MAC Layer Approaches for Mitigating the Spectrum Underutilization Due to Overlapping BSS Problem in WLAN

Abstract

Overlapping basic service set (OBSS) interferences cause significant reductions in the network throughput as seen at the medium access control (MAC) layer. When the frequency bands of multiple OBSSs are partially overlapped, the reduction of throughput are worsened because in addition to the OBSS interference problem there arises the OBSS-induced spectrum underutilization problem which has not been addressed so far. In this paper, two complementary contention-based MAC layer schemes are proposed to solve the OBSS-induced spectrum underutilization problem by setting up a virtual-primary channel for contention. The proposed time-limited (TL) scheme is designed for transmitting long data packets, which is suitable for applications requiring high throughputs. In contrast, the multi-contention (MC) scheme is designed for transmitting short data packets, which is suitable for applications requiring short delays. By efficiently exploiting the OBSS-induced underutilized spectrum, simulation results verify that the proposed TL scheme can increase the throughput greatly and the proposed MC scheme can reduce the delay for short data packets significantly.

Appendix

1.1 Time–Frequency Resource Ratio Analysis

The time–frequency resource is defined as the product of occupied/available time and bandwidth. Then, the time–frequency resource ratio (TFRR) of a MAC protocol can be defined as the ratio between the time–frequency resource needed to transmit some data frames using a direct link and the total time–frequency resource used to transmit these data frames under the MAC protocol.

1.2 TFRR of 802.11ac Protocol

Assume there are N_S STAs (including AP) contending for the channel access in a BSS and each STA has the same probability to access the channel. If only one data frame is transmitted for each channel access, then M frames are transmitted during a given time when there are M times of successful channel accesses for all STAs in the BSS. Let p_n be the probability that an STA in the BSS is a 20n megahertz capable STA, where n (n = 1, 2, …, N) is the number of 20 MHz channels that the STA can operate on. In 802.11ac, when the available bandwidth in the BSS is 80 MHz, the total number of available channels N is 4. Then, in average, Mp_n data frames are transmitted by the N_Sp_n 20n megahertz capable STAs in the BSS during the given time duration.

Let L_MPDU be the MAC protocol data unit length in bytes, T_S be the orthogonal frequency-division multiplexing (OFDM) symbol duration, m be the modulation and coding scheme (MCS) index and L_m,n be the number of data bits that can be transmitted per OFDM symbol over n 20 MHz channels corresponding to an MCS with index m. If the number of bits in the service field plus tail field of the physical layer convergence protocol data unit is n_tail, the transmission time for each data frame (without considering the transmission time of the preamble) is

$${\text{T}}_{{{\text{m}},{\text{n}}}} \left( {{\text{L}}_{\text{MPDU}} } \right) = {\text{T}}_{\text{s}} \frac{{{\text{n}}_{\text{tail}} {\text{ + 8L}}_{\text{MPDU}} }}{{{\text{L}}_{\text{m,n}} }}$$

(1)

where ⌈·⌉ represents round toward positive infinity. Then, the total time–frequency resource needed to transmit the M data frames (with MCS m) using a direct link is

$${\text{R}}_{{\rm m}} = \mathop \sum \limits_{{{{\rm n}} = 1}}^{{\rm N}} \left\{ {{\text{Mp}}_{{\rm n}} {\text{T}}_{{{{\rm m}},{\text{n}}}} \left( {{\text{L}}_{{\rm MPDU}} } \right)} \right\}\left\{ {20{\text{n}}} \right\}$$

(2)

However, the total time–frequency resource used to transmit these M data frames in 802.11ac is

$${\hat{\text{R}}}_{\text{m}} = \mathop \sum \limits_{{{\text{n}} = 1}}^{\text{N}} \left\{ {{\text{Mp}}_{\text{n}} \left[ {{\text{T}}_{\text{DIFS}} + {\bar{\text{T}}}_{\text{B}} + {\text{T}}_{\text{RTS}} + {\text{T}}_{\text{CTS}} + {\text{T}}_{\text{p}} + {\text{T}}_{{{\text{m}},{\text{n}}}} \left( {{\text{L}}_{\text{MPDU}} } \right) + {\text{T}}_{{{\text{ACK}},{\text{n}}}} + 3{\text{T}}_{\text{SIFS}} } \right]} \right\} \cdot 20{\text{N}}$$

(3)

where ${\text{T}}_{\text{p}}$ is the preamble duration, ${\text{T}}_{\text{DIFS}}$ is the distributed coordination function inter-framing spacing (DIFS) time and ${\text{T}}_{\text{SIFS}}$ is the short-inter-frame-spacing (SIFS) time. The average backoff time is assumed to be equal to ${\bar{\text{T}}}_{{\rm B}} = {\text{T}}_{{\rm slot}} {\text{CW}}_{{\rm min} } /2$, where CW_min is the minimum contention window size and T_slot is the slot duration. (Since we use CW_min to calculate the backoff time, the TFRR to be defined later will be denoted as maximum TFRR). The transmission times for RTS, CTS and ACK frames are given by T_RTS = T_P + T_m,n(L_RTS), T_CTS = T_P + T_m,n (L_CTS) and T_ACK,n = T_P + T_m,n(L_ACK). Here, L_RTS, L_CTS and L_ACK, denoting the number of bytes for RTS, CTS and ACK frames respectively. Then

$$\upeta_{\text{ac}} = \frac{{{\text{R}}_{\text{m}} }}{{{\hat{\text{R}}}_{{\rm m}} }} = \frac{{\mathop \sum \nolimits_{{{{\rm n}} = 1}}^{{\rm N}} {\text{p}}_{{\rm n}} {\text{nT}}_{{{{\rm m}},{\text{n}}}} \left( {{\text{L}}_{{\rm MPDU}} } \right)}}{{\left\{ {{\text{T}}_{{\rm DIFS}} + {\bar{\text{T}}}_{{\rm B}} + {\text{T}}_{{\rm RTS}} + {\text{T}}_{{\rm CTS}} + {\text{T}}_{{\rm p}} + 3{\text{T}}_{{\rm SIFS}} + \mathop \sum \nolimits_{{{{\rm n}} = 1}}^{{\rm N}} {\text{p}}_{{\rm n}} \left\{ {{\text{T}}_{{{{\rm m}},{\text{n}}}} \left( {{\text{L}}_{{\rm MPDU}} } \right) + {\text{T}}_{{{{\rm ACK}},{\text{n}}}} } \right\}} \right\}{\text{N}}}}$$

(4)

1.3 TFRR of the TL Scheme

Without loss of generality, we assume that MCS m₂ is adopted over the first i 20 MHz channels and MCS m₁ is adopted over the left N − i 20 MHz channel to transmit more data when the first i 20 MHz channels are occupied by the neighbor BSS in the TL scheme. Let L_SNI denote the number of bytes for SNI, T_SNI = T_P + T_m,n(L_SNI). Then, the total bytes transmitted over the left N − i 20 MHz channels is

$${\text{L}}_{\text{MPDU}}^{\text{TL}} = \frac{{{\text{L}}_{{{\text{m}}_{{1,{\text{N}} - {\text{i}}}} }} \frac{{{\text{T}}_{{{\text{m}}_{2} ,{\text{i}}}} \left( {{\text{L}}_{\text{MPDU}} } \right) - {\text{T}}_{\text{DIFS}} - {\bar{\text{T}}}_{\text{B}} - {\text{T}}_{\text{RTS}} - {\text{T}}_{\text{CTS}} - {\text{T}}_{\text{p}} - 2{\text{T}}_{\text{SIFS}} - {\text{T}}_{\text{SNI}} }}{{{\text{T}}_{\text{s}} }} - {\text{n}}_{\text{tail}} }}{8}$$

(5)

It is illustrated in Fig. 11. Then the total time–frequency resource needed to transmit the data frames (with MCS m₂) using a direct link is

$${\text{R}}_{{{{\rm m}}_{1} ,{\text{m}}_{2} }}^{1} = \mathop \sum \limits_{{{{\rm n}} = 1}}^{{\rm N}} \left\{ {{\text{Mp}}_{{\rm n}} {\text{T}}_{{{{\rm m}}_{2} ,{\text{n}}}} \left( {{\text{L}}_{\text{MPDU}} } \right)} \right\}\left\{ {20{\text{n}}} \right\} + \mathop \sum \limits_{{{\text{n}} = 1}}^{{\rm N}} {\text{p}}_{\text{n}} \left( {\mathop \sum \limits_{{{\text{i}} = 1}}^{{{\text{N}} - 1}} {\text{Mp}}_{{\rm i}} {\text{T}}_{{{\text{m}}_{1} ,{\text{m}}_{2} ,{\text{n}}}}^{\text{i}} \left( {{\text{L}}_{\text{MPDU}} } \right)\left\{ {20{\text{n}}} \right\}} \right)$$

(6)

where

$${\text{T}}_{{{\text{m}}_{1} ,{\text{m}}_{2} ,{\text{n}}}}^{\text{i}} \left( {{\text{L}}_{\text{MPDU}} } \right) = {\text{T}}_{\text{s}} \frac{{8{\text{L}}_{\text{MPDU}}^{\text{TL}} + {\text{n}}_{{\rm tail}} }}{{{\text{L}}_{{{\text{m}}_{2} ,{\text{n}}}} }}$$

(7)

Substituting (5) into (7),

$$\begin{aligned} {\text{T}}_{{{\text{m}}_{1} ,{\text{m}}_{2} ,{\text{n}}}}^{\text{i}} \left( {{\text{L}}_{\text{MPDU}} } \right) & = {\text{T}}_{\text{s}} \frac{{\frac{{{\text{L}}_{{{\text{m}}_{{1,{\text{N}} - {\text{i}}}} }} \frac{{{\text{T}}_{{{\text{m}}_{2} ,{\text{i}}}} \left( {{\text{L}}_{\text{MPDU}} } \right) - {\text{T}}_{\text{DIFS}} - {\bar{\text{T}}}_{\text{B}} - {\text{T}}_{\text{RTS}} - {\text{T}}_{\text{CTS}} - {\text{T}}_{\text{p}} - 2{\text{T}}_{\text{SIFS}} - {\text{T}}_{{\rm SNI}} }}{{{\text{T}}_{\text{s}} }} - {\text{n}}_{\text{tail}} }}{8}8 + {\text{n}}_{{\rm tail}} }}{{{\text{L}}_{{{\text{m}}_{2} ,{\text{n}}}} }} \\ & \approx \frac{{{\text{L}}_{{{\text{m}}_{{1,{\text{N}} - {{\rm i}}}} }} }}{{{\text{L}}_{{{\text{m}}_{2} ,{{\rm n}}}} }}{\text{T}}_{{{\text{m}}_{2} ,{\text{i}}}} \left( {{\text{L}}_{\text{MPDU}} } \right) - {\text{T}}_{\text{DIFS}} - {\bar{\text{T}}}_{\text{B}} - {\text{T}}_{\text{RTS}} - {\text{T}}_{\text{CTS}} - {\text{T}}_{\text{p}} - 2{\text{T}}_{\text{SIFS}} - {\text{T}}_{\text{SNI}} \\ \end{aligned}$$

(8)

The total time–frequency resource used to transmit these data frames using TL scheme is

$${\hat{\text{R}}}_{{{\text{m}}_{1} ,{\text{m}}_{2} }}^{\text{TL}} = \mathop \sum \limits_{{{\text{n}} = 1}}^{\text{N}} \left\{ {{\text{Mp}}_{\text{n}} \left[ {{\text{T}}_{\text{DIFS}} + {\bar{\text{T}}}_{\text{B}} + {\text{T}}_{\text{RTS}} + {\text{T}}_{\text{CTS}} + {\text{T}}_{\text{p}} + {\text{T}}_{{{\text{m}}_{2} ,{\text{n}}}} \left( {{\text{L}}_{\text{MPDU}} } \right) + {\text{T}}_{{{{\rm ACK}},{\text{n}}}} + 3{\text{T}}_{\text{SIFS}} } \right]} \right\} \cdot 20{\text{N}}$$

(9)

Then

$$\begin{aligned}\upeta_{\text{TL}} & = \frac{{{\text{R}}_{{{\text{m}}_{1} ,{\text{m}}_{2} }}^{1} }}{{{\hat{\text{R}}}_{{{\text{m}}_{1} ,{\text{m}}_{2} }}^{\text{TL}} }} \\ & = \frac{{\mathop \sum \nolimits_{{{\text{n}} = 1}}^{\text{N}} \left\{ {{\text{p}}_{{\rm n}} {\text{T}}_{{{\text{m}}_{2} ,{\text{n}}}} \left( {{\text{L}}_{\text{MPDU}} } \right)} \right\}{\text{n}} + \mathop \sum \nolimits_{{{\text{n}} = 1}}^{{\rm N}} {\text{p}}_{\text{n}} \left( {\mathop \sum \nolimits_{{{\text{i}} = 1}}^{{{\text{N}} - 1}} {\text{p}}_{\text{i}} \frac{{{\text{L}}_{{{\text{m}}_{{1,{\text{N}} - {{\rm i}}}} }} }}{{{\text{L}}_{{{\text{m}}_{2} ,{{\rm n}}}} }}{\text{T}}_{{{\text{m}}_{2} ,{\text{i}}}} \left( {{\text{L}}_{\text{MPDU}} } \right) - {\text{T}}_{\text{DIFS}} - {\bar{\text{T}}}_{\text{B}} - {\text{T}}_{\text{RTS}} - {\text{T}}_{\text{CTS}} - {\text{T}}_{\text{p}} - 2{\text{T}}_{\text{SIFS}} - {\text{T}}_{{\rm SNI}} {\text{n}}} \right)}}{{\mathop \sum \nolimits_{{{\text{n}} = 1}}^{\text{N}} \left\{ {{\text{p}}_{\text{n}} \left[ {{\text{T}}_{\text{DIFS}} + {\bar{\text{T}}}_{\text{B}} + {\text{T}}_{\text{RTS}} + {\text{T}}_{\text{CTS}} + {\text{T}}_{\text{p}} + {\text{T}}_{{{\text{m}}_{2} ,{\text{n}}}} \left( {{\text{L}}_{\text{MPDU}} } \right) + {\text{T}}_{{{{\rm ACK}},{\text{n}}}} + 3{\text{T}}_{\text{SIFS}} } \right]} \right\} \cdot {\text{N}}}} \\ \end{aligned}$$

(10)

Fig. 11

Illustration for TFRR calculation of TL scheme

1.4 TFRR of the MC Scheme

Assuming that there are N_MC times of successful channel accesses over the left N − i 20 MHz channels (with MCS m₂) during the transmission time ${\text{T}}_{{{\text{m}}_{2} ,{\text{i}}}} \left( {{\text{L}}_{\text{MPDU}} } \right)$ of the first i 20 MHz channels (with MCS m₁), the total bytes transmitted over the left N − i 20 MHz channels is

$${\text{L}}_{\text{MPDU}}^{\text{MC}} = \frac{{{\text{L}}_{{{\text{m}}_{{1,{\text{N}} - {\text{i}}}} }} \frac{{{\text{T}}_{{{\text{m}}_{2} ,{\text{i}}}} \left( {{\text{L}}_{\text{MPDU}} } \right) - {\text{T}}_{\text{SNI}} - {\text{N}}_{{\rm MC}} ({\text{T}}_{\text{DIFS}} + {\bar{\text{T}}}_{\text{B}} + {\text{T}}_{\text{RTS}} + {\text{T}}_{\text{CTS}} + {\text{T}}_{\text{p}} + 2{\text{T}}_{\text{SIFS}} ) - \left( {{\text{N}}_{\text{MC}} - 1} \right)({\text{T}}_{\text{SIFS}} + {\text{T}}_{{{\text{ACK}},{\text{N}} - {\text{i}}}} )}}{{{\text{T}}_{\text{s}} }} - {\text{n}}_{\text{tail}} }}{8}$$

(11)

It is illustrated in Fig. 12. Then the total time–frequency resource needed to transmit the data frames (with MCS m₂) using a direct link is

$${\text{R}}_{{{\text{m}}_{1} ,{\text{m}}_{2} }}^{2} = \mathop \sum \limits_{{{\text{n}} = 1}}^{\text{N}} \left\{ {{\text{Mp}}_{{\rm n}} {\text{T}}_{{{\text{m}}_{2} ,{\text{n}}}} \left( {{\text{L}}_{\text{MPDU}} } \right)} \right\}\left\{ {20{\text{n}}} \right\} + \mathop \sum \limits_{{{\text{n}} = 1}}^{{\rm N}} {\text{p}}_{\text{n}} \left( {\mathop \sum \limits_{{{\text{i}} = 1}}^{{{\text{N}} - 1}} {\text{Mp}}_{\text{i}} {\hat{\text{T}}}_{{{\text{m}}_{1} ,{\text{m}}_{2} ,{\text{n}}}}^{\text{i}} \left( {{\text{L}}_{\text{MPDU}}^{\text{MC}} } \right)\left\{ {20{\text{n}}} \right\}} \right)$$

(12)

where

$${\hat{\text{T}}}_{{{\text{m}}_{1} ,{\text{m}}_{2} ,{\text{n}}}}^{\text{i}} \left( {{\text{L}}_{\text{MPDU}}^{\text{MC}} } \right) = {\text{T}}_{\text{s}} \frac{{8{\text{L}}_{\text{MPDU}}^{\text{MC}} + {\text{n}}_{\text{tail}} }}{{{\text{L}}_{{{\text{m}}_{2} ,{\text{n}}}} }}$$

(13)

Fig. 12

Illustration for TFRR calculation of MC scheme

Substituting (11) into (13),

$$\begin{aligned} & {\hat{\text{T}}}_{{{\text{m}}_{1} ,{\text{m}}_{2} ,{\text{n}}}}^{\text{i}} \left( {{\text{L}}_{\text{MPDU}}^{\text{MC}} } \right) \\ & \quad = {\text{T}}_{\text{s}} \frac{{\frac{{{\text{L}}_{{{\text{m}}_{{1,{\text{N}} - {\text{i}}}} }} \frac{{{\text{T}}_{{{\text{m}}_{2} ,{\text{i}}}} \left( {{\text{L}}_{\text{MPDU}} } \right) - {\text{T}}_{\text{SNI}} - {\text{N}}_{{\rm MC}} ({\text{T}}_{\text{DIFS}} + {\bar{\text{T}}}_{\text{B}} + {\text{T}}_{\text{RTS}} + {\text{T}}_{\text{CTS}} + {\text{T}}_{\text{p}} + 2{\text{T}}_{\text{SIFS}} ) - \left( {{\text{N}}_{\text{MC}} - 1} \right)({\text{T}}_{\text{SIFS}} + {\text{T}}_{{{\text{ACK,N}} - {\text{i}}}} )}}{{{\text{T}}_{\text{s}} }} - {\text{n}}_{\text{tail}} }}{8}8 + {\text{n}}_{\text{tail}} }}{{{\text{L}}_{{{\text{m}}_{2} ,{\text{n}}}} }} \\ & \quad \approx \frac{{{\text{L}}_{{{\text{m}}_{{1,{\text{N}} - {\text{i}}}} }} }}{{{\text{L}}_{{{\text{m}}_{2} ,{\text{n}}}} }}{\text{T}}_{{{\text{m}}_{2} ,{\text{i}}}} \left( {{\text{L}}_{\text{MPDU}} } \right) - {\text{T}}_{\text{SNI}} - {\text{N}}_{\text{MC}} ({\text{T}}_{\text{DIFS}} + {\bar{\text{T}}}_{\text{B}} + {\text{T}}_{\text{RTS}} + {\text{T}}_{\text{CTS}} + {\text{T}}_{\text{p}} + 2{\text{T}}_{\text{SIFS}} ) - \left( {{\text{N}}_{\text{MC}} - 1} \right)({\text{T}}_{\text{SIFS}} + {\text{T}}_{{{\text{ACK}},{\text{N}} - {\text{i}}}} ) \\ \end{aligned}$$

(14)

However, the total time–frequency resource used to transmit these data frames using MC scheme is

$${\hat{\text{R}}}_{{{\text{m}}_{1} ,{\text{m}}_{2} }}^{\text{MC}} = \mathop \sum \limits_{{{\text{n}} = 1}}^{\text{N}} \left\{ {{\text{Mp}}_{\text{n}} \left[ {{\text{T}}_{\text{DIFS}} + {\bar{\text{T}}}_{\text{B}} + {\text{T}}_{\text{RTS}} + {\text{T}}_{\text{CTS}} + {\text{T}}_{\text{p}} + {\text{T}}_{{{\text{m}}_{2} ,{\text{n}}}} \left( {{\text{L}}_{\text{MPDU}} } \right) + {\text{T}}_{{{\text{ACK}},{\text{n}}}} + 3{\text{T}}_{\text{SIFS}} } \right]} \right\} \cdot 20{\text{N}}$$

(15)

Then

$$\begin{aligned}\upeta_{\text{MC}} & = \frac{{{\text{R}}_{{{\text{m}}_{1} ,{\text{m}}_{2} }}^{2} }}{{{\hat{\text{R}}}_{{{{\rm m}}_{1} ,{\text{m}}_{2} }}^{{\rm MC}} }} \\ & = \frac{{\mathop \sum \nolimits_{{{\text{n}} = 1}}^{\text{N}} \left\{ {{\text{p}}_{{\rm n}} {\text{T}}_{{{\text{m}}_{2} ,{\text{n}}}} \left( {{\text{L}}_{\text{MPDU}} } \right)} \right\}{\text{n}} + \mathop \sum \nolimits_{{{{\rm n}} = 1}}^{{\rm N}} {\text{p}}_{\text{n}} \left( {\mathop \sum \nolimits_{{{\text{i}} = 1}}^{{{\text{N}} - 1}} {\text{p}}_{\text{i}} {\hat{\text{T}}}_{{{\text{m}}_{1} ,{\text{m}}_{2} ,{\text{n}}}}^{\text{i}} \left( {{\text{L}}_{\text{MPDU}}^{\text{MC}} } \right){\text{n}}} \right)}}{{\mathop \sum \nolimits_{{{\text{n}} = 1}}^{{\rm N}} \left\{ {{\text{p}}_{\text{n}} \left[ {{\text{T}}_{\text{DIFS}} + {\bar{\text{T}}}_{\text{B}} + {\text{T}}_{\text{RTS}} + {\text{T}}_{\text{CTS}} + {\text{T}}_{\text{p}} + {\text{T}}_{{{\text{m}}_{2} ,{\text{n}}}} \left( {{\text{L}}_{\text{MPDU}} } \right) + {\text{T}}_{{{\text{ACK}},{\text{n}}}} + 3{\text{T}}_{{\rm SIFS}} } \right]} \right\} \cdot {\text{N}}}} \\ \end{aligned}$$

(16)

where ${\hat{\text{T}}}_{{{\text{m}}_{1} ,{\text{m}}_{2} ,{\text{n}}}}^{\text{i}} \left( {{\text{L}}_{\text{MPDU}}^{\text{MC}} } \right)$ is given by (14).

For convenience and without loss of generality, consider that the AP and the STAs in BSS1 and BSS2 are 80 MHz and 20 MHz capable devices respectively. Assuming the probability that the primary channel occupied by BSS2 is p, then p₁ = p, p₂ = 0, p₃ = 0 and p₄ = 1 − p₁ = 1 − p. The minimum and maximum contention window sizes are 15 and 1023, respectively. MCS index m₁ = m₂ = 1 is used for data transmission. T_S = 4 μs, n_tail = 22 bytes, ${\text{T}}_{\text{p}} = 40\;\upmu{\text{s}}$, ${\text{T}}_{\text{DIFS}} = 34\;\upmu{\text{s}}$, ${\text{T}}_{\text{SIFS}} = 16\;\upmu{\text{s}}$ and T_slot = 9 μs. We assume that the transmitted frame length is 3000 bytes when the channel is accessed over primary channel. Table 3 shows the TFRR of the proposed scheme and 802.11ac protocol. Obviously, the TFRR of TL scheme is highest which is in line with the intuition. Just considering the results for MC scheme, we observe that the TFRR decreases as N_MC increases. This is due to the fact that the bigger the N_MC is, the larger the proportion of the overhead is.

Table 3 TFRR of the proposed schemes and 802.11ac protocol