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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. $MN$

How is sequence number at receiver is distinct? Receiver is expecting one of frame with sequence number equal to the frame with sequence number sent by receiver. When sequence number of one of the frame matches to one which is expected by receiver then then receiver window slide further.

I am not able to understand this someone please explain
I have this doubt too.
This is from Kurose and Ross book

Direct

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.

please explain in detail I am not getting it if possible

According to the given question the communication between sender and receiver is unidirectional or bi directional?

If it is unidirectional then answer should be max (N,M) , as sender will send N frames and each frame should have unique sequence number and if it is bidirectional then it should be M+N as both will have their Frames In transition with unique sequence number.

Is my interpretation is correct? Please correct me if I am wrong!!

@vupadhayayx86, the main idea of sequence number comes from the need of keeping track of frames for ARQ. The provision of ARQ only comes in bidirectional connection.

In unidirectional connection(not practical) like Simplex Protocol etc., generally there is no Flow or Error control, so sequence number is not necessary.

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.

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For People who still have not understand.Just read it you will get it for sure.

https://stackoverflow.com/questions/28444055/go-back-n-window-size
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 .

$M+N$

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)

by

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yes musa you are correct.