Impact of deployment size on the asymptotic capacity for wireless ad hoc networks under Gaussian channel model

Abstract

We study the throughput capacity and transport capacity for both random and arbitrary wireless networks under Gaussian Channel model when all wireless nodes have the same constant transmission power P and the transmission rate is determined by Signal to Interference plus Noise Ratio (SINR). We consider networks with n wireless nodes $\{v_1,v_2,\ldots,v_n\}$ (randomly or arbitrarily) distributed in a square region B _a with a side-length a. We randomly choose n _s node as the source nodes of n _s multicast sessions. For each source node v _i , we randomly select k points and the closest k nodes to these points as destination nodes of this multicast session. We derive achievable lower bounds and some upper bounds on both throughput capacity and transport capacity for both unicast sessions and multicast sessions. We found that the asymptotic capacity depends on the size a of the deployment region, and it often has three regimes.

Appendix

1.1 Useful known results

Throughout this paper, we will repeatedly use the following results from probability theory literature.

Lemma 33

[Azuma’s inequality] Suppose that random variables $X_0, X_1, X_2, \ldots, X_n, \ldots$ are martingale and $|X_k -X_{k-1}| \le a_k$ almost surely for any $k \ge 1.$ Then for all positive integers N and all positive real number t , we have

$$ {{\bf Pr}\left({|X_N -X_0| \ge t}\right)} \le 2 \exp\left(-{\frac{t^2} {2\sum_{i=1}^N a_k^2}} \right) $$

Recall that here a sequence of random variables $X_i, 0 \le i,$ are called martingale if they satisfy that: $E(X_{N+1} \mid X_0, X_1, \ldots, X_N) = X_N.$

Lemma 34

[Chebyshev’s inequality] For a variable X, 

$$ {{\bf Pr}\left({|X - \mu|\geq A }\right)} \le {\frac{Var(X)} {A^2}}, $$

where μ = E(X), Var(X) is the variance of X , and A > 0.

Lemma 35 [Law of large numbers]

Considernuncorrelated variables$X_i, 1 \le i \le n$with same expected value μ = E(X _i ) and variance σ² = Var(X _i ). Let$\overline{X}= {\frac{\sum_{i=1}^{n} {X}_i} {n}}. \forall \epsilon >0,$

$$ {\bf Pr}\left({|\overline{X} - {\mu}| < \epsilon}\right) \ge 1 - {\frac{\sigma^2} {n \cdot \epsilon^2}}. $$

Lemma 36 [Binomial distribution]

Considernindependent variables$X_i \in\{0,1\}, p={{\bf Pr}\left({X_i=1}\right)},$ and $X=\sum_{i=1}^{n} X_i.$

$$ \left\{ \begin{array}{ll} {{\bf Pr}\left({X \leq \xi}\right)} \leq e^{{\frac{-2(n\cdot p-\xi)^{2}} {n}}}, & \hbox{when}\,0< \xi \le n \cdot p.\\ {{\bf Pr}\left({X>\xi}\right)} < {\frac{\xi (1-p)} {(\xi -n \cdot p)^2}}, & \hbox{when}\,\xi>n\cdot p. \end{array} \right. $$

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Lemma 37

[4] For a Poisson random variable X of parameter λ, 

$$ {{\bf Pr}\left({X \geq x}\right)} \leq {\frac{e^{-\lambda}(e \lambda)^x} {x^x}}, {\ for\ }\ x > \lambda $$

1.2 Proof of some technical lemmas

1.2.1 Proof for Lemma 5

Proof

We use the similar idea as Theorem 3 in [4] to prove this. We first partition the square into ${\frac{a^2} {c^2}}$ cells with side length c, here c can be rounded such that ${\frac{a^2} {c^2}}$ is an integer. Then we divide time into a sequence of q = (c ₃(d + 1))² successive mini-time-slots. Here, c ₃ is a constant no less than 2 and can be rounded such that q is an integer. Then based on a TDMA schedule, in each mini-time-slot, we let only one node in each of the disjoint sets (square with color grey) to transmit simultaneously.

First, we know the distance between u and v is at most $\sqrt{2}(d+1)c.$ Hence, the signal strength received by v is at least $S(u,v)\geq P\cdot \ell(\sqrt{2}(d+1)c)=P\cdot \min \{1, (\sqrt{2}(d+1)c)^{-\beta}\}.$ Next, we analyze the total interference received by v based on our TDMA scheduling described above. Given a transmitter u in one cell s _i , we know the receiver v is in some cell that is at most d-cells apart from s _i . The total interference caused by all other simultaneous transmissions can be computed as follows. First, any transmitter located in one of the eight closest cells is at least c ₃(d + 1) − (d + 1) cells away from v, then the Euclidean distance is at least $c\cdot (c_3(d+1)-(d+1)).$ Next, any transmitter of the sixteen next closest cells has Euclidean distance from v is at least $c\cdot(2c_3(d+1)-(d+1)),$ and so forth. Thus, the total interference caused by all simultaneous transmitters is

$$ \begin{aligned} I(u,v) &\leq \sum_{i=1}^\infty 8i\cdot P\cdot (c\cdot i\cdot (c_3 -1)d+i\cdot c_3-1)^{-\beta}\\ &\leq \sum_{i=1}^{\infty} {\frac{8P} {c^{\beta}}} {\frac{i} {(i(c_3-1)d)^\beta}} = {\frac{8P} {(c_3-1)^{\beta}(cd)^{\beta}}} \sum_{i=1}^{\infty} {\frac{1} {i^{\beta-1}}} \end{aligned} $$

Clearly, when β > 2, the summation in the above formula converges. So the total interference is at most $c_2 P \cdot (cd)^{-\beta},$ here $c_2={\frac{8} {(c_3-1)^{\beta}}} \sum_{i=1}^{\infty} {\frac{1} {i^{\beta-1}}}$ is a constant.

We can get the data rate R(u, v) is at least

$$ \begin{aligned} R(u,v) &=\log \left(1+{\frac{S(u,v)} {N_0+I(u,v)}}\right)\\ &\ge \log \left(1+{\frac{P\cdot \min \{1,(\sqrt{2}(d+1)c)^{-\beta}\}} {N_0+c_2 P (cd)^{-\beta}}}\right) \end{aligned} $$

which does not depend on n.

Clearly, when both c and d are constants, $R(u,v)=\Uptheta(1).$ When $c\cdot d\rightarrow \infty,$ by taking the limit for $c\cdot d\rightarrow \infty$ and by the fact that every transmitter can transmit once every $q^2=(c_3(d+1))^2$ mini-time-slots, the lemma follows.$\square$

1.2.2 Proof for Lemma 6

Proof

The proof follows from Lemma 37. Let A _n be the event that there is at least one cell with more than $\log {\frac{a} {c}} \times{\frac{nc^2} {a^2}}$ nodes. Since the number of nodes x in each cell of the partition is a Poisson random variable of parameter ${\frac{nc^2} {a^2}},$ by the union the Chernoff bounds, we have $\Pr(A_n) \leq {({\frac{a} {c}})}^2 \Pr(x>\log {\frac{a} {c}} \times{\frac{nc^2} {a^2}}) \leq {({\frac{a} {c}})}^2 e^{-{\frac{nc^2} {a^2}}}{({\frac{{\frac{nc^2} {a^2}}e} {\log {\frac{a} {c}}\times {\frac{nc^2} {a^2}}}})}^{{\frac{nc^2} {a^2}}\log {\frac{a} {c}}} .$ which tends to 0 as n tends to infinity. $\square$

1.2.3 Proof for Lemma 7

Proof

Let x be the number of nodes falling in one rectangle with size c ₁ × a and A _n be the event that there is at least one rectangle with more than nodes, by Lemma 37, we get $\Pr(A_n)\leq {\frac{\sqrt{n}} {c_1}}\times \Pr(x>c_1 \sqrt{n}) \leq {\frac{\sqrt{n}} {c_1}} e^{-c_1\sqrt{n}}{({\frac{ec_1\sqrt{n}} {c_1\sqrt{n}}})}^{2c_1\sqrt{n}} = {\frac{\sqrt{n}} {c_1}} e^{-c_1\sqrt{n}}{({\frac{e} {2}})}^{2c_1\sqrt{n}}.$ When n tends to infinity, it goes to 0. This finishes the proof. $\square$

1.2.4 Proof for Theorem 19

Proof

First, we partition B _a into cells with side length c, here c is some constant. Let C(P _i ) denote the number of cells a routing path P _i will use, i.e., the number of cells crossed by P _i . Let variable $L=\sum_{i=1}^{n_s} C(P_i),$ denoting the total load of all cells. Here the load of a cell by a routing method is the number of flows visiting the cell for the unicast path constructed. Then $L \ge \sum_{i=1}^{n_s} l_i/(\sqrt{2}{\frac{a} {m}}),$ where l _i denotes the Euclidian distance between the i-th source/destination pair.

Define random variables $X_q=\sum_{j=1}^{q} (l_j-E(l_j)).$ Then $E(X_{q+1} \mid X_1, \ldots, X_{q})=E(\sum_{j=1}^{q+1} (l_j-E(l_j))\mid X_1, \ldots, X_{q}) =E(\sum_{j=1}^{q} (l_j-E(l_j))+(l_{q+1}-E(l_{q+1})\mid X_1, \ldots, X_{q})=X_q+E(l_{q+1}-E(l_{q+1}))=X_q.$ In other words, variables X _i are martingale.

In addition, $|X_{q} - X_{q-1}| =\left| l_q - E(l_q) \right|\le {\sqrt 2 } a .$ Note that the last inequality holds for any l _q [12]. From Azuma’s Inequality, we have ${{\bf Pr}\left({|X_{n_s}-X_0| \ge t}\right)} \le 2 \exp(-{\frac{t^2} {2\sum_{i=1}^{n_s} 8 a^2 }}).$ Let $t= \epsilon {\sum_{i=1}^{n_s} E(|l_i|)} .$ Clearly, $\epsilon n_s c_3 a \le t \le \epsilon n_s \sqrt{2} a$ for some constant c ₃. Note that X ₀ = 0. Then, ${{\bf Pr}\left({\sum_{i=1}^{n_s} l_i \le \sum_{i=1}^{n_s} E(l_i) - t}\right)} \le {{\bf Pr}\left({|X_{n_s}| \ge t }\right)} \le \exp( -{\frac{t^2} {2\sum_{i=1}^{n_s} 8 a^2}} ) \le \exp( -{\frac{(\epsilon n_sc_3 a)^2} { 8n_s a^2}}) = \exp( -{\frac{n_s \epsilon^2 c_3^2} {8}}) .$ Thus, for a constant $\varepsilon$  ∈ (0, 1), 

$$ {{\bf Pr}\left({\sum_{i=1}^{n_s} l_i \le (1-\epsilon) n_s \sqrt{2} a }\right)} \le 2 e^{-{\frac{n_s \epsilon^2 c_3^2} { 8}}} $$

Thus, by letting $\epsilon={\frac{1} {2}},$ we have ${{\bf Pr}\left({\sum_{i=1}^{n_s} l_i \ge n_s \sqrt{2} a /2}\right)} \ge 1-2e^{-n_s c_3^2/32}.$ Then, ${{\bf Pr}\left({L \ge n_s m /2}\right)} \ge 1-2e^{-n_s c_3^2/32}.$

Recall that L denotes the total load of all cells. Then by pigeonhole principle, with probability at least $1-2e^{-n_sc_3^2/32},$ there is at least one cell, that will be used by at least ${\frac{n_sc_3 m} {m^2}}$ flows. According to Lemma 9, we know that the capacity of a cell with constant side length is O(1). Thus, with probability at least $1-2e^{-n_sc_3^2/32},$ the minimum data rate that can be supported using any routing strategy, due to the congestion in some cell, is ${\frac{1} {{\frac{n_s m} {c_3m^2}}}} = {\frac{c_3m} {n_s }} = {\frac{c_3 a} {c n_s}}=O({\frac{a} {n_s}}),$ since m = a/c for some constant c. $\square$

1.2.5 Proof for Lemma 26

Proof

We will prove this lemma using some existing results under protocol model, especially the area argument [12]. For the sake of our proof, assume that every node has an artificial “transmission radius” r such that each node v can only communicate with other nodes in its transmission range (a transmitting disk with its center at v and radius r). In addition, we define the area covered by a tree T as the union of its nodes’ transmitting disks. Then by showing a lower bound on the area of the region covered by any multicast tree T, we can give the desired lower bound on the number of cells it will cross.

Recall Lemma 11 in [11], it is proved that in protocol model, the area of the region D(T), w.h.p., is at least $\theta_0\sqrt{k}ar$ when $k<\theta_1\cdot{\frac{a^2} {r^2}}$ for some constant θ₀ and θ₁. Here r denotes the transmission range of each node in protocol model and D(T) denotes the region covered by all transmitting disks of all transmitting nodes (internal nodes of T) in the any multicast tree T. Unfortunately, this result can not help us directly, since in our model, each node has no fixed transmission range r. Instead, any pair of nodes can communicate with each other even though the data rate may be very small. Based on the original network under Gaussian channel model, we construct a new network under protocol model as follows.

• 1) Set each node’s transmission range as the side length of each cell g.

• 2) Add some artificial “additional relay nodes” V _ad such that any pair of nodes will have enough relay nodes along its link to make sure that the minimum number of cells the routing path crosses under protocol model is no more than the number of cells the direct link will cross in Gaussian channel model. Notice that V _ad cannot be selected as source or receivers, they can only act as relay nodes.

Let T be any multicast tree in original network under Gaussian channel model and T _p denote the corresponding multicast tree(spanning the same multicast session) constructed on this network under protocol model. We have two important observations here:

• 1) Our preceding two modifications will not affect the proof for Lemma 11 in [11]. In other words, the lower bound on |D(T _p )| still holds,

• 2) Furthermore, any link in Gaussian channel model can be simulated by using these artificial “additional relay nodes” in the protocol model such that the number of cells it will cross is not increased. So the lower bound of C(T) is no smaller than the lower bound of C(T _p ).

Together with Lemma 11 in [11], we get

$$ D(T_p)\geq \theta_0\sqrt{k}ar =\theta_0\sqrt{k}ag $$

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Since one transmitting disk can cover no more than 4 cells, we have $C(T_p)\geq \theta_0\sqrt{k}ag/4\times g^2 ={\frac{\theta_0} {4}}\cdot {\frac{\sqrt{k}a} {g}}.$ It follows that when $k<\theta_4\cdot{\frac{a^2} {g^2}},$ with high probability,

$$ C(T)\geq {\frac{\theta_0} {4}}\cdot {\frac{\sqrt{k}a} {g}}. $$

Since $|EMST|\leq 2\sqrt{2}\sqrt{k}a,$ if we set θ₃ as ${\frac{\theta_0} {4}}/2\sqrt{2},$ our lemma follows.$\square$

1.2.6 Proof for Theorem 28

Proof

Let variable $L=\sum_{i=1}^{n_s} C(T_i),$ denoting the total load of all cells. Here the load of a cell by a routing method is the number of flows visiting the cell for the multicast tree constructed. Then based on Lemma 26, we know that $L \ge \sum_{i=1}^{n_s} \theta_3| \hbox{EMST}(M_i)|/({\frac{a} {m}})$ with high probability. Notice that $E(\sum_{i=1}^{n_s} |\hbox{EMST}(M_i)|) = n_s c_7 a \sqrt{k}.$

Define random variables $X_q=\sum_{j=1}^{q} (|\hbox{EMST}(M_j)|-E(|\hbox{EMST}(M_j)|)).$ Then $E(X_{q+1} \mid X_1, \ldots, X_{q})=E(\sum_{j=1}^{q+1} (l_j-E(l_j))\mid X_1, \ldots, X_{q}) =E(\sum_{j=1}^{q} (l_j-E(l_j))+(l_{q+1}-E(l_{q+1})\mid X_1, \ldots, X_{q})=X_q+E(l_{q+1}-E(l_{q+1}))=X_q,$ so variables X _i are martingale. $|X_{q} - X_{q-1}| =\left| | \hbox{EMST}(M_q)| - E(|\hbox{EMST}(M_q)|) \right|\le E(|\hbox{EMST}(M_q)) |\le 2{\sqrt 2 }{\sqrt k} a .$ This inequality holds for any $\hbox{EMST}(M_q)$ [12].

From Azuma’s Inequality, we have ${{\bf Pr}\left({|X_{n_s}-X_0| \ge t}\right)} \le 2 \exp(-{\frac{t^2} {2\sum_{i=1}^{n_s} 8 k a^2 }})$ Let $t= \epsilon {\sum_{i=1}^{n_s} E(\hbox{EMST}(M_i))} .$ Clearly, $\epsilon n_s c_8{\sqrt k } a \le t \le 2 \sqrt{2} n_s \epsilon {\sqrt k} a.$ Note that X ₀ = 0. Then, ${{\bf Pr}\left({\sum_{i=1}^{n_s} |\hbox{EMST}(M_i)| \le \sum_{i=1}^{n_s} E(|\hbox{EMST}(M_i)|) - t}\right)} \le {{\bf Pr}\left({|X_{n_s}| \ge t }\right)} \le \exp( -{\frac{t^2} {2\sum_{i=1}^{n_s} 8 k a^2}} ) \le \exp( -{\frac{(\epsilon n_sc_8 {\sqrt k}a)^2} { 8n_s k a^2}}) = \exp( -{\frac{n_s \epsilon^2 c_8^2} {8}}).$ Thus, for a constant $\varepsilon$  ∈ (0, 1), 

$$ {{\bf Pr}\left({\sum_{i=1}^{n_s} |\hbox{EMST}(M_i)| \le (1-\epsilon) n_s c_9 \sqrt {k} a }\right)} \le 2 e^{-{\frac{n_s \epsilon^2 c_8^2} {8}}} $$

Then by letting $\epsilon = {\frac{1} {2}},$ we have

$$ {{\bf Pr}\left({\sum_{i=1}^{n_s} |\hbox{EMST}(M_i)| \ge n_s c_9 \sqrt {k} a /2}\right)} \ge 1-2e^{-n_s c_8^2/32}. $$

Based on Lemma 26, we get

$$ {{\bf Pr}\left({L \ge n_s \theta_3 c_9 \sqrt {k}m/2}\right)} \ge 1-2e^{-n_s c_8^2/32}. $$

It implies that

$$ {{\bf Pr}\left({L \ge n_s \theta_3 c_9 \sqrt {k} m/2}\right)} \ge 1-2e^{-n_sc_8^2/32}\quad{\ if\ } k\le \theta_1 \sqrt{n}. $$

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Recall that L denotes the total load of all cells. Then by pigeonhole principle, with probability at least $1-2e^{-n_sc_8^2/32},$ there is at least one cell, that will be used by at least ${\frac{n_s c_{10} \sqrt{k}m} {m^2}}$ flows where $c_{10}= \theta_3 c_9.$ Again, according to Lemma 9, the capacity of a cell with constant side length is O(1). Thus, when $n_s=\Uptheta(n),$ with probability at least $1-2e^{-n_sc_8^2/32},$ the minimum data rate that can be supported using cellular routing strategy is at most, for any routing strategy, due to the congestion in some cell, ${\frac{1}{{\frac{n_s c_{10} \sqrt{k}m} {m^2}}}} = {\frac{m} {c_{10} n_s \sqrt{k}}} = O( {\frac{a} {n_s \sqrt{k}}}).$ This finishes the proof of the theorem. $\square$

1.2.7 Proof for Theorem 29

Proof

Our proof again is to analyze the load of some cells. We use L to denote the total load of all cells. Then we get $L\geq \sum_{i=1}^{n_s} \theta_3 |EMST(M_i)|/ \left((\kappa \log m -\epsilon_m)\times {\frac{a} {m}} \right)$ based on Lemma 26. Since ${{\bf Pr}\left({\sum_{i=1}^{n_s} |\hbox{EMST}(M_i)| \ge n_s c_{9} \sqrt {k} a /2}\right)} \ge 1-2e^{-n_s c_8^2/32},$ from Lemma 26, we get

$$ {{\bf Pr}\left({L \ge n_s c_{10} \sqrt{k}{\frac{m} {\lg m}}}\right)} \ge 1-2e^{-n_s c_8^2/32}. $$

for some constant $c_{10}=c_9 \theta_3.$ Here we use ${\mathbb{L}}$ to denote the total number of flows crossing some super-cell. Notice that here “crossing” means visiting and leaving. We get,

$$ {{\bf Pr}\left({{\mathbb{L}} \ge L-n_s k=n_s c_{10} \sqrt{k}{\frac{m} {\lg m}}/2-n_s k}\right)} \ge 1-2e^{-n_s c_8^2/32}. $$

It is easy to show that, any multicast routing tree will cross at least $\lceil\delta \lg m\rceil$ quasi-closed cuts if it crosses three super-cells. Denoted by ${\mathbb{L}'}$ the total number of flows crossing some quasi-closed cut. We have ${\mathbb{L}}^{\prime}\geq {\frac{\mathbb{L}}{3}}\times \lceil\delta {\hbox{lg}}{m}\rceil.$

It follows that, with probability at least $1-2e^{-n_sc_8^2/32},$ the total load of all quasi-closed cell is at least

$$ {\frac{n_s c_{10} \sqrt{k}{\frac{m} {\lg m}}/2-n_s k} {3}}\times \lceil\delta \lg m\rceil. $$

Then by pigeonhole principle, with probability at least $1-2e^{-n_sc_8^2/32},$ there is at least one quasi-closed cell, that will be used by at least ${\frac{{\frac{n_s c_{10} \sqrt{k}{\frac{m} {\lg m}}/2-n_s k} {3}}\times \lceil\delta \lg m\rceil} {m^2}}$ flows which can be rewritten as

$$ \theta_2 {\frac{n_s \sqrt{k}} {\sqrt{n}}} $$

for some constant θ₂ when $k=O(({{\frac{m} {\lg m}}})^2).$ Then with probability at least $1-2e^{-n_sc_8^2/32},$ the minimum data rate that can be supported using any routing strategy, due to the congestion in some quasi-closed cell, is at most

$$ O\left({\frac{\left({\frac{a} {\sqrt{n}}}\right)^{-\beta} \sqrt n} {\theta_2 n_s \sqrt{k}}}\right) = O\left( ({\frac{a} {\sqrt{n}}})^{-\beta}\cdot {\frac{1} {n_s}}\cdot{\frac{\sqrt n} {\sqrt{k}}}\right), $$

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when the number of receivers k per-flow is $O({\frac{n} {\log^2 n}}).$ $\square$