Throughput and delay analysis for hybrid radio-frequency and free-space-optical (RF/FSO) networks

Abstract

In this paper the per-node throughput and end-to-end delay of randomly deployed (i.e. ad-hoc) hybrid radio frequency - free space optics (RF/FSO) networks are studied. The hybrid RF/FSO network consists of an RF ad hoc network of n nodes, f(n) of them, termed ‘super nodes’, are equipped with an additional FSO transceiver with transmission range s(n). Every RF and FSO transceiver is able to transmit at a maximum data rate of W ₁ and W ₂ bits/sec, respectively. An upper bound on the per node throughput capacity is derived. In order to prove that this upper bound is achievable, a hybrid routing scheme is designed whereby the data traffic is divided into two classes and assigned different forwarding strategies. The capacity improvement with the support of FSO nodes is evaluated and compared against the corresponding results for pure RF wireless networks. Under optimal throughput scaling, the scaling of average end-to-end delay is derived. A significant gain in throughput capacity and a notable reduction in delay will be achieved if $f(n) = \Upomega\left(\frac{1}{s(n)}\sqrt{\frac{n}{\log n}}\cdot \frac{W_1}{W_2} \right)$. Furthermore, it is found that for fixed W ₁, f(n) and n where f(n) < n, there is no capacity incentive to increase the FSO data rate beyond a critical value. In addition, both throughput and delay can achieve linear scaling by properly adjusting the FSO transmission range and the number of FSO nodes.

Appendices

Appendix 1

Proof of Lemma 1

To prove this lemma we need to show there exists constants c ₄ and c ₅ such that λ′ is upper bounded by

$$ \lambda' \leq \frac{c_{16}W_2f(n)s(n)}{n} $$

(78)

and there exists a spatial and temporal scheduling scheme such that a per session throughput

$$ \lambda' = \frac{c_{17}W_2f(n)s(n)}{n} $$

(79)

is achievable.

To derive the upper bound for λ′, we notice that the number of simultaneous FSO transmissions is no more than m, thus the total data rate served by the entire FSO network is no more than W ₂ f(n). Now let $\overline{L}$ denote the mean length of a line connecting two independently and uniformly distributed points on S ². Then the mean length of the path of packets is at least $\overline{L}-o(1)$. Thus the mean number of hops taken by a packet is at least $\frac{\overline{L}-o(1)}{s(n)}$. Then the total number of bits per second served by the entire network needs to be at least $\frac{(\overline{L}-o(1))n\lambda'}{s(n)}$. Then we have the following inequality:

$$ \frac{(\overline{L}-o(1))n \lambda'}{s} \leq W_2f(n) $$

(80)

Therefore we have the following upper bound on λ′:

$$ \lambda' \leq \frac{c_{16}W_2f(n)s(n)}{n} $$

(81)

To show that this upper bound is order tight, we design a routing scheme using similar techniques as in [2]. Let

$$ \rho'(n) := \hbox{radius of a disk of area } \frac{\pi s^2(n)}{64} $$

(82)

Then it is readily known that

$$ s(n)\geq 8\rho'(n) $$

(83)

According to Lemma 4.1 in [2], we can construct a Voronoi tessellation $\mathcal{V}'_n$ such that

• (A1) Every Voronoi cell contains a disk of radius ρ′(n);

• (A2) Every Voronoi cell is contained in a disk of radius 2 ρ′(n);

This setting allows direct communication within a cell and between adjacent cells, i.e., every node in a cell is within this distance s(n) from every other node in its own cell or adjacent cell. The routing strategy forwards the packets along the sequence of hops that approximate the straight line that connects the source and destination, i.e. packets are passing the cells that the straight-line intersects.

Now we show that the number of nodes in each cell is bounded from below with high probability. Let $\mathcal{F}$ denote the class of disks of area $\frac{\pi s^2(n)}{64}$. Then it is implied by Vapnik-Chervonenkis Theorem (see [2] and references therein) that

$$ \begin{aligned} &Pr\left\{ \sup_{D \in{\mathcal{F}}}\left|\frac{\hbox{number\,of\,nodes\,in\,D}} {f(n)}-\frac{\pi s^2(n)}{64}\right|\leq\epsilon(n)\right\}\\ &\quad >1-\delta(n) \\ \end{aligned} $$

(84)

whenever

$$ n>\max\left\{\frac{24}{\epsilon(n)}\log\frac{16e} {\epsilon(n)},\frac{4}{\epsilon(n)}\log\frac{2} {\delta(n)}\right\} $$

(85)

Note that s(n) is chosen such that the FSO network is asymptotically connected, and s(n) must be less than the maximum distance between any two points in S ² which is $\sqrt\pi/2$, then

$$ c_5\sqrt{\frac{\log f(n)}{f(n)}}\leq s(n)\leq\frac{\sqrt\pi} {2} $$

(86)

then Eq. 85 is satisfied by choosing $\epsilon(n)={s^2(n)}/{64}$ and δ(n) = 1/n when n is large enough.

Since each cell V in $\mathcal{V}'_n$ contains a disk of area $\frac{\pi s^2(n)}{64}$, then for large n we have

$$ Pr \left\{\hbox{number\,of\,nodes\,in }V \geq \frac{f(n)s^2(n)} {32}\right\} \geq 1-\delta(n) $$

(87)

Next we show that the traffic to be served by each cell V is bounded from above with high probability. Towards this objective we let i and j denote two randomly located points on S ², and L _ij denote the line segment connecting these two points. We claim that for any cell $V\in\mathcal{V}'_n$,

$$ Pr(L_{ij} \hbox{ intersects } V) \leq c_{24}s(n) $$

(88)

This is proved as follows². From property (A2), every cell $V\in\mathcal{V}'_n$ is contained in a disk of radius s(n)/4. If i lies at a distance x from the disk, then the angle θ subtended at i by the disk is no more than c ₁₈ s(n)/x. The area of the sector so formed is no more than $\frac{c_{19}\theta} {2\pi}$. If j does not lie in this sector, Then the line L _ij cannot intersect the disk containing the cell V. Hence for a point i at a distance x from the disk of radius s(n)/4 containing the cell V, the probability that the line L _ij intersects the disk is no more than c ₂₀ s(n)/x. Since i is uniformly distributed on S ², the probability density that it is at a distance x from the disk is bounded above by 2c ₂₁π(x + s(n)/4). Integrating, we obtain

$$ \begin{aligned} Pr(L_{ij} \hbox{ intersects } V) &\leq \int\limits_{s(n)/4}^{\sqrt\pi/2}\left( \frac{c_{20}s(n)}{x}\right)\\ & \cdot 2c_{21}\pi\left(x+\frac{s(n)}{4}\right)dx \\ &\leq c_{24}s(n) \end{aligned} $$

(89)

By a similar reasoning through Lemma 4.10–4.14 in [2] we have

$$ \begin{aligned} &Pr\{ \sup_{V \in {\mathcal{V}}'_m}(\hbox{traffic needing to be carried by cell V})\\ &\quad \leq c_{24} {\lambda'} n s(n) \} \geq 1-\delta_1(m) \end{aligned} $$

(90)

It can be implied by Eq. 87 that each cell $V\in\mathcal{V}'_n$ is able to transmit at least $\frac{f(n)s^2(n)}{32}W_2$ bits/sec. Then with high probability, the rate c ₂₄ λ′ n s(n) can be accommodated by all cell if

$$ c_{5} {\lambda'} n s(n) \leq \frac{f(n)s^2(n)}{32}W_2 $$

(91)

Then the per session achievable throughput is given by

$$\lambda' = \frac{W_2f(n)s(n)}{32c_{24}n} $$

(92)

which proves Lemma 1. □

Appendix 2

Proof of Lemma 4

Clearly the number of type-1 RF λ₂-traffic segments is (n − f). Let random variable $\mathcal{B}$ denote the number of type-2 RF λ₂-traffic segments. Then

$$ {\mathcal{N}}={\mathcal{B}}+n-f $$

(93)

For each source-destination pair, the destination node is a regular node with probability $\frac{n-f}{n}$. Then $\mathcal{B}$ follows binomial distribution $B(n,\frac{n-f}{n})$. It follows from the law of large numbers that given any $\epsilon>0$,

$$ \lim_{n\rightarrow\infty}Pr\left\{\left|\frac{{\mathcal{B}}} {n}-\frac{n-f}{n}\right|<\epsilon\right\} =1 $$

(94)

Let $\epsilon=c_{22}(n-f)/n$, then for large n and f < n, there is $\delta_2(n) \rightarrow 0$ such that

$$ Pr\left\{{\mathcal{B}}\leq (1+c_{22})(n-f)\right\} \geq 1-\delta_2(n) $$

(95)

Then Eq. 39 follows directly from Eqs. 93 and 95. □

Appendix 3

Proof of Lemma 5

This proof uses a similar technique introduced in the proof of Lemma 4.9 in [2].

According to the property (V2), there exists a disk of radius 2ρ that contains the Voronoi cell V. Let D denote such a disk. If i lies at a distance x from the disk D, then the angle $\alpha(\leq\pi)$ of the sector subtended at node i by the disk is no more than $\frac{c_{23}}{x}\sqrt{\frac{\log n} {n}}$. If the segment L _i joining node i and node g(i) intersects the disk D, then node g(i) should lie inside the sector and $|X_i-X_{g(i)}|\geq x$. Then the probability that L _i intersects the disk D is no more than $\frac{\alpha}{2\pi} Pr\{|X_i-X_{g(i)}|\geq x\}$. Note that $|X_i-X_{g(i)}|$ follows the probability distribution of random variable T defined in Section IV, which is given by Eq. 23. Since node i is uniformly distributed on S ², the probability density that it is at a distance x from the disk is bounded above by c ₂₄(x + 2ρ). Let x = z u ₁, then

$$ \begin{aligned} p_{V} = & Pr\{ L_{i} \,{\text{intersects}}V\} \\ \le & \int_{0}^{\infty } {{\frac{\alpha }{{2\pi }}}} Pr\{ |X_{i} - X_{{g(i)}} | \ge x\} c_{{24}} (x + 2\rho )dx \\ \le & \int_{0}^{\infty } {\min } \left\{ {\pi ,{\frac{{c_{{22}} }}{x}}\sqrt {{\frac{{\log n}}{n}}} } \right\}{\frac{{c_{{23}} }}{{2\pi }}}\cdot f^{{ - z^{2} }} (x + 2\rho )dx \\ = & \int_{0}^{{{\frac{{c_{{22}} }}{\pi }}\sqrt {{\frac{{\log n}}{n}}} }} \pi \cdot{\frac{{c_{{23}} }}{{2\pi }}}\cdot f^{{ - z^{2} }} (x + 2\rho )dx \\ & + \int_{{{\frac{{c_{{22}} }}{\pi }}\sqrt {{\frac{{\log n}}{n}}} }}^{\infty } {{\frac{{c_{{22}} }}{x}}} \sqrt {{\frac{{\log n}}{n}}} \cdot{\frac{{c_{{23}} }}{{2\pi }}}\cdot f^{{ - z^{2} }} (x + 2\rho )dx \\ \le & c_{{14}} \sqrt {{\frac{{\log n}}{{fn}}}} \\ \end{aligned} $$

(96)

Hence we have proved the lemma. □

Appendix 4

Proof of Lemma 6

Given $\mathcal{N}=N$, the number of RF λ₂-traffic segments $\mathcal{M}$ follows the binomial distribution B(N, p _V ).

If N = ω(1), i.e. N is large, then it follows from the law of large numbers that given any $\epsilon>0$,

$$ \lim_{n\rightarrow\infty}Pr\left\{\left|\frac{{\mathcal{M}}} {N}-p_V\right|<\epsilon\right\} =1 $$

(97)

Let $\epsilon=p_V$, then according to Lemma 4, there is a $\delta_4(n) \rightarrow 0$ such that

$$ Pr\left\{{\mathcal{M}} \leq 2c_{14} N \sqrt{\frac{\log n}{fn}}\right\} \geq 1-\delta_4(n) $$

(98)

If N = O(1), then for large n,

$$ E\{{\mathcal{M}}\} \leq c_{14}N\sqrt{\frac{\log n}{f n}} << 1 $$

(99)

Thus for large n,

$$ \begin{aligned} Pr\left\{{\mathcal{M}} \leq 2c_{14} N \sqrt{\frac{\log n}{f n}}\right\} &\geq Pr\left\{{\mathcal{M}} \leq E\{{\mathcal{M}}\}\right\} \\ &= Pr\left\{{\mathcal{M}}=0\right\} \\ &\geq \left(1-c_{14}\sqrt{\frac{\log n}{f n}}\right)^{N} \\ \rightarrow& 1 \end{aligned} $$

(100)

Note that Eqs. 98 and 23 indicate that for any ${N\in\mathbb{N}}$, there is a $\delta_5(n) \rightarrow 0$ such that

$$ Pr\left\{\left.{\mathcal{M}} \leq 2c_{14} {\mathcal{N}} \sqrt{\frac{\log n} {f n}}\right|{\mathcal{N}}=N\right\} \geq 1-\delta_5(n) $$

(101)

According to Lemma 3, we have

$$ \begin{aligned} Pr & \left\{{\mathcal{M}} \leq (2+c_{13})(n-f)\cdot 2c_{14} \sqrt{ \frac{\log n}{f n}}\right\} &\geq Pr\left\{{\mathcal{M}} \leq 2c_{14} {\mathcal{N}} \sqrt{\frac{\log n}{f n}}, {\mathcal{N}}\leq(2+c_{13})(n-f)\right\} & = Pr\left\{{\mathcal{N}}\leq(2+c_{13})(n-f)\right\} &\quad \cdot Pr\left\{\left.{\mathcal{M}} \leq 2c_{14} {\mathcal{N}} \sqrt{\frac{\log n}{f n}}\right|{\mathcal{N}}\leq(2+c_{13})(n-f)\right\} &\geq (1-\delta_2(n))(1-\delta_5(n)) \end{aligned} $$

(102)

Hence we have proved the lemma. □