Gains provided by multichannel transmissions Month Year doc.: IEEE 802.11-yy/xxxxr0 Gains provided by multichannel transmissions Date: 2010-01-19 Authors: John Doe, Some Company

Interests toward multi-channel Currently, the increase of bandwidth seems to be the only solution to increase the single user throughput. However, the probability of being allowed to transmit at 80MHz could be quite low in dense environment Interests of multi-channel Multi-channel is an efficient solution to increase that probability and ensure an increase of single-user throughput compared to 11n. In [1], we proposed 3 different methods for multichannel transmissions

Most favoured solution for multi-channel Non contiguous synchronous Aggregation of non contiguous channel wherever in the band Only one CSMA-CA on the primary channel Secondary, tertiary and quaternary can be on any channel of the band Primary Channel AIFS + backoff Ack time Secondary Channel Ack time PIFS Tertiary Channel Ack time Quaternary Channel PIFS Ack time PIFS SIFS 30 23 17 Primary Secondary Tertiary Quaternary

Most favoured solution for multi-channel Non contiguous synchronous Without channel bonding within 80MHz With channel bonding within 80MHz Primary Channel AIFS + backoff Ack time Secondary Channel Adj. BSS trans. SIFS time Tertiary Channel time Quaternary Channel PIFS time PIFS Primary Channel AIFS + backoff Ack Secondary Channel Adj. BSS trans. time time PIFS Tertiary Channel Ack time Quaternary Channel PIFS Ack time PIFS SIFS

Most favoured solution for multi-channel Non contiguous synchronous Synchronous non contiguous implementation allows easy implementation of channel bonding Seg 1 Seg 2 Transmitter Segment 1 Digital processing DAC I DAC 5GHz 6GHz Seg 1 Seg 2 DAC DAC Segment 2 IF RF Seg 1 Seg 2 Busy channel

Goal of this presentation Demonstrate the gains provided by multichannel to increase the probability to reach the desired bandwidth Clarify the advantages of multichannel transmissions  We ran some simulations for that purpose

Simulation scenarios Simulate different load configurations on the whole 5GHz band Select a number of busy channels among the 19 available in Europe: called density Restrict the set of scenarios by setting a unique load for each busy channel (fixed to 0.3, 0.6 and 0.9) For each density from 0 to 19, we simulate all possible configurations 30 23 36 40 44 48 52 56 60 64 100 104 108 112 116 120 124 128 132 136 140 Example: density=8, load =0.3, one configuration among all possible

Simulation scenarios Simulate different load configurations on the whole 5GHz band For each configuration, for each multichannel solution (non contiguous, bonding, nb of aggregated channels) and for each channel allocation, we calculate the equivalent bandwidth (bandwidth that you would be able to use 100% of the time) And select the best channel allocation and its corresponding optimal equivalent bandwidth Average the equivalent bandwidth for each density

Calculation of the equivalent bandwidth Assume we have 4 aggregated channels, each with a specific probability Pi of being idle. We can calculate the probability to transmit at 20, 40, 60 or 80MHz P1, P2, P3, P4 Primary P60 P60 Without channel bonding P80 = P1*P2*P3*P4. P60 = P1*P2*P3*(1-P4) P40 = P1*P2*(1-P3) P20 = P1*(1-P2) With channel bonding P80 = P1*P2*P3*P4. P60 = P1*P2*P3*(1-P4) +P1*P3*P4*(1-P2) +P1*P2*P4*(1-P3) …

Calculation of the equivalent bandwidth Assume we have 4 aggregated channels, each with a specific probability Pi of being idle. We can calculate the probability to transmit at 20, 40, 60 or 80MHz And calculate the equivalent bandwidth (bandwidth that you would be able to use 100% of the time) P1, P2, P3, P4 Primary

Simulations Compare the equivalent bandwidth vs density for Contiguous mode (with or without channel bonding) Ideal synchronous non contiguous mode (with or without channel bonding) Restricted* synchronous non contiguous mode (with or without channel bonding) *Restricted: only two non contiguous sets of contiguous channels Tested for different multichannel bandwidth (40, 60, 80MHz) Tested for different loads for busy channels: 0.3, 0.9 - 40MHz = 2*20MHz channels - 60MHz = 3*20MHz channels - 80MHz = 4*20MHz channels

Simulation results: 80MHz, load=0.9 Significant improvements with ideal non contiguous Still very good gains with restricted non contiguous Importance of the channel bonding

Simulation results: 80MHz, load=0.3 More compact results Still significative gains with restricted non contiguous Importance of the channel bonding

Simulation results: Restricted non contiguous gains, load=0 Simulation results: Restricted non contiguous gains, load=0.9, 40MHz up to 80MHz Leads to a potential reduction of the bandwidth used by a BSS Example: Density 12, Eq bandwidth required 50MHz: - contiguous required used bandwidth: 80 MHz - non contiguous required used bandwidth: 60 MHz Example: Density 13, - contiguous: saturated

Advantages of synchronous multi-channel (Conclusions) Clear increase of the probability to reach the desired bandwidth Increase of the equivalent bandwidth Or reduce the bandwidth occupied by a BSS for the same equivalent bandwidth Importance of the channel bonding The implementation with two front-end segments enables easy channel bonding, even in contiguous mode Better use of the whole band Assuming the knowledge of the 5GHz band occupancy, it offers a good flexibility in the use of the band: possible switch for the best channel allocation The two front-end segments implementation enables the possibility to sound the band while maintaining the link

References [1] Cariou, L. and Benko, J., Multichannel transmissions, IEEE 802.11-09/1022r0, Sep. 2010 Slide 16 Youhan Kim, Atheros