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+3 votes

Best answer

+1 vote

In SR protocol size of sender window and reciever window is same.

let their size be x

so sequence number >= sender window size + reciever window size

2^{m }>= x + x

2^{m} >= 2x

x <= 2^{m-1}

hence proved

0 votes

In case of selective repeat protocol the window size may be calculated as follows. Assume that the size of both the sender's and the receiver's window is w. So initially both of them contain the values 0 to (w-1). Consider that sender's data link layer transmits all the w frames, the receiver's data link layer receives them correctly and sends acknowledgements for each of them. However, all the acknowledgemnets are lost and the sender does not advance it's window. The receiver window at this point contains the values w to (2w-1). To avoid overlap when the sender's data link layer retransmits, we must have the sum of these two windows less than sequence number space. Hence, we get the condition

$Max Window Size = Sequence Number Space/2$

$Max Window Size = Sequence Number Space/2$

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