Efficient MAC protocol design and performance analysis for dense WLANs

Abstract

In this paper, we analytically study the dense basic service set network transmission problems in very high throughput (VHT, namely IEEE 802.11ac) wireless local area networks (WLANs) due to nervous bandwidth resources. Our contributions are threefold as follows. Firstly, we derive the closed-form expressions of throughput gains for primary channel establishment from multi-band selection using the optimal skipping rule, which balances the throughput gain from finding a good quality band with the overhead of measuring multiple bands. Secondly, in order to satisfy the quality of service of overlapping BSS users, we design a space interference avoidance mechanism, which can improve the system throughput for the whole dense WLANs. Thirdly, in order to further improve the transmission performance of dense BSS networks, we propose an unequal bandwidth transmission mechanism based on the VHT WLANs, which can not only clear the redundant network allocation vector duration timely but also use the limited bandwidth efficiently. The proposed protocols and mechanisms exploit both time and frequency diversity sufficiently, and are shown to result in typical throughput gains compared with the traditional IEEE 802.11 MAC protocol.

Appendices

Appendix 1

Here we will give the proof of the proposition 2. Due to the throughput gains definition and $E[P_u]=\lambda _s$, then we can rewritten the definition as

$$\begin{aligned} Throughput\_gains=\eta = \dfrac{\lambda _s\sum ^{\overline{\mathcal {N}}}_{u=1}\lambda _u}{T\_{redundant\_t}} =\dfrac{x}{y} \end{aligned}$$

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Then we can know the distribution of throughput gains can be calculated as follows.

$$\begin{aligned} f(\eta )&= \int\limits ^{t-\Delta }_{0}\dfrac{\lambda e^{-\lambda (t-\Delta -y)}[\lambda (t-\Delta -y)]^{m-1}}{(m-1)!} \dfrac{\lambda ^{\eta y}_{\overline{\mathcal {N}}}}{(\eta y)!} \cdot e^{-\lambda _{\overline{\mathcal {N}}}}\delta (\eta y-k)ydy\nonumber \\&= \sum ^{\lfloor \eta (t-\Delta ) \rfloor }_{k=0} \dfrac{\lambda e^{\lambda (t-\Delta )}}{(m-1)!}e^{-\lambda _{\overline{\mathcal {N}}}} \dfrac{\lambda ^{k}_{\overline{\mathcal {N}}}}{k!} \dfrac{k}{\eta }e^{\dfrac{\lambda k}{\eta }} [\lambda (t-\Delta -\dfrac{k}{\eta })]^{m-1} \end{aligned}$$

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Appendix 2

This appendix gives the proof of the corollary 1. $E_{SU\_gains}=E[\dfrac{x}{y}]=E[x]E[\dfrac{1}{y}]=\lambda _s\lambda _{\overline{\mathcal {N}}}E[\dfrac{1}{y}]$ Then

$$\begin{aligned} E[\frac{1}{y}]&= \int\limits ^{\infty }_{\frac{1}{t-\Delta }} \frac{\lambda ^m}{(m-1)!}y\frac{1}{y^2}e^{-\lambda (t-\Delta -\frac{1}{y})} \left(t-\Delta -\frac{1}{y}\right)^{(m-1)}dy \nonumber \\&\mathop{\longrightarrow}\limits^{{t-\Delta =a}}\frac{\lambda ^m}{(m-1)!}\int\limits ^{\infty }_{\frac{1}{a}} \frac{1}{y}e^{-\lambda (a-\frac{1}{y})}\left(a-\frac{1}{y}\right)^{(m-1)}dy\nonumber \\&\mathop {=}\limits ^{s=1/y}\frac{\lambda ^m}{(m-1)!}\int\limits ^{a}_{0} \frac{1}{s}e^{-\lambda (a-s)}(a-s)^{(m-1)}ds\nonumber \\&=\frac{1}{(m-1)!}\sum ^{\infty }_{k=1}(\frac{1}{a})^{k}(\frac{1}{\lambda })^{k-1} \gamma (m+k-1,\lambda a) \end{aligned}$$

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Then corollary 1 can be obtained.

Appendix 3

In this appendix we can give the proof of the proposition 3. Due to the redundant time of MU mode is $T_{MU\_redun}=t-\Delta -con\_t\sum ^{G}_{u=2}\lambda _u-X$ , then the distribution of the $T_{MU\_redun}$ contains the sum of Poisson variable $x$ and $m$-Erlang variable $y$. We firstly obtain the $z=x+y$ distribution as follows.

$$\begin{aligned} f(z)&= \int\limits ^{z}_{0}\dfrac{\lambda ^{x}_{G}}{(x)!}e^{-\lambda _G}\delta (x-k) \dfrac{\lambda e^{-\lambda (z-x)}[\lambda (z-x)]^{m-1}}{(m-1)!}dx\nonumber \\&= e^{-\lambda _G}\dfrac{\lambda ^m}{(m-1)!}\sum ^{\lfloor z \rfloor }_{k=0} \dfrac{\lambda ^{k}_{G}}{(k)!}e^{-\lambda (z-k)}(z-k)^{m-1} \end{aligned}$$

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Then the distribution of the $T_{MU\_redun}$ can be obtained as follows.

$$\begin{aligned} f_{MU\_redun}(x)=e^{-\lambda _G}\dfrac{\lambda ^m}{(m-1)!}\sum ^{\lfloor t-\Delta -x \rfloor }_{k=0}\dfrac{\lambda ^{k}_{G}}{k!}e^{-\lambda (t-\Delta -x-k)} \cdot (t-\Delta -x-k)^m \end{aligned}$$

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Appendix 4

In this appendix we can give the proof of the corollary2. $E_{MU\_gains}=E[\dfrac{x}{y}]=E[x]E[\dfrac{1}{y}]=\lambda _s\lambda _{\overline{\mathcal {N}}}E[\dfrac{1}{y}]$ Then

$$\begin{aligned} E[\frac{1}{y}]&= \int\limits ^{\infty }_{\frac{1}{t-\Delta }} \frac{e^{-\lambda _G}\lambda ^m}{(m-1)!}\frac{1}{y}\sum ^{\lfloor 1/y \rfloor }_{k=0}\frac{\lambda ^{k}_{G}}{k!} e^{-\lambda (\frac{1}{y}-k)}\left(\frac{1}{y}-k\right)^{m-1}dy\nonumber \\&\mathop{\longrightarrow}\limits_{[x=1/y]}^{{t-\Delta =a}}\frac{e^{-\lambda _G}\lambda ^m}{(m-1)!}\int\limits ^{a}_{0} \sum ^{\lfloor 1/y \rfloor }_{k=0}\frac{\lambda ^{k}_{G}}{k!}e^{-\lambda (x-k)}(x-k)^{m-1}\frac{1}{x}dx\nonumber \\&=\frac{e^{-\lambda _G}\lambda ^m}{(m-1)!} \sum ^{\lfloor 1/y \rfloor }_{k=0}\frac{\lambda ^{k}_{G}}{k!}\int\limits ^{a}_{k}e^{-\lambda (x-k)}(x-k)^{m-1}\frac{1}{x}dx \end{aligned}$$

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The integral term can be calculated as follows.

$$\begin{aligned} Integral &= \int\limits ^{a-k}_{0}e^{-\lambda t}t^{m-1}\frac{1}{t+k}dx\nonumber \\&\mathop{\longrightarrow}\limits^{u=a-k}\int\limits ^{u}_{0}e^{-\lambda t}t^{m-1}\frac{1}{t+k}dx \nonumber \\ &= \int\limits ^{\infty }_{0}e^{-\lambda t}\frac{t^{m-1}}{t+k}dx-\int\limits ^{\infty }_{u}e^{-\lambda t}\frac{t^{m-1}}{t+k}dx =I_1-I_2 \end{aligned}$$

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where $I_1$ can be calculated using [14, 3.353.5]

$$\begin{aligned} I_1&= \int\limits ^{\infty }_{0}e^{-\lambda t}t^{m-1}\dfrac{1}{t+k}dx =(-1)^{m-2}k^{m-1}e^{\lambda k}Ei(-\lambda k)+\sum ^{m-1}_{i=1}(i-1)!(-k)^{m-1-i}\lambda ^{-i}\end{aligned}$$

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$$\begin{aligned} I_2&= \int\limits ^{\infty }_{u+k}e^{-\lambda (s-k)}(s-k)^{m-1}\dfrac{1}{s}ds =\int\limits ^{\infty }_{b}e^{\lambda k}e^{-\lambda s}(s-b+b-k)^{m-1}\dfrac{1}{s}ds \nonumber \\&= e^{\lambda k}\sum ^{m-1}_{j=0}C^{j}_{m-1}(b-k)^{j}\int\limits ^{\infty }_{b}e^{-\lambda s}\dfrac{(s-b)^{m-1-j}}{s}ds \end{aligned}$$

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where $\int\limits ^{\infty }_{b}e^{-\lambda s}(s-b)^{m-1-j}\dfrac{1}{s}ds$ can be obtained through [14, 3.383.9]. then the results can be seen as follows.

$$\begin{aligned} I_2=e^{\lambda k}\sum ^{m-1}_{j=0}C^j_{m-1}(t-\delta -k)^j(t-\Delta )^{m-1-j}\cdot \Gamma (m-j)\Gamma (-(m-1-j),\lambda (t-\Delta )) \end{aligned}$$

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