Gains provided by multichannel transmissions

1. Gains provided by multichannel transmissions Date: 2010-01-19 Authors:

2. 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

3. 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 Secondary Quaternary Primary Tertiary

4. 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 Adj. BSS trans. Secondary Channel SIFS time Tertiary Channel time Quaternary Channel PIFS time PIFS Primary Channel AIFS + backoff Ack time Adj. BSS trans. Secondary Channel time PIFS Tertiary Channel Ack time Quaternary Channel PIFS Ack time PIFS SIFS

5. 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 DAC Digitalprocessing I Seg 1 Seg 2 5GHz 6GHz DAC DAC DAC Segment 2 Seg 1 Seg 2 RF IF Busy channel

6. 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

7. 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 23 100 104 108 112 116 120 124 128 132 140 136 56 64 52 60 44 36 40 48 Example: density=8, load =0.3, one configuration among all possible

8. 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

9. 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) • …

10. 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

11. 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

12. 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

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

14. 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, Eq bandwidth required 50MHz: - contiguous: saturated - non contiguous required used bandwidth: 60 MHz

15. 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

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