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19 votes
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The maximum window size for data transmission using the selective reject protocol with $n\text{-bit}$ frame sequence numbers is:

  1. $2^n$
  2. $2^{n-1}$
  3. $2^n-1$
  4. $2^{n-2}$
in Computer Networks
edited by
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4 Answers

44 votes
 
Best answer

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 $2^n-1$.

http://webmuseum.mi.fh-offenburg.de/index.php?view=exh&src=73


edited by
0
can anyone plzz explain this question
1
can i say reciever window size = senders windows size
4
Are selective repeat and selective reject same?
3
yes...
0
Link inactive
3
First time saw this girl actually explain her answers, else she just runs away after mentioning the option no.
12 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?

5 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)
1 vote

We know that,

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

Answer:

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