Coverage in Heterogeneous Sensor Networks: Loukas Lazos and Radha Poovendran
Coverage in Heterogeneous Sensor Networks: Loukas Lazos and Radha Poovendran
Abstract— In this paper we study the problem of coverage in according to a distribution, and is relevant in applications
heterogeneous planar sensor networks. Coverage as a perfor- where the sensors’ positions cannot be selected a priori.
mance metric, quantifies the quality of monitoring provided by
In this paper, we analyze the following stochastic coverage
the sensor network. We formulate the problem of coverage as a
set intersection problem arising in Integral Geometry, and derive problem. Given a planar field of interest and N sensors
analytical expressions for stochastic coverage. Our formulation deployed according to a known distribution, compute the
allows us to consider a heterogeneous sensing model, where fraction of the field of interest that is covered by at least
sensors need not have an identical sensing capability. In addition, k sensors (k ≥ 1). The problem can also be rephrased as,
our approach is applicable to scenarios where the sensing area
given a field of interest and a sensor distribution, how many
of each sensor has arbitrary shape and sensors are deployed
according to any distribution. We present analytical expressions sensors must be deployed in order for every point in the field
only for convex sensing areas, however, our results can be of interest to be covered by at least k sensors with a probability
generalized to non-convex areas. The validity of our expressions p (k-coverage problem) [20].
is verified by extensive simulations. In this paper we make the following contributions. We
formulate the problem of coverage in sensor networks as
I. I NTRODUCTION
a set intersection problem. We use results from integral
Sensor networks are projected to have a significant impact geometry to derive analytical expressions quantifying the
into our everyday lives, with applications to environmental coverage achieved by stochastic deployment of sensors into
monitoring, home health care, disaster relief operations, and a planar field of interest. Compared to previous analytical
ambient monitoring [1]. One of the primary tasks of sensor results [8], [12], [20], our formulation allows us to consider
networks is the collective monitoring of a field of interest. a heterogeneous sensing model, where sensors need not have
Sensors may monitor physical properties such as temperature, an identical sensing capability. In addition, our approach is
humidity, air quality, or track the motion of objects moving applicable to scenarios where the sensing area of a sensor
within the field of interest. In order for the sensor network is not an ideal circle, but has any arbitrary shape. To the
to sufficiently monitor the entire field of interest, one needs best of our knowledge, only [15] considers a heterogeneous
to ensure that every point of the field is covered by at least sensing model, though only incorporating the mean value of
one sensor. Furthermore, to provide the desired accuracy and the sensing range in the coverage computation. In addition, the
robustness against node failures, many applications require formulation in [15] considers only uniformly deployed sensors.
that each point of the field of interest is sensed by more In our approach, sensors can be deployed according to any
than one sensor. Hence, the problem of node deployment for distribution. We provide formulas for k-coverage in the case
the purpose of sensing can be viewed as a coverage problem, of heterogeneous sensing areas, as well as the simplified forms
defined below. in the case of identical sensing areas, and give an example for
The coverage problem is to quantify how well is the field of the computation of the number of sensors required to cover
interest sensed by the deployment of the sensor network. The a field of interest with a pre-specified probability. Finally, we
coverage problem can be studied under different objectives and validate our theoretical expressions via simulations and show
constraints imposed by the applications such as, worst-case an exact match between simulation and theory.
coverage [10], deterministic coverage [10], [12] or stochastic The rest of the paper is organized as follows. In Section
coverage [8], [10], [12], [15], [20]. The worst-case coverage III we formulate the coverage problem as a set intersection
problem quantifies coverage based on the parts of the field of problem. In Section IV we derive analytical expressions for
interest that exhibit the lowest observability from the sensors coverage. In Section V, we validate our theoretical results via
[10], and is relevant in applications where a desired threshold simulation. Section VI presents our conclusions.
number of sensors need to observe the field of interest.
The deterministic coverage problem [10], [12] quantifies the II. R ELATED W ORK
coverage achieved by deploying sensors in a deterministic
way, and is relevant in applications where one can select In this section we describe related work to the coverage
the positions where the sensors are placed. The stochastic problem in wireless sensor networks. The coverage problem
coverage problem [8], [10], [12], [15], [20], on the other hand, can be classified under different objectives and metrics. The
quantifies the coverage achieved when sensors are deployed different approaches to the coverage problem are, deterministic
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10
unclassified unclassified unclassified
x A0
O
k out of the N sets Si intersect. In figure 1(b), we show a set S, a randomly selected
reference point O ∈ S, and the axis of a coordinate system.
In the mapping of the stochastic coverage problem to the All rotations and translations for the set S are defined with
set intersection problem, the fixed closed set S0 corresponds to respect to the reference point O. Integrating the kinematic
the field of intersect A0 . The N closed sets dropped according density of a set A over a group of motions M in the plane,
to the distribution K(S0 ) correspond to the sensing areas of yields a measure for the set of motions M, which is called
the N sensors deployed according to the distribution K(A0 ). the kinematic measure [17], defined below.
By computing the fraction of the set S0 , where at least k out Definition 2: Kinematic measure–The kinematic measure
of N sets Si intersect, we equivalently compute the fraction m of a set of motions M in the plane is defined by the integral
of the field of interest that is k-covered3 . of the kinematic density dA over M :
The set intersection problem has been a topic of research of Z
Integral Geometry and Geometric Probability [7], [14], [17]– m= dA. (2)
M
[19]. Before we provide analytical coverage expressions based By measuring the motions of a set in the plane, we quantify
on our formulation, we present relevant background. the space of all possible positions of the set that correspond
to that motion. The quotient of the measure of any random
B. Background on Integral Geometry
motion path Z over the measure of all possible motions M in
In this section, we present relevant background on Integral the plane, yields the probability p(Z) for that random motion
Geometry that we use in Section IV for deriving analytical path Z to occur:
coverage expressions based on our formulation. Interested
reader is referred to [7], [14], [17]–[19], as reference to m(Z)
p(Z) = . (3)
Integral Geometry. m(M)
We first define the notion of the kinematic density for the The kinematic measure allows us to compute the geometric
group of motions of a set A in the plane, that is used to define probability for a specific set configuration to occur, as depicted
a measure that quantifies the possible positions of A, such that in (3). Equation (3), is used in our formulation to derive the
a specific event occurs [17]. The kinematic density expresses fraction of the field of interest covered by a sensor deployment,
the differential element of motion of a set in the plane and is as it is illustrated in the following section.
defined as follows.
Definition 1: Kinematic Density–Let M denote the group IV. C OVERAGE IN HETEROGENEOUS SENSOR NETWORKS
of motions of a set A in the plane. The kinematic density dA In this section, we derive analytical expressions for coverage
for the group of motions M in the plane for the set A, is by analyzing the coverage problem as a set intersection prob-
defined as the differential form: lem. We first illustrate the coverage computation when a single
sensor is randomly deployed to monitor the field of interest,
dA = dx ∧ dy ∧ dφ, (1) by studying the intersection of two sets in the plane. We then
where ∧ denotes the exterior product used in exterior calculus compute coverage when the sensor is deployed according to a
[5], [6], (x, y) denote the Cartesian coordinates, and φ denotes distribution K(A0 ). We extend our expressions to the general
the rotation angle of A with respect to the x axis of the case where N sensors are deployed at random. We compute
coordinate system4 [17]. the fraction of the field of interest covered by exactly k sensors
in the case of heterogeneous sensing areas and simplify the
3 Due to their equivalence, A and S as well as the terms sensing area
0 0 formula when the sensors have identical sensing areas. Finally,
and set are used interchangeably in the rest of the paper.
4 For every set A, one can randomly choose a reference point O, based on we compute the fraction of the field of interest covered by at
which all translations and rotation motions of A are defined. least k sensors.
T
A. Coverage Achieved by Random Deployment of a Single A0 A1 6= 0 is:
Sensor Z
\
m(A1 : A0 A1 6= ∅) = T
dA1
Let us consider the simple case where a single sensor s1
ZA0 A1 6=∅
is randomly deployed in such a way that it monitors some
part of the field of interest. The achieved coverage can be = T
dx ∧ dy ∧ dφ
A0 A1 6=∅
computed by considering the intersection of two sets in the = 2π(F0 + F1 ) + L0 L1 . (6)
plane. Let A0 , A1 denote two sets in a plane with A0 being
fixed, while A1 can move freely. A0 represents the field of Due to the length and complexity, the proof of (6) is omitted.
interest, while A1 represents the sensing area of node s1 . The Interested reader is referred to [17], [18] for details.
average size of the common area A01 between sets A0 , A1 ,
By combining (5) and (6) we can compute the probability
when A1 is randomly dropped in the plane, defines the area
p(P ∈ A1 ) as:
of A0 , covered by A1 . Normalizing A01 over A0 we obtain
the fraction f r(A0 ) of A0 covered by A1 . In figure 1)(c), we T
m(A1 : P ∈ A0 A1 )
show two sets A0 , A1 and the common area between them. p(P ∈ A1 ) = T
m(A1 : A0 A1 6= ∅)
To compute f r(A0 ), we randomly select a point P of A0 , 2πF1
and find the set of all positions of A1 that include P. Dividing = . (7)
2π(F0 + F1 ) + L0 L1
the measure of all the positions of A1 that include T P over the
measure of all the positions of A1 such that A0 A1 6= ∅
yields the probability p(P ∈ A1 ) that the randomly selected Note that p(P ∈ A1 ) is only dependent on the area and the
point P is covered by A1 [17], [18]. Integrating p(P ∈ A1 ) perimeter of the convex sets that intersect and not on the shape
over all P ∈ A0 and normalizing over the size of A0 yields of those sets.
f r(A0 ). The following theorem holds only for convex sets,
though it can be extended in the case of non-convex sets by Lemma 1: The fraction f r(A0 ) of a fixed convex set A0 of
appropriate computation of the kinematic measures [17], [18]. area F0 and perimeter L0 that is covered by a convex set A1
of area F1 and perimeter L1 , when A1 is randomly dropped
in the plane in such a way that it intersects with A0 is given
Theorem 1: Let A0 be a fixed convex set of area F0 and by:
perimeter L0 , and let A1 be a convex set of area F1 and 2πF1
perimeter L1 , randomly dropped in the plane in such a way f r(A0 ) = . (8)
2π(F0 + F1 ) + L0 L1
that it intersects with A0 . The probability that a randomly
selected point P ∈ A0 is covered by A1 is given by:
Proof: Equation (7) expresses the probability that a
2πF1 randomly selected point P ∈ A0 is covered by A1 . Integrating
p(P ∈ A1 ) = . (4) (7) over all points P ∈ A0 provides the size F01 of the
2π(F0 + F1 ) + L0 L1
common area A01 between A0 and A1 :
Z
Proof: The probability that P is covered by A1 is equal F01 = p(P ∈ A1 )dP
P ∈A0
to the measure of the set of motions of A1 such that P ∈ A1 Z
= p(P ∈ A1 ) dP
T by the measure of the set of motions of A1 such that
divided
P ∈A0
A0 A1 6= ∅. We now compute the two measures.
= p(P ∈ A1 )F0
Z 2πF0 F1
\ (i) = . (9)
m(A1 : P ∈ A0 A1 ) = T
dA1 2π(F0 + F1 ) + L0 L1
ZP ∈A0 A1
(ii) Normalizing F01 by F0 yields:
= dA1
P ∈A1
Z Z F01
2π f r(A0 ) =
= dx ∧ dy dφ F0
P ∈A1 0 2πF0 F1 1
= 2πF1 , (5) =
2π(F0 + F1 ) + L0 L1 F0
2πF1
=
where in 5(i)Twe integrate dA1 over all motions of A1 such 2π(F0 + F1 ) + L0 L1
that P ∈ A0 A1 . Since by assumption P ∈ A0 and A0 is = p(P ∈ A1 ). (10)
fixed, in 5(ii) we integrate dA1 over all motions of A1 such
that P ∈ A1 . The measure of all motions of A1 such that
B. Coverage Achieved by Deployment of a Single Sensor exactly k sensors is equal to the probability ¡that
¢ P is covered
According to an Arbitrary Distribution by exactly k specific sets. Let T denote a kx Nk matrix where
In the case where the A1 is not randomly deployed in each row j is a k-permutation
¡ ¢ of the vector [1 . . . N ], and let
the plane, but it follows an arbitrary distribution K(A0 ), the G denote a (N −k+1)x Nk matrix where each row j contains
measures in (5), (6) are calculated as weighted functions of the elements of [1 . . . N ], that do not appear in the j th row
the probability density function k(x, y, φ) of A1 . of T. Consider for example, T (1) = [1 . . . k] and G(1) =
Z [k + 1 . . . N ]. The probability p(T (1)) that P is covered by
\
m(A1 : A0 A1 6= ∅) = kdx ∧ dy ∧ dφ, (11) exactly the sets with indexes in the first row of T is given by:
ZZ
\ p(T (1))
(i)
= p(P ∈ A1 , . . . , P ∈
/ Ak+1 , . . . , P ∈
/ AN )
m(A1 : P ∈ A0 A1 ) = kdx ∧ dy ∧ dφ,(12)
(ii)
P ∈A1 = p(P ∈ A1 ), . . . , (P ∈ Ak )
T
where Z = A0 A1 6= ∅. Depending on the distribution p(P ∈
/ Ak+1 ), . . . , p(P ∈/ AN )
K(A0 ), the measures in (11), (12) may have a closed form. (iii) 2πF1
When A∞ is deployed according to the distribution K(A0 ), =
2π(F0 + F1 ) + L0 L1
we can calculate the probability p(P ∈ A1 ), by substituting 2πFk
the measures in (11), (12) into (7). The p(P ∈ A1 ), is the ...
2π(F0 + Fk ) + L0 Lk
basic building block for deriving expressions for coverage in
2πF0 + L0 Lk+1
the general case where N sensors are deployed, as we show
in the following section. 2π(F0 + Fk+1 ) + L0 Lk+1
2πF0 + L0 LN
...
C. Coverage in the Case of Multiple Sensors 2π(F0 + FN ) + L0 LN
Qk QN
In this section, we compute the probability p(S = k) that j=1 (2πFj ) z=k+1 (2πF0 + L0 Lz )
a randomly selected point P ∈ A0 is covered by k sensors = QN
when N sensors are randomly deployed. Using p(S = k), we r=1 (2π(F0 + Fr ) + L0 Lr )
Qk ¡ ¢ QN −k ¡ ¢
compute the probability that P is covered by at least k sensors, j=1 2πFT1,j z=1 2πF0 + L0 LG1,z
= QN .
as well as the fraction of A0 covered by at least k sensors.
r=1 (2π(F0 + Fr ) + L0 Lr )
Theorem 2: Let A0 be the field of interest of size F0 and (15)
perimeter L0 , and let N sensors with sensing area Ai of size
In (i), we show which k sets include point P. Due to the
Fi and perimeter Li be deployed over A0 . The probability
independence in the set deployment, in (ii), the intersection of
p(S = k) that a randomly chosen point P of A0 is covered
the events in (i) becomes a product of the individual events. In
by exactly k sensors when k ≥ 1 is given by:
(iii), we substitute the individual probabilities from (7), (14).
P(Nk ) ³Qk QN −k ´
In the general case, the probability that the sets with indexes
i=1 j=1 (2πFTi,j ) z=1 J (i, z)
p(S = k) = , (13) of the ith row of T cover point P is given by:
QN
r=1 (2π(F0 + Fr ) + L0 Lr ) Qk ¡ ¢ QN −k
j=1 2πFTi,j z=1 J (i, z)
where J (i, j) = (2πF0 + L0 LGi,z ), T is a matrix in which p(T (i)) = QN . (16)
each row j is a k-permutation of [1 . . . N ], and G is a matrix r=1 (2π(F0 + Fr ) + L0 Lr )
in which each row j contains the elements of [1 . . . N ], that Since we are not interested in a specific set permutation to
do not appear in the j th row of T. cover point P, the probability that p(S = k) is a summation
Proof: In order to prove Theorem 2, we map the problem of p(T (i)) for all possible k-permutations. Summing p(T (i))
of coverage to the set intersection problem, as illustrated in our over all i yields (13):
problem formulation in Section III-A. When a single sensor si (Nk )
is deployed, the probability that it covers a randomly selected X
p(S = k) = p(T (i))
point P ∈ A0 is given by Theorem 1. Hence, the probability i=1
p(P ∈/ Ai ) can be computed as:
(Nk ) Ã Qk ¡
X
¢ QN −k !
j=1 2πFTi,j z=1 J (i, z)
p(P ∈
/ Ai ) = 1 − p(P ∈ Ai ) = QN
2πFi i=1 r=1 (2π(F0 + Fr ) + L0 Lr )
= 1−
2π(F0 + Fi ) + L0 Li
2πF0 + L0 Li
= . (14) According to Lemma 1, (13) also expresses the fraction of A0
2π(F0 + Fi ) + L0 Li
that is covered by exactly k sensors. Equation (13) is valid for
Given that fact that the N sensors are independently deployed k ≥ 1. The fraction of the A0 that is not covered by any
in the plane so that they cover some part of A0 , the probability sensor, is given by the following corollary.
p(S = k) that a randomly selected point P ∈ A0 is covered by
Non−covered fraction of A0
1
Theoretical
Simulated
0.8
Fraction fr(A0)
0.6
0.4
0.2
0
0 100 200 300 400 500 600
Sensor deployed (N)
Fig. 2. Fraction f r(A0 ) of A0 , that remains non-covered as a function of the number of sensors N that are deployed to monitor the field of interest.
Corollary 1: The fraction of A0 that is not covered by any Theorem 3: Let A0 be the field of interest of size F0 and
sensor when N sensors are randomly deployed is given by, perimeter L0 , and let N sensors with sensing area Ai of size
N
Fi and perimeter Li be deployed over A0 . The probability
Y 2πF0 + L0 Li that a randomly selected point of A0 is covered by at least k
p(S = 0) = . (17)
i=1
2π(F 0 + F i ) + L0 Li sensors is given by:
Proof: Given that fact that the N sensors are indepen- k−1
X
dently deployed in the plane so that they cover some part of p(S ≥ k) = 1− p(S = h) (20)
A0 , the probability p(S = 0) that none of the Ai , i = 1 . . . N h=1
Fraction fr(A )
0
0
0.2
0.15 0.4
0.1
0.2
0.05
0 0
0 5 10 0 2 4 6 8 10 12 14
k (sensors) k (sensors)
(a) (b)
Probability density function p(S=k) for N=600 Fraction of A0 covered by at least k sensors for N=600
Theoretical Theoretical
0.16 Simulated Simulated
0.14 0.8
Fraction fr(A )
0.12
Fraction fr(A )
0
0
0.6
0.1
0.08
0.4
0.06
0.04 0.2
0.02
0 5 10 0 2 4 6 8 10 12 14
k (sensors) k (sensors)
(c) (d)
(e) (f)
Fig. 3. (a) The pdf of the fraction f r(A0 ) covered by exactly k sensors when N = 300 sensors with identical sensing area are randomly deployed. (b)
The fraction f r(A0 ) covered by at least k sensors when N = 300 sensors with identical sensing area are randomly deployed. (c) The pdf of the fraction
f r(A0 ) covered by exactly k sensors when N = 500 sensors with identical sensing area are randomly deployed. (d) The fraction f r(A0 ) covered by at
least k sensors when N = 500 sensors with identical sensing area are randomly deployed. (a) The pdf of the fraction f r(A0 ) covered by exactly k sensors
when N = 1000 sensors with identical sensing area are randomly deployed. (f) The fraction f r(A0 ) covered by at least k sensors when N = 1000 sensors
with identical sensing area are randomly deployed.
The theoretical formula that computes f r(A0 ) is obtained the theoretical formula in (22) conforms with the simulation
from Corollary 1 and is equal to: results. Since our method does not suffer from the border effect
µ ¶N problem, (22) is accurate despite the bounded size of the field
2πF0 + L0 L of interest.
f r(A0 ) = p(S = 0) = , (22)
2π(F0 + F ) + L0 L
In figure 3(a), we show the pdf of the fraction f r(A0 )
where F0 = πR2 , L0 = 2πR, F = πr2 , L = 2πr. In figure covered by exactly k sensors when N = 200 sensors with
2(a), we show the fraction f r(A0 ) of A0 , that remains non- identical sensing area are randomly deployed. The equivalent
covered as a function of the number of sensors N that are sensor density is equal to ρ = 0.0063 sensors/m2 . The
deployed to monitor the field of interest. We observe that same graphs for N = 600, N = 1000 (densities ρ = 0.019
Fig. 4. Fraction f r(A0 ) of A0 , that remains non-covered as a function of the number of sensors N that are deployed to monitor the field of interest.
sensors/m2 , ρ = 0.032 sensors/m2 ) are provided in figures sensor density is equal to ρ = 0.0095 sensors/m2 . The
3(c) and 3(e), respectively. The pdf of f r(A0 ) is equal to same graph for N = 500, N = 1000 (densities ρ = 0.019
the probability that a randomly selected point P is covered sensors/m2 , ρ = 0.032 sensors/m2 ) are provided in figures
by exactly k sensors. Our analytical derivation in Section IV, 5(c) and 5(e), respectively. The f r(A0 ) covered by exactly K
yields: sensors is equal to the pdf p(S = k) of the probability that
a randomly selected point P is covered by exactly k sensors.
f r(A0 ) = p(S = k)
¡N ¢ k N −k
Our analytical derivation in Section IV, yields:
k (2πF ) (2πF0 + L0 L)
= N
. (23) f r(A0 ) = p(S = k)
(2π(F0 + F ) + L0 L)
In figure 3(b), we show the fraction of A0 covered by For k = 0 :
N µ
Y ¶
at least k sensors when N = 200. The same graphs for 2πF0 + L0 Li
N = 600, N = 1000 are provided in figure 3(d) and f r(A0 ) = ,
i=1
2π(F0 + Fi ) + L0 Li
3(f), respectively. For both values of N we observe that our
theoretic formulas conform with the simulation results. For all while for k ≥ 1 :
graphs in figure 2, 3 we show the theoretical result according P(Nk ) ³Qk QN −k ´
to our expressions, and the simulation values. i=1 j=1 (2πFTi,j ) z=1 J (i, z)
f r(A0 ) = QN .
B. Coverage in Heterogeneous Sensor Networks r=1 (2π(F0 + Fr ) + L0 Lr )
In our second experiment, we considered a hierarchical In figure 5(b), we show the fraction of A0 covered by
(heterogeneous) sensor network, where two types of sensors at least k sensors when N = 300. The same graphs for
are deployed. Type A has a sensing area of disk shape with a N = 500, N = 1000 are provided in figures 5(d), and
sensing range rA = 10m, while type B has a sensing area of 5(f), respectively. We again verify that our theoretical formula
disk shape with a sensing range of rB = 15m. We randomly agrees with the simulation results.
deployed an equal number NA = NB = N2 of sensors of In the case of heterogeneous sensor networks where each
each type over a circular field of interest of size F0 = πR2 sensor has a different sensing area, the formula in (25)
where R = 100m. In figure 4, we show the fraction f r(A0 ) has an exponentially increasing computational cost, since an
of A0 , that remains non-covered as a function of the number exponentially increasing summation of terms must be com-
of sensors N that are deployed to monitor the field of interest. puted in order to derive the exact coverage achieved. Such a
The theoretical formula that compute that is equal to: computation may not be feasible for large networks. In such
a case, an approximation can be used for our formulas by
f r(A0 ) = p(S = 0) employing the expressions derived for a homogeneous sensor
YN
2πF0 + L0 Li network and substituting the size F and perimeter L of the
= , (24) sensing area of the sensors with the expected size E[F ] and
i=1
2π(F 0 + Fi ) + L0 Li
expected perimeter E[L]. The theoretical approximation for
where F0 = πR2 , L0 = 2πR, Fi = πri2 , L = 2πri . such a case is:
We observe that the simulation results verify the validity
f r(A0 ) = p(S = k)
of our theoretical expression. In figure 5(a), we show the pdf ¡N ¢ k N −k
of the fraction f r(A0 ) covered by exactly k sensors when k (2πE[F ]) (2πF0 + L0 E[L])
= N
.(25)
N = 300 sensors are randomly deployed. The equivalent (2π(F0 + E[F ]) + L0 E[L])
(a) (b)
(c) (d)
(e) (f)
Fig. 5. Heterogeneous sensor network, with the field of interest being a disk of radius R = 100m. An equal number of two types of sensors are deployed;
Type A has a sensing area of a disk shape with radius rA = 10m, while type B has a sensing area of a disk shape with rB = 15m. (a) The pdf of the
fraction f r(A0 ) covered by exactly k sensors when N = 300 sensors. (b) The fraction f r(A0 ) covered by at least k sensors when N = 300 sensors. (c)
The pdf of the fraction f r(A0 ) covered by exactly k sensors when N = 500 sensors. (d) The fraction f r(A0 ) covered by at least k sensors when N = 500
sensors. (e) The pdf of the fraction f r(A0 ) covered by exactly k sensors when N = 1000 sensors. (f) The fraction f r(A0 ) covered by at least k sensors
when N = 1000 sensors.
In figure 6(a) we show the pdf obtained via simulation for C. An Example of Computing the Coverage in a Sample
our heterogeneous sensor network experiment, for N = 500 Network
sensors, the theoretical values based on the exact formula in
(25), and the approximation in (25). In figure 6(b), we show In this section, we provide an example of applying our
the fraction of A0 covered by at least k sensors. We observe results to a sample sensor network. Consider an F oI of size
that for the case of heterogeneous sensor networks where F0 = 106 m2 and perimeter L0 = 4, 000m where sensors of
each sensor has a different sensing area, (25) provides a good identical sensing area F = 100π and perimeter L = 20π
approximation of the coverage achieved, without incurring the are randomly deployed. We want to compute the number of
computational cost of (25). sensors needed in order for a randomly selected point of the
F oI to be covered by at least one sensor with a probability
(a) (b)
Fig. 6. Heterogeneous sensor network, with the field of interest being a disk of radius R = 100m. An equal number of two types of sensors are deployed;
Type A has a sensing area of a disk shape with radius rA = 10m, while type B has a sensing area of a disk shape with rB = 15m. (a) The pdf of the
fraction f r(A0 ) covered by exactly k sensors when N = 500 sensors. (b) The fraction f r(A0 ) covered by at least k sensors when N = 500 sensors.