EFT: a high throughput routing metric for IEEE 802.11s wireless mesh networks

Abstract

In this paper, we present a throughput-maximizing routing metric, referred to as expected forwarding time (EFT), for IEEE 802.11s-based wireless mesh networks. Our study reveals that most of the existing routing metrics select the paths with minimum aggregate transmission time of a packet. However, we show by analyses that, due to the shared nature of the wireless medium, other factors, such as transmission time of the contending nodes and their densities and loads, also affect the performance of routing metrics. We therefore first identify the factors that hinder the forwarding time of a packet. Furthermore, we add a new dimension to our metric by introducing traffic priority into our routing metric design, which, to the best of our knowledge, is completely unaddressed by existing studies. We also show how EFT can be incorporated into the hybrid wireless mesh protocol (HWMP), the path selection protocol used in the IEEE 802.11s draft standard. Finally, we study the performance of EFT through simulations under different network scenarios. Simulation results show that EFT outperforms other routing metrics in terms of average network throughput, end-to-end delay, and packet loss rate.

Appendices

Appendix

1.1 A. Defer time with traffic priority

For simplicity of analysis, we consider only three classes of traffic, and according to [11], for j = 2,3, defer time (i.e., AIFS[j]) is equal to the duration of DIFS and is constant. However, for low-priority traffic (i.e., j = 1), AIFS[1] = DIFS + d _s . Because a node with high-priority traffic can initiate a transmission during the extra slot of the low-priority traffic, a node with low-priority traffic cannot defreeze its backoff counter. Thus, $d_d^1$ depends on the number of nodes associated with the higher-priority traffic and the probability at which they attempt to transmit. The defer time of traffic class 1 will include the transmission time of the high-priority packet initiated in the extra slot. Furthermore, the number of interruptions can be more than one. Therefore, the defer time of a packet of traffic class 1 is

$$ d_d^1 = {\rm AIFS}[1] + H \times T, $$

(15)

where, H is a random variable representing the number of interruptions in a single defer time and T is a random variable representing the amount of time of each interruption, given by

$$ T = {\rm AIFS}[2/3] + d_{\rm t} + {\rm SIFS} + d_{{\rm ACK}}. $$

(16)

The expected value of T is given by

$$ E[T] = {\rm AIFS}[2/3] + E[d_{\rm t}] + {\rm SIFS} + d_{{\rm ACK}}. $$

(17)

Let, μ ₂ and μ ₃ denote the probability at which a node with traffic classes 2 and 3, respectively, attempts to transmit in any randomly selected slot. The number of interruptions in a defer time is geometrically distributed with parameter μ, where μ is the probability that at least one high-priority packet transmits. Let n ₂ and n ₃ be the number of neighbors with traffic classes 2 and 3, respectively, and τ ₂ and τ ₃ be the probability of their transmissions, respectively. We therefore have

$$ \mu_2 = 1 - (1-\tau_2)^{n_2} $$

(18a)

$$ \mu_3 = 1 - (1-\tau_3)^{n_3} $$

(18b)

$$\begin{array}{lll} \mu &= 1 - (1-\mu_2)(1-\mu_3)\\ & = 1 - (1-\tau_2)^{n_2}(1-\tau_3)^{n_3}\\ & = 1 - (1 -\tau)^{n_2 + n_3} \end{array} $$

(18c)

The expected number of interruptions by a node with a higher-priority packet is therefore $E[H] = \frac{\mu}{1 - \mu}$.

B. Backoff delay

The backoff time at each transmission attempt depends on the size of the backoff counter (i.e., the size of the contention window), the number of interruptions in the backoff process (which depends on the number of contending neighbors) due to transmissions (either successful or unsuccessful) by the neighbors and the channel quality (transmission rate) of the transmitting neighbors, as well as the defer time after each freezing. Thus, the backoff delay in the i-th transmission attempt (where, i = 0,1,...,M ^′ − 1) of the j-th traffic class $d_{\rm b}^j(i)$ is given by

$$ d_{\rm b}^j(i) = \sum\limits_{k=1}^{w_i^j}d_s + \sum\limits_{k=1}^{b_i^j} (d_{\rm b} + d_d^j) $$

(19)

At the beginning of the backoff process of the i-th transmission attempt, a node uniformly chooses a backoff value $w_i^j$ in the range $(0, 2^i \times CW_{\rm min}(j))$, where $2^i \times CW_{\rm min}(j)$ defines the size of the current contention window for traffic class j. Therefore, the expected size of the backoff counter is $E[w_i^j] = \frac{w_i^j}{2}$ [20].

If the number of busy slots at the i-th transmission attempt is $b_i^j$, then a node has to wait $B_i^j = w_i^j + b_i^j$ slot times so that its backoff value $w_i^j$ reaches zero and it can start transmission. Let there be n neighbors of a tagged node (the node that is in backoff). Let p _t be the probability that there is a transmission in a slot. The probability of a slot being idle is the probability that none of the neighbors transmit in that slot, given by (1 − p _t). Now, the expected number of $B_i^j$ slots required to get $w_i^j$ idle slots is found by using Pascal distribution [20] with parameters $w_i^j$ and (1 − p _t). Therefore, the expected number of slots is $E[B_i^j]=\frac{E[w_i^j]}{1-p_{\rm t}}$. The expected number of busy slots can then be calculated as

$$ E[b_i^j]=E[B_i^j]-E[w_i^j]=\frac{w_i^j \times p_{\rm t}}{2(1-p_{\rm t})}. $$

(20)

The expected backoff delay for a packet of the j-th traffic class at the i-th transmission attempt can be found using Wald’s equation [21] as

$$ E[d_{\rm b}^j(i)]= E[w_i^j] \times d_{\rm s} + E[b_i^j]\times (d_{\rm b} + d_{\rm d}^j). \label{eq_backoff_single} $$

(21)

The number of transmission attempts (M) required for a successfully delivered packet has a truncated geometric distribution, with the PMF given by

$$ p[M=i] = \frac{p_{\rm s}(1-p_{\rm s})^i}{1-(1-p_{\rm s})^{M^{\prime}}} [i=0,1,2,...,M^{\prime}-1]. \label{eq_etx} $$

(22)

Thus, by combining Eqs. 21 and 22, the expected backoff delay for a packet of the j-th traffic class is

$$ \label{eq_exp_BO1111} E[d_{\rm b}^j] = \frac{p_{\rm s}}{1-(1-p_{\rm s})^{M^{\prime}}} \sum\limits_{i=0}^{M^{\prime}-1}(1-p_{\rm s})^i.E[d_{\rm b}^j(i)]. $$

(23)