Asymptotic throughput for large-scale wireless networks with general node density

Abstract

We study the asymptotic throughput for a large-scale wireless ad hoc network consisting of n nodes under the generalized physical model. We directly compute the throughput of multicast sessions to unify the unicast and broadcast throughputs. We design two multicast schemes based on the so-called ordinary arterial road system and parallel arterial road system, respectively. Correspondingly, we derive the achievable multicast throughput by taking account of all possible cases of n _s  = ω(1) and 1 ≤ n _d  ≤ n − 1, rather than only the cases of $n_s=\Uptheta(n)$ as in most related works, where n _s and n _d denote the number of sessions and the number of destinations of each session, respectively. Furthermore, we consider the network with a general node density $\lambda \in [1,n]$, while the models in most related works are either random dense network (RDN) or random extended network (REN) where the density is λ = n and λ = 1, respectively, which further enhances the generality of this work. Particularly, for the special case of our results by letting λ = 1 and $n_s=\Uptheta(n)$, we show that for some regimes of n _d , the multicast throughputs achieved by our schemes are better than those derived by the well-known percolation-based schemes.

Proofs of some lemmas and theorems

1.1 Proof of Lemma 4

We first recall a useful lemma about the tails of Chernoff bound from [34].

Lemma 14

([34]) Let X be a Poisson random variable with parameter μ. Then

$$ \Pr(X \ge x) \le { e^{- \mu }{\cdot}(e{\cdot}\mu)^x} /{x^x}, \quad{\rm for}\, x > \mu \\ $$

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$$ \Pr(X \leq x) \le { e^{- \mu }{\cdot}(e{\cdot}\mu)^x} /{x^x},\quad{\rm for}\, 0<x < \mu $$

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Next, we begin to prove Lemma 4.

Proof

Define the maximum and minimum number of nodes for all ${\frac{n}{\lambda {\cdot} S }}$ subregions as $\overline{\mu}$ and $\underline{\mu}$, respectively.

• Case 1 : When $S {\cdot} \lambda =\Upomega(1)$ and $S {\cdot} \lambda =o(\log n)$.

• According to Eq. (14) and union bound, it follows that

  $$ {\Pr}(\overline{\mu}\geq \Updelta_1 {\cdot} \log n) \leq {\frac{n}{\lambda {\cdot} S }} {\cdot} \left( {\frac{e^{1-{\frac{S \lambda}{\Updelta_1 \log n}}}{\cdot} S \lambda}{\Updelta_1 \log n}}\right)^{\Updelta_1 \log n} \to 0, $$

  for $\Updelta_1$ in Table 2, as $n \to \infty$, which proves the upper bounds. The lower bounds is straightforward.

• Case 2 : When $S {\cdot} \lambda =\omega( \log n)$.

• According to Eq. (14) and union bound, it follows that

  $$ {\Pr}(\underline {\mu}\leq (1-\Updelta_2) {\cdot} S \lambda) \leq {\frac{n}{\lambda {\cdot} S }} {\cdot} \left( {\frac{e^{-\Updelta_2}}{(1-\Updelta_2)^{(1-\Updelta_2)}}}\right)^{S \lambda} \to 0, $$

  for $\Updelta_2$ in Table 2, which proves the lower bounds. Similarly,

  $$ {\Pr}(\overline{\mu}\geq (1+\Updelta_3) {\cdot} S \lambda) \leq {\frac{n}{\lambda {\cdot} S }} {\cdot} \left( {\frac{e^{\Updelta_3}}{(1+\Updelta_3)^{(1+\Updelta_3)}}}\right)^{S \lambda} \to 0, $$

  for $\Updelta_3$ in Table 2, as $n \to \infty$, which proves the upper bounds.

• Case 3 : When $S{\cdot}\lambda : z {\cdot} \log n$, for any $z\in (0, \infty)$.

• Similarly, we have

  $$ {\Pr}(\underline {\mu}\leq (1-\Updelta_4) {\cdot} z {\cdot} \log n) \leq {\frac{n}{\lambda {\cdot} S }} {\cdot} \left( {\frac{e^{-\Updelta_4}}{(1-\Updelta_4)^{(1-\Updelta_4)}}}\right)^{z {\cdot} \log n} \to 0, $$

  for $\Updelta_4$ in Table.2, which proves the lower bounds; and

  $$ {\Pr}(\overline{\mu}\geq (1+\Updelta_5) {\cdot} z {\cdot} \log n) \leq {\frac{n}{\lambda {\cdot} S }} {\cdot} \left( {\frac{e^{\Updelta_5}}{(1+\Updelta_5)^{(1+\Updelta_5)}}}\right)^{z {\cdot} \log n} \to 0, $$

  for $\Updelta_5$ in Table.2, which proves the upper bounds.

1.2 Proof of Lemma 8

To simplify the description, we denote ${v_{\mathcal{S},k}}$, ${\mathcal{M}_{\mathcal{S},k}}$, ${\mathcal{U}_{\mathcal{S},k}}$, and ${\mathcal{D}_{\mathcal{S},k}}$ by v _k , $\mathcal{M}_{ k}$, $\mathcal{U}_{ k}$, and $\mathcal{D}_{ k}$, respectively, without confusion. Furthermore, we denote the set of all links in $\hbox{EST}(\mathcal{U}_{ k})$ as $\Uppi_k$.

Given a cell c ^* _t , we define the number of multicast sessions that are routed through the station inside c ^* _t as a random variable Z _t , and finally we consider the uniform upper bound of Z _t for every cell, denoted as Z. Define an event B(k, t): Multicast session $\mathcal{M}_{k}$ passes through the cell c ^* _t . For any link $v_iv_j \in \hbox{EST}(U_k)$, define an event B ^h _ij (k, t): $\mathcal{L}^h_{ij}$ passes through c ^* _t ; and define an event B ^v _ij (k, t): $\mathcal{L}^v_{ij}$ passes through c ^* _t . Then,

$$ {\Pr}(B(k,t) )\le \sum\nolimits_{e_{ij} \in \Uppi_k} (\Pr(B_{ij}^h (k,t)) + \Pr(B_{ij}^v (k,t)) ) $$

Based on c ^* _t , we construct the region $\mathcal{D}^h_{ij}(k,t)$ as illustrated in Fig. 6, then the following proposition is true.

Fig. 6

Construction of the region $\mathcal{D}^h_{ij}(k,t)$. Here $|{\cdot}|$ represents the Euclidean length of a line segment or the Euclidean distance between two nodes

Proposition 3

The Poisson node v _i is located in the region $\mathcal{D}^h_{ij}(k,t)$ if the event B ^h _ij (k, t)happens.

Similarly, we can construct the region $\mathcal{D}^v_{ij}(k,t)$, and we have

Proposition 4

The Poisson node v _j is located in the region $\mathcal{D}^v_{ij}(k,t)$ if the event B ^v _ij (k, t)happens.

Define the number of nodes in the region of area D(k, t) as a random variable ϑ _t , where $D(k,t) = \min\{n/\lambda, \tilde{D}(k,t)\}$, and

$$ \tilde{D}(k,t) = \sum\limits_{e_{ij} \in \Uppi_k}(\|{\mathcal{D}}^h_{ij}(k,t)\|+\|{\mathcal{D}}^v_{ij}(k,t)\|) \le 2\left( {\frac{n_d {\cdot} 8\log n}{\lambda}}+ 2\sqrt {{\frac{2n_d n}{\lambda}}}{\cdot} 2\sqrt{{\frac{2\log n}{\lambda}}}\right) $$

Hence, $D(k,t)\leq D= \min \{{\frac{16}{\lambda}}{\cdot}(n_d {\cdot}\log n + \sqrt{n_d{\cdot} \log n {\cdot} n}, \,{\frac{n}{\lambda}}\}.$ Let $\Upgamma_o:=D {\cdot} \lambda$. Then, Obviously, ϑ _t follows a Poisson distribution with the mean of at most $\mu_o = {\frac{n_s }{n}} {\cdot} \Upgamma_o$. From Proposition 3 and Proposition 4, we have that Z _t  ≤ ϑ _t .

By Lemma 4, according to different cases of ${\frac{n_s }{n}} {\cdot} \Upgamma_o$, we prove this lemma.

1.3 Proof of Lemma 12

From Lemma 10, we get that the rate along the arterial roads can be achieved of order

$$ {\rm R}(n)=\left\{ \begin{array}{lll} \Upomega\left({\frac{\lambda^{{\frac{\alpha}{2}}}}{(\log n)^{{\frac{\alpha}{2}}}}}\right) & {when} & \lambda :[1, (\log n)^{1-{\frac{2}{\alpha}}}] \\ \Upomega\left({\frac{1}{\log n}} \right) & {when} & \lambda :[(\log n)^{1-{\frac{2}{\alpha}}}, n] \\ \end{array}\right.. $$

Thus, we only need prove that for every station the maximum relay burden is w.h.p. of order

$$ {\rm L}(n)=\left\{ \begin{array}{ll} O\left({\frac{n_s{\cdot}\Upgamma_{p}}{n}}\right) & {\rm when}\,n_s{\cdot} \Upgamma_{p} = \Upomega(n\log n) \\ O(\log n) & {\rm when}\,n_s{\cdot} \Upgamma_{p} = O(n\log n)\\ \end{array}\right. $$

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where $\Upgamma_{p}=\left\{ \begin{array}{ll} \Uptheta({\frac{\sqrt{n n_d}}{\sqrt{\log n}}}) & {when\,} n_d=O({\frac{n}{\log n}})\\ \Uptheta(n_d)& {when\,} n_d=\Upomega({\frac{n}{\log n}}).\\ \end{array}\right.$

Given a station v ^* _t passed by a horizontal arterial road ${\mathbf h}$ or a vertical arterial road ${\mathbf v}$, we define the number of multicast sessions routed through v ^* _t during Phase 2, 3 and 4 as a random variable X _t . Note that we finally need the uniform upper bound of X _t for all stations, denoted by X.

Define an event A(k, t): During Phase 2, 3 and 4, multicast session $\mathcal{M}_k$ passes through v ^* _t . Furthermore, for each edge $e_{ij} \in \Uppi_k$, define the event A ^h _ij (k, t) (or A ^v _ij (k, t)): During Phase 2 (or Phase 4), $\mathcal{M}_k$ passes through v ^* _t . Hence, we have

$$ {\Pr}(A(k,t) )\le \sum\nolimits_{e_{ij} \in \Uppi_k} (\Pr(A_{ij}^h (k,t) ) + \Pr(A_{ij}^v (k,t)) ) $$

For each link $v_iv_j \in \Uppi_k$, construct the region $\mathcal{S}^h_{ij}(k,t)$ as in Fig. 7, where $|{\cdot}|_{\rm h}$ and $|{\cdot}|_{\rm v}$ represent the horizontal and vertical Euclidean distance, respectively, that are defined as: In the 2-dimension plane, for any two nodes u ₁ and u ₂ with the coordinates (x ₁, y ₁) and (x ₂, y ₂) respectively, |u ₁ u ₂|_h = |x ₁ − x ₂| and |u ₁ u ₂|_v = |y ₁ − y ₂|. Therefore,

Fig. 7

Construction of the region $\mathcal{S}^h_{ij}(k,t)$. Here $(\Uppsi^h)^{-1}({\mathbf h})$ is the horizontal slice that corresponds to the arterial road ${\mathbf h}$. $\mathcal{S}^h_{ij}(k,t)$ is the shadowed region of sides l × 2L ^h _ij , where $l ={\frac{\sqrt{2}}{\sqrt{\lambda {\cdot} \log n}}}$ and $L_{ij}^h = | v_iv_j |_{\rm h} + 2\sqrt{2\log n /\lambda}$

Proposition 5

If A ^h _ij (k, t) happens, then the Poisson point v _i is located in the region $\mathcal{S}^h_{ij}(k,t)$.

By a similar method, based on the arterial road ${\mathbf v}$ and the vertical slice $(\Uppsi^v)^{-1}({\mathbf v})$, we construct the region $\mathcal{S}^v_{ij}(k,t)$ of height l and width 2L ^v _ij , where

$$ L_{ij}^v = | v_iv_j |_{\rm v} + 2\sqrt{2\log n /\lambda}. $$

Hence, we have the following result.

Proposition 6

If A ^v _ij (k, t) happens, then the Poisson point v _j is located in the region $\mathcal{S}^v_{ij}(k,t)$.

Define Y _t as the number of Poisson nodes in the region of area S(k, t) according to a p.p.p of density ${\frac{n_s}{n}}{\cdot} \lambda$, where

$$ S(k,t) = \min\{n, \tilde{S}(k,t)\}, \, \tilde{S}(k,t) = \sum \nolimits_{e_{ij} \in \Uppi_k}(\|{\mathcal{S}}^h_{ij}(k,t)\|+\|{\mathcal{S}}^v_{ij}(k,t)\|)\} $$

Then, Y _t follows Poisson distribution with $\mu_p = {\frac{n_s}{n}}{\cdot} \lambda {\cdot} S(k,t)$. According to Proposition 5 and Proposition 6, we can obtain that X _t  ≤ Y _t . Then, we have

$$ \tilde{S}(k,t) = 2\sum\nolimits_{e_{ij} \in \Uppi_k}( | v_iv_j |_{\rm h} + | v_iv_j |_{\rm v} + 2\sqrt{2\log n /\lambda}){\cdot}{l} $$

Since $| v_iv_j |_{\rm h} + | v_iv_j |_{\rm v} \le \sqrt 2 | {v_i v_j } |$ and by Lemma 6,

$$ \tilde{S}(k,t) \leq {\frac{2\sqrt{2}}{\sqrt{\lambda\log n}}} {\cdot} \sum\nolimits_{e_{ij} \in \Uppi_k}( \sqrt{2}{\cdot}| v_iv_j |+ 2\sqrt{{\frac{2\log n}{\lambda}}})\leq {\frac{8}{\lambda}} {\cdot} n_d +{\frac{8\sqrt{2}}{\lambda}}{\cdot} \sqrt {{\frac{n_d n}{\log n}}} $$

Then, we have

$$ S(k,t)\leq {\frac{8}{\lambda}}{\cdot} (n_d + \sqrt {{\frac{2 {\cdot} n_d {\cdot} n}{\log n}}}):= {\frac{8}{\lambda}}{\cdot} \Upgamma_{p} $$

Then, $\mu_p = {\frac{n_s}{n}}{\cdot} \lambda {\cdot} S(k,t)\leq {\frac{n_s}{n}}{\cdot} 8{\cdot} \Upgamma_{p}$.

From Lemma 4, according to different cases of ${\frac{n_s}{n}}{\cdot} 8{\cdot} \Upgamma_{p}$, we get the Eq. (15). Hence, combining Lemma 10, we prove this lemma.

1.4 Proof of Lemma 13

For any given multicast session $\mathcal{M}_k$, denote the set of leaf nodes of $\hbox{EST}(\mathcal{U}_k)$ by $\mathcal{\tilde{U}}_k$. Then, in Phase 1, only the nodes in the set $\mathcal{U}_k-\mathcal{\tilde{U}}_k$ intend to access into the corresponding arterial roads via a single hop; while in Phase 5, only the nodes in the set $\mathcal{U}_k-\{v_k\}$ need to receive the data from the corresponding arterial roads via a single hop. Since it is not always true that $|\mathcal{U}_k-\mathcal{\tilde{U}}_k|=|\mathcal{U}_k-\{v_k\}|$, where $|{\cdot}|$ represents the cardinality of a discrete set, we learn the fact that unlike the unicast case, Phase 1 is not symmetrical with Phase 5 due to the tree structure of each EST. However, for $|\mathcal{U}_k-\mathcal{\tilde{U}}_k|\leq|\mathcal{U}_k-\{v_k\}|$, the load in Phase 5 is no more than that in Phase 1. According to Lemma 11, by parallel transmission scheduling, each cell can sustain a total rate of order

$$ \left\{ \begin{array}{lll} \Upomega\left({\frac{\lambda^{{\frac{\alpha}{2}}}}{(\log n)^{{\frac{\alpha}{2}}-1}}}\right) & {when} & \lambda :[1, (\log n)^{1-{\frac{2}{\alpha}}}] \\ \Upomega(1) & {when} & \lambda :[(\log n)^{1-{\frac{2}{\alpha}}}, n] \\ \end{array}\right.. $$

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By bottleneck-principle, we only need to examine the throughput in Phase 5 out of those two phases. In Phase 5, we only need to prove that the bandwidth as in Eq. (16) bears the load of order

$$ \left\{ \begin{array}{ll} O({\frac{n_s {\cdot} n_d {\cdot} \log n}{n}}) & {\rm when}\,n_s {\cdot} n_d = \Upomega(n) \\ O(\log n) & {\rm when}\,n_s {\cdot} n_d = O(n)\\ \end{array}\right. $$

(17)

Similar to the proof of Lemma 12, since

$$ |\cup_{v_k\in {\mathcal{S}}}{({\mathcal{U}}_k-\{v_k\})}|\leq n_s{\cdot} (n_d-1), $$

an upper bound of the load of each group of $\Uptheta(\log n)$ arterial roads, with a total rate of $\Upomega((\log n)^{1-{\frac{\alpha}{2}}})$, follows a Poisson distribution of the mean

$$ \mu'_p= {\frac{n_s}{n}}{\cdot} \lambda {\cdot} (n_d-1)(2\sqrt{2\log n/\lambda})^2 \leq {\frac{8 {\cdot} n_s {\cdot} n_d {\cdot} \log n}{n}}, $$

where the area of each cell is $(2\sqrt{2\log n/\lambda})^2$. Hence, according to Lemma 4, we prove that Eq. (17). Then, we can complete the proof.