Answer:
see explaination
Explanation:
For GBN:
A sends will 9 segments in total. These are initially sent segments 1, 2, 3, 4, 5 and later re-sent segments 2, 3, 4, and 5.
B will send 8 ACKs. They are 4 ACKS with sequence number 1, and 4 ACKS withsequence numbers 2, 3, 4, and 5.
For SR:
A sends 6 segments in total. They are initially sent segments 1, 2, 3, 4, 5 and later re-sent segments 2. B sends 5 ACKs. They are 4 ACKS with sequence number 1, 3, 4, 5. And there is oneACK with sequence number 2.
For TCP:
A sends 6 segments in total. They are initially sent segments 1, 2, 3, 4, 5 and later re-sent segments 2. B sends 5 ACKs. They are 4 ACKS with sequence number 2. There is one ACK withsequence numbers 6.
Let it be observed that TCP always send an ACK with expected sequencenumber.
b). TCP. This is because TCP uses fast retransmit without waiting until time out.
