1. Understanding TCP fairness over Wireless LAN  班級：碩士在職專班（一）  學號： 492515045  姓名：呂國銓  日期： 92.11.18

2. - 2 - O UTLINE Introduction Experiment Simulation Mathematical Analysis Solution Conclusion

3. - 3 - I NTRODUCTION WLAN 盛行－ IEEE 802.11standard Private Area ： homes and offices Public Area ： airports, hotels, cafes,…

4. - 4 - W LAN M ODEL 對等式無線網路 (Ad-Hoc) 【點對點模式】 主從式無線網路 (Infrastructure) 【共用模式】 Ad-Hoc Infrastructure

5. - 5 - U NFAIRNESS OF 802.11 If the mobile hosts are all senders or all receivers, then they each have equal share of the total available bandwidth. === There is one mobile sender and the rest are all mobile receivers. This mobile sender, therefore gets half of the channel bandwidth and the remaining half is equally shared by all the mobile receivers. = ==

6. - 6 - E XPERIMENT In order to illustrate the subtle interactions of TCP with an unfair 802.11 MAC protocol. We conducted a series of performance tests on a commercial 802.11b network consisting of one base station and three mobile users. The ratios presented in the table are the average of 5-10 runs. In order to test the sensitivity of this ratio to the base station buffer size, use background UDP traffic. Install sniffers on the wireless interface.

7. - 7 - T ABLE OF E XPERIMENT Ru ： The average TCP uplink throughput R d ： The average TCP downlink throughput SD ： Standard Deviation

8. - 8 - C ONCLUSION F ROM T ABLE  Ru / Rd ＞ 1 ＝＞ upstream > downstream Number of flows ↑ ＝＞ Ru / Rd ↑ Use background UDP traffic ＝＞ Ru / Rd ↑ MTU ↓ ＝＞ Ru / Rd ↑↑

9. - 9 - I LLUSTRATION OF E XPERIMENT Upstream flow finished its upload and terminated Packets lost Congestion avoidance region : 9K – 18K First 150 sec throughput is very low

10. - 10 - S IMULATION Factors impact the throughput ratio in a test-bed  Wireless link interface  Base station buffer size  Implementation details of the 802.11 MAC layer  …… Simulation study using the NS2 simulator  One upstream and one downstream flow  Multiple flows

11. - 11 - S IMULATION ： up/down ratio Total throughput is stable Region I : 84 < buffer Up/down = 1 Region II : 42 < buffer < 84 Up/down = 10 → 1 Region III : 6 < buffer < 42 Up/down = {9,12} Region IV : buffer < 6 very noisy

12. - 12 - S IMULATION ： RTT  RTT increases monotonically with the base station buffer size without any significant rate changes  The RTT of downstream is almost equal to the upstream’s.  One upstream and one downstream flow  5 simulation runs  Each simulating 100 seconds

13. - 13 - S IMULATION ： D ata & A CK L oss  Data packet loss rate is always higher than the ACK loss rate  The dependency on the buffer size is not liner.  One upstream and one downstream flow  5 simulation runs  Each simulating 100 seconds

14. - 14 - S IMULATION ： Multiple Flows  The ratio is almost linear  All the downstream flows share the same resources while the total throughput remains stable.  One upstream and multiple downstream flows  buffer size = 100 packets  5 runs for each data point  Lasting for 100 seconds

15. - 15 - S IMULATION ： Multiple Flows  The ratio is high (up to 800)  Total throughput is low  ACKs of the upstream flows clutter the base station buffer and downstream packets are dropped.  Equal number of multiple upstream and downstream flows  buffer size = 100 packets  5 runs for each data point  Lasting for 100 seconds

16. - 16 - B : buffer size of the base station ω: TCP receiver window size α: ACK packet / Data packet window size between and Average window size A NALYSIS : 1 U P & 1 D OWN (?)

17. - 17 - A NALYSIS : 1 U P & 1 D OWN ρ: arrival rate / service rate p : drop rate [6] [7] 6 < B < 42 42 < B [7]

18. - 18 - A NALYSIS vs. S IMULATION

19. - 19 - A NALYSIS : M ULTIPLE F LOWS n2n2

20. - 20 - S OLUTION Modify the receiver window field of the ACK packets flowing through the base station.  The 16-bit receive window field is used for flow control.  n flows, buffer = B → receiver window = B/n »Assume 1 upstream, n-1 downstream upstream = B/n,and every downstream = B/n »Assume m upstream, n-m downstream every upstream = B/n,and every downstream = B/n

21. - 21 - T CP S EGMENT S TRUCTURE source port # dest port # 32 bits application data (variable length) sequence number acknowledgement number Receive window Urg data pnter checksum F SR PAU head len not used Options (variable length) URG: urgent data (generally not used) ACK: ACK # valid PSH: push data now (generally not used) RST, SYN, FIN: connection estab (setup, teardown commands) # bytes rcvr willing to accept counting by bytes of data (not segments!) Internet checksum (as in UDP)

22. - 22 - S IMULATION F OR S OLUTION  Throughput ratio of upstream and downstream = 1 ＝＞ resulting in fair allocation of bandwidth  Without the solution,the ratio up to 800. (P.15 Fig.7)  Set receiver window = 100/n  Buffer size = 100 packets  5 simulation runs for each n  Each simulating 100 seconds

23. - 23 - E XPERIMENT F OR S OLUTION Receiver windowRatio of up/downStandard deviation 65000 bytes ( default ) 7.94.57 2000 bytes ( modified by solution ) 1.0070.0005  2 upstream flows  2 downstream flows  MTU = 500 bytes  450 / 1ms UDP background

24. - 24 - C ONCLUSION & D ISCUSSION CONCLUSION  The buffer size at the base station plays a key role.  Modifying the receiver window size can provide fair TCP throughput for any buffer size or number of flows. The other ways of research  Channel losses  TCP flows with different RTT  Interaction with IPSec

25. - 25 - R EFERENCES