Hi all, Tony P. asked for a counter-example to why neighbor-only algorithm information is sufficient. This email attempts to articulate just such an example.
Suppose that we have a bi-partite network, with two halves, A and B. Part A contains nodes A1, A2, A3, …. Part B contains nodes B1, B2, B3, …. The two halves are connected by three links (A1, B1), (A2, B2), and (A3, B3). The correct flooding topology in this situation is to select exactly two of the three links. Selecting only one of the links would create a single point of failure. Selecting all three links leads to unacceptable and unnecessary flooding. Suppose that A1 and B1 are running algorithm X. All other nodes are running algorithm Y. Suppose that under algorithm X, links 2 and 3 are selected. Therefore, A1 and B1 choose to prune (A1, B1). Further, suppose that under algorithm Y, links 1 and 2 are selected. Therefore, nodes A3 and B3 choose to prune (A3, B). Now, only (A2, B2) is selected, creating a single point of failure. The key points here are simple: - An algorithm makes assumptions about how other nodes in the topology are going to behave. If multiple algorithms are in play, those assumptions may not hold. - Two concurrent algorithms, while each individually correct, can still produce a flawed flooding topology if they are asked to interoperate. - Full flooding at the boundary between the algorithms is not sufficient to correct the situation. Regards, Tony _______________________________________________ Lsr mailing list -- [email protected] To unsubscribe send an email to [email protected]
