文献阅读交流:COPE(Xors in the Air: Practical Wireless Network Coding)

COPE是无线网络inter-flows网络编码的开创性文章,本文列出我在读COPE论文的几个困惑,期待与您交流。

COPE论文

COPE论文如下:

S. Katti, H. Rahul, W. Hu, D. Katabi, M. Medard, and J. Crowcroft, “Xors in the Air: Practical Wireless Network Coding,” in ACM SIGCOMM, 2006.

困惑1:Coding+MAC gain为怎么达到2?

对于无线中继网络(假设为:A — R — B),coding gain是1.33很好理解。但Coding+MAC gain怎么能达到2,摘抄原文如下:

The mismatch between the traffic the router receives from the edge nodes and its MAC-allocated draining rate makes the router a bottleneck; half the packets transmitted by the edge nodes are dropped at the router’s queue. COPE allows the bottleneck router to XOR pairs of packets and drain them twice as fast, doubling the throughput of this network. Thus, the Coding+MAC gain of the Alice-and-Bob topology is 2.

因为A和B不能同时占用信道,否则R同时收到两个信号造成干扰,即隐藏终端问题,这样R从AB接收消息还是需要2个time slot,怎么Coding+MAC gain可以达到2?

以下是我想的两种可能:

  • A与B占用不同的信道,如wifi信道1,6,12中的2个,这样R就可以同时接收来自AB的消息,即用一个time slot。
  • R做物理层网络编码,R将从AB收到的两个信号做异域,再广播出去,这样交换AB的一个消息只需要2个time slot。

但这样的解释明显不是原文的意思。求指点。

困惑2:对于不同速率,性能提升怎么算?

以无线中继网络(假设为:A — R — B)为例,AB每单位时间发送数据包,得到的coding gain是1.33,提升的性能是25%。那么,如果AB以不同速率发送数据包,网络编码带来的性能提升怎么算?

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