Hans Journal of Wireless Communications
Vol.08 No.02(2018), Article ID:24588,6 pages
10.12677/HJWC.2018.82007

Analysis of Throughput of IEEE 802.11 Protocol Based on Exponential Backoff Algorithm

Qiao Xiang, Suoping Li

School of Science, Lanzhou University of Technology, Lanzhou Gansu

Received: Apr. 3rd, 2018; accepted: Apr. 20th, 2018; published: Apr. 27th, 2018

ABSTRACT

In the actual network communication environment, data packets cannot be efficiently transmitted due to channel interference, so it is necessary to improve the system performance impact under non-ideal channels. This paper takes the wireless LAN as the communication background, considering the impact of non-ideal channels on WLANs such as signal fading and distortion, adds the freezing probability based on the previous literature exponential backoff algorithm to reduce the data collision in the channel and combine them. The two types of access methods are used to obtain the saturation throughput formula under the IEEE 802.11 protocol. Finally, simulation and analysis of the model are performed.

Keywords:IEEE 802.11, Backoff Algorithm, Nonideal Channel, Throughput

1. 引言

2. 改进后的模型分析

(1)

$\left\{\begin{array}{l}P\left\{i,k|i,k+1\right\}=1-P\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}k\in \left(0,{W}_{i}-\text{3}\right),\text{\hspace{0.17em}}i\in \left(0,m\right)\\ P\left\{0,k|i,0\right\}=\frac{\left(1-{P}_{r}\right)P}{{W}_{\text{0}}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}k\in \left(0,{W}_{0}-\text{2}\right),\text{\hspace{0.17em}}i\in \left(0,m\right)\\ P\left\{i,k|i-1,0\right\}=\frac{{P}_{r}}{{W}_{\text{0}}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}k\in \left(0,{W}_{0}-\text{2}\right),\text{\hspace{0.17em}}i\in \left(0,m\right)\\ P\left\{m,k|m,0\right\}=\frac{{P}_{r}}{{W}_{m}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}k\in \left(0,{W}_{m}-\text{2}\right)\\ P\left\{-1,0|i,0\right\}=\left(1-P\right)\left(1-{P}_{r}\right)\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}k\in \left(0,{W}_{i}-\text{2}\right)\end{array}$ (2)

3. 饱和吞吐量分析

${T}_{S}^{bas}=H+E\left[L\right]+\delta +{T}_{SIFS}+{T}_{ACK}+\delta +{T}_{DIFS}$ (3)

RTS/CTS接入方式：

${T}_{S}^{rts}={T}_{RTS}+\sigma +{T}_{SIFS}+{T}_{CTS}+\sigma +{T}_{SIFS}+H+E\left[L\right]+\delta +{T}_{SIFS}+{T}_{ACK}+\delta +{T}_{DIFS}$ (4)

 (5)

$P=1-{\left(1-\tau \right)}^{n}$ (6)

$\tau =\sum _{i=-1}^{m}{b}_{i,0}=\frac{\left(1-P{P}_{r}^{m+2}-\left(P+{P}_{r}\right){P}_{r}^{m+1}\right)\cdot \left(2-2P\right)\left(1-2{P}_{r}\right)}{2{\left(1-P\right)}^{2}\left(1-2{P}_{r}\right)\left(1-{P}_{r}\right)+\left(\left(1-{P}_{r}\right)P+{P}_{r}\right)\left(1-2{P}_{r}\right)\left({W}_{0}+1\right)+\left(\left(1-{P}_{r}\right)P+{P}_{r}\right)\left(1-{\left(2{P}_{r}\right)}^{m}\right){W}_{0}{P}_{r}}$ (7)

$\begin{array}{c}{P}_{r}^{bas}\\ {P}_{r}^{rts}\end{array}\right\}=\left\{\begin{array}{l}1-{\left(1-\tau \right)}^{n}{\left(1-BER\right)}^{H+E\left[L\right]+{T}_{ACK}}\\ 1-{\left(1-\tau \right)}^{n}{\left(1-BER\right)}^{{T}_{RTS}+{T}_{CTS}+H+E\left[L\right]+{T}_{ACK}}\end{array}$ (8)

${P}_{S}^{bas}=\frac{n\tau {\left(1-\tau \right)}^{n}{\left(1-BER\right)}^{H+E\left[L\right]+{T}_{ACK}}}{P}$ (9)

RTS/CTS访问方式成功传输概率：

 (10)

$S=\frac{E\left[数据成功传输的平均分组长度\right]}{E\left[数据成功传输的平均时延\right]}$ (11)

$\begin{array}{c}\begin{array}{c}{S}^{bas}\\ {S}^{rts}\end{array}\right\}=\frac{E\left[d\right]}{E\left[\sigma \right]}=\frac{{P}_{S}\cdot P\cdot E\left[L\right]}{\left(1-P\right)\sigma +{P}_{S}\cdot {T}_{S}\cdot P+\left(1-{P}_{S}\right)\cdot {T}_{C}\cdot P}\\ =\left\{\begin{array}{l}\frac{n\tau {\left(1-\tau \right)}^{n}{\left(1-BER\right)}^{{T}_{RTS}+{T}_{CTS}+H+E\left[L\right]+{T}_{ACK}}\cdot E\left[L\right]}{\left(1-P\right)\sigma +{P}_{S}\cdot {T}_{S}\cdot P+\left(1-{P}_{S}\right)\cdot {T}_{C}\cdot P}\\ \frac{n\tau {\left(1-\tau \right)}^{n}{\left(1-BER\right)}^{H+E\left[L\right]+{T}_{ACK}}\cdot E\left[L\right]}{\left(1-P\right)\sigma +{P}_{S}\cdot {T}_{S}\cdot P+\left(1-{P}_{S}\right)\cdot {T}_{C}\cdot P}\end{array}\end{array}$ (12)

4. 数值分析

Figure 1. Saturation throughput graph for basic access method

Figure 2. Saturation throughput graph for RTS/CTS access method

5. 结束语

Analysis of Throughput of IEEE 802.11 Protocol Based on Exponential Backoff Algorithm[J]. 无线通信, 2018, 08(02): 64-69. https://doi.org/10.12677/HJWC.2018.82007

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