Chapter 3 P37. Compare GBN, SR, and TCP (no delayed ACK). Assume that the timeou

Question

Chapter 3
P37. Compare GBN, SR, and TCP (no delayed ACK). Assume that the timeout values for all
three protocols are sufficiently long such that 5 consecutive data segments and their
corresponding ACKs can be received (if not lost in the channel) by the receiving host (Host B)
and the sending host (Host A) respectively. 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.
a. How many segments has Host A sent in total and how many ACKs has Host B sent in
total? What are their sequence numbers? Answer this question for all three protocols.
b. If the timeout values for all three protocol are much longer than 5 RTT, then wh

Explanation / Answer

a)

Go-Back-N (GBN):

The Go-Back-N protocol retransmit all the segments from the lost segments.

The given lost segment = 2nd segment

Initially, the five segments with sequence numbers 1, 2, 3, 4, and 5 are transmitted to the Host B. The 2nd segment is lost during the transmission. In the Go-Back-N protocol, all segments from the 2nd segment shoud be retransmitted. The retransmitted segments sequence numbers are 2, 3, 4, and 5.

Therefore, the total number of segments transmitted from Host A to Host B = 9

The total number of ACKs sent by Host B = 8

The first 4 ACKs contains the Sequence number 1 and the remaining 4 ACKs contains the sequence numbers 2, 3, 4, and 5.

Here, The Acknowledgement number is the number of received segment.

Selective Repeat (SR):

In the Selective Repeat mechanism, there is no need to resend the N frames when a single frame is lost. Instead of resending the N frames, only the damaged frame is retransmitted.

Initially, all the five frames with sequence numbers 1, 2, 3, 4, and 5 should be transmitted. After the damage of frame 2, only frame 2 should be retransmitted.

Therefore, the total number of frames sent = 6

The total number of ACKs sent by Host B = 5

The first 4 ACKs contains sequence numbers 1, 3, 4, 5 and the remaining 1 ACK contains the sequence number 2

Here, The Acknowledgement number is the number of received segment.

TCP:

Initially, the segments with sequence numbers 1, 2, 3, 4, and 5 are sent to the Host B. After the damage of frame 2, only frame 2 is retransmitted.

Therefore, The total number of segments sent from Host A to Host B = 6 segments

The total number of ACKs sent from Host B = 5

The first 4 ACKs contains the sequence numbers 2 and the remaining 2 ACKs contains the sequence number 6.

Here, The Acknowledgement number is the expected segment number.

b)

TCP delivers all the five data segments in shortest time interval when the timeout values are much longer than 5 RTT. The other protocols GBN and SR retransmit the lost segment only after the time-out. But the TCP uses the fast retransmit mechanism to retransmit the lost segments. It does not wait for the timeout to retransmit the lost segments.

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