hcip 第二天

需求：在R1-R5运行RIPV2

[r1]rip 100
[r1-rip-100]version 2
[r1-rip-100]network 1.0.0.0
[r1-rip-100]network 12.0.0.0
[r1-rip-100]network 14.0.0.0
[r1-rip-100]network 172.16.0.0

[r2]rip 100
[r2-rip-100]version 2
[r2-rip-100]network 2.0.0.0
[r2-rip-100]network 12.0.0.0
[r2-rip-100]network 23.0.0.0

[r3]rip 100
[r3-rip-100]version 2
[r3-rip-100]network 3.0.0.0
[r3-rip-100]network 23.0.0.0
[r3-rip-100]network 34.0.0.0

[r4]rip 100
[r4-rip-100]version 2
[r4-rip-100]network 4.0.0.0
[r4-rip-100]network 14.0.0.0
[r4-rip-100]network 34.0.0.0
[r4-rip-100]network 45.0.0.0
[r4-rip-100]network 46.0.0.0

[r5]rip 100
[r5-rip-100]version 2
[r5-rip-100]network 45.0.0.0

在R6-R7上运行RIPV1

[r6]rip 100
[r6-rip-100]version 1
[r6-rip-100]network 6.0.0.0
[r6-rip-100]network 46.0.0.0
[r6-rip-100]network 67.0.0.0

[r7]rip 100
[r7-rip-100]version 1
[r7-rip-100]network 7.0.0.0
[r7-rip-100]network 67.0.0.0

1.使用合理IP地址规划网络，各自创建环回接口

[r1]int loopback 0
[r1-LoopBack0]ip add 1.1.1.1 24
[r1-LoopBack0]int g0/0/0
[r1-GigabitEthernet0/0/0]ip add 12.1.1.1 24
[r1-GigabitEthernet0/0/0]int g0/0/1
[r1-GigabitEthernet0/0/1]ip add 14.1.1.1 24



[r2]int loopback 0
[r2-LoopBack0]ip add 2.2.2.2 24
[r2-LoopBack0]int g0/0/0
[r2-GigabitEthernet0/0/0]ip add 12.1.1.2 24
[r2-GigabitEthernet0/0/0]int g0/0/1
[r2-GigabitEthernet0/0/1]ip add 23.1.1.1 24

[r3]int loopback 0
[r3-LoopBack0]ip add 3.3.3.3 24
[r3-LoopBack0]int g0/0/0
[r3-GigabitEthernet0/0/0]ip add 23.1.1.2 24
[r3-GigabitEthernet0/0/0]int g0/0/1
[r3-GigabitEthernet0/0/1]ip add 34.1.1.1 24

[r4]int loopback 0
[r4-LoopBack0]ip add 4.4.4.4 24
[r4-LoopBack0]int g0/0/0
[r4-GigabitEthernet0/0/0]ip add 14.1.1.2 24
[r4-GigabitEthernet0/0/0]int g0/0/1
[r4-GigabitEthernet0/0/1]ip add 34.1.1.2 24
[r4-GigabitEthernet0/0/1]int g0/0/2
[r4-GigabitEthernet0/0/2]ip add 45.1.1.1 24
[r4-GigabitEthernet0/0/2]int g0/0/3
[r4-GigabitEthernet0/0/3]ip add 46.1.1.1 24

[r5]int loopback0 
[r5-LoopBack0]ip add 5.5.5.5 24
 [r5-LoopBack0]int g0/0/0 
[r5-GigabitEthernet0/0/0]ip add 45.1.1.2 24

[r6]int loopback0
[r6-LoopBack0]ip add 6.6.6.6 24
[r6-LoopBack0]int g0/0/0
[r6-GigabitEthernet0/0/0]ip add 46.1.1.2 24
[r6-GigabitEthernet0/0/0]int g0/0/1
[r6-GigabitEthernet0/0/1]ip add 67.1.1.1 24

[r7]int loopback0
 [r7-LoopBack0]ip add 7.7.7.7 24
 [r7-LoopBack0]int g0/0/0
 [r7-GigabitEthernet0/0/0]ip add 67.1.1.2 24

2.r1创立环回172.16.1.1/24 172.16.2.1/24 172.16.3.1/24

[r1]int loopback 1
[r1-LoopBack1]ip add 172.16.1.1 24
[r1-LoopBack1]int loopback 2
[r1-LoopBack2]ip add 172.16.2.1 24
[r1-LoopBack2]int loopback3
[r1-LoopBack3]ip add 172.16.3.1 24

3.要求r3使用r2访问r1环回

[r3]acl 2000
[r3-acl-basic-2000]rule 	
[r3-acl-basic-2000]rule permit source 1.1.1.0 0
[r3-acl-basic-2000]rule  permit source 172.16.0.0 0
[r3-GigabitEthernet0/0/0]rip metricin 2000   3

4、减少路由条目数量，增加路由传递安全性

[r1]int g0/0/0
[r1-GigabitEthernet0/0/0]rip summary-address 172.16.0.0 255.255.252.0（路由汇总）
[r1]int g0/0/1
[r1-GigabitEthernet0/0/1]rip summary-address 172.16.0.0 255.255.252.0（路由汇总）

[r1]rip 100
[r1-rip-100]timers rip 15  60  100

5、R5创建一个环回模拟运营商，不能通告

R5上虚拟运营商5.5.5.5/24未通告到RIP协议内，想与其通信，则要在R5路由器RIP协议内下发缺省路由

[r5-rip-100]default-route originate 

6、R1 telnet R2环回实际telnet 到R7上

7、R6-R7路由器不能学习到达R1换回路由

[r6]acl 2000
[r6-acl-basic-2000]rule deny source 1.1.1.0 0
[r6-acl-basic-2000]rule deny source 172.16.0.0 0
[r6-acl-basic-2000]rule permit source any
[r6]rip 100
[r6-rip-100]filter-policy 2000 import g0/0/0—调用acl2000从入方向过滤R1的两个环回路由

8、全网可达

查看R1-R7的rip路由表：
[r1]dis ip routing-table protoco rip
.
.
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[r7]dis ip routing-table protoco rip（查看是否有缺省路由）