Question

Explain the solution with every single detail like 888888855555555 why there is 6 8 and 8 5. why did put some numbers add and some even. also graph it with details

3) (25 points) Consider TCP slow start. Suppose a new TCP connection is just starting up, i.e., sends one maximum size segment (MSS) in the first transmission round. Suppose the Slow Start Threshold is initially 8 MSSs. Suppose the connection has an infinite number of MSSs to send. Further, suppose that a triple-duplicate ACK occurs at the end of transmission round 6. No other triple-duplicate ACKs or time-outs occur. Tabulate or plot the congestion window (in units of MSSs) as a function of time (in units of transmission rounds) for TCP Tahoe and for TCP Reno from transmission round 1 up to and including transmission round 14. Include the Slow Start Threshold in your table or plot. Tahaz me lvona

Answer 1

Given that In the first transmission round = 1 mss threshold is initially 20 mss End of transmission round J, time out occurs No other triple - duplicate Acks or time out occurs Infinite number of Mss's to send 1^st transmission round = 1mss 2^nd transmission round = 2mss 3^rd transmission round = 3mss 4^th transmission round = 4mss 5^th transmission round = 8mss 6^th transmission round = 16mss respectively Threshold changes to half of the previous congestion window size. Threshold = 16mss/2 = 8mss \r\n 7^th transmission round = 2mss 8^th transmission round = 4mss 9^th transmission round = 8mss 8mss is obtained of threshold linear increase the congestion window for every transmission round 10^th transmission round = 9mss 11^th transmission round = 10mss 12^th transmission round = 11mss 13^th transmission round = 12mss 14^th transmission round = 13mss 15^th transmission round = 14mss 16^th transmission round = 15mss \r\n plotting the congestion window: - \r\n There is no difference between