Route selection question: Feasible successors in EIGRP

[mpeppers]

This may be a stupid question, but I am having trouble working out how a feasible successor is selected. As I understand it:

In a situation where there are multiple routes to a given destination, EIGRP computes the cost of each link (based on constants that weight the following: bandwidth, delay, link status, reliability, and MTU, with only the first two being defaults) to determine the best route(s), this route(s) is named the successor and its cost is the feasible distance/advertised distance/reported distance (Are these synonymous?). In order to avoid recalculations, EIGRP also looks for feasible successors, which replace the successor route if it becomes unavailable. Any other route may be a feasible successor if its cost is less then the feasible distance/advertised distance/reported distance. I am confused about how those feasible successors are chosen. If another route has a cost that is less then the feasible distance, wouldn't it be the successor? I am obviously missing something here.

Mike

(another CCNP hopefull)

[steve.barlow]

Feasible distance (FD) is the lowest metric/best path to a destination. Advertised/reported distance (AD) is the metric sent to the local router by the advertising/next hop router. FD = AD + the cost of the local router to get to the next hop (who advertises the AD). Remember cost/metric is calculated by the bandwidth/delay leaving an interface, so that must be taken into account in the FD (ie the cost to get to the router that sent the AD).

For a route to enter the eigrp toplogy table, the AD must be less than the FD.

Hope it helps.

Steve

[thisisshanky]

HI,

Advertised distance is distance from the neighbor to the destination.

Feasible distance is distance from you to the destination.

A----B----C.

A , B, C are 3 routers connected serially. A Network X on C, will be advertised by B to A. Distance (Total) from A to network X is feasible distance.

Distance of B to network X is called advertised distance.

Now if another router D is connected to A and C (not to B), that is A has a parallel link to C through D, D will also advertise the same route X to A. NOw A is in a dilemma to find out which route to use.

To solve this, comes Feasibility condition. It states, that feasible distance to network X through any other neighbor (this distance is called Reported distance or Advertised distance) should be less than the current feasible distance. See the following example.

A----B----C-x

|-----D-------|

|------E-------|

A has 3 neighbors B, D, E.

I am assuming, A has equal distance of 30 to neighbors B, D, E.

Assume B reports (reported or advertised) its FD to X as 20. and A's FD to network X be 20+A's distance to B = 50.

Now D reports an FD to X as 30 which is less than 50 (A's current FD), D wil be eligible as a Feasible successor. (Note that A's distance to X through D would be 30+30=60, which is greater than 50 and hence wont be there in routing table, but since its eligible as Feasible successor, it will be there in topology table.

If another neighbor E reported FD to X as 60, which is greater than 50, E wont be eligible to become a feasible successor.

Hope its clear.

[mpeppers]

Thanks for the help, I knew that I was missing something obvious.