0

I am going through one video on BGP with IBGP neighbors having multiple paths between them. The author says that ordinarily we will need so many IBGP session between the routers as shown below

enter image description here

Then he says that the solution is to form neighbor ship over loopbacks as shown below. My question is, in this case also, full IBGP neighbor ship will still be required, otherwise how will R2 and R2 (the physical paths) will know about the routes in the ASs connected to R1 and R4 and hence the packets will be dropped. Please let me know if I am correct, if not what's the correct concept.

enter image description here

Akshay J
  • 5,362
  • 13
  • 68
  • 105

1 Answers1

0

Whether or not you have a full IBGP mesh and whether you do physical interface peering or loopback peering are separate ("orthogonal") questions.

In both figures in your question, there is only BGP peering between routers R1 and R4. R2 and R3 do not have any BGP sessions. This design is a so-called "BGP free core" (see http://bgphelp.com/2017/02/12/bgp-free-core/). This is a common design in service providers. In a BGP free core, routers R2 and R3 do not have a full BGP routing table. As a result they do not know how to route traffic to other autonomous systems. The purple autonomous system in the figures is still able to able to forward "transit traffic" because R1 and R2 have some sort of tunnel to each other (typically an MPLS LSP).

PS 1: Do not ask people to contact you via email or twitter or facebook. The purpose of StackOverflow is to have the answer(s) and discussion along with the question on the site.

PS 2: The StackExchange "Network Engineering" community is a better place for networking related questions.

Bruno Rijsman
  • 3,715
  • 4
  • 31
  • 61