Question

Assume a TCP sender is continuously sending 1,369-byte segment. If a TCP receiver advertises a window size of 7,001 bytes, and with a link transmission rate 34 Mbps an end-to-end propagation delay of 30.7 ms, what is the utilization? Assume no errors, no processing or queueing delay, and ACKs transmit instantly. Also assume the sender will not transmit a non-full segment. Give answer in percentages, rounded to one decimal place, without units (e.g. for an answer of 10.43% you would enter "10.4" without the quotes).

Answer 1

Sol) Given data Transmission rate = 34 Mbps Window size 7001 bytes size = 1369. bytes Segment The propagation delay a 30.7 ms Window size Total segments = segment size 7001 - 5.113 ~ 5 = 1369 = 5 Round trip time = 2 x propagation delay 2 X 30.7 = 61.4 ms Throughput Segment size Round trip time 1369x8 3 S = ims = 10 -3 61.4 x 10 Ibyte = 8 bits - 178,37X 100 bps

Through put Utilization = Transmission rate 178.37x103 [: ! IMbps =lobps 6 34 X10 = 0.005 The percentage value of utilization = 0.005 X 100 = 0.5% The overall utilization for total segments (5) = 5 x 0.5 = 2.5%, .. utilization needed = 2.5%