[Network Computing] Suppose that in TCP, the sender window is of size N, the bas

Question

[Network Computing]

Suppose that in TCP, the sender window is of size N, the base of the window is at sequence number x, and the sender has just sent a complete window’s worth of segments. Let RTT be the sender-to-receiver-to-sender round-trip time, and let MSS be the segment size.

1. Is it possible that there are ACK segments int eh reciever-to-sender channel for segments with sequence number lower than x? Justify your answer.

2. Assuming no loss, what is the throughput (in packets/sec) of the sender-toreceiver connection?

3. Suppose TCP is in its congestion avoidance phase. Assuming no loss, what is the window size after N segments are ACKed?

Explanation / Answer

CONCEPT:understanding Network Computing.

Answers:

1.No

Explanation:

Because with the size less than sequence number we are not able to pass segments.

2.Throughput of the sender to receiver is MSS/RTT

Explanation:

as there are no data loss over the communication so as per basic calculation it is set to MSS/RTT ,here RTT is round trip time between sender and receiver.

3 .It set to MSS-N of the current congestion window size.

Explanantion:

there are already N segments ACKed so we need to set the size to MSS-N segments.Here MSS is the total size of segment.

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