understanding tcp fairness over wireless lan ieee infocom 2003 saar pilosof, ramachandran ramjee,...
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Understanding TCP fairness over Wireless LAN IEEE INFOCOM 2003 Saar Pilosof, Ramachandran Ramjee, Danny Raz, Yuval Shavitt, Prasun Sinha. Presented by Yixin Hua. Agenda. Introduction Problem Overview Simulation Study Modeling TCP Access Our Solution Related Work Conclusion & Discussion. - PowerPoint PPT PresentationTRANSCRIPT
Understanding TCP fairness over Wireless LAN
IEEE INFOCOM 2003Saar Pilosof, Ramachandran Ramjee, Danny Raz, Yuval Shavitt, Prasun
Sinha
Presented by Yixin Hua
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Agenda Introduction Problem Overview Simulation Study Modeling TCP Access Our Solution Related Work Conclusion & Discussion
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Introduction A typical 802.11 installation
CISCOSYSTEMS
802.11 Base Station
802.11 Mobile host
802.11 Mobile host
802.11 Mobile host
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IntroductionScenarios (Assumption)
All senders or receivers: Share bandwidth equally.
One sender and two receivers: Sender get half BW, and receivers share other half.
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Problem OverviewReal Experiment Setup
CISCOSYSTEMS
802.11 Base Station
802.11 Mobile host
802.11 Mobile host
802.11 Mobile host
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Problem OverviewReal Experiment Setup
Ru: Average TCP uplink throughput Rd: Average TCP downlink throughput Ru/Rd: Ratio MTU: Maximum Transmission Unit – Varied Background UDP: to reduce buffer
available to TCP flows – varied by packet size and arrival interval
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Problem OverviewResult
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Problem OverviewResult
In basic case, Ru/Rd = 1.44 Does commercial system give high
priority to downstream? Since most applications involve download rather than upload.
With UDP flows, ratio Ru/Rd increase With smaller MTU, ratio reaches 8
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Problem OverviewFurther investigation with sniffers
Upstream TCP window size reaches its maximum in all cases
Downstream TCP window size changes
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Problem OverviewFurther investigation with sniffers
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Problem OverviewConcerns
Wireless link interference Base station buffer size Implementation details of 802.11
MAC layer Difficult to vary and isolate
parameters, and trace their impacts
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Simulation StudyExperiment 1: 1 TCP sender and 1 TCP receiver
TCP receiver window size w = 42 MTU = 1500 Base station buffer size B = 6 ~ 85 packets Number of ACK per data packet = 1 Data packet size 1024 bytes 5 runs, each lasts 100 second Nodes don’t move
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Simulation Study Experiment 1: Observation
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Simulation Study Experiment 1: Observation
Region 1: over 84, ratio is 1 Region 2: 42 to 84, ratio decreases
from 10 to 1 Region 3: 6 to 42, ratio varies
between 9 and 12 Region 4: below 6, data points wide
spread (too noisy)
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Simulation Study Experiment 1: More observations
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Simulation Study Experiment 1: More observations
RTT increases monotonically with base station buffer size w/o significant rate changes
Data packet loss rate is always higher than ACK loss rate, not linear with base station buffer size
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Simulation Study Experiment 1: Further investigation
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Simulation Study Experiment 1: Further investigation
When buffer size is smaller than 42, sharing result is 1:10
When buffer size becomes larger, sharing ratio increases
When buffer size is larger than 84, Base Packet is equal to the difference between Down ACK and the Up Packet
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Simulation StudyExperiment 2: Multiple flows
Case 1: One upstream and multiple downstream flows
Case 2: Equal number of multiple upstream and downstream flows
Base station buffer size is 100 packets
5 runs, each lasts 100 seconds
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Simulation Study Experiment 2: Observation
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Simulation Study Experiment 2: Observation
Case 1: Ratio is linear All downstream flows share bandwidth equally Total throughput stay stable
Case 2: Average ratio goes up to 800, since upstream
flows’ ACK clutter base station buffer Upstream flows maintain maximum window size Downstream flows struggle with a window of 0-2
packets
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Modeling TCP Access Scenario 1: One upstream and one downstream flow
Base station buffer size: B TCP receiver window size: w All packet loss due to buffer
overflows at base station
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Modeling TCP Access Scenario 1: Analysis
Upstream flow window behavior When sender window is large, a loss of
an ACK has no effect on the window size due to TCP cumulative acknowledgement nature.
Sender window will reach w.
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Modeling TCP Access Scenario 1: Analysis
Downstream flow window behavior It changes depending on B and w, since loss of
data packet will cause sender half window size. If B ( + 1)w, all packets have room in base
station buffer, no drops. Assume BS buffer is full of w ACKs.(?) If B
(+1)w, B - w buffer available for downstream. Sender window will vary between (B - w)/2 and B - w, average window size is 3(B - w)/4. (Simplified)
Ratio: R = 4w/(3(B - w))
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Modeling TCP Access Scenario 1: Further Analysis
Using bounded size queuing system (M/M/1/K) Arrival rate Rd + Ru , = 1. The probability of K packets in a buffer in a stable
state, Pk = (1-) k/(1- k+1) (1) is ratio between arrival rate and service rate,
= (Rd + Ru)/ Ru= 1+R, where R = Rd/Ru (2) Drop rate approximate to p = (1+BR)/(B+1) (3) Using Rd= sqrt(3/(2p))/RTTd, and Ru=w/RTTu R = RTTu/RTTd*sqrt(3/(2w2p)) (4)
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Modeling TCP Access Scenario 1: Further Analysis
Using (3) and (4), We get (1+BR)/(B+1) = 3/(2w2R2) Finally
Using 1+B B and 1+BR BR, R = 1/10.56, it gives an approximation for region 6 to 42.
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Modeling TCP Access Scenario 1: Further Analysis
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Modeling TCP Access Scenario 1: Further Analysis
When B > w and only loss is due to buffer overflow, window size is composed from a fixed part B - w, and a part of interaction with acknowledgements in the BS buffer.
Effective average window size is sqrt(3/(2p)) + 3(B - w)/4
R=RTTu/(w*RTTd)*sqrt(3/(2p))+3(B-w)/4 (6) It gives an approximation to region 42 to 85. When w=42, it matches with Eq. 4
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Modeling TCP Access Scenario 1: Validation
TCP doesn’t provide a nice arrival behavior like M/M/1/K
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Modeling TCP Access Scenario 2: Small Buffer Upstream flow with small buffer size, using discrete
time Markov chain
State i represents a state where TCP window size is 2i On state i, go to state 0 if a timeout occurs with
probability p2i
Otherwise double the window size and move into state i+1 with probability 1- p2i
With ns2, the upstream flow always end up with maximum window size
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Modeling TCP Access Scenario 3: Multiple Flows
From eq. 1, = (Rd + nRu)/ Ru= 1+nR (7)
Drop rate, p = (1+nBR)/(B+1)(8)
R=sqrt(3/(2nw2p))+3(B-w)/(4nw)(9)
It fits figure 6 very well!
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Modeling TCP Access Scenario 3: Multiple Flows
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Our solution
1. Separately queue for TCP data and ACK packets at base station - Doesn’t wok
2. Fake duplicate ACK packets or discard data packets to force TCP to reduce the upstream window size – Waste BW
3. Using advertised receiver window field in the ACK packets towards TCP sender, BS manipulates the receiver window
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Our solution Solution 3
Keep a counter for numbering current TCP flow in the system
If n flows in system, BS set receiver window to B/n
? Web traffic, bursty flow, UDP ! An XCP way
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Our solutionA simulation for solution 3
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Related Work Lu et al. [2] first identified the problem under a UDP
model. They proposed a centralized scheduling algorithm performed at BS.
Nandagopal et al. [3] suggests a fairness model that identify the different node fairness and flow fairness.
Research [9] suggests employ BW reservation over MA channels to support QoS.
Sobrinho and Krishnakumar [10] suggests blackburst to find the the real-time sender with longest waiting time( and thus the highest priority).
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Related Work Deng and Chang [1] suggested to change the backoff
period according to a station priority. The lower the priority the higher is the maximum backoff period a station can draw.
Berry et al. [11] follower the line and use two distinct backoff periods for two priority classes.
Vaidya et al. [4] suggests a distributed algorithm that calculates the backoff period for the stations that resulted access to the channel will closely match the Self-Clocked Fair Queueing scheduling.
Ada and Castelluccia suggests three differential mechanism based on scaling of the congestion window, modifying the IFSs, and changing the maximum frame length.
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Conclusion & Discussion Buffer size at base station plays a key role in the
observed unfairness. Based on simulation, the unfairness in TCP throughput
ration could be as high as 800. Using bounded size queuing system (M/M/1/K), authors
explained TCP’s behavior and interaction with MAC layer. The analysis identified four regions of unfairness that
depend on the buffer availability at base station. Proposed solution using advertised window manipulated
by base station alleviates the problem in simulation and testbed.
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Conclusion & Discussion Open discussion:
Channel losses TCP with different RTT Providing higher share of the media to
the base station Interaction with IPSec