# Number of Retransmission in case of Go Back N

While Going through Exercise problem of " Computer Networking A Top-Down Approach" by Kurose and Ross, i encountered this problem.I am giving my approach and the point where i am stuck at.

## Question

Suppose Host A sends 5 data segments to Host B, and the 2nd segment (sent from A) is lost. In the end, all 5 data segments have been correctly received by Host B.How many segments has Host A sent in total and how many ACKs has Host B sent in total for Go Back N

GoBackN: $A$ sends $9$ segments in total. They are initially sent segments $1, 2, 3, 4, 5$and later resent segments $2, 3, 4,$and $5$. $B$ sends $8$ ACKs. They are $4$ ACKS with sequence number $1,$ and $4$ ACKS with sequence numbers $2, 3, 4,$ and $5.$

## My Approach /Doubt

I agree with the number of transmission of data segment by Host $A$.But i have doubt regarding the Acknowledgement by Host $B$.

### Why?

We know that GBN has sender side window size=$N$ while reciever side as only$1$ which is the reason that it cannot recieve Out of order packet.

Now When Host $A$ sends entire packet $1,2,3,4,5$ where $2^{nd}$ gets lost then Host $B$ which is expecting Sequence number $1$.On recieving sequence number $1$, it will send ack as $2$i.e expecting sequence number $2$.As Sequence number gets lost , host $B$ will recieve sequence $3,4,5$ for which it will discard the packet as it is not the packet it is expecting.

On recieving again sequence number $2,3,4,5$(retransmitted packet), Host $B$ will send the Acknowledgement.

Total Acknowledgement i am getting is $5$ i.e ACk no for sequence number $1,2,3,4,5$ not $8$.