Assume a TCP sender is continuously sending 1,118-byte segment. If a TCP receiver advertises a window size of 6,046 bytes, and with a link transmission rate 40 Mbps an end-to-end propagation delay of 32.3 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: 1.7%

Explanation:

As TCP is running in a sliding window mode, it can't send more segments in sequence than the window size, so the packet length, will be equal to the integer number of full segments within the window, times the segment length, as follows:

Number of Segments = (Window Size / Segment Size) = 6,046 / 1,118 = 5.4

This means that, as a máximum, the sender will be able to send 5 segments in sequence, so the packet length will be as follows.

Packet Length = 1,118 bytes * 5 = 5,590 bytes.

If we know that the end-to-end delay is of 32.3 mseg, the RTT will be exactly the double, i. e., RTT = 64.6 mseg.

The actual bit rate, taking into account the packet length, and the RTT, will be as follows:

Tb = 5,590 bytes. 8 bits/byte / 64.6. msec = 692.3 Kb/s

The link utilization, will be the proportion between this effective rate, and the link transmission rate, as follows:

η = 692.3 Kb/s / 40,960 Kb/s = 0.0169 = 1.7%