Sliding Window protocol.

Question

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)?

Answer 1

$BW =  512∗10^3$

$V = 10^8m/s$

$D = 45000 * 10^3m$ 

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

For 2a : https://gateoverflow.in/107261/optimal-window-size

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

              https://gateoverflow.in/144462/cn-optimal-windows-size