Multiple ABRs in stub area

Unanswered Question
Aug 3rd, 2007
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Hi all,


Attached is my topology.

I am injecting default route into area 1 from R1 with cost 10 and from R2 with cost 40.

In that case, will both default route be in area1's database ?

OR let me know, which default route does R5 takes and which one does R4 takes to reach external destination .


Thanks,

Vijaybabu



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cjnwodo Fri, 08/03/2007 - 04:34
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Hi,


Both default routes should be in the OSPF database, however only the one from R1 [with the lower cost] should make it into the routing table. So both R4 and R5 should use R1.

GillieLucent Fri, 08/03/2007 - 07:00
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Hi,

Can someone explain bit more ?

It would be better if you explain how R2's default route reaches R5 and vice versa.


Thank,

Vijaybabu

cjnwodo Fri, 08/03/2007 - 07:10
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Hi,


Before I do so, can I ask a question? Are the physical links in your diagram the only ones, in the network? or do R4 and R5 have other connectivity to each other or do they [R4 & R5] both connect to R1 and R2?

cjnwodo Fri, 08/03/2007 - 07:15
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Hi,


Also what is the physical link between R1 and R2 and what area is that link in?

GillieLucent Fri, 08/03/2007 - 09:35
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Hi,


Thanks for your prompt response.

There is no connectivity between R4 & R5. R4 is connected only to R2 and R5 is connected only to R1. The connection between R1 and R2 is broadcast network and its in Area 0.

royalblues Fri, 08/03/2007 - 11:43
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Is R3 connected to only R2?


Narayan

sundar.palaniappan Fri, 08/03/2007 - 13:31
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R5 would use R1. R4 would use R2.


Both R5 and R4 would see only one ABR each. You can verify that by issuing the command 'show ip ospf border-routers'. There will be only one option in your setup for R5 and R4 and thus choosing the route based on OSPF metric doesn't arise.


HTH


Sundar

GillieLucent Fri, 08/03/2007 - 21:52
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Hi Sundar,



But both R5,R5 will receive Router LSAS from both R1 and R2, so both R4 and R5 will come to know there are two ABRs in the area. In that case, it should choose ABR which advertises least default cost, isn't ?

Please explain elaborately.


Thanks,

Vijaybabu


sundar.palaniappan Sat, 08/04/2007 - 04:45
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Vijaybabu,


R5 will not receive LSAs from R2. The reason for that is you have a partitioned area1. What you are saying would be true if the link between R1 and R2 is in area 1. As far as R1 and R2 is concerned they consider area1 as two separate areas.


HTH


Sundar

GillieLucent Sat, 08/04/2007 - 09:11
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Hi Sundar,


Thanks for your clearing my doubt.

I really didn't think about that, I just enabled neighbourship without thinking in which area its needed.

So, there is no use of the link between R1 and R2 in area 0.

Can you explain me if area 1 is not a stub area ?


Thanks,

Vijaybabu

sundar.palaniappan Sat, 08/04/2007 - 09:21
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Area 1 is a normal area and not OSPF stub area. If you want to make area1 as stub you would have to configure all the routers in area 1 as stub area. As I said before if you configure the link between R1 and R2 to be area 1 then both R4 and R5 would receive LSAs from both ABRs.


HTH


Sundar

GillieLucent Sat, 08/04/2007 - 10:09
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Hi Sundar,


Actually, in my topology area 1 is a stub area.

Now, I want you to explain the behavior if area 1 is not a stub area .


Thanks,

Vijaybabu

sundar.palaniappan Sat, 08/04/2007 - 11:31
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It shouldn't make any difference in the current behavior whether it's normal or stub area. If you desire you can make R5 and R4 a stub , a totally stub would even be a better option, in their own area.


HTH


Sundar

GillieLucent Sat, 08/04/2007 - 13:25
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Hi Sundar,


You are answering well, but not trying to explain.

We are not experts, we are learners.

Explain me in the below scenario :

In the topology diag, I have made a connection between R4 and R5, then what could happen ?


Thanks,

Vijaybabu

sundar.palaniappan Sat, 08/04/2007 - 19:01
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Vijaybabu,


I was under the impression that you were only looking for a solution to address a problem in your network. I have no problems explaining the reasoning behind my suggestion.


Area 1 is stub area. Now you have added a connection between R4 and R5, two stub routers in area 1. There are two ABRs namely R1 & R2. ABRs inject a IA (type 3 LSA) default route into the stub area. Both R4 and R5 would learn the default route from both ABRs. At this point R4 and R5 would use the default route with the best metric (lower cost). The OSPF cost calculation is cumulative as the route traverses every link and therefore you have to configure the link cost(s) appropriately to route the traffic as you desire.


HTH


Sundar

GillieLucent Sun, 08/05/2007 - 00:41
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Hi Sundar,


Thanks for the explanation.

If I advertise a default route from R1 with default cost 10 and from R2 with default cost 40, in that case R4 would choose R1 to reach external destinations though R2 is connected to ASBR, am I correct ?


Thanks

Vijaybabu

GillieLucent Sun, 08/05/2007 - 04:36
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Hi Sundar,


I don't have configs now.

Actually, I am understanding the concepts. But you could get that from my topology, only modification, there is a connection between R4 and R5. So if I advertise a default routes from R1 and R2 with stub default cost 10 and 40 respectively, in that case which ABR would R4 choose to reach external destinations ?

I need answer with explanation.


Thanks,

Vijaybabu

sundar.palaniappan Sun, 08/05/2007 - 05:38
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ABRs in a stub network inject a default route automatically into the stub domain. How are you setting the metric of 10 & 40 for the default route?


In a normal OSPF area you can redistribute a default route from static/another routing process or use the default information originate command to inject a default route and every one of these options give you the option to set the metric. When the route is redistributed it's advertised as E2 (External type2) route by default and the metric stays the same as the route traverses through the area. You can optionally configure the redistributed route as E1 (External type1) which would cause the cost to increment as the route traverses more links.


HTH


Sundar

GillieLucent Sun, 08/05/2007 - 08:20
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Hi,


Isn't this possible with the below command ?


'area 2 default cost 10'


Thanks,

Vijaybabu

sundar.palaniappan Mon, 08/06/2007 - 13:50
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Vijaybabu,


Yes you can use this command to set the default cost.


With that said here's my response to your previous question.


""But you could get that from my topology, only modification, there is a connection between R4 and R5. So if I advertise a default routes from R1 and R2 with stub default cost 10 and 40 respectively, in that case which ABR would R4 choose to reach external destinations ?""



Whether R4 would choose R1 or not depends on the cost of the link between R4 and R5. If the link cost is less than 30 (difference in metric set by R1 and R2) then R4 would choose R1's default route as it's gateway of last resort.


I shall try to explain this with an example.


R4 currently would have two type 3 LSA for default route, from R1 with cost of 10 and from R2 with a cost of 40.


Let's assume the following cost;


R1 - cost 10 - R5

R2 - cost 10 - R4

R4 - cost 20 - R5


R4 would have two type 3 LSAs for default route, metric of 10 from R1 and metric of 40 from R2.


The final tally would look like this.


via R1 - 10 (advertised metric)+ 10 (cost of link between R1 and R5) + 20 (cost of link between R4 and R5) = 40 (final Metric).


via R2 - 40 (advertised metric) + 10 (cost of link between R2 and R4) = 50 (final metric).


As you can see in this example R4 would choose R1 as it's default route.


HTH


Sundar

GillieLucent Mon, 08/06/2007 - 22:13
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Hi,


I tried this scenario in the lab. In my setup, R4 has two default routes, one through R5 and another R2. I shall try to explain .

R1 advertises with default cost 10 and R4 receives this LSA both through R5 and R2.

I have used Gig cards all through the setup, so interface cost is 2.


First route : 10 (R1's advertised metric) + 2 (Link between R1 and R5) + 2 (Link between R5 and R4) = 14

Second route : 10 (R1's advertised metric) + 2 (Link between R4 and R2) + 2 (Link between R2 and R1) = 14


Hope, I am correct.



Thanks,

Vijaybabu

balajee Mon, 08/06/2007 - 22:43
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How the cost will be 2 for the link between R5 and R4? From R5, R4 is two hops away. In that case, cost should be 3. Correct me if I am wrong.


Thanks,

Balajee

sundar.palaniappan Tue, 08/07/2007 - 04:13
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Vijaybabu,


Do you have the link between R1 and R2 in area0 as stated in your earlier post. If you do then the cost between R1 and R2 (cost of 2) shouldn't be added to the metric calculation for the second route via R2. Moreover the cost set by R2 (40) doesn't show up on R4. It sounds that you might have the link between R1 and R2 in area 1.


Second route : 10 (R1's advertised metric) + 2 (Link between R4 and R2) + 2 (Link between R2 and R1) = 14 --> The link cost of 2 shouldn't be there.


Balajee,


Vijaybabu has established direct link between R5 and R4 and it's part of area 1 (stub area). You will see that in one of his above posts.


HTH


Sundar

GillieLucent Tue, 08/07/2007 - 04:49
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Hi Sundar,


Yes, I have link between R1 and R2 in area 1 and area 0.

Since R4 receives R1,R2 router LSAs in area 1, then the second route is valid, am I correct



Thanks,

Vijaybabu

sundar.palaniappan Tue, 08/07/2007 - 11:33
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Is this how it's setup now?


R1 - area 0 - R2

R1 - area 1 - R5

R2 - area 1 - R4

R4 - area 1 - R5



GillieLucent Tue, 08/07/2007 - 21:57
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Hi Sundar,


Setup is like :


R1-area 0-R2

R2-area 0-R3

R1-area 1-R2

R1-area 1-R5

R2-area 1-R4

R4-area 1-R5


Thanks,

Vijaybabu



sundar.palaniappan Wed, 08/08/2007 - 07:04
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Vijaybabu,


Why do you have R1 and R2 talking to each other via both area 0 and area 1?


Remove the link between R1 and R2 from area 1. As per your earlier post you stated the link was in area 0.


Try that and let us know the outcome.


HTH


Sundar

GillieLucent Wed, 08/08/2007 - 08:30
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Hi Sundar,


I don't want area 1 to be partitioned. Hence I connected R1 and R2 in area 1.


I tried with no link between R1 and R2 in area 1. In that case R4 chose the default route advertised by R1.


Route advertised by R1:

10 ( R1's advertised metic) + 2 (cost of link between R4 and R5) + 2 (cost of link between R5 and R1) = 14


Route advertised by R2:

40 ( R2's advertised metic) + 2 (cost of link between R4 and R2) = 42



Thanks,

Vijaybabu

sundar.palaniappan Wed, 08/08/2007 - 16:18
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What you are seeing is the normal behavior. Now you've seen the OSPF behavior when the area is partitioned and not partitioned. Hope your concerns have been addressed.


HTH


Sundar

GillieLucent Wed, 08/08/2007 - 21:36
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Hi Sundar,


yes, I have understood well.

Thanks for all your explanation.


Thanks,

Vijaybabu

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