Maximum window size can be = 1+2a
here: a = Tt / Tp
Tt = $\frac{Packet Size}{Bandwidth} = \frac{65*8 bits}{40*10^{6}bps}$
Tp = 30 msec = $30*10^{-3}$
So Maximum window size =$1 + \frac{2*30*10^{-3}*40*10^{6}}{65*8}$
= 4616.38
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