GATE CSE 2005 | Question: 25

Question

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

• A. $2^n$
• B. $2^{n-1}$
• C. $2^n-1$
• D. $2^{n-2}$

Comments

Selective Repeat is sometimes calles as Select Reject.

Source : Forouzan, 4^th Edition.

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Are selective repeat and selective reject the same thing?

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GO to the Root of Concept: Video Explanation with timestamp Efficiency,Throughput| Relation Between WindowSize & Sequence No. Sliding Window | With NOTES

Answer 1

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 2

In selective repeat or selective reject Sender window size = receiver window size = x (let)

and we know that

available sequence number $\leq$  sender window size + receiver window size            ${\color{Red} \rightarrow }$ 1

for n bits sequence number available sequence number = $2^{n}$

put in equation 1

 $2^{n}$ = x + x              (since sender window size= receiver window size = x (let))

$2^{n}$ = 2x

then,

x= $2^{n}$ /2

x= $2^{n-1}$

therefore maximum window size is  $2^{n-1}$

therefore answer is B

Answer 3

In Selective Reject (or Selective Repeat), maximum size of window must be half of the maximum

sequence number = 2$^n$/ 2 = 2$^n$-1
For Go-back N, the maximum window size can be 2$^n$ - 1.

Answer 4

Because for selective repeat reciver also need window.

So total window is divided by two

Here total window is 2^n so sender window is 2^n/2