Consider the TCP procedure for estimating RTT. Suppose that = 0.1. Let SampleRTT
ID: 3678544 • Letter: C
Question
Consider the TCP procedure for estimating RTT. Suppose that = 0.1. Let SampleRTT 1 be the most recent sample RTT, let SampleRTT 2 be the next most recent sample RTT, and so on. (a) For a given TCP connection, suppose four acknowledgments have been returned with corresponding sample RTTs: SampleRTT 4 , SampleRTT 3 , SampleRTT 2 , and SampleRTT 1 . Express EstimatedRTT in terms of the four sample RTTs. (b) Generalize your formula for n sample RTTs. (c) For the formula in part (b) let n approach infinity. Comment on why this averaging procedure is called an exponential moving average.Explanation / Answer
Please provide me tym so that i ll try to do bit (c)
a)
Denote EstimatedRT T(n) for the estimate after the nth sample.
EstimatedRT T(1) = SampleRT T
EstimatedRT T(2) = x × SampleRT T1 + (1 x) × SampleRT T2
EstimatedRT T(3) = x × SampleRT T1 + (1 x) [x × SampleRT T2 + (1 x) × SampleRT T3]
= x × SampleRT T1 + (1 x)x × SampleRT T2 + (1 x)2 × SampleRT T3
EstimatedRT T(4) = x × SampleRT T1 + (1 x) × EstimatedRT T(3)
= x × SampleRT T1 + (1 x)x × SampleRT T2 + (1 x)2x × SampleRT T3+(1 x)3 × SampleRT T4
b)
EstimatedRT T(n) = xSIGMA(base)j=1 n 1(1 x)j × SampleRT Tj + (1 x) n × SampleRT Tbase n