Question

Assume a TCP sender is continuously sending 1162-byte segment. If a TCP receiver advertises a window size of 7873 bytes, and with a link transmission rate 23 Mbps an end-to-end propagation delay of 32 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

Answer 1

Answer & Explanation:

Given:

Propagation delay =
`32ms`

Segment size =
`1162bytes`

Link transmission rate =
`23Mbps`

Round trip time =
`2*`Propagation delay =
`2*32ms = 64ms`

Calculating the total segment:
`total\:segment = (Receiver\:window \:size)/(sender\:segment)=(7873)/(1162)=6.7753\approx 6`

`through put = (segment)/(Round-trip\:time)=(1162byte)/(64ms)=\frac{1162*8bits}{64*{10^(-3)}s}=145.25*{10^(3)}bps`

`utilization=(throughout)/(bandwidth)=(145.25* 10^(3)bps)/(23Mbps)=(145.25* 10^(3)bps)/(23* 10^(6) bps)=6.315* 10^(3-6)=6.315* 10^(-3)=0.0063`

Utilization in percentage is 0.63%

Over all Utilization =
`6* 0.63\%=3.78\%`