# Routing, Anycast, in addition to Multicast as long as Mesh in addition to Sensor Networks RAM Thank you! Questions / Comments

## Routing, Anycast, in addition to Multicast as long as Mesh in addition to Sensor Networks RAM Thank you! Questions / Comments

Willsey, Les, High School Sports Reporter has reference to this Academic Journal, PHwiki organized this Journal Routing, Anycast, in addition to Multicast as long as Mesh in addition to Sensor Networks Rol in addition to Flury Roger Wattenhofer RAM Distributed Computing Group All-In-One Solution

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Routing in Mesh in addition to Sensor Networks Unicast Routing (send message to a given node) Multicast Routing (send message to a given set of nodes) Routing in Mesh in addition to Sensor Networks (2) Anycast Routing (send message to any node of a given set) Unicast Multicast Anycast Unicast Anycast Multicast 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 in addition to 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, in addition to Multicast Labeled routing scheme (1+)-approximation as long as Unicast Routing Constant approximations to Multicast in addition to 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 as long as Routing Schemes of Stretch Factor Three. No labeling in addition to stretch < 3 requires routing tables of size (n) Not only Unicast Routing, but also Multicast, Anycast, in addition to 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) > Dominance Net Hierarchy Build -nets as long as 2 {1, 2, 4, , 2d log e } = 2 = 4 = 8 = 1 Level 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 as long as the enumeration Name of net-center obtained by concatenation of enum values Name at most 2 log bits long Root 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 as long as m a tree Level 0 Level 1 Level 2 Level 3 Per level at most 22 ¢ 2 bits d log e levels ) Label size of O(log ) as long as 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 routing Recall: 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 as long as at most 22(8/ + 6) net-centers per level If a node needs to store a routing entry as long as net-center c, then it also needs to store a routing entry as long as the parent of c. The routing table can be stored as a tree For each net-center, we need to store its enumeration value, in addition to the next-hop in as long as mation, 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 in addition to 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 in as long as mation in addition to as long as wards 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 c2 c3

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 in addition to 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, in addition to 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, in addition to send the message to u. Main idea: Our distance estimates may be wrong up to a factor 6, in addition to message routing on the tree has stretch (1+), which results in a 6(1+)-approximation. Summary Unicast Multicast Anycast Unicast Anycast Multicast (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 Rol in addition to Flury Roger 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, in addition to each net-center u covers Bu(2), then any network node is covered by at most 22 net-centers. 2-net Bu(4)

Distance Queries Problem: Determine the distance between two nodes u in addition to v, given their labels L(u) in addition to L(v) Solution: Determine the smallest level i as long as which the labels of u in addition to 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) in addition to L(v) have a common net-center on level i-1. Upper bound: By distinction of cases in addition to applying triangle inequality (see paper) Constant Doubling Metrics

## Willsey, Les High School Sports Reporter

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