Hidden Terminal Decoding and Mesh Network Capacity Y. Richard Yang 2/5/2009

Recap: ZigZag Decoding Exploits 802.11’s behavior Senders use random jitters Collisions start with interference-free bits Retransmissions Same packets collide again Pa Pa ………… How do we leverage these interference free bits to decode the collisions? ∆1 Pb ∆2 Pb Interference-free Bits

ZigZag Technical Issues 1 1 ∆1 ∆2 2 Collision detection Chunk subtraction Backward ACK compatibility

Outline 802.11 hidden terminal decoding Overall idea Technical issues Collision detection

Collision Detection: How does the AP know it is a collision and where the second packet starts? Time ∆

Detecting Collisions and the Value of ∆ Time AP received signal Correlate Packets start with known preamble AP correlates known preamble with signal ∆ Correlation Time

Correlation

Matching Collision Given (Pa + Pb()) and (Pa’, Pb’(’)), how to determine that Pa = Pa’ and Pb = Pb’ Determine offset first Correlation of Pb() and Pb’(’) Pa P’a  Pb ’ P’b

Outline 802.11 hidden terminal decoding Overall idea Technical issues Collision detection Subtracting chunks

How Does the AP Subtract the Signal? 1 2 Channel’s attenuation or phase may change between collisions Can’t simply subtract a chunk across collisions signal in first collision signal in second collision Now that the AP knows where the second packet starts, it is going to take the interference-free chunk from at the start of the first collision and subtract it from the second. In practice, it is not that simple…………..

Re-modulate bits to get channel-free signal 1 2 Subtracting a Chunk Decode chunk into bits Removes effects of channel during first collision Re-modulate bits to get channel-free signal Apply effect of channel during second collision Correlation estimates channel despite interference (how?) The solution to this problem however is simple. …….. Now the AP can subtract the first chunk. The AP iterates this process to decode both the whole packet.

Subtracting a Chunk: Error Propagation 1 3 1 ∆1 ∆2 2 2 For example, if the channel was noisy it can cause a decoding error in chunk 1. Since we subtract chunks across collisions, errors in the first collision can propagate to the second collision For example, Now when we subtract, the error in chunk 1 can lead to a decoding mistake in chunk 2, which itself could propagate further.

Subtracting a Chunk: Error Propagation Temporal Diversity : A bit is unlikely to be affected by noise in both collisions ∆1 ∆2 The good new is that we have temporal diversity. Each bit appears twice on the channel. It is unlikely that the same bit is in error in both collisions. We can leverage this property by decoding each bit twice using two independent decodings. But how do we get these two independent decodings? I have already described to u … one way of decoding. Get two independent decodings

AP Decodes Backwards as well as Forwards ∆1 ∆2 2 2 3 1 1 Errors propagate differently in the two decodings Which decoded value should the AP pick? But in addition to this forward decoding method we can also decode backwards. For each bit, AP picks the decoding that has a higher PHY confidence

Outline Admin. and recap 802.11 hidden terminal decoding Overall idea Technical issues Collision detection Subtracting chunks ACK for backward compatibility

802.11 Acknowledgement Problem: Use as much synchronous acknowledgements as possible for backward compatibility Pa Pa Pb Pb

802.11 Ack Backward Compatibility

Outline Admin 802.11 hidden terminal decoding Mesh network capacity

Infrastructure Mode Wireless 802.11 by default operates in infrastructure mode. AP wired network AP: Access Point AP AP Problems of infrastructure mode wireless networks?

Mesh Networks infrastructure mode ad-hoc (mesh) mode AP wired network AP: Access Point AP AP ad-hoc (mesh) mode

Mesh Networks Advantages Potential problems fast and low-cost deployment no central point of failure no APs overhead for users who can reach each other Potential problems low overall capacity

Recap: Capacity of Mesh Networks The question we study: how much traffic can a mesh wireless network carry, assuming an oracle to avoid the potential overhead of distributed synchronization (MAC)? Why study capacity? learn the fundamental limits of mesh wireless networks separate the spatial reuse perspective and system design perspective gain insight for designing effective wireless protocols

Recap: Two Constraints transmission successful if there are no other transmitters within a distance (1+D)r of the receiver receiver sender r (1+D)r Interference constraint a single half-duplex transceiver at each node: either transmits or receives transmits to only one receiver receives from only one sender Radio interface constraint

Recap: Model Domain is a disk of unit area There are n nodes in the domain The transmission rate is W bits/sec

Capacity of Mesh Wireless Network Consider two types of networks arbitrary networks: place nodes optimally to derive overall upper bound random network: nodes are placed randomly

Outline Admin 802.11 hidden terminal decoding Mesh network capacity arbitrary networks: place nodes optimally to derive overall upper bound

Transmission Model: Bit-time Perspective Chop time into a total of WT bit-times in T seconds The transmission decision is made for each bit bit time 1 bit time 2 bit time t bit time WT 1 2 3 5 4 1 2 3 5 4

Transmission Model: End-to-end Perspective Assume the network sends a total of T end-to-end bits in T seconds Assume the b-th bit makes a total of h(b) hops from the sender to the receiver Let rbh denote the hop-length of the h-th hop of the b-th bit 1 2 T 2

Hop-Count Constraint 2 Since there are a total of WT bit-times, and during each bit-time there are at most n/2 simultaneous transmissions, we have

Area Constraint m k (1+)r’ r’ i j (1+)r r ½ r’ ½ r Consider two simultaneous transmissions at a bit-time i j (1+)r r ½ r’ ½ r Djm + r >= Dim >= (1+)r’ Djm + r’ >= Djk >= (1+)r  Djm >= (r+r’)  /2

Area Constraint For each transmission with distance r from sender to receiver, we draw a circle with radius ½  r These circles do not overlap

Area Constraint: Global Picture 2

Area Constraint: Global Picture 2 sum over all circles, since each circle has at least ¼ of its area in the unit disk,

Summary: Two Constraints Radio interface constraint Interference constraint a single half-duplex transceiver at each node transmission successful if there are no other transmitters within a distance (1+D)r of the receiver

Discussion: what does the result mean? Capacity Bound Note: Let L be the average (direct-line) distance for all T end-to-end bits. Discussion: what does the result mean?

Discussion L depends on traffic pattern (who needs to talk to whom) and positions of the end-to-end (application-level) senders and (application-level) receivers If end-to-end senders and receivers are spread out throughout the network, L is large per node capacity Otherwise, L will be small as network becomes denser

Achieving Capacity: Example n/2 senders and receivers: L=

Results: Arbitrary Networks Protocol Model

Outline Admin 802.11 hidden terminal decoding Mesh network capacity arbitrary networks: place nodes optimally to derive overall upper bound random networks: uniform distribution of nodes and senders/receivers

Uniform Random Networks Uniform distribution of n nodes n origin-destination (OD) pairs Each node chooses same power level P, and thus equal radius r(n) Equal throughput (n) bits/sec for all OD pairs

Random Networks: Required Bits Assume: average length of each OD pair is L Average number of hops: Total required bit transmissions per second to support l(n) :

Random Networks: Offered Bits Required bit transmissions per second: What is the maximum number of transmissions (of bits) in one second? space used per transmission (interference limited): at least ¼ p [Dr(n)/2]2 = pD2r2(n)/16 number of simultaneous transmissions at most (interference limited): total bits per second

Random Networks: Capacity Required ≤ offered 

Connectivity Constraint Need routes between origin-destination pairs - places a lower bound on transmit range r(n) A A D D Connected Not connected To maintain connectivity with a high probability, requires r(n) on the order:

Random Networks: Capacity Required ≤ offered 

Measurement Measured scaling law: throughput declines worse with n than theoretically predicted: 1/n1.68 Remaining story line mesh networks with wide-area traffic may have low scalability, and need techniques to increase capacity

Backup Slides

Uniform Random Networks