2 Replies Latest reply: Jan 3, 2018 10:47 AM by Christian

# OSPF number of adjacencies on VPLS

Hi Team,

I just started to read something about “how to deploy OSPF (IS-IS) on VPLS”.

VPLS offers you a broadcast service and you can deploy OSPF in broadcast mode. The following thread is discussing to use P2MP to avoid problems in cases of PW outages or RP failures:

When looking from the design perspective the number of adjacency states is relevant. But I don’t understand the calculation

Quote from the mail thread: <With the p2mp network approach, the number of adjacencies is O(N^2) (vs O(N) in the broadcast network case)>

From my point of view the overall number of adjacency states (example network with 3 routers) in case of OSPF:

Number of full adjacencies with LSA processing …

DR/BDR = O{N-1}

other routers = O{N} (DR and BDR)

total adjacencies = O{(N-1)+N}

*** P2MP mode:

total adjacencies = 2+2+2 = O(N-1)^N

Where did I miss something?

Thanks for your support

Christian

Nachricht geändert durch Christian

• ###### 1. Re: OSPF number of adjacencies on VPLS

Hi Christian,

My understanding in this discussion is that the routers in the broadcast domain establish adjacencies with DR and BDR only. So, it will be 2 adjacencies per router. For DR and BDR, each will have an adjacency to all the remaining routers.

In the P2MP scenario, this means many point-to-point connections, so you do not have the DR/BDR election. So, basically the number of adjacencies will be the number of the 'point-to-point links' in this VPLS domain. If you have a full mesh 'circuits' it means you would have 3 'circuits' when you have in total 3 routers.

RouterA --- RouterB

RouterB --- RouterC

RouterC --- RouterA

Regards,

Liz

• ###### 2. Re: OSPF number of adjacencies on VPLS

Hi Liz,
thanks for your feedback.
I have the some understanding in both scenarios (broadcast and P2MP).

In a VPLS network (that is emulating a switched environment) every router peers with the others.

Router A has OSPF Full state (=transmitting LSA packets) with router B and C.
Router B with A and C ; Router C with A and B

So the overall number of adjacencies (O) is 6.

Here I made a mistake in the generic formular.    It should be number of states O{(N-1)*N}
Thanks for the feedback
Christian