Symptotics: a framework for estimating the scalability of real-world wireless networks

Abstract

We present a framework for non-asymptotic analysis of real-world multi-hop wireless networks that captures protocol overhead, congestion bottlenecks, traffic heterogeneity and other real-world concerns. The framework introduces the concept of symptotic scalability to determine the number of nodes to which a network scales, and a metric called change impact value for comparing the impact of underlying system parameters on network scalability. A key idea is to divide analysis into generic and specific parts connected via a signature—a set of governing parameters of a network scenario—such that analyzing a new network scenario reduces mainly to identifying its signature. Using this framework, we present the first closed-form symptotic scalability expressions for line, grid, clique, randomized grid and mobile topologies. We model both TDMA and 802.11, as well as unicast and broadcast traffic. We compare the analysis with discrete event simulations and show that the model provides sufficiently accurate estimates of scalability. We show how our impact analysis methodology can be used to progressively tune network features to meet a scaling requirement. We uncover several new insights, for instance, on the limited impact of reducing routing overhead, the differential nature of flooding traffic, and the effect real-world mobility on scalability. Our work is applicable to the design and deployment of real-world multi-hop wireless networks including community mesh networks, military networks, disaster relief networks and sensor networks.

Appendix: Shortest paths through degree-4 grid center

We derive an approximate formula for the expected number of shortest paths between uniformly randomly selected nodes that go through the center of an m by m degree-4 grid containing $N = m^2$ nodes.

Consider the route from source s to a destination d. For the path from s to d to go through the center, they must clearly lie in the opposite quadrant inclusive of the nodes parallel and perpendicular to the center. Let $P_c(s,d)$ denote the probability that the path from s to d goes through the center, and Q(d) denote the probability that d is in the opposite quadrant. Then the expected number of paths through the center is given by

$$\begin{aligned} E[P_c] = Q(d_q)\cdot P_c(s,d)\cdot (N-1) \end{aligned}$$

(17)

Since we don’t count the center node c and s,

$$\begin{aligned} Q(d) = \frac{\left( \frac{m+1}{2}\right) ^2 - 1}{m^2 - 2} \approx \frac{1}{4} + \frac{1}{2m} \end{aligned}$$

(18)

$P_c(s,d)$ depends upon the exact location of s and d. Instead of iterating over all values of s and d (which gets very complicated very quickly), we approximate it by assuming the average to be equal to when s and d are in the centers of their respective quadrants, and denote these nodes by $s_q$ and $d_q$. Let n(x, y) denote the number of shortest paths from x to y. Then

$$\begin{aligned} P_c(s,d) = P_c(s_q,d_q) = \frac{n(s_q,c)\cdot n(c,d_q)}{n(s_q,d_q)} \end{aligned}$$

(19)

Fig. 12

Reference figure for derivation

Given a square z by z grid, the number of shortest paths from one corner to another is ${2z \atopwithdelims ()z}$[37]. Let $b = m/4$. Then, from Fig. 12 and the above, we have

$$\begin{aligned} P_c(s,d) = \frac{{2b \atopwithdelims ()b}^2}{{4b \atopwithdelims ()2b}} \end{aligned}$$

(20)

The above is related to the concept of Catalan numbers [37]. The $n{\hbox {th}}$ Catalan number and an approximate formula thereof [38] are given by

$$\begin{aligned} C_n = \frac{1}{n+1}\cdot {2n \atopwithdelims ()n} \approx \frac{4^n}{n^{1.5}\sqrt{\pi }} \end{aligned}$$

(21)

Using Eq. 21 in 20, we have

$$\begin{aligned} P_c(s,d) = \frac{((b+1)C_b)^2}{(2b+1)C_{2b}} \approx \frac{0.8(b+1)}{b^{1.5}} \end{aligned}$$

(22)

Re-substituting $b = m/4$ in the above, and using this and Eq. 18 in 17, we have

$$\begin{aligned} E[P_c] &\approx \left( 0.25 + \frac{1}{2m}\right) \cdot m^2\cdot \frac{1.6(m+4)}{m^{1.5}} \\ &\approx 0.4\left( 1 + \frac{2}{m}\right) (m^{1.5} + 4m^{0.5})\\ &\approx 0.4\left( 1+ \frac{2}{\sqrt{N}}\right) (N^{\frac{3}{4}} + 4N^{\frac{1}{4}}) \end{aligned}$$