Answer:
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.
