(21 pts: 5 for a, 2 each for b-j) We are using TCP Reno. ssthresh is set to 16.

Question

(21 pts: 5 for a, 2 each for b-j) We are using TCP Reno. ssthresh is set to 16. Assume no errors 3. (5 pts) Starting from a window size of 1, calculate how many packets can be sent as the gwnd window size grows in each round. Subtract that from your remaining total to show how many packets remain to be sent in the last column: a. Round # Packets Sent Packets Remaining 2 4 b. If you had 30 packets to send instead of 31, would anything change? Assume a packet is determined to be lost when cwnd-32 and ssthresh-16. Will ssthresh be changed? To what value? c. d. If you are in slow start, will çwnd pass through each integer value? In other words, when we move in one round from gwnd-8 packets to the next round when cwnd-16, will cwnd jump to 16 or will it grow one by one? Why? When TCP sends an entire window's worth of many packets at a time, how many RTT's (round-trip times) is that considered to take? e.

Explanation / Answer

Solution:

a)

below is the table for how the cwnd will grow

b)

No, nothing will change

c)

·         So if the packet is lost in case of TCP Reno, then there will be a timeout and slow start will happen after this so the value of cwnd will again be set to 1 or 2.

·         then again we will start to increase the size of the cwnd, by cwnd = cwnd + 1, at each recieved ACKs.

·         In this case the value of ssthresh will be changed and it will become Flight size/2.

·         Flight size=> The size of the data which was sent the last time when timeout happended.

d)

the value of cwnd will be growing one by one, this is how TCP Reno works. to avoid congestion.

f)

then expected RTT is 32 unit of time.

I hope this helps if you find any problem. Please comment below. Don't forget to give a thumbs up if you liked it. :)

#Round Packets #packets sent packets remaining 1 1 1 30 2 2 3 28 3 3 6 25 4 4 10 21 5 5 15 16

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