One difficulty with the original TCP SRTT estimator is the choice of an initial value. In the absence of any special knowledge of network conditions, the typical approach is to pick an arbitrary value, such as 3 s, and hope that this will converge quickly to an accurate value. If this estimate is too small, TCP will perform unnecessary retransmisions. If it is too large, TCP will wait a long time before retransmitting if the first segment is lost. Also, the convergence may be slow, as this problem indicates. (a) Choose f = 0.85 and SRTT(0) = 3 s, and assume all measured RTT values = 1 s and no packet loss. What is SRTT(19)?
Hint: Eq 22.3 can be rewritten to simplify the calculation, using the expression (1n)/(1). SRTT1K + 12 = a * SRTT1K2 + 11 - a2 * RTT1K + 12 (20.3) (b) Now let SRTT(0) = 1 s and assume measured RTT values = 3 s and no packet loss. What is SRTT(19)?