Alcatel Unleashed

Hi,

I am preparing for Nokia's IS-IS exam and encountered a challenging question about IS-IS asymmetric routing in the practice test. I believe I'm overlooking something and would appreciate guidance from experts.

In the attached network topology, the route from R2 to R7 is R2 > R5 > R7, but the return route from R7 to R2 is R7 > R8 > R6 > R5 > R4 > R3 > R1 > R2.
It makes sense that R7 chooses the path through R8 > R6 > R5 because of the lower cost, with R5 subsequently routing through R4 to reach area 49.001 due to the cost advantage. However, I am confused as to why R2 takes the direct path through R5 to R7, which has a higher metric, instead of using the lower-cost route through R1 > R3 > R4 > R5 > R6 > R8 > R7. I would greatly appreciate it if someone could help me understand this routing behaviour

Thank you.

Without giving it too much thought, I believe it will prefer an L2 route over an L1 route since its destination is in an other area. Hence R2 -> R5 is the most appropriate route to take.

Thank you Stoffen

No prob.
Giving it more thought, the leftmost router R1, is a L1 only router.
Hence it will only have a default route towards the other areas.
R2 is L1/L2 and have the other area routes in its routing table, and will forward using longest-prefix-match.
R2 will never use R1 as next hop for inter-area transport as long as adjacency with R5 is established.