Ssume 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 (B) and the sending host (A) respectively.
Suppose host A sends 5 data segments to host B, and the second 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 number? Answer this question for all three protocols.
(b) If the timeout values for all three protocols are much longer than 5 RTT, then which protocol
successfully delivers all five data segments in the shortest time interval?

The explanation of the question is described in the section below.

Explanation:

(a)

Go Back N :

A gives in a maximum of 9 pieces. Typically they will be sent back sections 1, 2, 3, 4, 5 and then re-sent segments 2, 3, 4 and 5.

B sends out 8 ACK's. They are 4 ACKS including 1 series and 4 ACKS with 2, 3, 4 and 5 series amounts.

Selective Repeat :

A sends in such max of 6 bits. Subsequently, segments 1, 2, 3, 4, 5 and earlier re-sent Segments 2 will be sent.

B assigns five ACKs. We are 4our ACKS with numbers 1, 3, 4, 5. However there is one sequence quantity 2 ACK.

TCP :

A assigns in a total of 6 bits. Originally, segments 1, 2, 3, 4, 5 and future re-sent Segments 2 have always been sent.

B sends five ACKs. There are 4 ACKS with either the number 2 series. There has been one ACK with a sequence of numbers 6. Remember that TCP always needs to send an ACK with a sequence number you anticipate.

(b)

This is although TCP utilizes convenient retransmission without searching for the time out.

So, it's the right answer.

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no, it just either gets automatically 5 points to give away, or u can choose how many points u want to give but, the answerer only gets half