Which statement is correct about IS-IS when it performs the Dijkstra algorithm?
Which statement is correct about IS-IS when it performs the Dijkstra algorithm?
When performing the Dijkstra algorithm in IS-IS routing, new neighbor IDs that appear in the tree database and match a router ID in the Link-State Database (LSDB) are moved to the candidate database. This step ensures that all reachable nodes are considered for the shortest path calculations.
Based on AJSPR course; Not A. because the local router moves its own local tuple into the tree database (and not into candidate database) Not C. because SPF algorithm moves the tuple with the lowest cost from the candidate database into the tree database. (and not into the LSB) Not D. because Step 1 (router determining cost from root to each neighbor ID) is repeated until the candidate database is empty. (and not until the tree database is empty) Therefore the answer is B. because if a new neighbor ID appears in the tree database, any tuples in the LSDB with a router ID equal to the new tree entry’s neighbor ID are moved into the candidate database.
B is correct
candidate database is not the same as tree database
There is a part on "Advanced Junos Service Provider Routing On-Demand " that says: "The candidate database is not empty, so the algorithm continues". Well, we can presume if the database is empty, algorithm stops.
candidate database is not the same as tree database