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+13 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. $2^{n-2}$
asked in Computer Networks by Veteran (59.6k points)
edited by | 2.4k views

4 Answers

+34 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

answered by Loyal (5.3k points)
edited by
0
can anyone plzz explain this question
+1
can i say reciever window size = senders windows size
0
Are selective repeat and selective reject same?
+2
yes...
0
Link inactive
+7 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

answered by Boss (15k points)
+4 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)
answered by Active (1.3k points)
0 votes

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

answered by (219 points)
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