Chapter 6
86 Time 0 1 2 3 4 5 6 7 8 9
A recvs Ack0 Ack1 Ack2 Ack3 Ack3/timeout4 Ack3/timeout5&6 Ack5/timeout7&8 Ack7 Ack8
A sends Data0 Data1,2 Data3,4 (4 dropped) Data5,6 (6 dropped) Data7,8 (8 dropped) Data4 Data6 Data8 Data9,10 (slow start)
R sends Data0 Data1 Data2 Data3 Data5 Data7 Data4 Data6 Data8 Data9
cwnd size 1 2 3 4 5 1 1 1 1 2
22. We follow the conventions and notation of the preceding exercise. Although the first packet is lost at T=4, it wouldn’t have been transmitted until T=8 and its loss isn’t detected until T=10. During the final few seconds the outstanding losses in the existing window are made up, at which point slow start would be invoked. A recvs Ack # T=0 T=1 T=2 T=3 T=4 T=5 T=6 T=7 T=8 T=9 T=10 T=11 T=12 T=13 T=14 T=15 T=16 T=17
1 2 3 4 5 6 7 8 8 8 8 8 10 12 14 16 17
cwnd size 1 2 3 4 5 6 7 8 9 9 9
2
A sends Data 1 2,3 4,5 6,7 8,9 10,11 12,13 14,15 16,17 9
11 13 15 17 18,19
R sending/R’s queue 1/ 2/3 3/4,5 4/5,6,7 5/6,7,8 6/7,8,10 7/8,10,12 8/10,12,14 10/12,14,16 12/14,16 14/16,9 16/9 9/ 11/ 13/ 15/ 17/ 18/19
9 lost 11 lost 13 lost 15 lost 17 lost 2nd duplicate Ack8
B gets 9
slow start
23. R’s queue size is irrelevant because the R-B link changed from having a bandwidth delay to having a propagation delay only. That implies that packets leave R as soon as they arrive and hence no queue can develop. The problem now becomes rather trivial compared to the two previous questions. Because no queue can develop at the router, packets will not be dropped, so the window continues to grow each RTT. In reality this scenario could happen but would ultimately be limited by the advertised window of the connection. Note that the question is somewhat confusingly worded—it says that 2 packets