Question

A host starts a TCP transmission with an EstimatedRTT of 36.3ms (from the “handshake”). The host then sends 3 packets and records the RTT for each:

SampleRTT1 = 49.5 ms

SampleRTT2 = 15.8 ms

SampleRTT3 = 27.1 ms

(NOTE: SampleRTT1 is the “oldest”; SampleRTT3 is the most recent.)

Using an exponential weighted moving average with a weight of 0.4 given to the most recent sample, what is the EstimatedRTT for packet #4? Give answer in miliseconds, rounded to one decimal place, without units, so for an answer of 0.01146 seconds, you would enter "11.5" without the quotes.

Answer 1

Given weight = w = 0.4

For Packet #1

Initial EstimatedRTT1 = 36.3 ms

Time out value = 36.3 * 2 = 72.6 ms

SampleRTT1 = 49.5 ms (Less than timeout value)

For Packet #2

EstimatedRTT2 = 0.4 * SampleRTT1 + (1-0.4) * EstimatedRTT1

EstimatedRTT2 = 0.4 * 49.5 + 0.6 * 36.3 = 41.58 ms

Time out value = 41.58 * 2 = 83.16 ms

SampleRTT2 = 15.8 ms (Less than timeout value)

For Packet #3

EstimatedRTT3 = 0.4 * SampleRTT2 + (1-0.4) * EstimatedRTT2

EstimatedRTT3 = 0.4 * 15.8 + 0.6 * 41.58 = 31.27 ms

Time out value = 31.27 * 2 = 62.54 ms

SampleRTT3 = 27.1 ms (Less than timeout value)

For Packet #4

EstimatedRTT4 = 0.4 * SampleRTT3 + (1-0.4) * EstimatedRTT3

EstimatedRTT4 = 0.4 * 27.1 + 0.6 * 31.27 = "29.6 ms"