edited by
412 views
0 votes
0 votes
Host A is sending data to host B over a full duplex link. A and B are using the sliding window protocol for flow control. The sender and receiver window sizes are 5 packets each. Data packets (sent only from A to B) are all 1000 bytes long and the transmission time for such a packet is 50 μsec. Acknowledgement packets (sent only from B to A) are very small and require negligible transmission time. The propagation delay over the link is 200 μsec. What is the total time required in this communication?

a) 250 μsec

b) 200 μsec

c) 275 μsec

d) 450 μsec
edited by

1 Answer

0 votes
0 votes
It is based on the concept of pipelining.See there are five packets.Assume A,B,C,D,E.

A transmits, TT+ journey on link = 50μsec + 200μsec = 250μsec                                .  .................(1)

Now , during this journey of A more packet are introduced on the link.Like when A finished 50 on link,B is introduced on link.Similarly other packets are also introduced.

NOw ,When when A is reached to the end,E has finished landing on link.

So only E 's propagation delay is left out which is 200                                                      ..................(2)

So, overall (1)+(2)=450μsec

Related questions

1 votes
1 votes
1 answer
3
sripo asked Jan 5, 2019
568 views
I am unable to understand their explanation,can anyone explain it in a better way?