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

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 or archive

Comments

@rfzahid 😂

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@rfzahid Clicking on edit history can tell you why :) https://gateoverflow.in/revisions/5654

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Arjun sir 😂🤣

Answer 2

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

Comments

@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?

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Sender and Receiver should always have unique(different) sequence no. for prevention of acceptance of duplicate packets as a next window of packets.

Total available seq no. = 4

Lets say SWS=4 and RWS=4

Sender having sequence no.( 0,1,2,3)

Receiver seq no also (0,1,2,3)

Say after sending first windows, all the acks are lost. Now sender thinks all packets are lost(how it will know that acks are lost) and receiver thinks all acks are sent and received.

Now sender will again retransmit all the 4 packets thinking all were lost but receiver will accept them thinking that all are from next set of windows. So here duplicate packets are received.

Same is the case for any common seq. no. between Sender and Receiver.

Now lets take total Seq. no. =8

SWS = RWS = 4

First window is sent of (0,1,2,3) and all acks are send but not received by sender.

But now Receiver shifts its window to (4,5,6,7) after acknowledging the previous window.

Now if any duplicate packets comes it will discard( but will send acks to confirm sender)

Answer 3

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)

Comments

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

correct me if i am wrong anywhere

Answer 4

n bits in sequence no means 2^n sequence numbers are there
we know that in SR ,sender and reciever's window size are same...so divide these 2^n sequence no's equally..
so sender can have maximum 2^(n-1) sequence nos...so maximum possible window size is 2^(n-1)...

Comments

@sudsho i hav edoubt that 
in general SWS+RWS<=N

SWS: sender window size=2^n-1

RWS: receiver window size=2^n-1
so 2^n-1+ 2^n-1<=N
2ⁿ <=N

and my question is that it is not asking about the sender window size  but it asks about window size ?? correct me

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sequence nos are greater than or equal to sender window size+reciever window size

here N is sequence no bits...so sequence nos are 2^N..means 2^N >= sender+reciever...

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why is this so is important...to avoid duplicate packets with same sequence nos...if total u have 100 sequence nos than it depends on you what u wanna use..say u want to use all..then in SR u have to give 50 to sender and 50 to reciever...if u want to use only 80..no problem..give 40 to sender and 40 to reciever...
in GBN if u have 100 sequence nos..then u can give 99 to sender and 1 different to reciever,...so on

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question is saying maximum window size for data tranmission....if u have 100 sequence nos and u divided them 50-50 to sender and reciver..than at any time u can transfer atmost 50 bytes data only..so 2^(n-1)