Questions and Answers
1
a] The data sent to host H3 is forwarded through interface 3, and the forwarding table of router A is:
target address link interface.
H3 3.
b) No, because the forwarding rule is only based on the destination address
c) One possible configuration is:
access port incoming VC number output port output VC number
1 12 3 22
2 63 4 18
d) one possible configuration is:
router B:
Access port Entering VC number Output port Output VC number
1 22 2 24
Router C:
Access port Entering VC number Output port Output VC number
1 18 2 50
Router D:
Access port entering VC number Output port output VC number.
1 24 3 70。
2 50 3 76。.
2
Destination host address range link interface
00000000 to 00111111 0
01000000 to 01011111 1
01100000 to 01111111 2
10000000 to 10111111 2
1100000 to 11111111 3
The number of addresses for interface 0 is: 2^6=64.
The number of addresses for interface 1 is: 2^5=32.
The number of addresses for interface 2 is: 2^6+2^5=64+32=96
The number of addresses for interface 3 is: 2^5=32.
subnet 2 223.1.17.0/25
subnet 1 223.1.17.128/26
subnet 3 223.1.17.192/28
3 I feel that there is a problem with the standard answer
23.1.17.0/26
223.1.17.128/25
223.1.17.192/28
4
a. Since all IP packets are sent to the outside, we can use a packet sniffer to record all IP packets generated by the hosts behind the NAT. Since each host generates a sequence of IP packets with a sequence number and a different (probably, since they were chosen randomly from a large space) initial identification number (ID) sequence, IP packets with consecutive IDs can be grouped together group into --series clusters. The number of clusters is the number of hosts behind NAT.
b. If these identification numbers are assigned randomly instead of sequentially, then the technique suggested in part a. will not work because there will be no clusters in the sniffed data.
