|
R5：
- access-list 2 permit 10.1.201.0 0.0.0.255
-
- route-map LEAK permit 10
- match ip address 1
-
- router eigrp QYT
- address-family ipv4 unicast autonomous-system 666
- af-interface Serial1/1
- summary-address 10.2.0.0 255.255.0.0 leak-map LEAK
验证：
- R2#show ip eigrp topology 10.1.201.0/24 | include from
- 10.12.25.5 (Serial1/1), from 10.12.25.5, Send flag is 0x0
- 10.2.12.1 (Serial1/0), from 10.2.12.1, Send flag is 0x0
-
-
- R5#show ip eigrp topology 10.2.201.0/24 | include from
- 10.12.25.2 (Serial1/1), from 10.12.25.2, Send flag is 0x0
测试：
- R2#traceroute 10.1.201.100 source 10.2.201.100 numeric
- Type escape sequence to abort.
- Tracing the route to 10.1.201.100
- VRF info: (vrf in name/id, vrf out name/id)
- 1 10.12.25.5 9 msec * 9 msec
-
-
- R5#traceroute 10.2.201.100 source 10.1.201.100 numeric
- Type escape sequence to abort.
- Tracing the route to 10.2.201.100
- VRF info: (vrf in name/id, vrf out name/id)
- 1 10.12.25.2 9 msec * 9 msec
七、配置NAT
此部分不作为本案例研究重点，配置部分仅作示例
- R3(config)#int e0/2
- R3(config-if)#ip nat outside
-
- R3(config)#int e0/3
- R3(config-if)#ip nat inside
-
- access-list 1 permit 10.0.0.0 0.0.0.255
-
- ip nat inside source list 1 interface 【XXX】 overload
八、合理分配R3和R4的上行链路资源
在R3和R4上配置静态默认路由，并以不同度量值引入EIGRP，尽量体现网络上行链路实际情况
- R3(config)#ip route 0.0.0.0 0.0.0.0 36.1.1.6
- R4(config)#ip route 0.0.0.0 0.0.0.0 s1/0
R3：
- router eigrp QYT
- address-family ipv4 unicast autonomous-system 666
- topology base
- redistribute static metric 10000 100 255 1 1500
R4：
- router eigrp QYT
- address-family ipv4 unicast autonomous-system 666
- topology base
- redistribute static metric 1544 2000 255 1 1500
九、在R5上实现去往互联网的业务流量非等价负载均衡
查看拓扑表，仅有一个后继，没有发现可行后继，说明要么只收到一个路径，要么就是有其它路径，但并不符合可行条件;
- R5#show ip eigrp topology
-
- P 0.0.0.0/0, 1 successors, FD is 196608000
- via 10.1.35.3 (196608000/131072000), Ethernet0/1
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