题目:
思路:
分成三个部分。需要运用到IPV6地址来解题,左边是IPV4和IPV6右边只有IPV6,中间是IPV4。三个部分需要用到不同的协议。此题与之前最大的区别在用IP地址的不同,此题运用IPV6地址与之前大不相同,所以要考虑地址的转换。其余的思路就是配置好各个路由器的IP地址,而后通过每段应有的协议宣告该路由器的协议。
地址转换:
23.1.1.1
0001 0111
2002:1701:101::/48
2002:1701:101:0000::/64 --2002:1701:101:FFFF::/64
2002:1701:101:0000:8000::/65
34.1.1.2
2002:2201:0102::/48 AS1
2002:2221:0102::、64 – 2002:2201:0102:7FFF::/64
2002:2201:0102:8000::/49 AS2
2002:2201:0102:8000::/64 – 2002:2201:0102:FFFF::/64
路由器配置:
R1:
[r1]int lo0
[r1-LoopBack0]ip add 192.168.1.1 25
[r1-LoopBack0]int lo1
[r1-LoopBack1]ip add 192.168.1.129 25
[r1-LoopBack1]int g0/0/1
[r1-GigabitEthernet0/0/1]ip add 192.168.0.1 30
[r1]ip route-static 0.0.0.0 0.0.0.0 192.168.0.2
[r1]ipv6
[r1]int lo0
[r1-LoopBack0]ipv6 enable
[r1-LoopBack0]ipv6 address 2002:1701:101::1/65
[r1-LoopBack0]int lo1
[r1-LoopBack1]ipv6 enable
[r1-LoopBack1]ipv6 address 2002:1701:101:0000:8000::1 65
[r1-LoopBack1]int g0/0/1
[r1-GigabitEthernet0/0/1]ipv6 enable
[r1-GigabitEthernet0/0/1]ipv6 address 2002:1701:101:1::1 64
[r1]rip
[r1]ripng 1
[r1-ripng-1]q
[r1]int lo0
[r1-LoopBack0]ripng 1 enable
[r1-LoopBack0]int lo1
[r1-LoopBack1]ripng 1 enable
[r1-LoopBack1]int g0/0/1
[r1-GigabitEthernet0/0/1]ripng 1 enable
[r1-GigabitEthernet0/0/1]ripng summary-address 2002:1701:101:: 64
[r1]ipv6 route-static 2002:1701:101:: 64 NULL 0
[r1]ipv6 route-static 2002:1701:101:: 64 NULL 0
R2:
[r2]int g0/0/1
[r2-GigabitEthernet0/0/1]ip add 23.1.1.1 24
[r2]ip route-static 0.0.0.0 0.0.0.0 23.1.1.2
[r2]int g0/0/0
[r2-GigabitEthernet0/0/0]ip add 192.168.0.2 30
[r2-GigabitEthernet0/0/0]int lo0
[r2-LoopBack0]ip add 192.168.2.1 24
[r2]acl 200