TCP-Fast Retransmit and Fast Recovery Technique

Question

Comments

is it 800ms. i think sender has to wait for one rtt before advancing window

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Is answer 1100ms? 800 for rtt and 300 for successive 3 duplicate acks.

Answer 1

We wait 300 ms initially to detect the third duplicate ACK, and then one full 800 ms RTT as the sender waits for the ACK of the retransmitted segment.

Suppose pakcet number P is sent at t=0 by the sender's clock. Suppose P is lost. 7 packets – P+1 to P+7 - will be sent at t=100, 200, ..., 700, since the window size is 8. Sender stops sending at t=800, since window size is 8 and ACK for P has not arrived. ACK for P+1, P+2 and P+3 will be duplicate ACKs of P-1, and will arrive at t=900, 1000 and 1100.

P is re-transmitted at t=1100. ACK arrives at 1900.

Hence time lost = 1900-800 = 1100.( in msec)

Comments

Thankyou Sir for correcting ,

receiver also generates acknowledge at rate of once in every 100 ms

initially if packet reached successfully it will take 800 or 900

bz receiver also take 100 ms to generate .