What is the optimal window size when the one way delay is 30 msec, operates on............. #techtud

Question

What is the optimal window size when the one way delay is 30 msec, operates on 40 Mbps bottleneck bandwidth and packet size is 65 bytes?

Answer 1

Maximum window size can be = 1+2a

here: a = T_t / T_p

T_t = $\frac{Packet Size}{Bandwidth} = \frac{65*8 bits}{40*10^{6}bps}$

T_p = 30 msec = $30*10^{-3}$

So Maximum window size =$1 +  \frac{2*30*10^{-3}*40*10^{6}}{65*8}$

                                           = 4616.38