Sentry Selection in Sensor Networks Radha Krishna Ganti and Martin Haenggi University of Notre Dame

In the case when the number of partitions is less than the maximum node degree of disk graph, the optimization function becomes more complex. Let Dj (Rt, xk ) = {xi : p(xi) = j, xi ∈ Nk (Rt)}

Sentry Selection C(xi) = α

• Sensor Nodes have limited power resources.

Y

[1 − δ(p(xi) − p(xj ))]

xj ∈Ni(R)

– The peak power constraint limits the sensor’s communication range +

• Maximize coverage, connectivity (between nodes) and the lifetime of the network. – Maximizing the sensor network life is important, especially if the sensors cannot be replaced or are inaccessible. • The typical power consumption of a sleeping node is about 3 orders of magnitude smaller than for a node that is awake sensing, listening, or transmitting.

N X

[

δ(p(xi) − j) · area(B(xi, R) \

j=1

xk ∈Dj (Rt,xi)

N X δ(p(xi) − j) B(xk , R)) + β |Dj (Rt, xi)| + 1

Simulation Results For the purpose of simulation, a square of 15 × 15 was used. The coverage metric was evaluated by numerically calculating the area of the union of disks by using a fine grid of resolution 0.01. The positions of the nodes were generated by using a Poisson point process of intensity λ = 1. The transmission radius Rt is taken to be 1.5 times the coverage radius, for all simulations. For the distributed algorithm, α = 100, β = πR2, γ = 1 were chosen. In the following table, the average percentage of area not covered by subsets as generated by two methods is compared

j=1

Phase 1 Phase 2 Phase 3 Distributed 1.58% 1.61% 1.61% Random 4.32% 4.32% 4.32%

• The first term tries to select a partition not chosen by the neighbors in a disk of radius R. • The second term tries to maximize the coverage, in case such a free partition does not exist. • The third term tries to give more weight to the partition with least number of nodes.

In the following figure, the variance of the normalized number of nodes in the partitions is plotted when • Since the node has knowledge only about the neighbors in a disk Rt, the number of points in a the number of partitions is 4. partition is estimated as the number of points of the partition in a Rt-radius disk about the node.

– Put the nodes to sleep as aggressively as possible. • The active nodes (Sentries) should provide acceptable coverage and be connected. – Switch the active set of nodes from time to time.

• α and β are parameters to be chosen.

Variance of number of points in a subset versus radius for N= 4

• The sentry selection problem is to divide the nodes in a sensor network into subsets The total cost function to be maximized is given by with good coverage and minimum overlap.

Variance of number of points in each subset

−3.3

|φ| X

– No global knowledge can be assumed.

C2 = C(xi) 2. The nodes are Each node of the sensor network is considered as a point in a bounded space B ⊂ R S i=1 assumed to form a point process φ in B. Let ρ(A) = volume( x∈A B(x, R)) denote the coverage  of set A for some sensing radius R. Let N denote the number of sets into which the nodes have to be The nodes are initially uniformly assigned a partition from 1 to N . The update rule for each node xi is divided. The problem in a very general form can be stated as follows: as follows. For some maximum overlap D, choose sets A1, A2, . . . , AN ⊂ φ, such that |Ai∩Aj | • |φ| ≤ D, i 6= j

X

p(xi) = argmaxk=1,...,N {C(xi|p(xi) = k) + γ

10

−3.4

10

−3.5

10

0.5

0.6

0.7

0.8 0.9 1 Coverage Radius: R

C(xj |p(xi) = k)}

1.1

1.2

1.3

xj ∈Ni(R)

• mini ρ(Ai) is maximized • | ∪ Ai| ≈ |φ| • |Ai| = |Aj | ∀ i, j or minimize maxi6=j ||Ai| − |Aj || • The algorithm should be local, i.e., each node need not have the knowledge about every other node in the network.

• C(xi|p(xi) = k) denotes the individual node cost function given that the node xi belongs to partition k. • γ denotes the relative weight of the node’s cost function to that of its neighbors’. • Making β large makes the number of nodes in each set almost equal.

Figure 2: Variance of the number of points in a subset versus the coverage radius for N = 4 The mean intensity of each set is 0.25, and the variance is of the order of 10−3.5. This implies that the number of points in the sets are almost equal.

• By construction C2(k) ≥ C2(k − 1) and C2 ≤ |α + πR2 + β||φ|. Number of points in all sets with internode distance< R for all sets versus Set Number N

• Convergence of the algorithm depends on the initial choice of the partition.

Distributed Optimization

80 R=1 R=1.5

An example of the resulting partition is shown below. No of points with internode distance < R in all sets

70

• The problem can be viewed as a distributed optimization problem . • If N is greater than chromatic number of the disk graph, then the partitioning can be found by optimizing |φ| X C1 = {

Y

[1 − δ(p(xj ) − p(xi))]}

(1)

15

10

60

50

40

30

20

10

j=1 xj ∈Ni(R)

0

Where δ(.) is the Kronecker delta function, the neighborhood Ni(R) = {xj ∈ B(xi, R), j 6= i}, and p(xi) denotes the sentry set to which the node placed at xi belongs.

• The only constraint is that this should happen sequentially, i.e. one node does its optimization only after another node finishes.

3.5

4

4.5

5

5.5 6 Number of Sets N

6.5

7

7.5

8

5

Figure 3: Average no. of nodes with internode distance < R in all subsets for different N and R

• The maximum value of C1 is achieved only when there is a proper partitioning of the nodes. • The product term inside the sum can be evaluated by any node with the information available from its neighbors and it can choose its partition so as to maximize the product.

3

0 0

2

4

6

8

10

12

14

16

References Figure 1: Sentry selection with 4 phases

 Peter Hall, Introduction to The Theory of Coverage Processes, Wiley, 1988.  M. Rabbat and R. Nowak, “Distributed optimization in sensor networks”, in International Workshop on Information Processing in Sensor Networks (IPSN’04), 2004.

## Sentry Selection in Sensor Networks

Sensor Nodes have limited power resources. â The peak ... The typical power consumption of a sleeping node is about 3 orders of magnitude smaller than for a ... divided. The problem in a very general form can be stated as follows: For some maximum overlap D, choose sets A1. ,A2. ,...,AN. â Ï, such that. â¢. |Aiâ©Aj|. |Ï|.

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