Connectivity of multihop wireless networks with log-normal shadowing

Abstract

In this paper, we study the connectivity of multihop wireless networks with the log-normal shadowing model by investigating the critical transmission power required by each node for asymptotic vanishing of the isolated nodes, and the precise distribution of the number of isolated nodes. The vanishing of isolated nodes is not only a prerequisite but also a good indication of network connectivity. Most of the known works on network connectivity under such a shadowing model were obtained only based on simulation studies or ignoring the important boundary effect to avoid the challenging technical analysis, and thus hardly applied to practical wireless networks. It is extremely challenging to take the complicated boundary effect into consideration under such a realistic shadowing model because the transmission area of each node is an irregular region other than a circular area. Assume the wireless nodes are represented by a Poisson point process with density $n$ over a unit-area disk. With the boundary effect taken into consideration, we first obtain an explicit formula for the expected number of isolated nodes, we then derive an upper and a lower bounds of the critical transmission power for asymptotic vanishing of the isolated nodes. The tightness of the upper and lower bounds for the critical transmission power are analyzed via numerical analysis by using the software engineering approach. When a wireless network consists of $n$ nodes distributed independently and uniformly over a unit-area disk, we derive the precise distribution of the number of the isolated nodes under such a realistic shadowing model with the linear power assignment, taking the boundary effect into consideration.

Appendix

This appendix is dedicated to the proof of Theorem 3 in Sect. 4.

Proof of Theorem 3

Let $X$ be a random point with uniform distribution over ${{\mathbb{D}}}$ and independent of ${\mathcal {P}}_{n}$. For any finite planar set $V$ and any subset $U$ of $V,$ define $h\left( U,V\right)$ as follows: $h(U,V)$ $=1$ if $\left| U\right| =1$ and the node in $U$ is isolated from all the nodes in $V,$ and $h(U,V)$ $=0$ otherwise. Note that ${{\mathbb{D}}}$ is partitioned into two sub-regions ${\mathbb{D}}(1)$ and ${\mathbb{D}}(2)$ (see Fig. 2).

Let $N$ denote the number of isolated nodes in the network, then by the Palm Theory,

$$\begin{aligned} {\mathbf{E}}\left[ N\right]&=\underset{Y\in {\mathcal {P}}_{n}}{ {\sum } }{\mathbf{E}}\left[ h(\left\{ Y\right\} ,{\mathcal {P}}_{n})\right] ={\mathbf{E}}\left[ \underset{Y\in {\mathcal {P}}_{n}}{ {\sum } }h(\left\{ Y\right\} ,{\mathcal {P}}_{n})\right] \nonumber \\&=n{\mathbf{E}}\left[ h(\left\{ X\right\} ,\left\{ X\right\} \cup {\mathcal {P}}_{n})\right] \nonumber \\&=n\Pr \left( X\ {\text {is isolated in }}\left\{ X\right\} \cup {\mathcal {P}}_{n}\right) \nonumber \\&=n {\iint \limits _{x\in {{\mathbb{D}}}}} \Pr (X\ {\text {is isolated in }}\left\{ X\right\} \cup {\mathcal {P}} _{n}\vert X=x)dA.\nonumber \\&=n\left( {\iint \limits _{x\in {\mathbb{D}}(1)}} + {\iint \limits _{x\in {\mathbb{D}}(2)}} \right) \Pr (X\ {\text {isolated in }}\left\{ X\right\} \cup {\mathcal {P}}_{n} \vert X=x)dA. \end{aligned}$$

(18)

To compute the integral, we equally partition the interval $[0,R_{n}]$ into $k$ subintervals with $r_{0}=0<$ $r_{1}<$ $r_{2}<\cdots <r_{k}=R_{n}.$ Let $\triangle r_{i}=r_{i}-r_{i-1}=\frac{R_{n}}{k},$ where $i=1,2,\ldots ,k.$

First we calculate the integral on ${\mathbb{D}}(1)$. For any $x\in {\mathbb{D}}(1),$ the disk $D(x,R_{n})$ is divided into $k-1$ concentric annuli and the center disk $D(x,r_{1}).$ We consider the center disk $D(x,r_{1})$ as an annulus with inner radius $r_{0}=0$ and outer radius $r_{1}.$ For each $1\le i\le k,$ the $i$th annulus is the one with inner radius $r_{i-1}$ and outer radius $r_{i}.$ The area of the $i$th annulus can be written as $2\pi r_{i}\triangle \widetilde{r_{i}},$ where

$$\begin{aligned} \triangle \widetilde{r_{i}}=\triangle r_{i}+o(\triangle r_{i})=\frac{R_{n}}{k}+o\left( \frac{1}{k}\right) \end{aligned}$$

as $k\rightarrow \infty .$ Let $N_{i}$ denote the number of nodes in the $i$th annulus, $1\le i\le k$, then $N_{i}$ is a Poisson random variable with mean $n(2\pi r_{i}\triangle \widetilde{r_{i}}).$ For every $1\le i\le k,$ we have

$$\begin{aligned}&\Pr \left( X\; has\; no\; link\; with\; nodes\; in\; i{\text{-}}th\; annulus | X=x\right) \nonumber \\&\quad = {\sum \limits _{j=0}^{\infty }} \Pr \left( \left( \begin{array}{l} X\; has\; no\; link\ \\ with\; nodes\; in\ \\ the\; i{\text{-}}th\; annulus \end{array} | X=x\right) \mid N_{i}=j\right) \cdot \nonumber \\&\Pr \left( N_{i}=j\mid X=x\right) \nonumber \\&\quad = {\sum \limits _{j=0}^{\infty }} \left( 1-f(r_{i},n)\right) ^{j}\left( \frac{(2n\pi r_{i}\triangle \widetilde{r_{i}})^{j}}{j!}e^{-2n\pi r_{i}\triangle \widetilde{r_{i}}}\right) \nonumber \\&\quad =e^{-2n\pi r_{i}\triangle \widetilde{r_{i}}} {\sum \limits _{j=0}^{\infty }} \frac{(2n\pi r_{i}\triangle \widetilde{r_{i}}\left( 1-f(r_{i},n)\right) )^{j}}{j!}\nonumber \\&\quad =e^{-2n\pi r_{i}\triangle \widetilde{r_{i}}f(r_{i},n)}. \end{aligned}$$

(19)

For any $x\in {\mathbb{D}}(1),$ we have

$$\begin{aligned}&\Pr \left( \hbox {X isolated in }\left\{ X\right\} \cup {\mathcal {P}} _{n}\mid X=x\right) \nonumber \\&\quad =\lim _{k\rightarrow \infty }\,\Pr \left( \begin{array}{l} For\; all\; 1\le i\le k,\ \\ X\; has\; no\; link\; with\\ nodes\; in\; i{\text{-}}th\; annulus \end{array} \mid X=x\right) \nonumber \\&\quad =\lim _{k\rightarrow \infty } {\prod \limits _{i=1}^{k}} \Pr \left( \begin{array}{l} X\; has\; no\; link\; with\ \\ nodes\; in\; the\; \\ i{\text{-}}th\; annulus \end{array} | X=x\right) \nonumber \\&\quad =\lim _{k\rightarrow \infty } {\prod \limits _{i=1}^{k}} e^{-2n\pi r_{i}\triangle \widetilde{r_{i}}f(r_{i},n)}\nonumber \\&\quad =e^{\lim _{k\rightarrow \infty } {\sum \nolimits _{i=1}^{k}} \left( -2n\pi r_{i}\right) \left( \frac{R}{k}+o\left( \frac{1}{k}\right) \right) f(r_{i},n)}\nonumber \\&\quad =e^{\lim _{k\rightarrow \infty } {\sum \nolimits _{i=1}^{k}} \left( -2n\pi r_{i}\right) \frac{R}{k}f(r_{i},n)+ {\sum \nolimits _{i=1}^{k}} \left( -2n\pi r_{i}\right) o\left( \frac{1}{k}\right) f(r_{i},n)}\nonumber \\&\quad =e^{\lim _{k\rightarrow \infty }\left( {\sum \nolimits _{i=1}^{k}} \left( -2n\pi r_{i}\right) \frac{R}{k}f(r_{i},n)\right) }\nonumber \\&\quad =e^{-\int _{0}^{R_{n}}2n\pi rf(r,n)dr}\nonumber \\&\quad =e^{-M_{n}}. \end{aligned}$$

(20)

Therefore,

$$\begin{aligned}&{\iint \limits _{x\in {\mathbb{D}}(1)}} \Pr (X\ isolated\ |\ X=x)dA\nonumber \\&\quad =e^{-M_{n}}\left| {\mathbb{D}}(1)\right| \nonumber \\&\quad =\pi e^{-M_{n}}\left( \frac{1}{\sqrt{\pi }}-R_{n}\right) ^{2}. \end{aligned}$$

(21)

Next we calculate the integral on ${\mathbb{D}}(2)$. For any $x\in {\mathbb{D}}(2),$ the region $D(x,R_{n})\cap {{\mathbb{D}}}$ is divided into $k-1$ concentric circular belt regions and the center disk $D(x,r_{1}).$ For each $1\le i\le k,$ the $i$th circular belt region (annulus if it is contained in ${{\mathbb{D}}}$) is the one with inner radius $r_{i-1}$ and outer radius $r_{i}.$ Let $l_{i}$ denote the arc length of the $i$th circular belt region and $l_{i}^{\prime }=2\pi r_{i}-l_{i}$, then its area can be written as $l_{i}\triangle \widetilde{r_{i}},$ where

$$\begin{aligned} \triangle \widetilde{r_{i}}=\triangle r_{i}+o(\triangle r_{i})=\frac{R_{n}}{k}+o\left( \frac{1}{k}\right) \end{aligned}$$

as $k\rightarrow \infty .$ (If the $i$th annulus is entirely contained in ${{\mathbb{D}}}$, then $l_{i}=2\pi r_{i}.$) Let $N_{i}$ denote the number of nodes in the $i$th circular belt region, $1\le i\le k$, then $N_{i}$ is a Poisson random variable with mean $n(l_{i}\triangle \widetilde{r_{i}}).$

For any $1\le i\le k,$

$$\begin{aligned}&\Pr \left( \begin{array}{l} X\ has\ no\ link\ with\ nodes\ in\\ the\ i{\text{-}}th\ circular\ belt\ region \end{array} \mid X=x\right) \\&\quad = {\sum \limits _{j=0}^{\infty }} \Pr \left( \left( \begin{array}{l} X\ has\ no\ link\ with\ \\ nodes\ in\ i{\text{-}}th\ \\ circular\ belt\ region \end{array} \ |\ X=x\right) \mid N_{i}=j\right) \cdot \\&\Pr \left( N_{i}=j\mid X=x\right) \\&\quad = {\sum \limits _{j=0}^{\infty }} \left( 1-f(r_{i},n)\right) ^{j}\left( \frac{(nl_{i}\triangle \widetilde{r_{i}})^{j}}{j!}e^{-nl_{i}\triangle \widetilde{r_{i}}}\right) \\&\quad =e^{-nl_{i}\triangle \widetilde{r_{i}}} {\sum \limits _{j=0}^{\infty }} \frac{(nl_{i}\triangle \widetilde{r_{i}}\left( 1-f(r_{i},n)\right) )^{j}}{j!}\\&\quad =e^{-nl_{i}\triangle \widetilde{r_{i}}f(r_{i},n)}. \end{aligned}$$

So, for any $x\in {\mathbb{D}}(2),$

$$\begin{aligned}&\Pr \left( X\ is\ isolated\mid X=x\right) \nonumber \\&\quad =\lim _{k\rightarrow \infty }\,\Pr \left( \begin{array}{l} For\ all\ 1\le i\le k,\ X\ has\\ no\ link\ with\ the\ nodes\ in\\ \ the\ i{\text{-}}thth\ circular\ belt\ region \end{array} |\ X=x\right) \nonumber \\&\quad =\lim _{k\rightarrow \infty } {\prod \limits _{i=1}^{k}} \Pr \left( \begin{array}{l} X\ has\ no\ link\ with\ nodes\\ in\ i{\text{-}}th\ circular\ belt\ region \end{array} |\ X=x\right) \nonumber \\&\quad =\lim _{k\rightarrow \infty } {\prod \limits _{i=1}^{k}} e^{-nl_{i}\triangle \widetilde{r_{i}}f(r_{i},n)}\nonumber \\&\quad =e^{\lim _{k\rightarrow \infty } {\sum \nolimits _{i=1}^{k}} \left( -nl_{i}\right) \left( \frac{R_{n}}{k}+o\left( \frac{1}{k}\right) \right) f(r_{i},n)}\nonumber \\&\quad =e^{\lim _{k\rightarrow \infty }\left( -n {\sum \nolimits _{i=1}^{k}} \left( 2\pi r_{i}-l_{i}^{\prime }\right) \frac{R_{n}}{k}f(r_{i},n)\right) }\nonumber \\&\quad =e^{-\int _{0}^{R_{n}}2n\pi rf(r,n)dr\ +\ \lim _{k\rightarrow \infty }n {\sum \nolimits _{i=1}^{k}} l_{i}^{\prime }\frac{R_{n}}{k}f(r_{i},n)}\nonumber \\&\quad =e^{-M_{n}\ +\ n {\iint \nolimits _{D(x,R_{n})\backslash {{\mathbb{D}}}}} f(\rho ,n)\rho d\rho d\theta }, \end{aligned}$$

(22)

where the double integral over the region $D(x,R_{n})\backslash {{\mathbb{D}}}$ in the last equation is based on the polar coordinate system ($\rho ,\theta$) with the point $x$ as the origin and ray $ox$ as the fixed direction.

Fig. 4

When $x\in {{\mathbb{D}}}(2)$, a part of the disk $D(x,R_{n})$ is outside of the unit-area disk ${{\mathbb{D}}}$

Next we calculate the double integral over the region $D(x,R_{n} )\backslash {{\mathbb{D}}}$. Extend $ox$ to $c$ on the boundary $\partial {{\mathbb{D}}}$ (see Fig. 4). Let $a$ and $b$ be the two intersection points of $\partial D(x,R_{n})$ and $\partial {{\mathbb{D}}}$ such that $a$ is above $oc.$ Under the polar coordinate system ($\rho ,\theta$) mentioned above, the equation of the circle $\partial D(x,R_{n})$ is $\rho =R_{n}$ and the equation of the arc $\partial {\mathbb{D}} \cap D(x,R_{n})$ is given by

$$\begin{aligned} \left( \rho \cos \theta +\left\| x\right\| \right) ^{2}+\left( \rho \sin \theta \right) ^{2}=\left\| a\right\| ^{2}. \end{aligned}$$

That is,

$$\begin{aligned} \rho ^{2}+2\left\| x\right\| \rho \cos \theta -(\left\| a\right\| ^{2}-\left\| x\right\| ^{2})=0. \end{aligned}$$

Solving this quadratic equation for $\rho ,$ we have

$$\begin{aligned} \rho =\sqrt{\frac{1}{\pi }-\left\| x\right\| ^{2}\sin ^{2}\theta }-\left\| x\right\| \cos \theta . \end{aligned}$$

Let $\beta =\angle axc.$ By the law of cosine, we have

$$\begin{aligned} \cos (\pi -\beta )=\frac{\left\| x\right\| ^{2}+R_{n}^{2}-\left( \frac{1}{\sqrt{\pi }}\right) ^{2}}{2R_{n}\left\| x\right\| }. \end{aligned}$$

Thus,

$$\begin{aligned} \beta =\pi -\arccos \frac{\left\| x\right\| ^{2}+R_{n}^{2}-\frac{1}{\pi } }{2R_{n}\left\| x\right\| }. \end{aligned}$$

Hence,

$$\begin{aligned}&{\iint \nolimits _{D(x,R_{n})\backslash {{\mathbb{D}}}}} f(\rho ,n)\rho d\rho d\theta \\&\quad =\int _{-\beta }^{\beta }d\theta \int _{\sqrt{\frac{1}{\pi }-\left\| x\right\| ^{2}\sin ^{2}\theta }-\left\| x\right\| \cos \theta }^{R_{n} }f(\rho ,n)\rho d\rho \\&\quad =2\int _{0}^{\beta }d\theta \int _{\sqrt{\frac{1}{\pi }-\left\| x\right\| ^{2}\sin ^{2}\theta }-\left\| x\right\| \cos \theta }^{R_{n}}f(\rho ,n)\rho d\rho . \end{aligned}$$

Therefore,

$$\begin{aligned}&{\iint \limits _{x\in {\mathbb{D}}(2)}} \Pr (X\; isolated\; |\ X=x)dA\nonumber \\&\quad = {\iint \limits _{x\in {\mathbb{D}}(2)}} e^{-\int _{0}^{R_{n}}2n\pi rf(r,n)dr\ +\ n {\iint \nolimits _{D(x,R)\backslash {{\mathbb{D}}}}} f(\rho ,n)\rho d\rho d\theta }dA\nonumber \\&\quad =e^{-M_{n}\ }\int _{\sqrt{\frac{1}{\pi }}-R_{n}}^{\sqrt{\frac{1}{\pi }}}2\pi e^{2n\int _{0}^{\beta }d\theta \int _{\sqrt{\frac{1}{\pi }-r^{2}\sin ^{2}\theta }-r\cos \theta }^{R_{n}}f(\rho ,n)\rho d\rho }rdr. \end{aligned}$$

(23)

The last equation holds by setting $r=\left\| x\right\| .$

Thus, Theorem 3 follows by combining Eqs. (21) and (23). $\square$