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

Solution:

Given,

=>EstimatedRTT1 = 36.3 ms

=>Host sends 3 packets with the following SampleRTT's

SampleRTT1 = 49.5 ms

SampleRTT2 = 15.8 ms

SampleRTT3 = 27.1 ms

=>Weight($\alpha$) = 0.4

Explanation:

=>We know that, new EstimatedRTT = (1-$\alpha$)*Previous EstimatedRTT + $\alpha$ *SampleRTT

Calculating EstimatedRTT for packet 2:

=>EstimatedRTT2 = (1-$\alpha$)*EstimatedRTT1 + $\alpha$ *SampleRTT1

=>EstimatedRTT2 = (1-0.4)*36.3 ms + 0.4*49.5 ms

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

=>EstimatedRTT2 =21.78 ms + 19.8 ms

=>EstimatedRTT2 = 41.58 ms

Calculating EstimatedRTT for packet 3:

=>EstimatedRTT3 = (1-$\alpha$)*EstimatedRTT2 + $\alpha$ *SampleRTT2

=>EstimatedRTT3 = (1-0.4)*41.58 ms + 0.4*15.8 ms

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

=>EstimatedRTT3 = 24.948 ms + 6.32 ms

=>EstimatedRTT3 = 31.268 ms

Calculating EstimatedRTT for packet 4:

=>EstimatedRTT4 = (1-$\alpha$)*EstimatedRTT3 + $\alpha$ *SampleRTT3

=>EstimatedRTT4 = (1-0.6)*31.268 ms + 0.4*27.1 ms

=>EstimatedRTT4 = 0.4*31.268 ms + 0.4*27.1 ms

=>EstimatedRTT4 = 12.5072 ms + 10.84 ms

=>EstimatedRTT4 = 23.3472 ms

Rounding off the EstimatedRTT for packet 4 to one decimal place:

=>EstimatedRTT4 = 23.3 ms

=>Hence answer = 23.3

I have explained each and every part with the help of statements attached to it.