Suppose a TCP connection has a window size of 8 segments and an RTT of 800ms. Th

Question

Suppose a TCP connection has a window size of 8 segments and an RTT of 800ms. The sender sends segments at a constant rate of one every 100ms, and the receiver sends ACKs back at the same rate without delay. A segment is lost, and the fast retransmit algorithm on the receipt of the third duplicate ACK detects the loss. At the point when the ACK of the retransmitted segment finally arrives, how much total time has the sender lost (compared to lossless transmission) if the sender waits for the ACK from the retransmitted lost packet before sliding the window forward again? Assume that the packet lost is not the first data packet of the connection

Explanation / Answer

Ans: 1100 ms.

Explanation:
Initially we wait 300 ms time to detect the APK of third duplicate.
As per given one full 800 ms RTT of links as sender waits for retransmitted segment.
For example, if pakcet P is sent at t=0 ,then assume P is lost.
For 7 packets = P+1 to P+7 will be sent at intervals of time t=100, 200, ..., 700.
Because of window size is 8,sender stops at t=800, and ACK for P has not recieved.
So that,ACK for P+1, P+2 and P+3 will be duplicate ACKs of P-1,so that they will arrive at t=900, 1000 and 1100 respectively.
Hence p is re-transmitted at t=1100 and ACK arrives at 1900.

Hence,time lost will be = 1900-800 = 1100

Explanation:
Initially we wait 300 ms time to detect the APK of third duplicate.
As per given one full 800 ms RTT of links as sender waits for retransmitted segment.
For example, if pakcet P is sent at t=0 ,then assume P is lost.
For 7 packets = P+1 to P+7 will be sent at intervals of time t=100, 200, ..., 700.
Because of window size is 8,sender stops at t=800, and ACK for P has not recieved.
So that,ACK for P+1, P+2 and P+3 will be duplicate ACKs of P-1,so that they will arrive at t=900, 1000 and 1100 respectively.
Hence p is re-transmitted at t=1100 and ACK arrives at 1900.

Hence,time lost will be = 1900-800 = 1100

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