Bgp 4 peer problems, 2 peer stay in connect

[Grant Taylor]

On 03/07/2018 09:20 AM, Mattia Milani wrote:
> all external interface of the peer belong to the same network, and the 
> address of the network is 10.0.0.0/8 <http://10.0.0.0/8> that is unic.

Okay.  The addresses are in the same network.  Are the routers connected 
to one common broadcast domain?

Your diagram looks more like point to point links between the routers. 
These point to point (like) links will need to be in different networks.

If you have two different networks (broadcast domains) connected to a 
router with addresses in the same (sub)network, the router will get 
confused and not work as expected.

(You can do some crazy things at layer 2, but that's not traditional 
routing.)

> H0 belong to AS2 and have the interface eth1 with the address 10.0.0.1/8 
> and it is connected with H1
> H1 belong to AS4 and have two interfaces:
>       -eth1 with the address 10.0.1.1/8 that is 
> connected with H0.eth1
>       -eth2 with the address 10.0.1.2/8 that is 
> connected with H2.eth2
> H2 belong to AS3 and have two interfaces:
>       -eth1 with the address 10.0.2.1/8 that is 
> connected with H3.eth1
>       -eth2 with the address 10.0.2.2/8 that is 
> connected with H1.eth2
> H3 belong to AS1 and have the interface eth1 with the address 10.0.3.1/8 
> and it is connected with H2

To use addresses like that, you will need to have all of the following 
interfaces in the same network (broadcast domain).

  - H0.eth1
  - H1.eth1
  - H1.eth2
  - H2.eth1
  - H2.eth2
  - H3.eth1

+----------+--------+---+--------+------------+
|          |        |   |        |            |
| +----+   | +----+ |   | +----+ |     +----+ |
+-+ H0 |   +-+ H1 +-+   +-+ H2 +-+     | H3 +-+
   +----+     +----+       +----+       +----+

eth1 is on the left and eth2 is on the right

All interfaces are in the SAME broadcast domain and (sub)network.  Thus 
all IP addresses will be able to communicate directly with each other.

If you want to do something like the following:

+----+   +----+   +----+   +----+
| H0 AAAAA H1 BBBBB H2 CCCCC H3 |
+----+   +----+   +----+   +----+

AAAAA and BBBBB and CCCCC are three completely separate networks, 
physically, and addresses.

H0.eth2 = A0
H1.eth1 = A1
H1.eth2 = B1
H2.eth1 = B2
H2.eth2 = C2
H3.eth1 = C3

So, if you want to use 10/8 for A, you *MUST* use something different 
for B and C.

You can re-use any IP addresses you want in your lab as long as they 
don't conflict with each other.  Your original example has conflicting 
IP networks.

I personally would use the reserved Test-Net-1 (192.0.2.0/24) for A, 
Test-Net-2 (198.51.100.0/24) for B, and Test-Net-3 (203.0.113.0/24) for C.

I think a lot of people would use different sub-nets of 10.0.0.0/8 for 
each of the networks.  I.e. 10.0.1.0/24 for A, 10.0.2.0/24 for B, and 
10.0.3.0/24 for C.  The key being that they are completely separate /24 
networks.

> every interace is on the same network address so do you mean that every 
> bgp session between two peer need to have different network address?

Are the interfaces in the /same/ physical broadcast domain?  Based on 
your diagram they are on *separate* physical broadcast domains.



-- 
Grant. . . .
unix || die


P.S.  Here's hoping my ASCII art is up to snuff /and/ that it survives 
email.

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