BGP综合实验

一.实验要求：

1、AS1中存在两个环回，一个地址为192.168.1.0/24，该地址不能在任何协议中宣告， AS3中存在两个环回，一个地址为192.168.2.0/24，该地址不能在任何协议中宣告，最终要求这两个环回可以ping通

2、R1-R8的建邻环回用x.x.x.x/32表示

3、整个A2的Ip地址为172.16.0.0/16，R3-R7上各有一个业务网段，请合理划分;并且其内部配置osPF协议

4、AS间的骨干链路IP地址随意定制，

5、使用BGP协议让整个网络所有设备的环回可以互相访问，

6、减少路由条目数量，避免环路出现;

二.实验拓扑图

三.实验过程

1.R1配置

$R1$ int g 0/0/0

$R1-GigabitEthernet0/0/0$ ip address 12.1.1.1 24

$R1-GigabitEthernet0/0/0$ int l 0

$R1-LoopBack0$ ip address 1.1.1.1 32

$R1-LoopBack0$ int l 1

$R1-LoopBack1$ ip address 192.168.1.1 24

2.R2配置

$R2$ int g 0/0/0

$R2-GigabitEthernet0/0/0$ ip address 12.1.1.2 24

$R2-GigabitEthernet0/0/1$ int g 0/0/1

$R2-GigabitEthernet0/0/1$ ip address 172.16.1.1 30

$R2-GigabitEthernet0/0/1$ int g 0/0/2

$R2-GigabitEthernet0/0/2$ ip address 172.16.1.21 30

$R2-GigabitEthernet0/0/0$ int l 0

$R2-LoopBack0$ ip address 2.2.2.2 32

3.R3配置

$R3$ int g 0/0/0

$R3-GigabitEthernet0/0/0$ ip address 172.16.1.2 30

$R3-GigabitEthernet0/0/0$ int g 0/0/1

$R3-GigabitEthernet0/0/1$ ip address 172.16.1.5 30

$R3-GigabitEthernet0/0/1$ int l 0

$R3-LoopBack0$ ip address 3.3.3.3 32

$R3-LoopBack0$ int l 1

$R3-LoopBack1$ ip address 172.16.3.3 24

4.R4配置

$R4$ int g 0/0/0

$R4-GigabitEthernet0/0/0$ ip address 172.16.1.6 30

$R4-GigabitEthernet0/0/0$ int g 0/0/1

$R4-GigabitEthernet0/0/1$ ip address 172.16.1.9 30

$R4-GigabitEthernet0/0/1$ int l 0

$R4-LoopBack0$ ip address 4.4.4.4 32

$R4-LoopBack0$ int l 1

$R4-LoopBack1$ ip address 172.16.4.4 24

5.R5配置

$R5$ int g 0/0/0
$R5-GigabitEthernet0/0/0$ ip address 172.16.1.22 30
$R5-GigabitEthernet0/0/0$ int g 0/0/1
$R5-GigabitEthernet0/0/1$ ip address 172.16.1.17 30
$R5-GigabitEthernet0/0/1$ int l 0
$R5-LoopBack0$ ip address 5.5.5.5 32
$R5-LoopBack0$ int l 1
$R5-LoopBack1$ ip address 172.16.5.5 24

6.R配置

$R6$ int g 0/0/0

$R6-GigabitEthernet0/0/0$ ip address 172.16.1.18 30

$R6-GigabitEthernet0/0/0$ int g 0/0/1

$R6-GigabitEthernet0/0/1$ ip address 172.16.1.13 30

$R6-GigabitEthernet0/0/1$ int l 0

$R6-LoopBack0$ ip address 6.6.6.6 32

$R6-LoopBack0$ int l 1

$R6-LoopBack1$ ip address 172.16.6.6 24

7.R7配置

$R7$ int g 0/0/0

$R7-GigabitEthernet0/0/0$ ip address 172.16.1.14 30

$R7-GigabitEthernet0/0/0$ int g0/0/1

$R7-GigabitEthernet0/0/1$ ip address 172.16.1.10 30

$R7-LoopBack1$ int g 0/0/2

$R7-GigabitEthernet0/0/2$ ip address 34.1.1.7 24

$R7-GigabitEthernet0/0/1$ int l 0

$R7-LoopBack0$ ip address 7.7.7.7 32

$R7-LoopBack0$ int l 1

$R7-LoopBack1$ ip address 172.16.7.7 24

8.R8配置

$R8$ int g 0/0/0

$R8-GigabitEthernet0/0/0$ ip address 34.1.1.8 24

$R8-GigabitEthernet0/0/0$ int l 0

$R8-LoopBack0$ ip address 8.8.8.8 32

$R8-LoopBack0$ int l 1

$R8-LoopBack1$ ip address 172.16.8.8 24