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+41 votes

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**Answer is b)**

In $\text{selective reject protocol},$ the maximum window size must be half

the Sequence number space $=\dfrac{2^n}{2}=2^{n-1}$.

For **Go-back n**, the maximum window size can be $2^n-1$.

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+9 votes

**NUMBER OF SEQUENCE NUMBERS AVAILABLE >= SENDER WINDOW SIZE+RECEIVER WINDOW SIZE.**

Thus with n bits 2^n sequence numbers are available.

**In SR protocol SENDER WINDOW SIZE = RECEIVER WINDOW SIZE..**

thus window size= 2^n/2= 2^n-1

0

@sushmita plz verify this

Ws+Wr=2*N for SR and Ws +Wr =N+1 for gobackN

2x=2^n so x=2^(n-1)

sender and receiver side have unique sequence no? why so?

+6 votes

Here n bit has been used for sequence no.. so total possible sequence no is = 2^n

As we know sender window size= receiver window size (Selective repeat / Reject protocol)..

So window size= (2^n)/2 => 2^(n-1)

As we know sender window size= receiver window size (Selective repeat / Reject protocol)..

So window size= (2^n)/2 => 2^(n-1)

0 votes

**Sender window size + Receiver window size <= Available Sequence No.**

i.e. Ws + Wr <= ASN .....(1)

In Selective Repeat/Reject Protocol, **Ws = Wr**

Say, Ws = Wr = N.....(2)

Also as in question it is given that n-bits have been used for frame sequence no.

Therefore, Available Sequence No. (ASN) = 2^n

Now from equation (1) and (2),

N + N <= 2^n

=>N <= 2^n-1

Therefore, **The maximum window size for data transmission using the selective reject protocol
with n-bit frame sequence numbers is 2^n-1**

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