380 views
2 votes
2 votes

1 Answer

Best answer
5 votes
5 votes

Suppose first packet is sent at t=0 and lost...next 7 packets will be sent at t=100, 200, ...700(window=8)
now stop at t=800 (window size is 8 and ACK for first packet has not arrived)
ACK for 2nd,3rd,4th packet will be duplicate ACKs, and will arrive at t=900, 1000 and 1100
now,after recieving all 3 duplicate acknowledgements first packet is again re-transmitted at t=1100
ACK will arrive at 1900
Time lost = 1900-800 = 1100

selected by

Related questions

0 votes
0 votes
1 answer
1
Madhu M Pandurangi asked Dec 2, 2021
306 views
At a transmission rate of 20Mbps and propagation speed of 2*10^8m/s. How many meters of the cable is one-bit delay in an Ethernet interface equivalent?
0 votes
0 votes
1 answer
2
0 votes
0 votes
1 answer
3
1 votes
1 votes
1 answer
4