A High-Throughput Path Metric for Multi Hop Wireless Routing

Douglas S. J. De Couto, Daniel Aguayo, John Bicket, Robert Morris MIT Computer Science and Artificial Intelligence Laborotory Presented by Tray Cooper, Feb 27, 2007

Background

 The most commonly used metric is minimum hop-count. Why?

Throughput = 1 Throughput = 1/2 Throughput = 1/3  Why do we need a new metric?

Problems with Min Hop-Count:

 A min hop-count protocol weighs each link the same.    Links have 2 values: 0 or 1 Routing packets sent, but data lost Lower hop count ~ lower SINR due to distance  Same hop count links 

How does the protocol pick the best path?

Possible Solutions

 Resend Lost Packets  Bad, only leads to lower bandwidth  Use a minimum SINR threshold value  Could lead to node/network partitioning

ETX?

 Find paths with the fewest expected number of total transmissions  “predict” # of transmissions (including retransmissions)  Lowest # = highest throughput = best route

A test

 A test was setup to see how the minimum hop count metric REALLY works.

Note this was an experimental test, not a simulation.

 During the test each packet sent contained 193 bytes (134 of data)  A “best” route was determined by trying 10 different routes and seeing which was best.

Results of the Test

 2 Regions   Above 250 PPS: 1 hop links Below 250 PPS: Multihop links  Note the 0 values for 1/5 of the packets, even though a route exists

Other Link Problems

 Poor links  Asymmetric links

A New Metric?

 SINR threshold?

 Node partitioning  Multiply link ratios?

1 1 .95

.95

1 = 1 = .90

ETX: Minimize the Expected Transmission Count

Link throughput  1/ Link ETX Delivery Ratio 100% 50% Link ETX 1 2 Throughput 100% 50% 33% 3 33%

Calculating ETX

 Assuming 802.11 link-layer acknowledgments (ACKs) and retransmissions:  P(TX success) = P(Data success)  P(ACK success) measured fwd delivery ratio

r

fwd measured rev delivery ratio

r

rev  Link ETX = 1 / P(TX success) = 1 / [ P(Data success)  P(ACK success) ]   Link ETX  1 / (

r

fwd 

r

rev)

Why measure both ACK and Data Success?

Measuring Delivery Ratios

 Each node broadcasts small link probes (134 bytes), once per second  Nodes remember probes received over past 10 seconds   Reverse delivery ratios estimated as

r

rev  pkts received / pkts sent Forward delivery ratios obtained from neighbors (piggybacked on probes)

A little practice

Route ETX Throughput 1 100% 2 2 3 50% 50% 33% 5 20%

ETX Good (and Bad)

  The Good:   ETX predicts throughput for short routes (<3) ETX accounts for    asymmetric links lossy links long links The Bad:    ETX probes are susceptible to loads  hidden terminals, heavy congestion, etc.

ETX always uses the same size packets (134), this causes loss estimates for data packets to be low and ack packets to be high ETX does not take into account a variable bit rate

Test Specifics:

 Indoor network, 802.11b, ‘ad hoc’ mode  1 Mbps, 1 mW, small packets (134 bytes), RTS/CTS disabled  DSDV + modifications to respect metrics  Packets are routed using route table snapshot to avoid route instability under load.

 DSR + modifications to respect metrics

ETX and DSDV

DSDV+hop-count DSDV overhead DSDV+ETX ‘Best’

ETX and DSR

DSR+hop-count DSR+ETX ‘Best’

A Quick Summary

 ETX aims to increase throughput by determining how many times it will take to transmit a message.

 This seems to work well, especially over multi-hop links.

Discussion:

 ETX is traffic independent…Why?

  Oscillations Traffic could be included in cases of consistent traffic (access point networks?)  Size of packets?

 Probing Period?