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GATE IT 2004 | Question: 81

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46 46 votes

In a sliding window $ARQ$ scheme, the transmitter's window size is $N$ and the receiver's window size is $M$. The minimum number of distinct sequence numbers required to ensure correct operation of the $ARQ$ scheme is

  1. $\min (M, N)$
  2. $\max (M, N)$
  3. $M + N$
  4. $MN$

6 Answers

Best answer
52 52 votes

C) M+N

Because $\large W_s+W_r \leq \text{Sequence numbers}$ (as the maximum number
of unacknowledged packets at sender will be $\large W_{s}$ and at the receiver it will be $\large W_{r},$
similar to the sequence numbering in Selective Repeat)

where $\large W_{s}$ is size of sender window and $\large W_{r}$ is receiver window size.

• edited by
13 13 votes
In general sliding window ARQ scheme , the sending process sends a number of frames without worrying about receiving an ACK(acknowledgement) packet from the receiver. The sending window size in general is N and receiver window is 1. This means it can transmit N frames to its peer before requiring an ACK. The receiver keeps track of the sequence number of the next frame it expects to receive and sends that number with ever ACK it sends. But in case of the question the sender window size is N and receiver is M so the receiver will accept M frames instead of 1 frame in general. Thus sending M sequence numbers attached with the acknowledgement.
Hence, for such a scheme to work properly we will need a total of M+ N distinct sequence numbers.
6 6 votes

Answer C

$M+N$

seq no.s > $W_s + W_r$

Sliding Window has two implementations

  • SR
  • GBN

In SR sender window size is equal to receiver window size

but in GBN sender window size is N and Receiver Window size is 1

we need minimum sequence number = sender window size + receiver window size

i.e. in SR sq no.s = 2N

and in GBN sq. no.s = N+1

if we use less seq.no.s  than these numbers then following problem can occur

if acknowledgments are lost. Due to Time out timer on receivers side Retransmission of those packets will be started but receiver has already received them so receiver may receive duplicate packets so extra seq. no.s are required to reject packets if they were accepted.(as we reuse seq.no.s)

2 2 votes
Always try to apply this :
  
    WS (recieving window)+ WS (sending window) <= ASN (available sequence numbers)  --------(1)

So , applying  (1)
 
M+N <= ASN

Hence minimum number of distinct  sequence numbers required is  , M+N .
1 1 vote
We all know that sender window is taken to be equal to reeciver window

or you can say number of sequence numbers are divided equally to both

So if sender is M

and receiver is N

to maintain the equilibrium, M+N is a constant
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