in non increasing ordermerge the pair of tours starting from the greatest saving as long as the vehicle capacity allowed it and the savings is positive


Heuristics for the CARPConstructive heuristics: The Augment-Insert Algorithm

Initialize:Set RU=R, and the list L of edges (i, j) in RU in descending order of SP(0, i) + SP(j,0). Find the least cost cycles C(i, j) that serves the required edges

IterationAugment:

Select the first (farthest) edge in RU from L. If no edge is found go to the following step. Augment C(i, j) by assigning edges in RU on the cycle following L until the vehicle capacity is exhausted. Remove the serviced edges from RU

Insert:Following L insert in the current cycle edges in RU as long as the insertion cost does not exceed the upper bound CLIM (parameter) or the vehicle capacity is exhausted.

100


Heuristics for the CARPConstructive heuristics: The Augment-Insert Algorithm

In a version II of the algorithm the priority is given to the edge with smaller demand

The value assigned to CLIM influence the length of the cycles, and larger values allows greater detours


Heuristics for the CARPConstructive heuristics: The Construct-Strike Algorithm

IterationConstruct:

By sequentially adding edges form a capacity-feasible cycle Csuch that G-C remains connected

Strike: remove the cycle from G and repeat the construct step until all the required edges are served or no more cycles are found.

Remove any non required edges added in previous iterationObtain an eulerian circuit by solving a matching problem for the odd degree vertices remaining in G and two copies of the depot (that must have even degree)Iterate until all the required edges are served

101


Heuristics for the CARPConstructive heuristics: The Construct-Strike Algorithm

Criteria similar to the Path-Scanning can be used for edge selection in the construct phaseA good quality alternative is to select the (i, j) edge that maximize the total demand on the path from j back to the depot with the smallest total demand

The heuristics behaving better:Augment-InsertConstruct-Strike (with the above selection rule)Path Scanning with random insertion

No constructive algorithm is dominant


Heuristics for the CARPTwo-Phase heuristics: The Route First - Cluster SecondAlgorithm (Ulusoy)

Route: Determine a single tour C that include all the required edges but disregards the vehicles’ capacity (solve a RCPP)

Balance&

Connect

102



Cluster: Build an auxiliary directed graph T = (V,A) as follows:

the vertices in V correspond to the edges in R, plus a final vertex feach arc (i, j)∈A connects a pair of vertices associated with two edges in R such that i is visited before j and the partial tour from ito j-1 can be feasibly served by vehicle. An arc (i, f)∈A means that all the edges from i to the end of tour C can be feasibly served by a vehiclea cost is associated with each arc (i, j)∈A which includes the cost from depot to edge i, from edge i to edge j-1 and from this latter to the depotFind the min cost path on T from the vertex associated with the first edge of C to vertex f. Each arc on such path corresponds to a vehicle route on the original network



fa

b

cd

ea b c d e

3

2Vehicle capacity = 5

2

2

3

cac ⇒ cost of cycle =

from depot to edge a

+ from edge a to edge b

+ from edge b to depot

cac

103


BibliographyHandbooks in Operations Research and Management Science, Vol. 8: Network. Routing, eds M. O. Ball, T. L. Magnanti, C. L. Monma, and G. L. Nemhauser, Elsevier, Amsterdam, NL (1995).

The vehicle routing problem, eds P. Toth, D. Vigo, SIAM, Philadelphia, USA (2001).

Scienze delle decisioni per i trasporti, eds S. Pallottino, A. Sciomachen, Franco Angeli, Milano, I (1999).

Arc Routing : Problems, Applications and Algorithms. Eiselt,H.A., Gendreau,M., Laporte,G., Centre for Research in Transportation Edition, University of Montreal, (1992).

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Vehicle Routing Problems - Massimo Paolucci Home Page · 2 VRP – Massimo Paolucci 3 Vehicle Routing Problems VRPs include: Static problems: the service de mand is fixed and a priori