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9月14日

Back to Research: Network Coding and a Small Riddle for You

Last year, I was investigating the scaling laws of ad hoc wireless networks, that is how different parameters, mainly throughput and end-to-end delay scale (asymptotically) with the number of nodes in the network. I've already written a very brief description of the mathematical model, the problem statement and the design challenges here.

Moving on with ad hoc wireless networks, I am currently investigating, along with another student and two faculty professors at UT, how network coding can improve (if it does) the throughput (or decrease delay) of a given ad hoc wireless network. Our research is supported by Cisco. If you haven't heard of the term "network coding before", here's a very simplistic example to illustrate the concept. Say you have two wireless nodes A and B and a wireless router R as shown in the configuration below:


image


Let's assume that A and B each has a package to transmit to the other, and that each transmission must happen through the router R, i.e. there's no direct channel between A and B. The trivial solution would require 4 transmissions:

  1. A sends a message to R.
  2. R forwards the message to B.
  3. B sends a message to R.
  4. R forwards it to A.

But with network coding, one can achieve the same result in 3 transmissions. Can you find a way to do so? I'll give you some time to think about it and will post the solution later as a comment. Feel free to post your guesses as comments below. [Hint: The solution is simple.]

Although the example above demonstrates the key idea of network coding, the setting itself is very unrealistic. In an actual ad hoc wireless network, we have much more than two nodes and one router. As interference comes into play, it's not obvious how to extend the solution that works for few nodes efficiently to one that works for hundreds or thousands of nodes. Moreover, the network topology can be very complicated (in the above example, it's a simple linear chain) which significantly increases the complexity of the problem. Finally, in an ad hoc network, there are no routers but rather other nodes that act as routers. This tremendously increase the number of possible routing schemes as the number of nodes in the network increases. Depending on the network topology, optimizing over all possible routing schemes can be very challenging.

- Rayyan

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Rayyan撰寫:
Heyyy! German! Are you talking about Windows Live Ad Hoc Networks here?! :P~
10 月 4 日
MunozGerma​n撰寫:
I totally missed this discussion, but I got wind of it thanks to some "new" stuff coming along. Expect surprises soon! ;-)
10 月 2 日
Rayyan撰寫:
I'm always glad to discuss my work (and thoughts in general) with others. It's great you enjoyed thinking all the way about network coding and figured out the solution. You already pointed out a key point along the way: the importance of modeling in analysis. Although the problem sounded very wierd to you at first, when you thought it's a trick question, and later thought it was a "specialized setup", it is nothing more than defining the communication model in a wireless setting. The broadcast model I assumed above is more or less standard in the wireless setting, but I still agree that that modeling is often a challenging task indeed and has a significant impact on the results. The challenge lies in the fact that a very complciated model is very hard to analyze analyitcally and an overly simplified model is often unrealistic and thus does not yield useful results that can be put to practice. Thanks for your interest in my research!
9 月 23 日
It was fun to figure it out.  Thanks for your help.  As for the thousand-node problem . . . you're the grad student, not I!
9 月 21 日
Rayyan撰寫:
Yey, we've got a winner :) Let's now solve a "similar" problem for a thousand nodes! :)
9 月 20 日
I think I got it now.  If we assume this specialized setup then I think that the following will work:

1) R receives the messages from both A and B
2) R XORs the messages together and  re-broadcasts.
3) Both A and B XOR the broadcast with their original message to obtain the other's message.
9 月 18 日
Rayyan撰寫:

Oh.. You're totally right. The case where both nodes have the same message is definitely not interesting. As you pointed out, there is no need to communicate then. But even with distinct messages, there exists a solution so that the whole process finishes in exactly three transmissions and as pointed out already, the last transmission of which will be a broadcast. Here's another hint: the nodes can do some simple processing on the packets sent and received as long as:
1. The total bits sent in a single transmission, whether be it a broadcast or not, is always fixed (one packet).
2. The scheme is lossless, i.e., the nodes should be able to recover the original messages.

PS. A message in this context is assumed to be a "packet" of certain fixed size, i.e. some binary sequence of bits of fixed length such as {11010100111}. I hope this helps :)

9 月 17 日
Maybe that would help if both A and B had the same information to transmit to each other.  But in that case they could both hang on to their information in the first place.
9 月 17 日
Rayyan撰寫:
No, it's not meant to be a trick question.The solution is simple but not obvious. It exploits the fact that the channels are all wireless. In particular, assume that the router R can broadcast a message, that is the same message, to both A and B with a single transmission. Well, this assumption depends on the physical model of the transmissions for sure but it's not an unreasonable one in wireless networks.  In fact, it's very typical to model the propagation of a wireless signal in communication systems as being symmetric in all directions with its strength (amplitude) diminishing exponentially as it travels away from the source. That's one possible reason why the wireless nodes A and B above are not able to communicate directly. I hope the problem setup makes more sense now. Can you see how this can be helpful in reducing total number of transmissions?
9 月 16 日
Is this a trick question?  How could it be done in three unless A and B have some sort of a direct connection?
9 月 14 日

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