CODEWITHSUNDEEP
network-programmingtcpcongestion-control
schnauzer
schnauzer

Reputation: 33

Why there is a TCP congestion window inflation during fast recovery?

TCP fast recovery algorithm is described as follows(from TCP illustrated vol. 1). What I can't understand is in step 1, why there's a CWD window inflation by three times the segment size?

  1. When the third duplicate ACK is received, set ssthresh to one-half the current congestion window, cwnd. Retransmit the missing segment. Set cwnd to ssthresh plus 3 times the segment size.
  2. Each time another duplicate ACK arrives, increment cwnd by the segment size and transmit a packet (if allowed by the new value of cwnd).
  3. When the next ACK arrives that acknowledges new data, set cwnd to ssthresh. This should be the ACK of the retransmission from step 1, one round-trip time after the retransmission. Additionally, this ACK should acknowledge all the intermediate segments sent between the lost packet and the receipt of the first duplicate ACK. This step is congestion avoidance, since we're slowing down to one-half the rate we were at when the packet was lost.

Upvotes: 2

Views: 877

Answers (1)

PaoloJ42
PaoloJ42

Reputation: 152

From [RFC 2001][1]

When the third duplicate ACK in a row is received, set ssthresh

to one-half the current congestion window, cwnd, but no less than two segments. Retransmit the missing segment. Set cwnd to ssthresh plus 3 times the segment size. This inflates the congestion window by the number of segments that have left the network and which the other end has cached

So, when you receive 3 duplicate ACKs in a row you cut cwnd to half and perform a fast retransmit, from now on you're trying to not just idle while waiting for the next new ACK (1 RTT at best). Once you enter fast recovery, you send new data with

cwnd= original cwnd + # of duplicate ACKs received

until either you receive the ACK you were waiting for or the timer for that ACK expires.

Basically, that "+3" takes account for those 3 acks received that made you enter fast recovery in the first place so that you transmit a number of new bytes equal to the lost bytes + the ones that got to the receiver but were discarded. [1]: https://www.rfc-editor.org/rfc/rfc2001

Upvotes: 2

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