Static routing comprehensive experiment

The experimental requirements ：

One 、 First, build the topology diagram

Two 、 After the topology is built, proceed to IP To configure . It's going on IP Before configuration , First, be reasonable IP Address planning .
The backbone area only needs two ip Network segment , So you can borrow from the Internet 6 position ,
IP The address division is as follows ：

To configure IP The operation of is not repeated here .
3、 ... and 、 The computer needs to automatically obtain IP Address , To configure DHCP service .

    [r3]dhcp enable 
    [r3]ip  pool aa
    [r3-ip-pool-aa]network 192.168.1.96 mask 27
	[r3-ip-pool-aa]gateway-list 192.168.1.97
	[r3]interface GigabitEthernet0/0/2
	[r3-GigabitEthernet0/0/2]dhcp select global

Then it will belong to r3 Of the two hosts DHCP Service open , Input ipconfig Check whether you can get IP Address

As shown in the figure, it can be obtained normally IP Address
Four 、IP After the address is configured , Then the whole network can reach .
r1：
1）r1 Default point of r2,r3

    [r4]ip route-static 0.0.0.0 0 192.168.1.2
	[r4]ip route-static 0.0.0.0 0 192.168.1.6

2）r1 To r2 The loopback of ( because r2 The loopback of can be summarized , So you just need to write a summarized route ）

 	 [r1]ip route-static 192.168.1.64 27 192.168.1.2

3）r1 To r3 The loopback of ( ditto ,r3 The loopback addresses of can be summarized , I'm not going to go over it ）

    [r1]ip route-static 192.168.1.96 27 192.168.1.6	

4）r1 To 192.168.1.8 30 Network segment

   [r1]ip route-static 192.168.1.8 30 192.168.1.2

5）r1 To 192.168.1.12 30 Network segment

   [r1]ip route-static 192.168.1.12 30 198.168.1.6

r2
1）r2 The default point of r4

 [r2]ip route-static 0.0.0.0 0 192.168.1.10

2）r2 To r3 There are two ways to go around , Do a load balancing

[r2]ip route-static 192.168.1.96 27 192.168.1.10
[r2]ip route-static 192.168.1.96 27 192.168.1.1

3）r2 To 192.168.1.4 30 Network segment .( The cost of taking the default route is large , So choose to go through r1）

[r2]ip route-static 192.168.1.4 30 192.168.1.1 

4）r2 To r1 The loopback of

[r2]ip route-static 192.168.1.32 27 192.168.1.1

The following routers are similar to the previous method ......

r3

ip route-static 0.0.0.0 0.0.0.0 192.168.1.14
ip route-static 192.168.1.0 255.255.255.252 192.168.1.5
ip route-static 192.168.1.32 255.255.255.224 192.168.1.5
ip route-static 192.168.1.64 255.255.255.224 192.168.1.5
ip route-static 192.168.1.64 255.255.255.224 192.168.1.14

r4

Writing r4 The default route for , Because the upper and lower bandwidths are different , Floating static configuration is possible

ip route-static 0.0.0.0 0.0.0.0 192.168.1.18
ip route-static 0.0.0.0 0.0.0.0 192.168.1.22 preference 61

The next configuration is similar to the above

ip route-static 0.0.0.0 0.0.0.0 192.168.1.18
ip route-static 0.0.0.0 0.0.0.0 192.168.1.22 preference 61
ip route-static 192.168.1.0 255.255.255.252 192.168.1.9
ip route-static 192.168.1.4 255.255.255.252 192.168.1.13
ip route-static 192.168.1.32 255.255.255.224 192.168.1.9
ip route-static 192.168.1.32 255.255.255.224 192.168.1.13
ip route-static 192.168.1.64 255.255.255.224 192.168.1.9
ip route-static 192.168.1.96 255.255.255.224 192.168.1.13

r5

ip route-static 0.0.0.0 0.0.0.0 12.0.0.2
ip route-static 192.168.1.0 255.255.255.0 192.168.1.17
ip route-static 192.168.1.0 255.255.255.0 192.168.1.21 preference 61

because 192.168.1.0/24 There are many unopened network segments , So it's written that the next hop will lead out of the loop , So we are r4 Write an empty interface route

[r4]ip route-static 192.168.1.0 24 NULL 0

At this time, you will find the intranet ping no r5 The loopback of , Because to r5 What's going back is r4 Default , The matching length of the empty interface is longer than that of the missing province , So I'm going to r5 The loopback route is thrown to the empty interface, resulting in different .

So we need to write another one to r5 Ring back to the startling address

 ip route-static 192.168.1.160 255.255.255.224 192.168.1.18
 ip route-static 192.168.1.160 255.255.255.224 192.168.1.22 preference 61

At this point ping through r5 Your loopback address

Considering that when it comes to loopback , All use the summary address , So we should write the interface address in themselves

[r1]ip route-static 192.168.1.32 27 NULL 0
[r2]ip route-static 192.168.1.64 27 NULL 0

Four 、r1-r5 All accessible r6 The loopback of

[r5]acl 2000
[r5-acl-basic-2000]rule permit source 192.168.1.0 0.0.0.255
[r5]int GigabitEthernet  0/0/1
[r5-GigabitEthernet0/0/1]nat outbound 2000