Re: resending: slow start triggered if recv window full on packet loss?

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Hi Peter

I think I didn't explain properly

my question is not about what it is slow-start but regarding where (if it) is specified
that after filling the receive window, slow-start process has to start when packet loss has occurred (?)

In the graph I sent as a link, fast-retransmit occurs, but afterwards I don't see fast-recovery but slow-start instead.

am I right?

Thanks

On Sun, Nov 9, 2008 at 3:30 PM, Peter Teoh <htmldeveloper@xxxxxxxxx> wrote:
this is from the RFC

http://www.faqs.org/rfcs/rfc2001.html

 (noticed how the sender/receiver window control the flow in different ways):

  The algorithm to avoid this is called slow start.  It operates by
  observing that the rate at which new packets should be injected into
  the network is the rate at which the acknowledgments are returned by
  the other end.

  Slow start adds another window to the sender's TCP:  the congestion
  window, called "cwnd".  When a new connection is established with a
  host on another network, the congestion window is initialized to one
  segment (i.e., the segment size announced by the other end, or the
  default, typically 536 or 512).  Each time an ACK is received, the
  congestion window is increased by one segment.  The sender can
  transmit up to the minimum of the congestion window and the
  advertised window.  The congestion window is flow control imposed by
  the sender, while the advertised window is flow control imposed by
  the receiver.  The former is based on the sender's assessment of
  perceived network congestion; the latter is related to the amount of
  available buffer space at the receiver for this connection.

  The sender starts by transmitting one segment and waiting for its
  ACK.  When that ACK is received, the congestion window is incremented
  from one to two, and two segments can be sent.  When each of those
  two segments is acknowledged, the congestion window is increased to
  four.  This provides an exponential growth, although it is not
  exactly exponential because the receiver may delay its ACKs,
  typically sending one ACK for every two segments that it receives.


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