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

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OUTLINE

Introduction

Experiment

Simulation

Mathematical Analysis

Solution

Conclusion

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INTRODUCTION

WLAN 盛行－ IEEE 802.11standard

Private Area ： homes and offices

Public Area ： airports, hotels, cafes,…

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WLAN MODEL

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

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UNFAIRNESS 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.

=

= =

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EXPERIMENT

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.

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TABLE OF EXPERIMENT

Ru： The average TCP uplink throughput

Rd ： The average TCP downlink throughput

SD ： Standard Deviation

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CONCLUSION FROM TABLE

Ru / Rd ＞ 1 ＝＞ upstream > downstream

Number of flows ↑ ＝＞ Ru / Rd ↑

Use background UDP traffic ＝＞ Ru / Rd ↑

MTU ↓ ＝＞ Ru / Rd ↑↑

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ILLUSTRATION OF EXPERIMENT

Upstream flow finished its upload and terminated

Packets lost

Congestion avoidance region : 9K – 18K

First 150 sec throughput is

very low

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SIMULATION

Factors impact the throughput ratio in a test-bedWireless link interfaceBase station buffer sizeImplementation details of the 802.11 MAC

layer……

Simulation study using the NS2 simulatorOne upstream and one downstream flowMultiple flows

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SIMULATION ： 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 < 6very noisy

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SIMULATION ： RTT

RTT increases monotonically with the base station buffer size without any significant rate changes

The RTT of downstream is alm

ost equal to the upstream’s.

One upstream and one

downstream flow

5 simulation runs Each simulating 100 seconds

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SIMULATION ： Data & ACK Loss

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

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SIMULATION ： 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 packets5 runs for each data pointLasting for 100 seconds

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SIMULATION ： 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 packets5 runs for each data pointLasting for 100 seconds

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B : buffer size of the base station

ω: TCP receiver window size

α: ACK packet / Data packet

window size between and

Average window size

ANALYSIS : 1 UP & 1 DOWN

B2

B

4

)(3 B

)(3

4

BupstreamTP

TPdownstreamR (?)

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ANALYSIS : 1 UP & 1 DOWNρ: arrival rate / service rate

p : drop rate

[6]

[7]

6 < B < 42

42 < B [7]

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ANALYSIS vs. SIMULATION

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ANALYSIS : MULTIPLE FLOWS

n2

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SOLUTION

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

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TCP SEGMENT STRUCTURE

source port # dest port #

32 bits

applicationdata

(variable length)

sequence number

acknowledgement numberReceive window

Urg data pnterchecksum

FSRPAUheadlen

notused

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 willingto accept

countingby bytes of data(not segments!)

Internetchecksum

(as in UDP)

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SIMULATION FOR SOLUTION

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 packets5 simulation runs for each nEach simulating 100 seconds

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EXPERIMENT FOR SOLUTION

Receiver window Ratio of up/down Standard deviation

65000 bytes

( default )7.9 4.57

2000 bytes( modified by solution )

1.007 0.0005

2 upstream flows 2 downstream flows

MTU = 500 bytes 450 / 1ms UDP background

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CONCLUSION & DISCUSSION

CONCLUSIONThe buffer size at the base station plays a key role.Modifying the receiver window size can provide fair T

CP throughput for any buffer size or number of flows.

The other ways of researchChannel lossesTCP flows with different RTT Interaction with IPSec

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REFERENCES