H12-821_V1.0 · Question #1126
See the following figure. All routers on the network run IS-IS and are in an area 49.0001. By referring to the LSDB of R1, the Level-2 DIS is. (the device name, for example, R1.) A. R1 B. R2 C. R3…
The correct answer is A. R1. R1 is the Level-2 DIS because, in IS-IS DIS election on a broadcast segment, the router with the highest interface priority wins; if priorities are equal, the highest SNPA (MAC address) is the tiebreaker. In the figure, R1 holds the highest election metric on that LAN segment…
Question
Options
- AR1
- BR2
- CR3
- DR4
How the community answered
(61 responses)- A75% (46)
- B7% (4)
- C3% (2)
- D15% (9)
Explanation
R1 is the Level-2 DIS because, in IS-IS DIS election on a broadcast segment, the router with the highest interface priority wins; if priorities are equal, the highest SNPA (MAC address) is the tiebreaker. In the figure, R1 holds the highest election metric on that LAN segment, and its LSDB confirms this: you will see a pseudonode LSP with R1's System ID and a non-zero circuit ID (e.g., R1.01-00), which only the DIS generates.
Why the distractors are wrong: R2, R3, and R4 each have a lower interface priority or a lower MAC address than R1, so they lose the election. A non-DIS router does not originate a pseudonode LSP - it only has its own node LSP - so their System IDs will not appear as a pseudonode entry in R1's LSDB.
Memory tip: "The DIS owns the pseudonode - look for the .01 LSP in the LSDB to identify who won the election. Highest priority → highest MAC → winner."
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