On the Stable Throughput in Wireless LANs

Abstract

The throughput stability is concerned with how much traffic load can be sustained in a network, and has been a research hotspot. Recent studies on the stability in 802.11 networks have arrived at contradictory conclusions. This paper delves into the reasons behind these contradictions. Our study manifests that the maximum stable-throughput is not simply larger than, less than, or equal to the saturation throughput as argued in previous works. Instead, there exists two intervals, over which the maximum stable-throughput follows different rules: over one interval, it may be far larger than the saturated throughput; over the other, it is tightly bounded by the saturated throughput. Most existing related research fails to differentiate the two intervals, implying that the derived results are inaccurate or hold true partially. Finally, we verify our study results via extensive simulations.

This work is partially supported by the Macao Science and Technology Development Fund under Grant (No. 081/2012/A3, No. 104/2014/A3, and No. 013/2014/A1), NNSF of China for Contract 61373027, NSF of Shandong Province for Contract ZR2012FM023, and Strategic Priority Research Program of the Chinese Academy of Sciences (No. XDA06040101). Qinglin Zhao is the corresponding author.

Appendix

Proof of Theorem 1 : We first prove that $0<\eta <1$ . Note that $T_{c}$ $=\max (T_{b},T_{b_{0}})$ $\gg \sigma$ and $0<C_{0}<1$. If $T_{b}\ge T_{b_{0}}$, then $0<\eta =\frac{(T_{b}-T_{b_{0} })+C_{0}(T_{b_{0}}-\sigma )}{T_{b}}<\frac{T_{b}-\sigma }{T_{b}}<1$. If $T_{b}<T_{b_{0}}$, then $0<\eta =\frac{C_{0}(T_{b}-\sigma )}{(1-C_{0})T_{b_{0} }+C_{0}T_{b}}<\frac{C_{0}T_{b}}{(1-C_{0})T_{b_{0}}+C_{0}T_{b}}<1$. Next we prove (a) and (b).

(a) When $n\rightarrow \infty$, noting $k=\lim \limits _{n\rightarrow \infty }(n-1)\beta$ if k exists, and applying the Poisson distribution to approximate the binomial distributions in (1), we have

$$\begin{aligned} \lim \limits _{n\rightarrow \infty }P_{e}&=e^{-k}C_{0},\text { }\lim \limits _{n\rightarrow \infty }P_{s}=ke^{-k}C_{0},\\ \lim \limits _{n\rightarrow \infty }P_{b}&=(1-e^{-k})C_{0},\text { } \lim \limits _{n\rightarrow \infty }P_{b_{0}}=e^{-k}(1-C_{0}),\\ \lim \limits _{n\rightarrow \infty }P_{c}&=(1-C_{0})(1-e^{-k}),\\ \lim \limits _{n\rightarrow \infty }\varOmega&=T_{c}+C_{0}(T_{b}-T_{c} )+\frac{C_{0}(\sigma -T_{b})+(1-C_{0})(T_{b_{0}}-T_{c})}{e^{k}}. \end{aligned}$$

Then, (4) is derived from $\varGamma (k)$ $=\lim \limits _{n\rightarrow \infty }P_{s}L/\varOmega$.

(b) To maximize $\varGamma (k)$, we set the first derivative of $\varGamma (k)$ to zero and have $k=1-\eta e^{-k}$, and hence $(k-1)e^{k-1}=-\eta e^{-1}$. Then $k-1=\mathscr {W}_{0}(-\eta e^{-1})$ or $\mathscr {W}_{-1}(\frac{-\eta }{e})$. We have $k_{opt}=\mathscr {W}_{0}(-\eta e^{-1})+1\ge 0$, since only $\mathscr {W} _{0}(-\eta e^{-1})>-1$ for $-\eta e^{-1}\in (-1/e,0).$     $\blacksquare$