Maintaining Connectivity in Sensor Networks Using Directional Antennae

Evangelos Kranakis,Danny Krizanc,Oscar Morales

Presented by Tal Beja

Outline

Introduction. The main problem. Definitions. Solving the main problem.

Introduction Connectivity in wireless sensor networks may be

established using either omnidirectional or directional antennae.

Each antennae has a transmission range (radios). Directional antennae has also a spread (angle).

Omnidirectional Antennae

Directional Antennae

rr

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Introduction

Given a set of sensors positioned in the plane with antennae (directional or omnidirectional), a directed network is formed as follows: The edge (u, v) is in the network if v lies within u’s

sector. It is easy to see that with omnidirectional

antennae with the same radius all network is unidirectional, because u in the range of v iff v in the range of u.

Introduction Examples:

Introduction

Why to use directional antennae? The coverage area of omnidirectional antennae

with range r is: The coverage area of directional antennae with

range r and spread is: The energy is proportional to the coverage area. With the same energy E the omnidirectional

antennae can reach a range of , while the directional antennae can reach a range of .

The main problem

Consider a set S of n point in the plane that can be identified with sensors having a range .

For a given angel and an integer , each sensor is allowed to use at most directional antennae each of angle at most .

Determine the minimum range required so that by appropriately rotating the antennae, a directed, strongly connected network on S is formed.

Definitions and notations – the minimum range of directed antennae of angular spread at most so

that every sensor in S uses at most such antennae (under appropriate rotation) a strongly connected network on S results.

– the set of all strongly connected graphs on S with out-degree at most k. – the maximum length of an edge in G. . – a set of all MSTs on S. - the maximum length of an edge in T. . .

MST and out degrees

If the set S of n points is on two-dimensional plane there is an MST with a maximum degree six.

This can be improved to an MST with max degree of five.

Definition: for is a spanning tree with max degree of k.

Solving the main problem When k = 1, and We have a problem of finding a Hamiltonian cycle than

minimize the maximum length of an edge. This known as the bottleneck traveling agent problem (BTSP). Parker and Rardin study the BTSP where the lengths satisfy

the triangle inequality. They found 2-approximation algorithm for this problem. They also proved that there is no better approximation

algorithm with polynomial time to the problem, unless P = NP.

Solving the main problem

Sensors on a line, k = 1: Where we have the same problem as omnidirectional

antennae. And r is the maximum length between a pair of two adjacent sensors.

When and k = 0, there exist an orientation of the antennae where the graph is strongly connected if and only if the distance between points I and i+2 is at most r, for any .

Solving the main problem Proof:

Assume , for some . Consider the antenna at - there are two cases to consider:

The antenna directed left – then the left side of the graph cannot be connected to the rest.

The antenna directed right – then the right side of the graph cannot be connected to the rest.

If we direct the odd label antennae to the right and the even label antennae to the left we will get a strongly connected graph.

Solving the main problem

Sensors on a plane, k= 1. Where we have the BTSP problem. When there exist a polynomial algorithm

that when given MST on S computes an orientation of the antennae with radius of

Solving the problem The algorithm:

Solving the problem

Proof:Lemma – for each and for each neighbor of

either or , G (the transmission graph) contains two opposite directed edge between and , and it contains a directed edge between either or to .

Solving the problem

Proof (of the lemma): – the range of the antennae. Since Because and are neighbors and their

antennae directed at each other they both in each other sector.

So we have in G two directed edges between and .

Solving the problem

Proof (of the lemma): Assuming is a neighbor of (if he is a neighbor of we

have a symmetrical case). If the counter clockwise angel then is in the sector of ,

since . If the counter clockwise angel then by the low of cosines

in the triangle define by , and :

So is in the sector of . So we have in G and edge between either or to .

Solving the problem Proof :

We need to prove now that for any edge in T we have a direct path from to and from to in G.

Without loss of generality, assume that is closer to the root of T (the first node we selected in the algorithm).

If the edge than we have two directed edges between them in G(from the lemma).

Otherwise let be the node where Since is a neighbor of there are a directed edge form or from to and a directed

edge from to in G (from the Lemma) – so we have a directed path from to in G. If (lead) there are a directed edge between and in G (from the algorithm). Otherwise let be the node where Since is a neighbor of there are a directed edge form or from to and a directed

edge from to in G (from the Lemma) – so we have a directed path from to in G.

Solving the main problem

Theorem: Consider a set S of n sensors in the plane and suppose each sensor has , directional antennae. Then the antennae can be oriented at each sensor so that the resulting spanning graph is strongly connected and the range is at most .

Moreover, given an MST on S the spanner can be compute with additional overhead.

Solving the main problem

When k = 1, and r > 0. Determining whether there exists an

orientation of the antennae so the transmission graph is strongly connected is NP-complete.

Proof: by reduction from an NP-hard problem of finding Hamiltonian cycles in degree three planar graphs.

Solving the main problem

When k = 2, and the angular sum of the antennae is then it is NP-hard to approximate the optimal radius to within a factor of x, where .

Proof: this is also proven with reduction from finding Hamiltonian cycles in degree three planar graphs.