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Is it really exponential?

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  • Raghu Mallena
    I was looking at some traces and noticed that the Slow Start algorithm opens the window size by *one* segment size everytime an ACK is received. This is how
    Message 1 of 1 , Oct 29, 1998
      I was looking at some traces and noticed that the Slow Start
      algorithm opens the window size by *one* segment size everytime
      an ACK is received. This is how it was described in Van Jacobson's
      "Congestion Avoidance and Control" SIGCOM'98 paper. Is there
      any reason why it is designed this way as opposed to opening
      the window by the amount of data that got acked? With Delayed
      Ack in effect, the one-segment growth scheme would increase the
      window in roughly 2^(x/2) way as opposed to a true 2^x growth.
      Basically, RTT*logW (where W is window size in segments) being
      the time to reach steady state as computed in VJ's paper will
      not be true for Delayed Ack. And this will be even worse
      for the stacks that are lazy about sending Acks i.e send one
      Ack for more than 2 packets (please... i dont mean to re-start
      last weeks discussion on this subject)

      However, a simple one-line fix to open the window by the amount of
      data acked (ack - snd_una) should work well always. The window would
      still open exponentially and the time to reach full window size would
      still be in the order of RTT*logW, approximately.

      Any reason why it shouldnt be done this way?

      - raghu

      ---
      Raghu Mallena
      Silicon Graphics
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