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If m number of bits are used for representing a sequence number in selective-repeat ARQ,then the maximum size of sliding window is 2m-1 .Justify this statement using a boundary case example.What will happen if the size of sliding window exceeds 2m-1 ?

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If  same sequence no is used but window size is increased , then we fail to find duplication of data. so we have to increase sequence no also such that it satisfy

## sequence no >= sender window size + receiver window size

in order to detect duplication of data

by Active (1.2k points)
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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

2m  >= x + x

2m >= 2x

x <= 2m-1

hence proved

by Loyal (9.2k points)
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
$Max Window Size = Sequence Number Space/2$
by (63 points)