Part of mooc exercises:
First question:
Due to message exchange, file A of size M is sent first, so dA = M/10 + M/20 + M/10 = 0.5 (sec)
B is sent at t = 0.1 + e (sec). Since only the link between the two routes is a shared link, in a non-shared link, the file transfer of A and B does not affect each other, and then the file of B It will lag behind A’s file e to reach the first router, and it takes 0.1 (sec), so A’s file is sent first, because the middle link is a shared link, so B can only wait for 0.1 (sec) , And then after A’s file arrives at the second router, B starts to transmit, which takes 0.05 (sec), and then the non-shared link, B then transmits to the destination host, which takes 0.1 (sec), so B’s Total time consumption dB = 0.1 + 0.1 + 0.05 + 0.1 = 0.35 (sec)
Second question:
The point is: in the shared link segment, the transmission bandwidth is equally divided
dA = 0.0001 * 2002 = 0.2002 （sec）
dB = 0.0001 * 1002 = 0.1002（sec）
Third question:
Packet switching appears to be fair, because after the rules for grouping are established, files with a large amount of data or files with a small amount of data will be grouped according to the regulations, but the transmission time of files with a large amount of data will be longer. In the message exchange, even if 2Mbits is larger than 1Mbits, if 1Mbits arrives at the shared link before 2Mbits, then files with small data will be sent first, and files with large data can only be waited, which is very inefficient and unfair!