Refer to the exhibit. AS65510 iBGP is configured for directly connected neighbors. R4 cannot ping or traceroute network 192.168.100.0/24. Which action resolves this issue?
A.
Configure R1 as a route reflector server and configure R2 and R3 as route reflector clients.
B.
Configure R4 as a route reflector server and configure R2 and R3 as route reflector clients.
C.
Configure R4 as a route reflector server and configure R1 as a route reflector client.
D.
Configure R1 as a route reflector server and configure R4 as a route reflector client.
All answers are wrong. R1 and R4 do not see each other's routes because they are two hops apart. They need a route reflector between them, either R2 or R3.
Indeed the quality of the question is low. To have any solution work, you need to add neighbor config on top of the specified "directly connected". Spend you time well and understand why all are wrong.
Dont think it possible to be D as the question clearly states there is ONLY BGP relationship between directly connected devices, D is only possible if you stand a BGP relationship between R1 & R4, then you have a full mesh and dont need route reflector at all.
If R2 & R3 are RR clients and R4 The RR. R1 advertises to R2 & R3 they then pas the route to the reflector.
Ehm... wrong, if you have connectivity between loopback via an IGP (or static routing), you can configure peering between two routers that are not directly connected... in my opinion D is not correct, but only because R1 is a border router and it's better not to configure it as a RR
Explanation of the Issue:
R1 learns the external BGP route to 192.168.100.0/24 from R5 (eBGP peer).
R1, R2, R3, and R4 are part of iBGP in AS 65510.
The iBGP is configured only between directly connected neighbours, meaning not a full mesh.
Problem with iBGP:
iBGP does not re-advertise routes learned from one iBGP peer to another iBGP peer by default (RFC 4271).
R1 learns 192.168.100.0/24 via eBGP and advertises it to R2 and R3.But R2 and R3 will NOT forward that route to R4, because it's iBGP-learned.
That means R4 never learns about 192.168.100.0/24, and can't reach it.
Solution: Use a Route Reflector (RR)
To fix this without building a full mesh (i.e., iBGP between every router), use route reflection.
Make R1 a route reflector — it already has the eBGP route.
Make R4 a route reflector client — so R1 will reflect the route to it.
D. Configure R1 as a route reflector server and configure R4 as a route reflector client.
Yes, if a route reflector client learns a route from an iBGP peer, it will advertise that route to the route reflector. Route reflectors are designed to overcome the iBGP loop prevention mechanism, allowing for routes learned from iBGP peers to be propagated within the same autonomous system (AS).
D makes sense if you look at the wording of the answer
Configure R4 as route reflector client. This doesn't mean your configuring RR client on R4 it means your configuring RR client for R4 from R2 and R3s perceptive
Perfect question Cisco , well done !! Professional level certification lol.
All questions are wrong , A and B for sure.
C and D , also. If you go and place the command lets say for example for D, to make the R1 server and R4 client , there is an error command : "% Specify remote-as or peer-group commands first." So you need to make R4 neighbor, so full mesh . So RR config is not needed and used. Same for C. In the end , we need to chose which wrong question is more correct/less wrong. So C ? cause edge router is RR clients ??
GUYS , SWITCH your brains off. The answers just saying the role, it doesn't really say make a router client with the other router as the server ( for example D , R1 RR and his client is R4) . C and D can work. So lets pick C :
1) R1 is RR-client of R2 ---> R2 RR (for R1)
2) R4 is RR (for R2) ---> R2 is RR-client for R4
(so R2 is server for R1 and client for R4 ) !
And it can work with the same logic with D , but I chose C only because , of RR-client placement one the edge of AS.
I think that is B, because 1st rule of router reflector: if a RR recive a NLRI from NON_RR_CLient, the RR advertise the NLRI to RR_Client and not advertise to another NON_RR_Client.
R1 is connected to the external network (AS 65520) via eBGP.
R1, R2, R3, and R4 are in the same iBGP system (AS 65510).
R5 is advertising the 192.168.100.0/24 network to R1 over eBGP.
However, R4 is unable to reach the 192.168.100.0/24 network. This is likely because iBGP requires a full-mesh configuration by default, meaning that R4 doesn’t have a direct iBGP session with R1, and so it isn't learning the route to 192.168.100.0/24.
Solution:
To fix this without needing full-mesh peering, R1 can be configured as a route reflector server, and R2 and R3 can be configured as route reflector clients. This way, R1 can reflect the route learned from R5 to the other routers in AS 65510, including R4.
The correct answer is: A
A. Configure R1 as a route reflector server and configure R2 and R3 as route reflector clients.
This setup will ensure that R4 can learn the route to 192.168.100.0/24 from R1 through the route reflector mechanism.
R1 as RR and R4 as client means we establish a full mesh, not R4 gets the advertisement of R1, as it didnt get advertisement from iBGP. Now as long as R4 has route to next-hop to eBGP AS, D solves it
B: R1 is out, Can't set reflector at the edge of BGP, Must be centrally located to avoid issues. R4 makes sense to be reflector and the opposite, attached routers to be clients.
None of the answers are correct. Lab'd it up. Only way for R4 (based on the question's requirement for directly connected iBGP neighbors) is if R2 or R3 or both to have R4 config'd as a route-reflector-client.
R4 can only be a route reflector client of either R2 or R3 for the ping to work. This means for this scenario question there is no correct answer, all the answers are incorrect.
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