Generic stationary backoff distributions for distributed multiple access control

Abstract

In this paper, we study the characteristics of two different backoff schemes: one that maximizes the channel utilization and one that maximizes the probability of a successful transmission. Our results indicate that while the latter provides slightly lower channel utilization, its shape is much less sensitive to the number of contending nodes. That is, the channel utilization is kept almost constant for a very wide range of node densities if the backoff distribution has increasing shape. This motivates us to propose a simple heuristic backoff scheme called the Truncated Geometric Backoff Distribution (TGBD). We provide simple analytical expressions for the probability of a successful transmission and the channel utilization. We also show that the TGBD can easily be extended to support service differentiation by adjusting the window lengths, and we provide a simple approximation that gives the relative share of the capacity for a node in a priority class compared to nodes in other classes. This extended backoff scheme easily outperforms the much more complex Quality of Service (QoS) standard, IEEE802.11e EDCA. Finally, a two-stage backoff model, based on the TGBD, is proposed that further increases the probability of a successful transmission. Results indicate that the channel utilization is almost independent of the number of contending nodes.

Appendix

1.1 8.1 Expected idle time in the priority extension

Here we give the full derivation of the expected number of idle slots before the winner in any priority class transmits.

To find the expected number of idle slots we first consider the distribution of Z,

$$\begin{aligned} P(Z > z) = &\prod_{l=1}^M\sum _{n_l=0}^{\infty} \frac{\lambda_l^{n_l}}{n_l!}e^{-\lambda_l} \bigl(1 - q^{\max{(L_l-z,0)}}\bigr)^{n_l} \\ = & \exp\Biggl(- \sum_{l=1}^M \lambda_l q^{\max{(L_l - z,0)}}\Biggr). \end{aligned}$$

We are only interested in the cases when there is at least one priority class having more than one node. Similarly as before, one solution is to condition on that there is at least one priority class with a positive number of nodes, i.e. N ₁+N ₂+⋯+N _M >0, we get

$$\begin{aligned} &{P(Z > z | N > 0)} \\ &{\quad = \frac{P(Z > z) - P(Z > z | N_1 =\cdots = N_M = 0)}{P(N > 0)}} \\ &{\quad = \frac{\exp(-\sum_{l=1}^M \lambda_l q^{\max{(L_l - z,0)}}) - \exp{(-\sum_{l=1}^M \lambda_l)}}{1 - \exp{(-\sum_{l=1}^M \lambda_l)}},} \end{aligned}$$

where N=N ₁+N ₂+⋯+N _M . Then it follows that the expected length until a winner transmits, with L _mx =max(L ₁,…,L _M ), is

$$\begin{aligned} &{E(Z | N > 0) } \\ &{\quad = \sum_{z=0}^{L_{mx} - 1} P(Z>z | N > 0)} \\ &{\quad = \frac{\sum_{z=0}^{L_{mx} - 1} ( \exp( - \sum_{l =1}^M \lambda_l q^{\max{(L_l - z,0)}}) - \exp{( - \sum_{l =1}^M \lambda_l)} )}{1-\exp(\sum_{l=1}^M \lambda_l)}.} \end{aligned}$$

1.2 8.2 Model of the two-stage backoff

In this section we derive the channel utilization of the two-stage backoff model. We start by deriving the full distribution of the number of winners in the TGBD stage (first stage) and then continue with the uniform backoff stage (second stage).

1.2.1 8.2.1 First stage backoff

To analyze the two-stage backoff, we first need to find the full distribution of the number of nodes from each priority class that selects the minimal backoff. This distribution is easily obtained in the non-Poisson case, we have due to independence between classes and nodes,

$$\begin{aligned} &{P(Y_1 = y_1, \dots, Y_M = y_M)} \\ &{\quad = \sum_{x=0}^{\infty}\prod _{l=1}^M \binom{N_l}{y_l} \bigl(P \bigl(X_{j,l}' = x\bigr)\bigr)^{y_l} \bigl(P \bigl(X_{j,l}' > x\bigr)\bigr)^{N_l-y_l}.} \end{aligned}$$

Now assume that the number of nodes in each priority class is Poisson distributed, i.e. N _l ∼Po(λ _l ) then the probability for y ₁,y ₂,…,y _M winners is

$$\begin{aligned} &{ P(Y_1 = y_1, \dots, Y_M = y_M) } \\ &{\quad = \sum_{x=0}^{\infty} \prod_{l=1}^M \sum _{n_l=y_l}^{\infty} \frac{\lambda_l^{n_l}}{{n_l}!}e^{-\lambda_l} \binom{n_l}{y_l}} \\ &{\qquad {}\times \bigl(P\bigl(X_{j,l}' = x\bigr) \bigr)^{y_l} \bigl(P\bigl(X_{j,l}' > x\bigr) \bigr)^{n_l-y_l},} \end{aligned}$$

and with $P(X_{j,l}' = x) = pq^{\max(L_{l} - x,0)}$ for x=1,2,…,L _l , $P(X_{j,l}' = x) = 0$ for x>L _l , $P(X_{j,l}' = 0) = q^{L_{l}}$ and $P(X_{j,l}' > x) = 1 - q^{\max(L_{l} - x,0)}$. By careful standard manipulations and using that $(\lambda_{l}^{n_{l}}/{n_{l}}!)\binom{n_{l}}{y_{l}}=\lambda_{l}^{n_{l}}/(y_{l}!(n_{l}-y_{l})!)$ and summing over n _l −y _l instead of n _l , this expression simplifies to, if y ₁+⋯+y _M >0,

$$\begin{aligned} &{P(Y_1 = y_1,\dots, Y_M = y_M)} \\ &{\quad = \sum_{x=0}^{L_{bnd}} \prod_{l=1}^M \frac{(\lambda_l\;P(X_{j,l}'=x))^{y_l}}{y_l!\; \exp(\lambda_l q^{\max(L_l-x,0)})}} \\ &{\quad = \sum_{x=1}^{L_{bnd} } \Biggl(\prod_{l=1}^M \frac{{(\lambda_lpq^{\max(L_l-x,0)})}^{y_l}}{y_l!\; \exp{(\lambda_l q^{\max{(L_l-x,0)}} ) } } \Biggr) } \\ &{\qquad {}+ \prod_{l=1}^M \frac{{(\lambda_l q^{L_l})}^{y_l}}{y_l! \exp{(\lambda_l q^{L_l})}}.} \end{aligned}$$

We also have P(Y ₁=0,…,Y _M =0)=0. The summation bound, L _bound , is defined as follows

$$ L_{bound} = \min_{l;y_l > 0}{(L_l)}. $$

If y ₁≥1, then the winners must select one of the slots 0,1,…,L ₁. However, if y ₁=0,y ₂≥1 then the winners may also select the additional slots L ₁,L ₁+1,…,L ₂, provided that priority class 1 has zero nodes. We finally get

$$\begin{aligned} &{P(Y_1=y_1,\dots,Y_M=y_M|N > 0)} \\ &{\quad = \frac{P(Y_1 = y_1, \dots, Y_M = y_M | N > 0)}{P(N > 0)}} \\ &{\quad = \frac{P(Y_1 = y_1, \dots, Y_M = y_M)}{1 - \exp{(-\sum_{l=1}^M \lambda_l)}}.} \end{aligned}$$

1.2.2 8.2.2 Second stage backoff

Now we continue with the analysis of the second backoff stage. Let $m_{l} = \#\{j, X_{j,l}' = Z\}$, i.e. the number of nodes selecting the minimal backoff time in priority class l, in the first backoff stage. First we will look at the probability of a single winner in the second stage, given that m ₁,m ₂,…,m _M nodes remain from the first backoff stage. Let Q _j,l ∼U{0,1,…,K−1}, j=1,2,…,m _l and $V_{l} = \min{(Q_{1,l},Q_{2,l},\dots,Q_{m_{l},l})}$. Furthermore, let V=min(V ₁,V ₂,…,V _M ) and W=#{(j,l),Q _j,l =V}, i.e. the number of nodes selecting the minimal backoff in the second backoff stage. There can be w winners on slot 0,2,…,K−1 and the total probability becomes

$$\begin{aligned} &{P(W = w)} \\ &{\quad = \sum_{x=1}^K \binom{m_1+ \dots + m_M}{w} \biggl(\frac{1}{K} \biggr)^w \biggl(1 - \frac{x}{K} \biggr)^{\sum_{l=1}^M m_l - w}.} \end{aligned}$$

If w winners select the minimal backoff time, then the remaining $\sum_{l=1}^{M} m_{l} -w$ nodes must select a larger backoff. These w winners can be chosen in $\binom{m_{1}+\cdots+m_{M}}{w}$ different ways. We are only interested in the case of a single winner and that probability is,

$$ P(W = 1) = \frac{\sum_{l=1}^M m_l}{K} \sum_{x=1}^K \biggl(1 - \frac{x}{K} \biggr)^{\sum_{l=1}^M m_l-1}. $$

Let p _l (m ₁,…,m _M ) denote the probability to have a single winner from priority class l and this probability is,

$$\begin{aligned} &{p_l(m_1,\dots,m_M)} \\ &{\quad = P(\text{winner} \in l | W = 1)} \\ &{\quad = \biggl(\frac{m_l}{\sum_{l=1}^M m_l} \biggr) \biggl(\frac{\sum_{l=1}^M m_l}{K} \biggr) \sum_{x=1}^K \biggl(1 - \frac{x}{K} \biggr)^{\sum_{l=1}^M m_l-1}} \\ &{\quad = \frac{m_l}{K}\sum_{x=1}^K \biggl(1 - \frac{x}{K} \biggr)^{\sum_{l=1}^M m_l-1},} \end{aligned}$$

where $m_{l}/{\sum_{l=1}^{M} m_{l}}$ is the fraction of nodes belonging to priority class l. The expected length of the second backoff stage can easily be computed through the distribution function of V. We have,

$$\begin{aligned} F_V(v) =& P(V \leq v) = 1 - P(V > v) \\ =& 1 - \biggl( 1 - \frac{v + 1}{K} \biggr)^{\sum_{l=1}^M m_l}. \end{aligned}$$

Let ω(m ₁,…,m _M ) denote the expected length of the second backoff stage and is given by

$$\begin{aligned} &{ \omega(m_1,\dots,m_M) } \\ &{\quad = E(V) = \sum _{v=0}^{\infty}P(V > v)} \\ &{\quad =\sum_{v=0}^{\infty}\bigl(1 - F_V(v)\bigr) = \sum_{v=0}^{K-2} \biggl(1 - \frac{v+1}{K} \biggr)^{\sum_{l=1}^M m_l}.} \end{aligned}$$

To compute the success probability of having a single winner in the second backoff stage then every combination of winners from the priority classes must be considered. However, theoretically there can be a very large number of winners from each priority class since the number of nodes is Poisson distributed, i.e. we have

$$\sum_{m_1=0}^{\infty} \dots \sum_{m_M=0}^{\infty}\frac{P(Y_1 = m_1 , \dots, Y_M = m_M)}{1 - \prod_{l=1}^M e^{-\lambda_l}} = 1. $$

However, the probability of having a large number of winners is very small, i.e. most of the probability mass is found in the region where y ₁,…,y _M are relatively small.

Let p _s,l denote the success probability of having a single winner from priority class l. This probability is given by

$$\begin{aligned} p_{s,l} &\approx \sum_{m_1=0}^{R_1} \cdots \sum_{m_M=0}^{R_M}\frac{P(Y_1 = m_1, \dots, Y_M = m_M)}{1 - \prod_{l=1}^M e^{-\lambda_l}} \\ &\quad {}\times p_l(m_1,\dots,m_M), \end{aligned} $$

where R _l ⪆10. Let μ denote the expected length of the second backoff stage. This length is given by

$$\begin{aligned} \mu &\approx \sum_{m_1=0}^{R_1} \cdots \sum _{m_M=0}^{R_M} \frac{P(Y_1 = m_1, \dots, Y_M = m_M)}{1 - \prod_{l=1}^M e^{-\lambda_l}} \\ &\quad {}\times \omega(m_1,\dots,m_M). \end{aligned} $$

Note that the expressions for p _s,l and μ quickly become unwieldy when the number of priority classes increases. However, in this work we are only considering a few number of priority classes. Now we can derive the channel utilization for the two-stage backoff model, we have

$$ \rho_l \approx \frac{p_{s,l} U}{E(Z|N > 0) + \mu + S + B}, $$

(20)

where B is the time for transmitting a short burst.