Performance Anomaly of 802.11b February 11, 2010 Paper By: Martin Heusse, Franck Rousseau, Gilles BergerSabbatel, Andrzej Duda Presentation By: Daniel Mitchell, Brian Shaw, Steven Shidlovsky CS4516 C10 1 Outline
802.11b Basics The Anomaly Simulation Verification Experimental Verification CS4516 C10 2 Abstract When one host on a IEEE 802.11b network is forced to transmit at less than the maximum bit rate of 11 Mbps then all other hosts are forced to also transmit at this lower rate CS4516 C10
3 Behind the Problem Access method Distributed Coordination Function (DCF) Uses CSMA/CA Low quality radio transmissions will result in a decrease in bit rate 5.5, 2, or 1 Mbps
Performance anomaly caused Privileges low speed hosts, penalize high speed hosts CS4516 C10 4 DFC Performance Overall transmission time T = ttr + tov Each packet has constant overhead time tov = DIFS + tpr + SIFS + tpr + tack DIFS = Time wait between senses of channel
SIFS = Period access point waits to send ACK tpr = PLPC Transmission time tack = MAC acknowledgement transmission time CS4516 C10 5 Throughput Efficiency Equation to determine useful throughput P = (Ttr/T) * (1500/1534)
Result: 70% useful throughput Thus 11 Mbps has 7.74 Mbps useful data CS4516 C10 6 Multiple Hosts Increases overall transmission time Decreases the proportion of useful throughput P(N) = ttr/T(N)
T(N) = Overall transmission time due to multiple hosts CS4516 C10 7 The Anomaly Since the slow hosts need more time to transmit the same data, all the hosts slow down to roughly the same speed The slow host holds the channel for a proportionately longer amount of time!
This anomaly occurs regardless of how many fast hosts are present Collisions and contention affect all hosts proportionally CS4516 C10 8 Why the Anomaly Exists
sd: Amount of data to be transmitted Time to transmit data = sd/(data rate) Over the long term, CSMA/CA provides each host with an equal probability of accessing the channel Therefore, all hosts will have the opportunity to transmit the same amount of data Fast hosts have a lower channel utilization CS4516 C10 9 Further Discussions
See Heuse, et. al. page 3 for the full mathematics. Contention periods and collisions are accounted for. UDP is expected to obey the mathematical models as generally no ACK packets are sent. TCP behaves as though there is 2 slow hosts, but can be shown mathematically to behave similarly. The second slow host is the ACK packets returning to it. TCP also incorporates congestion control, so a host may stop transmitting when its data rate is substantially below the 1Mbps minimum.
CS4516 C10 10 Verification A simulation was conducted to verify the mathematical results. Simulator is targeting a worst-case scenario: The channel is always busy the first time a node wants to transmit. All nodes configured to use 802.11b with exponential backoff Simulation showed the mathematics are
good, though not perfect The error: Other factors, besides the proportion of collisions, affect the average time spent in collisions CS4516 C10 11 Proportion of Collisions Figure 3 The mathematics assume a greater number of collisions than the simulation shows, particularly for very large numbers of hosts CS4516 C10
12 Throughput Figure 4 The performance anomaly is observed. Note that no configuration gets acceptable throughput for very large numbers of hosts. This triggers TCP congestion control algorithms and may force hosts to stop transmitting. CS4516 C10 13 Experimental Verification
Measure Throughput Four notebooks(Marie, Milos, Kea, Bali) RedHat 7.3, 802.11b cards Access Point is not the bottleneck CS4516 C10 14 Tool Used
Netperf:: generates TCP or UDP traffic and measures throughput Tcpperf:: generates TCP traffic and measures the throughput Udpperf:: generates UDP traffic and measures the throughput Measurements done with netperf, compared to results of tcpperf and udpperf CS4516 C10 15
Test 1: No Mobility All hosts near access point Force one to use degraded bit rate One test run with TCP, the other with UDP Using 2 hosts, 3 hosts, and 4 hosts, at bit rates 11, 5.5, 2, and 1 for Bali (slow host) For TCP, hosts are competing with the access point, which is sending TCP ACKs
on behalf of the destination For UDP, hosts compete with each other CS4516 C10 16 CS4516 C10 17 Test 1: Discussion
Measured values correspond well to analytical values (better for UDP) TCP traffic pattern more complex, due to Access Point competing with hosts (TCP ACKs), dependence on overall RTT and bottleneck link Pattern can become correlated with data segment traffic, since TCP ACK is sent upon arrival of 18 CS4516 C10 Test 2: Mobile Hosts
Bali (slow host) is a mobile host, bit rate automatically adapts to varying transmission conditions Other hosts located near access point with good conditions CS4516 C10 19 CS4516 C10 20 Test 2: Discussion
For TCP, when transmission conditions are bad (300-380) the throughput of Marie increases. This is due to Bali limiting its sending rate in adverse conditions Note, at 380, Bali stops sending completely even if its bit rate is not 0, and Marie gains almost all available throughput UDP shows similar results, although Maries gains during adverse conditions are not quite as large, unless Bali stops sending CS4516 C10
21 Related Work There have been many other papers studying 802.11 WLANs, but no prior papers use varying bit rates for hosts Most other papers use simulations, rather than analysis, which can give complex results Short-term unfairness of CSMAbased medium access protocols is also a topic of interest
CS4516 C10 22 Conclusions Throughput much lower than nominal bit rate Proportion of useful throughput depends strongly on number of hosts If a host degrades its bit rate due to bad transmission conditions, other hosts throughputs will drop roughly to the rate of
the slower host However, in real conditions using TCP, the slow host will be subject to packet loss, limiting its sending rate, allowing other hosts to take advantage of the unused capacity CS4516 C10 23
