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Multiple ABRs in stub area

GillieLucent
Level 1
Level 1

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

32 Replies 32

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

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

Can you post R1 and R2's OSPF configuration?

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

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

Hi,

Isn't this possible with the below command ?

'area 2 default cost 10'

Thanks,

Vijaybabu

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

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

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

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

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

Is this how it's setup now?

R1 - area 0 - R2

R1 - area 1 - R5

R2 - area 1 - R4

R4 - area 1 - R5

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

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

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

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