R outing, A nycast, and M ulticast for Mesh and Sensor Networks

Routing, Anycast, and Multicastfor Mesh and Sensor Networks

Roland FluryRoger Wattenhofer

RAM

DistributedComputing Group

2Roland Flury, ETH Zürich @ Infocom 2007


All-In-One Solution


Routing in Mesh and Sensor Networks

• Unicast Routing (send message to a given node)

• Multicast Routing (send message to a given set of nodes)


Routing in Mesh and Sensor Networks (2)

• Anycast Routing (send message to any node of a given set)

Unicast MulticastAnycastUnicast

AnycastMulticast

ALL IN ONE


Modeling Wireless Networks

Routing in limited, wireless networks– Limited Storage– Limited Power– No central unit, fully distributed algorithms

Description of network topology– Undirected Graph G=(V,E)– Vertices V: Set of network nodes– Edges E: present between any two connected nodes

Wireless Networks– Nodes tend to be connected to other nodes in proximity– Connectivity graphs: constant doubling


Constant Doubling Metrics

Ball: Bu(r) := { v | v 2 V and dist(u, v) · r }8 u 2 V, r > 0 : 9 S µ Bu(r) s.t.

8 x 2 Bu(r) : 9 s 2 S : dist(x, s) · r/2

|S| = 2 = O(1) doubling dimension


Routing, Anycast, and Multicast

• Labeled routing scheme• (1+)-approximation for Unicast Routing• Constant approximations to Multicast and Anycast• -approximate Distance Queries

• Label size: O(log ) (bits)– is the diameter of the network

• Routing table size: O(1/) (log ) (O() + log ) (bits)– is the doubling dimension of the graph (2..5)– is the max degree of any node


Some Related Work

• Gavoille, Gengler. Space Efficiency for Routing Schemes of Stretch Factor Three.No labeling and stretch < 3 requires routing tables of size (n)

Routing Table (bits) Label (bits)

Talwar, 2004 O(1/( )) log2+ O( log )Chan et al. 2005 ( / )O() log log O( log(1/)) log

Slivkins, 2005 -O() log log O( log(1/)) log

Slivkins, 2005 -O() log log log log n 2O() log n log(-1 log )

Abraham et al. 2006 -O() log n log(min(,n)) d log n eThis work O(1/) log (O() + log ) 2O() log

n Number of nodes in network Diameter of network Doubling dimension of network Max. degree of any node Approximation factor for unicast routing

Not only Unicast Routing, but also Multicast, Anycast, and distance queries

& distributed construction


Outline

Introduction

Related Work

Construction of Labels

Construction of Routing Tables

Unicasting

Multicasting

Anycasting


Node Labeling: -net

• Given a graph G=(V,E)• U ½ V is a -net if

a) 8 v 2 V: 9 u 2 U : d(u,v) · b) 8 u1, u2 2 U : d(u1, u2) > Net centers of the -net


Dominance Net Hierarchy

• Build -nets for 2 {1, 2, 4, …, 2d log e }

= 2

= 4

= 8

= 1Level 0

Level 1

Level 2

Level 3


Naming Scheme

• Select parent from next higher level• Parent enumerates all of its children

– At most 22 children– 2 bits are sufficient for the enumeration

• Name of net-center obtained by concatenation of enum values– Name at most 2 log bits long

1 2 3 4 5 6

3212121111

1 2 1 1 2 1 2 1 2 3 1 1 2 1 1 2

Root

R:6:3:2


Node Labeling

• Each net-center c of a -net advertises itself to Bc(2)• Any node n stores ID of all net-centers from which it receives

advertisements– Per level at most 22 net-centers to store– If net-center c covers n, then also the parent of c covers n– The set of net-centers to store form a tree

1 2 3 4 5 6

3212121111

1 2 1 1 2 1 2 1 2 3 1 1 2 1 1 2Level 0

Level 1

Level 2

Level 3– Per level at most 22 ¢ 2 bits– d log e levels) Label size of O(log ) for a

constant

n


Outline

Introduction

Related Work

Construction of Labels

Construction of Routing Tables

Unicasting

Multicasting

Anycasting


Routing Tables

• Routing tables to support (1+) stretch routingRecall: Routing table size of O(1/) (log ) (O() + log ) bits

• Every net-center c 2 -net of the dominance net advertises itself to Bc( (8/ + 6))

• Every node stores direction to reach all advertising net-centers

Bc( (8/ + 6))

c


Routing Tables – Analysis

• Each node needs to store direction for at most 22(8/ + 6) net-centers per level

• If a node needs to store a routing entry for net-center c, then it also needs to store a routing entry for the parent of c.– The routing table can be stored as a tree

• For each net-center, we need to store its enumeration value, and the next-hop information, which takes at most 2 + log bits

• Total storage cost is 22(8/ + 6) log (2 + log ) bits.


Unicast Routing

Problem: From a sender node s, send a message to a target node t, given the ID and label L(t).

Algorithm: From all net-centers listed in L(t), s picks the net-center c on the lowest level to which it has routing information and forwards the message towards c.

Main idea: Once we reach a first net-center of t, we are sure to find a closer net-center on a lower layer. The path to the first net-center causes only little overhead as the net-centers advertise themselves quite far.

s t

c1

c2c3


Multicast Routing

Problem: From a sender node s, send a message to a set of target nodes U.

Algorithm: Build a Minimum Spanning Tree (MST) on s U and send the message on this tree.

Main idea: The MST is a 2-approximation to the Minimum Steiner Tree problem.Our distance estimates may be wrong up to a factor 6, and message routing on the tree has stretch (1+), which results in a 12(1+)-approximation.


Anycast Routing

Problem: From a sender node s, send a message to an arbitrary target chosen from a set U.

Algorithm: Determine u 2 U with minimal distance to s, and send the message to u.

Main idea: Our distance estimates may be wrong up to a factor 6, and message routing on the tree has stretch (1+), which results in a 6(1+)-approximation.


Summary

Unicast

Multicast

Anycast

Unicast

AnycastMulticast

(1+) approximation

12 (1+) approximation

6 (1+) approximation

ALL IN ONE

Label size: O(log )

Routing table size: O(1/) (log ) (O() + log )


Questions / Comments

Thank you!Questions / Comments?

Roland FluryRoger Wattenhofer


-nets on Doubling Metrics

Property 1 (Sparseness): Any ball Bv(2x ) covers at most 2(1+x) nodes from an arbitrary -net.

2-net

v

Bv(4)


-nets on Doubling Metrics (2)

Property 2 (Dominance): Given a -net, and each net-center u covers Bu(2), then any network node is covered by at most 22 net-centers.

u

2-netBu(4)


Distance Queries

Problem: Determine the distance between two nodes u and v, given their labels L(u) and L(v)

Solution: Determine the smallest level i for which the labels of u and v have at least one common net-center…

Theorem: The distance query is a -approximation.Proof: Lower bound: dist(u, v) > 2i -1, otherwise L(u) and L(v) have

a common net-center on level i-1. Upper bound: By distinction of cases and applying triangle inequality… (see paper)


Constant Doubling Metrics

R outing, A nycast, and M ulticast for Mesh and Sensor Networks

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