Connectivity properties of large-scale sensor networks

Abstract

In wireless sensor networks, both nodes and links are prone to failures. In this paper we study connectivity properties of large-scale wireless sensor networks and discuss their implicit effect on routing algorithms and network reliability. We assume a network model of n sensors which are distributed randomly over a field based on a given distribution function. The sensors may be unreliable with a probability distribution, which possibly depends on n and the location of sensors. Two active sensor nodes are connected with probability p _e (n) if they are within communication range of each other. We prove a general result relating unreliable sensor networks to reliable networks. We investigate different graph theoretic properties of sensor networks such as k-connectivity and the existence of the giant component. While connectivity (i.e. k = 1) insures that all nodes can communicate with each other, k-connectivity for k > 1 is required for multi-path routing. We analyze the average shortest path of the k paths from a node in the sensing field back to a base station. It is found that the lengths of these multiple paths in a k-connected network are all close to the shortest path. These results are shown through graph theoretical derivations and are also verified through simulations.

Appendix

1.1 Proofs

1.1.1 Proof of Theorem 2

Proof

Define $\omega(n):=n {\pi}r^{2}(n) p_e(n)-\ln(n),$ thus $\pi r^{2}(n)=\frac{\ln n+ \omega(n)} {n p_{e}(n)}.$ Let $S_1=\overline{S(\overline{O},1-2r(n))}.$ We now obtain

$$ \begin{aligned} EZ_n & =n\int \limits_{S_0} \left(1-\nu(B(\overline{X},r(n)))p_{e}(n)\right)^{n-1}dm(\overline{X}) \\ & \geq n\int \limits_{S_1} \left(1-\nu(B(\overline{X},r(n)))p_{e}(n)\right)^{n-1}dm(\overline{X})\\ & = n \int \limits_{S_1} \left(1-\frac{\ln n+ \omega(n)} {n}\right)^{n-1}dm(\overline{X})\\ & = n \left(1-\frac{\ln n+\omega(n)} {n}\right)^{n-1} m(S_1)\\ & =e^{-\omega(n)}(1+o(1)). \end{aligned}.$$

(46)

Therefore, we conclude that $\lim \limits_{n\rightarrow\infty} EZ_n(r(n))= \infty$ if $\lim \limits_{n\rightarrow\infty} \omega(n)=-\infty.$ Moreover, $\lim \limits_{n\rightarrow\infty} EZ_n(r(n))>0$ if $\lim \limits_{n\rightarrow\infty} \omega(n)\le\infty.$ Now assume that $\lim \limits_{n\rightarrow\infty} \omega(n)>-\infty.$ Let Y _3,n be the number of isolated vertices in S ₃. Then we get

$$ \begin{aligned} EY_{3,n}\leq&n r^{2}(n) \left(1-\frac{\pi r^{2}(n)} {4}p_{e}(n)\right)^{n-1}\\ \leq&n r^{2}(n) e^{-\frac{\pi r^{2}(n)} {4}p_{e}(n)(n-1)}. \end{aligned} $$

(47)

Using $p_e(n)\geq \frac{c} {\ln n}$ and $\pi r^{2}(n)=\frac{\ln n+ \omega(n)} {n p_{e}(n)},$ we conclude

$$ EY_{3,n}=O\left (\frac{\ln n(\ln n+\omega(n))e^{-\omega(n)/4}} {n^{\frac{1}{4}}}\right)=o(1). $$

(48)

Therefore, there is no isolated vertex in S ₃ with high probability. Next, let Y _2,n be the number of isolated vertices in S ₂. Then

$$ EY_{2,n}=n\int \limits_{S_2} \left(1-\nu(B(\overline{X},r(n)))p_{e}(n)\right)^{n-1}dm(\overline{X}) $$

(49)

Using the Laplace method for integrals and Lemma 1, it can be shown that

$$ EY_{2,n}=O\left(\frac{e^{-\frac{\omega(n)} {2}}} {r(n)p_e(n) \sqrt{n}}\right) $$

(50)

Using $p_e(n)\geq \frac{c} {\ln n}$ and $\pi r^{2}(n)=\frac{\ln n+ \omega(n)} {n p_{e}(n)},$ we conclude

$$ EY_{2,n}=O\left (\frac{e^{-\frac{\omega(n)} {2}}}{\sqrt{\left(c+\frac{c \omega(n)} {\ln(n)}\right)}}\right).$$

(51)

Thus if $\lim \limits_{n\rightarrow\infty} \omega(n)=\infty$ then Y _2,n  = 0 asymptotically almost surely. Moreover, if $0\le\lim \limits_{n\rightarrow\infty} \omega(n)\le\infty$ then Y _2,n is finite asymptotically almost surely. Combining with (46) we conclude the theorem.□

1.1.2 Proof of Theorem 3

Proof

By Theorem 2, when $\lim \limits_{n\rightarrow \infty} \big[n \pi r^{2}(n) p_e(n)- \ln(n)\big]=\infty,$ we have $\lim\limits_{n\rightarrow\infty} EZ_n(r(n))=0.$ Thus, by Markov’s inequality there is no isolated vertex with high probability. Then, by Theorem 1 the graph is connected asymptotically almost surely. Hence, we focus on the proof of the other direction. That is if $0 < \lim \limits_{n\rightarrow \infty} \big[n \pi r^{2}(n) p_e(n)- \ln(n)\big] < \infty$ (or equivalently $0 < \lim\limits_{n\rightarrow\infty} EZ_n(r(n)) < \infty$), then there exists δ > 0 such that $\liminf \limits_{n\rightarrow\infty} p^{disc}_{n}> \delta > 0,$ where $p^{disc}_{n}$ is the probability that g _n is disconnected. The proof is as follows. Let A _n,j be the event that the vertex v _j is isolated. Then we want to prove

$$ \limsup \limits_{n \rightarrow \infty} \hbox{Pr}\left\{\bigcap_{i=1}^{n}\overline{A_{n,i}}\right\} \le 1. $$

(52)

To prove the above, we use Lemma 2. Let $\Updelta_n=\sum \limits_{i=1}^{n} \sum \limits_{j\neq i} \hbox{Pr}\{A_{n,i} \cap A_{n,j}\}.$ We show that under the condition $0\le\mu<\infty,$ we have $\lim \limits_{n \rightarrow \infty} \Updelta_n= \Updelta<\infty.$ Thus by applying Lemma 2 we conclude the theorem. It remains to prove Δ < ∞. We note that

$$ \begin{aligned} \Updelta_n \leq& n(n-1)\int \limits_{S_0\times S_0} \left(1-\nu(B(\overline{X},r(n)))p_{e}(n)\right.\\ &- \nu(B(\overline{X},r(n)))p_{e}(n)\\ &+ \nu(B(\overline{X},r(n))) \cap B(\overline{Y},r(n)))p_{e}^2(n) \left. \right)^{n-2}d m(\overline{X})\times m(\overline{Y}) \end{aligned} $$

(53)

We have $S_0\times S_0=(S_1\times S_1)\cup(S_0\times S_0 \setminus S_1\times S_1).$ It suffices to show that the integral over the set $S_1\times S_1$ and $S_0\times S_0 \setminus S_1\times S_1$ is finite. Let $\Updelta_n^{1}$ and $\Updelta_n^{2}$ be the two integrals respectively. For example, for S ₁ × S ₁ we have

$$ \begin{aligned} \Updelta_n^{1} & = n(n-1)\int \limits_{S_1\times S_1} \left(1-\nu(B(\overline{X},r(n)))p_{e}(n)\right.\\ & \quad -\nu(B(\overline{X},r(n)))p_{e}(n) \\ & \quad \left.+\nu(B(\overline{X},r(n))) \cap B(\overline{Y},r(n)))p_{e}^2(n) \right)^{n-1}d(m\times m)\\ & = n(n-1)\int \limits_{S_1\times S_1} \left(1-\frac{\ln n+ \omega(n)} {n}\right.\\ & \quad \left.-\frac{\ln n+ \omega(n)} {n}+\nu(B(\overline{X},r(n))) \cap B(\overline{Y},r(n)))p_{e}^2(n) \right)^{n-1}d(m\times m)\\ & \leq e^{-2\omega(n)}\int \limits_{S_1\times S_1} e^{\nu(B(\overline{X},r(n))) \cap B(\overline{Y},r(n)))p_{e}^2(n)(n-1)}d(m\times m)\\ & = e^{-2\omega(n)}\int \limits_{S_1} e^{\nu(B(\overline{O},r(n))) \cap B(\overline{Y},r(n)))p_{e}^2(n)(n-1)}dm(Y)\\ & = e^{-2\omega(n)}\int \limits_{S_1\setminus \quad B(\overline{O},2r(n))} e^{\nu(B(\overline{O},r(n))) \cap B(\overline{Y},r(n)))p_{e}^2(n)(n-1)}dm(Y)\\ & \quad + e^{-2\omega(n)}\int \limits_{ B(\overline{O},2r(n))} e^{\nu(B(\overline{O},r(n))) \cap B(\overline{Y},r(n)))p_{e}^2(n)(n-1)}dm(Y)\\ & = e^{-2\omega(n)}+e^{-2\omega(n)} \int \limits_{ B(\overline{O},2r(n))} e^{\nu(B(\overline{O},r(n))) \cap B(\overline{Y},r(n)))p_{e}^2(n)(n-1)}dm(Y). \end{aligned} $$

(54)

Using the Laplace method for integrals and Lemma 1 we obtain

$$ \Updelta_n^{1} = e^{-2\omega(n)}+O \left(\frac{e^{-(2-p_e(n)) \omega(n)}} {n^{(2-p_e(n))} p_e(n)^4 r^2(n)}\right) $$

(55)

Using $p_e(n)\geq \frac{c} {\ln n}$ and $0<\lim \limits_{n\rightarrow \infty} \omega(n)<\infty,$ we conclude

$$ \lim \limits_{n \rightarrow \infty} \Updelta_n^{1} <\infty. $$

(56)

Similarly, we can show $\lim \limits_{n \rightarrow \infty} \Updelta_n^{2} <\infty.$ Therefore, $\lim \limits_{n \rightarrow \infty} \Updelta_n=\Updelta<\infty,$ which concludes the theorem.□

1.1.3 Proof of Theorem 6

By a simple coupling argument, we find that the probability of having at least one isolated vertex is a decreasing function of r(n). If α < 1, then for any constant c and large enough n, we have

$$ r(n)<\sqrt{\frac{\ln n+ c} {\pi n p_{e}(n)}}. $$

(57)

Thus, by Theorem 4, the probability that g = g(n, r, p _e ) has at least one isolated vertex is asymptotically greater than or equal to $e^{-e^{-c}}$ for any real number c. Thus, if α < 1, the graph g = g(n, r, p _e ) has an isolated vertex with high probability, and thus it is not k-connected for any positive integer k.

Now, by Theorem 1, it suffices to prove that if α > 1, for any fixed $k\in \{0,1,2,\ldots\},$ g(n, r, p _e ) does not have any vertices of degree k with high probability. Let α > 1 and Y _j,k,n be the number of vertices of degree k in S _j , for j = 1, 2, 3. It suffices to show Y _j,k,n  = 0 asymptotically almost surely for j = 1, 2, 3.

We first consider Y _1,k,n . We have

$$ \begin{aligned} EY_{1,k,n}=& n\int \limits_{S_1} \binom{n}{k} [\nu(B(\overline{X},r(n)))p_{e}(n)]^k\\ &\left(1-\nu(B(\overline{X},r(n)))p_{e}(n)\right)^{n-k-1}dm(\overline{X}). \end{aligned} $$

(58)

But for $\overline{X}\in S_1,$ we have $\nu(B(\overline{X},r(n)))=\pi r^2(n).$ Thus

$$ EY_{1,k,n}= O\left( \frac{(\ln n)^k} {n^{\alpha-1}}\right)=o(1). $$

(59)

Therefore, Y _1,k,n  = 0 asymptotically almost surely. We now consider Y _2,k,n . We have

$$ \begin{aligned} EY_{2,k,n}=& n\int \limits_{S_2} \binom{n}{k} [\nu(B(\overline{X},r(n)))p_{e}(n)]^k\\ &\left(1-\nu(B(\overline{X},r(n)))p_{e}(n)\right)^{n-k-1}dm(\overline{X}). \end{aligned} $$

(60)

Using the Laplace method for integrals, Lemma 1, and $p_e(n)\geq \frac{c} {\ln n}$ we can write

$$ EY_{2,k,n}=O\left(\frac{(\ln n)^{2k+1}} {n^{\frac{\alpha}{2}-\frac{1}{2}-o(1)}}\right)=o(1) $$

(61)

This implies Y _2,k,n  = 0 asymptotically almost surely. We now prove Y _3,k,n  = 0 asymptotically almost surely. We note that

$$ \begin{aligned} EY_{3,k,n}\leq & n r^{2}(n) \binom{n}{k} (\pi r^2(n)p_e(n))^k \left(1-\frac{\pi r^{2}(n)} {4}p_{e}(n)\right)^{n-k-1}\\ \leq& n^{k+1} r^{2k+2}(n)p_e(n) e^{-\frac{\pi r^{2}(n)} {4}p_{e}(n)(n-k-1)}. \end{aligned} $$

(62)

Using $p_e(n)\geq \frac{c} {\ln n}$ and $\lim \limits_{n\rightarrow\infty} \left(\frac{n \pi r^2(n) p_e(n)} {\ln n} \right)=\alpha,$ we conclude

$$ EY_{3,k,n}=O\left (\frac{(\ln n)^{k+2}} {n^{\frac{1}{4}-o(1)}}\right)=o(1) $$

(63)

This implies that Y _3,k,n  = 0 asymptotically almost surely.□