OSPF Totally Stub area

[banlan.chen]

Hi All,

Could you please tell me what's the meaning of cisco's proprietary Totally Stub Area?

My understand is other vendors' router can't put into the Totally Stub Area whoes ABR is cisco router. Am I right? But from LSA view in Totally Stub Area your router only accept LSA1, LSA2 and default route LSA3, there is nothing with cisco proprietary LSA.

Thanks in adervance

[Faraz Shamim]

There is no such thing as "Cisco" Totally stub area. The RFC says that you have an "option" to leak "limited" type 3 LSA into the stub area so all we are doing is leaking type 3 default thats all. That does not make it a Cisco proprietry. As far as operations are concern, you can run Cisco router with totally stubby while other vendors router as stub and it will work just fine. But totally stub only make sense on ABR though.

Faraz

[j-bliss]

Let me quote the good book on this one.(from the Cisco Building Scalable Cisco Networks book).

"Totallyt stubby Area-This is an area that does not accept external autonomous system (As) router or summary routes from other areas internal to the autonomous system. Instead, if the router needs to send a packet to a network external to the area, it sends it using a default route. Totally stubby areas are Cisco propriatary."

BSCN Cisco press.

Catherine Paquet and

Diane Teare

ISBN 1-57870-228-3

Page 182

[Faraz Shamim]

Thanks for the info. Let me take this back to Cisco press. But aside from what is written in this book there is nothing proprietry about totally stub and it will work with any vendors' router without any issues. If you read RFC 2328 section 12.4.3.1, it clearly says that the algorithm mentioned in 12.4.3 about originating sumnmary LSA is optional if the area is stub. Here you have a choice of originating "fewer" summary LSA into stub area. So on Cisco router with totally stub area we generate only one summary LSA which is a default route. I do not see anything in the RFC that make totally stub behavior a proprietry to any vendor.

Faraz