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

2. Recap: ZigZag Decoding
Exploits ’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

3. ZigZag Technical Issues1 1 ∆1 ∆2 2
Collision detection
Chunk subtraction
Backward ACK compatibility

4. Outline 802.11 hidden terminal decoding Overall idea Technical issues
Collision detection

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

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

7. Correlation

8. 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

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

10. 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…………..

11. 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.

12. Subtracting a Chunk: Error Propagation1 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.

13. 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

14. 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

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

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

17. 802.11 Ack Backward Compatibility

18. Outline
Admin
hidden terminal decoding
Mesh network capacity

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

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

21. 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

22. 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

23. 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

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

25. 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

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

27. 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

28. 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

29. 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

30. 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

31. 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

32. Area Constraint: Global Picture2

33. Area Constraint: Global Picture2
sum over all circles, since each circle has at least ¼ of its area in the unit disk,

34. 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

35. 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?

36. 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

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

38. Results: Arbitrary Networks
Protocol Model

39. 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

40. 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

41. 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) :

42. 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

43. Random Networks: Capacity
Required ≤ offered 

44. 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:

45. Random Networks: Capacity
Required ≤ offered 

46. 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

47. Backup Slides

48. Uniform Random Networks