Question

Compare GBN, SR, and TCP (no delayed ACK). Assume that the timeout values for all three protocols are sufficiently long such that 4 consecutive data segments with sequence number 1, 2, 3, 4 can be received by the receiving host (Host B) and their corresponding ACKs can be received by the sending host (Host A). Suppose Host A sends data segments to Host B, and the 2nd segment (sent from Host A) is lost. In the end, all 4 data segments have been correctly received by Host B.

Answer 1

1.

First send all 4 segment.

1 2 3 4

When 2nd segment is lost, resend all segment starting from 2.

So again send 2 3 4.

So total # of segment sent = 7.

Total # of ACK received by Host B will be :

2 2 2 2 5 5 5  

Acknowledgement number always specify the next expected sequence number. So when seq number 2 packet is lost, it will repeatedly sent Acknowledgement packet with ACK=2 and then 5.

So total # of ACKs send by B is 7.

2.

The packets will be send in order :

1 2 3 4 2

Only send that packet which was lost.

The acknowledgement received will be ;

2 2 2 2 5

# of packet sent = # of ACKs received = 5

3.

The packet will be sent in groups. Like send first 1 packet, then 2 packet then 4 packet and so on since TCP follow AIMD

So the packets will be sent like

1 ( 2 , 3) ( 4) ( 2)

Now when 2nd packet is lost, the window size will be dropped to one and it will send only one packet.

The acknowledgement received will be in order :

2 ( 2 , 2) ( 2) ( 5)

# of segment sent = 5

# of Acknowledgement received = 5

Even in TCP, ACKNOWLEDGEMENT is sent for each packet received. Thus when 2nd packet is lost, it will continuously send Acknowledgement# = 2 stating 2nd packet is lost.

When it successfully receives the lost segment, give acknowledgement= 5 stating all the in between packet have been successfully received.

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