Computer network third time operation

 1-10  try to compare circuit switching and packet switching under the following conditions. The message to be sent is a total of x (bit). From the source station to the destination station a total of K-segment link, the transmission time delay for each link is D (s), the data rate is B (S/b). In circuit switching, the time of the circuit is set to S (s). In packet switching, the length is P (bit), and the queue waiting time of each node is negligible. Q Under what conditions, packet switched SHIENBI circuits are smaller. Answer: Packet switching:               Send delay = x/b (s)                propagation delay = k * D (s)                A total of k-1 packet switches, forwarding time = (k-1) *p/b                Total time delay = x/b + K * d + (k-1) *p/b (s) circuit switching:       total time Delay = s + k * d + x/b (s) so when (x/b + k * d + (k-1) *p/b)-(s + k * d + x/b) <o (k-1) p/b (s), the SHIENBI circuit of the packet switching is exchanged less.   1-21  assumes that the transmission rate of the signal on the media is 2.3x10 8 m/s. Media Length:      (1) 10cm (NIC)      (2) 100m (LAN)      (3) 100km (metropolitan area Network)      (4) 5000km (WAN)       trial calculation when the data rate is 1mb/s and 10The number of bits that are being propagated in the above media on GB/s. A: When the data rate is 1mb/s: (1) The number of bits on the NIC =0.1/(2.3x10 8) * (1x10 6) ≈0.000435b    (2) The number of bits on the LAN =100/(2.3x10 8) * (1x10 6) ≈0.435b &N bsp;  (3) Number of bits =100000/(2.3x10 8) * (1x10 6) ≈434.78b    (4) Bit number =5000000/(2.3x10 8) * (1x10 6) ≈21739. 13b when the data rate is 10gb/s: (1) The number of bits on the NIC =0.1/(2.3x10 8) * (1x10 10) ≈4.35b    (2) The number of bits on the LAN =100/(2.3x10 8) * (1x10 10) ≈4347.8 3b    (3) Number of bits =100000/(2.3x10 8) * (1x10 10) ≈4347826.09b    (4) Number of bits =5000000/(2.3x10 8) * (1x10 10 ≈217391304.3b   1-22  100-byte application layer data to transport layer transfer, plus 20 bytes of TCP header. And then to the network layer to transfer, plus 20 bytes of IP header. Finally to the data link layer of Ethernet transmission, plus the first and the tail a total of 18 bytes, trial data transmission efficiency. If the application layer data length is 1000 bytes, the transmission efficiency of the data is much. A: When the application layer data is 100 bytes:     data transfer efficiency = 100/(100+20+20+18) ≈63.3% when the application layer data is 1000 bytes: data transfer efficiency = 1000/(1000+20+20+18) ≈94 .5%   2-12  has 600MB (megabytes) of data that needs to be transferred from Nanjing to Beijing. One way is to write the data to disk and then send the disks by train. Another approach is to use the computer to transfer this data over a long-distance telephone line (the rate at which information is transmitted is 2.4KB/S). Try to compare the pros and cons of these two methods. If the information transmission rate is 33.6kb/s, the result is how. A: The train will take the disk, according to the speed of the motor, from Shanghai to Beijing also only 8 hours. When the information transmission rate is 2.4kb/s &Nbsp;   time =600x2 20x8÷2400=2097153 s = 582.54h when the information transmission rate is 33.6kb/s     time required =600x2 149796.57 s = 41.61h so there is no way to use the train in these two kinds of information transmission rates.