Redistribute routes into OSPF from BGP

[EAleg]

Hi guys, can u tell please, what i must type, that Server-Router know about this routes: 30.30.30.0, 60.60.60.0, 70.70.70.0, 10.10.10.0, 20.20.20.0? What i must specify in OSPF configuration on R3?

Thank u for help n attention

Server_Router knows about  40.40.40.0, 80.80.80.0, 50.50.50.0.

====R3=====

R3#sh run | s router bgp 300
router bgp 300
bgp log-neighbor-changes
network 40.40.40.0 mask 255.255.255.252
network 50.50.50.0 mask 255.255.255.252
network 80.80.80.0 mask 255.255.255.252

redistribute ospf 1
neighbor 40.40.40.1 remote-as 300
neighbor 50.50.50.1 remote-as 300
neighbor 80.80.80.2 remote-as 300

R3#sh run | s router ospf 1
router ospf 1
redistribute bgp 300 subnets
passive-interface default
no passive-interface GigabitEthernet0/2
network 10.11.12.0 0.0.0.3 area 1

====Server_Router===

router ospf 1
passive-interface default
no passive-interface GigabitEthernet0/0
network 10.11.12.0 0.0.0.3 area 1

[balaji.bandi]

If you not looking for Mutual redistribution,  If the Server router has only one path is to reach R3, R3 aware of all the routes you looking then the Default route should work here.

router ospf 1
 default-information originate

You can also add the always keyword to the default-information originate command to make a router originate a 0.0.0.0 type 5 LSA even if the router does not have a default route in its routing table.

Test and advise.

[Georg Pauwen · Accepted solution]

Hello,

on R3, try and add:

router bgp 300
bgp log-neighbor-changes
--> bgp redistribute-internal
network 40.40.40.0 mask 255.255.255.252
network 50.50.50.0 mask 255.255.255.252
network 80.80.80.0 mask 255.255.255.252

[paul driver]

Hello
As Rtr3 is an ibgp rtr then you would need to redistribute ibgp routes into ospf by means of bgp redistribute-internal as already suggested however looking at your physical design it seems to suggest all rtrs are in the same ASN domain with a central rtr as a RR.

If so then R1 -R2 rtrs don’t require a physical connection or ibgp peering to each other, And as R3 is connecting to those two rtrs it also doesn’t require a direct connection to RR rtr either.