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How does having an EIGRP feasible successor speed up convergence?

How does having an EIGRP feasible successor speed up convergence?

PrepAway - Latest Free Exam Questions & Answers

A.
EIGRP sends queries only if there is a feasible successor, which decreases the number of routers
that are involved in convergence.

B.
EIGRP sends queries only if there is not a feasible successor, which causes less control traffic to
compete with data.

C.
EIGRP immediately installs the loop-free alternative path in the RIB.

D.
EIGRP preinstalls the feasible successor in the RIB in all cases, which causes traffic to switch more
quickly.

2 Comments on “How does having an EIGRP feasible successor speed up convergence?

  1. TripleA says:

    From the CCIE R&S v5 Official Certification Guide Vol. 1, page 393:

    “It is a widespread belief about EIGRP that if the current Successor fails, a Feasible Successor (if one exists) will always be promoted to the Successor role. This statement is not entirely correct, however.

    “[Specific example omitted]

    “What really happens in EIGRP is the following:

    “■ Whenever EIGRP detects a topology change, it first records the change into the topology table and updates the RD and CD of the neighbor that advertised the
    change (in case of a received EIGRP message) or was influenced by it (in case of a link metric change).

    “■ From among all neighbors that advertise the network, EIGRP identifies the one that provides the least CD, taking into account the updated CDs. Note that the FC is not invoked at this step.

    “■ Only after identifying the neighbor offering the least CD, EIGRP verifies whether this neighbor meets the FC and is therefore a Feasible Successor. If it is, EIGRP will promote it to the Successor and start using it right away. If, however, that neighbor does not meet the FC, EIGRP will put the route into the Active state and send out Queries, asking its neighbors to assist in locating the best route.

    “In other words, EIGRP—just like any other routing protocol—always tries to choose the shortest path toward a destination, but before using it, EIGRP verifies whether it meets the FC to be loop-free. If it does, EIGRP will use it. If it does not, EIGRP puts the destination into the Active state”.

    So the correct answer is actually B.




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