Feasible successor in EIGRP

Unanswered Question
Aug 20th, 2007

to become feasible,

AD of route must small than FD of best route

OR

AD of route must small or equal to FD of best route.

(AD<FD) or (AD <= FD)

thanks

I have this problem too.
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glengregory Mon, 08/20/2007 - 10:27

Hi,

The AD (advertised Distance) for EIGRP will always be smaller than FD since it is the distance as reported by the neighbor; and has got nothing to do with being a feasible successor.

To be a feasible successor, there should only be more routes to the same destination.

Hope this answers you question...

Glen

tcheumassi Mon, 08/20/2007 - 20:28

thanks for you help.

Then if I well understand, if a have 2 routes to reach network X,

A-B-C--X AD 10 FD 20

A-E-C--X AD 20 FD 25

I will have just one succesor (route A-B-C--X) and no feasible successor? (because AD '20' is no small than FD '20')?

thanks

glengregory Mon, 08/20/2007 - 23:24

Hi,

If a router has 2 routes for X with different metrics (FD 20 & FD 10), then you will have 1 SUCCESSOR route FD 10 & 1 FEASIBLE SUCCESSOR FD 20 route. Both the routes will be in the topology table (#show ip eigrp topology). But, only the SUCCESSOR route will be in the routing table (#show ip route).

Just forget about the AD.

Glen

glengregory Wed, 08/22/2007 - 04:05

Hi,

Take for example 2 routers connected to each other with serial links R1 <--> R2. Each has an ethernet port also. R2 will advertise the (RD) of its e0 link to R1, for e.g. RD 2180. Now R1 has to compute FD for R2 e0 (its metric to reach R2 + RD) which will always be higher than RD.

There's an error in the cisco doc you mentioned:-

"There are two routes to Network A from Router One: one through Router Two with a metric of 46789376"

"The reported distance from Router Two is 46277376, which is higher than the feasible distance"

RD 46277376 is NOT HIGHER than FD 46789376.

I have not yet read the "loop free path" fully yet. But, I think that without Routing loops, the FD will always be higher than RD simply because the router has to compute the metric to the neighbor.

Please give me 1 scenario where in a perfectly working network, the RD > FD.

Thanks for the link. It was very insightful.

Glen

CSCO10892433 Thu, 08/23/2007 - 00:34

Hi, Glen

I think you get wrong with the definition of FD. Here is the correct definition of FD:

FD(network A) = Minimum(Metric-1,Metric-2,...,Metric-n)

where Metric-1 = metric to network A via neighbor 1

Metric-2 = metric to network A via neighbor 2

...

Metric-n = metric to network A via neighbor n

In this case, metric via neighbor TWO = 46789376 and metric via neighbor FOUR = 20307200. So, the FD = min(46789376,20307200)= 20307200.

Because AD of neighbor TWO = 46277376, which is greater than FD(20307200), so neighbor TWO cannot be a feasible successor for network A.

HTH

SSLIN

glengregory Fri, 08/24/2007 - 02:06

Hi,

Now, I got it!!!

I misunderstood the RD should be < FD for FS.

Even though reading the doc, I still did'nt understand until I read your txt file for the calculation.

Thanks a lot for the correction. I really appreciate it. I know more of EIGRP now for the CCNA.

Thanks again,

Glen

tcheumassi Thu, 08/23/2007 - 10:38

yes is correct that to become the Feasible Succesor the router must have AD small than FD of best route to a given destination.

AD

thanks for all

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