Protocols for Secure Communication in Wireless Sensor Networks

6.4. Security Evaluation 181
a sensor network. In terms of hop count, we get an average path length of
L = ∆(2)W/D ≈ .52W/D
hops.
Now we can estimate the number of paths going through a single node as
follows. The total number of paths in the network is N(N − 1) and the average
path length is denoted as L. To each position on each path, a node needs to
be assigned, thus the total number of such assignments is LN(N − 1). These
assignments are assumed to be equally distributed over all nodes, disregarding
border effects, so there are
LN(N − 1)
N
= L(N − 1) (6.3)
assignments per node, which equals the number of paths that each node is part
of. We observe that border nodes are part of fewer paths than central ones.
Since we are interested in an average case analysis only, we will not deal with
such effects.
Under the Canvas authentication regime, a path with non-compromised endpoints
is compromised only if there are two adjacent compromised nodes somewhere
on the path. Whenever a new node is being compromised, all of its paths
that are shared with compromised neighbours will also be compromised. The
number of newly compromised paths is thus dependent on the number of compromised
neighbours. We can assume that a fraction of x/N of the neighbours
are already compromised. Recall from Section2.5.3 that d denotes the density
of the network. Thus, L(N − 1)/d · xd/N = L(N − 1)x/N new paths are
compromised. But we must be careful not to count already compromised paths
since a good deal of them might already be compromised by other pairs of
nodes. Therefore we introduce another factor, c, which is the fraction of so far
non-compromised paths:
c = 1 − BC(x)
N(N − 1) .
Through the multiplication by c we ensure that we count only a certain fraction
of the newly compromised paths. Finally, this yields the following equation
that describes the number of compromised paths when there are x compromised
nodes:
L(N − 1)x
BC(x + 1) = BC(x) + c 2(N − 1) + (6.4)
N
Figures. 6.11 and 6.12 show the approximation given in equation 6.4 compared
to the simulation of Canvas on the standard square. N = 500 and N =