# 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 $$8$$.