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A selective repeat ARQ  uses a window size of $26$ to send the data. The number of bits needed to define the sequence number is _____.
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IN selective repeat ARQ , sender window <= (2^m)/2  where m is no. of bits

so ,in this case it is given 26 <= 2^m / 2 which gives  .... m-1 = 5 , so m= 6 bits
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in selective arq --->  sender window size = receiver window size.

min seq no  >= window size + sender size

min seq no >=52. For 52 seq no, 6 bits are required.

Is this approach correct ?

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@Bikram

Sir, can you confirm why answer is 6 ? or how

26 <= 2^m / 2 which gives  .... m-1

gives you m-1 ?

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In Selective Repeat whatever the size of Sender's window is, same is the size of the Receiver's Window.

It is given that the Sender's Window size is 26 (26 numbers not bits)

=> Total numbers = 26 + 26 = 52.

 

To uniquely identify 52 numbers, we need 6 bits.

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