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# Same direction flow

#1

If there is a flow that is only one direction in our file (this was found in top 3 duration), then should we take this flow into account as it would give us empty graphs for RTT portions of this assignment. There is no SYN/FIN/RST packets, and all of these ACK packets are Len=0, so I think we can assume that any data transferred or any connection establishment or finish is before and after the file.

the flow looks like this:
56182 > 5900 [ACK] Seq=1 Ack=126 Win=65535 Len=0 TSval=763111061 TSecr=2134609349
56182 > 5900 [ACK] Seq=1 Ack=449 Win=65535 Len=0 TSval=763111064 TSecr=2134609350
56182 > 5900 [ACK] Seq=1 Ack=1017 Win=65535 Len=0 TSval=763111065 TSecr=2134609350

if the source ports are placed into a set, it would be 1 number, likewise for destination ports.

#2

In this case, you probably observed a portion of a large file transfer. This is not unusual. What it means is that you can estimate on one direction very good because you have lots of sampled RTT, while there is no information to estimate it in other direction.

#3

if this is the case, how would we calculate the RTT for each of these packets? as stated in a previous question,

[quote=“shirali, post:4, topic:676, full:true”]Yes. If there is no data in a packet (i.e., len=0), then you will not receive any ACK for that.[/quote] , how would we be expected to get the rtt as there is only 1 time value per packet?

#4

On one direction, you see lots of packets that are sent and the other side sent back an acknowledgement. For all of these packets you can assume the RTT is the difference between the time the packet was sent and the time the ack is received. However, on the other side, you will not have any sample because you did not send any packet to receive an acknowledgement for that and use their time difference as sampled RTT.