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There are two stations A and B connected by a $512 * 10^3$ bps network using a sliding window protocol. The speed of the signal is $10^8$ meter/sec and the distance between two stations is $45,000$ km. If the packet size is $256$ B, then what will be the optimal window size(in packets)?

$BW = 512∗10^3$

$V = 10^8m/s$

$D = 45000 * 10^3m$

pkt size = 256B = 256 * 8 b = 2048b$Full duplex channel, so total data send from x to y is = (2* Tp) * Band Width =$2 * D/V * BW$=$2 * 450*512 b\$

Number of window require  =  Total Send Data / Pkt size

=  ( 2*450 * 512 )/ (256 * 8 )

=  225

so optimal window size is N = 1+2a = 1+225 = 226

Similar :: https://gateoverflow.in/1820/gate2006-44