# Connectivity in Sensor

Mobile - Wireless

Nov 21, 2013 (4 years and 6 months ago)

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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

r

r

𝛼

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:
𝜋

2

The coverage area of directional antennae with range
𝛼

is:
𝛼
𝑟
2
2

The energy is proportional to the coverage area.

With the same energy E the omnidirectional antennae
can reach a range of

𝜋

, while the directional
antennae can reach a range of
2

𝛼
.

The main problem

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

>
0
.

For a given angel
0

𝛼

2
𝜋

and an integer

1
, each sensor is allowed to use at most

directional antennae each of angle at most
𝛼
.

Determine the minimum range

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

Definitions and notations

𝑘
(

,
𝛼
)

the minimum range of directed antennae of angular
𝛼

so that every sensor in S uses at most

such
antennae (under appropriate rotation) a strongly connected network
on S results.

𝑘

=

𝑘

,
0

𝐷
𝑘
(

)

the set of all strongly connected graphs on S with out
-
degree at most k.

𝑘

,


𝐷
𝑘
(

)

the maximum length of an edge in G.

𝑘

=
min
𝐺

𝑘

𝑘

.

 
(

)

a set of all MSTs on S.


,


 
(

)

-

the maximum length of an edge in T.

𝑀 

=
min


𝑀 

(

)
.

𝑀 

𝑘
(

,
𝛼
)
.

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

2
,
𝐷



is a spanning
tree with max degree of k.

Solving the main problem

When k =
1
, and
𝛼
=
0

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
𝛼

𝜋

we have the same problem as
omnidirectional antennae. And r is the maximum
length between a pair of two adjacent sensors.

When
0

𝛼
<
𝜋

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
1

𝑖



2
.

Solving the main problem

Proof:

Assume


𝑖
,

𝑖
+
2
>

, for some
𝑖



2
.

Consider the antenna at

𝑖
+
1

-

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
𝛼
=
0

we have the BTSP problem.

When
𝜋

𝛼
<
8
𝜋
5

there exist a polynomial
algorithm that when given MST on S
computes an orientation of the antennae
2
sin

(
𝜋

𝛼
2
)

1

,
𝛼

Solving the problem

The algorithm:



,
Λ





𝑉






=



,


𝑖










.



𝑖



𝐵




𝑖

𝑖

𝑖  


Λ

Λ





𝑖


,
𝜋





=



,


𝑖















𝑖



𝑖 
.

  


,




 𝑖















 









 







𝑖



 𝑖
.

  



Λ

 𝑖









𝜋

.

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):


𝛼
=
2
𝑖
(
𝜋

𝛼
2
)

the range of the antennae.

Since
𝜋

𝛼
<
8
𝜋
5
,
d
𝛼
=
2
𝑖
𝜋

𝛼
2

2
𝑖
𝜋
5
>
2
𝑖
𝜋
6
=

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


𝛼

then


is in the
sector of

, since


.

If
the counter clockwise angel

>
𝛼

then by the low of
cosines in the triangle define by

,


and

:


=

2
+

2

2




2

2

=
2
𝑖
𝑥
2

2
𝑖
𝜋

𝛼
2
=

(
𝛼
)

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

1

be the node where

1
,



.

Since


is a neighbor of


there are a directed edge form


or from

1

to


and a directed edge from


to

1
in G (from the Lemma)

so we have
a directed path from


to


in G.

If


Λ

(lead) there are a directed edge between


and


in G (from the
algorithm).

Otherwise let

2

be the node where

2
,



.

Since


is a neighbor of


there are a directed edge form


or from

2

to


and a directed edge from


to

2
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

,
1

5
,
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
2
sin
𝜋
𝑘
+
1

𝑘
(

,
𝛼
)
.

Moreover, given an MST on S the spanner can be
Ο
(

)

Solving the main problem

When k =
1
,
𝛼
<
2
𝜋
3

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
𝛼

then it is NP
-
hard to
approximate the optimal radius to within a
factor of x, where

=
2
sin

(
𝛼
)
.

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