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Webcast-Catalyst9k
New Member

please help with EIGRP

Hello Everyone!

I am reading the CCNA 3 and 4 book, the EIGRP chapter. There is a bit that I just do not understand. Is there a mistake in the book? I have attached the diagram from the book and the text to the diagram, word to word, from the book is below:

"Note in the example in Figure 4-6 that the router D does not have a feasible successor identified. The FD (or calculated cost) for the router D to router A is 2 and the advertised distance (AD), or reported distance, via router C is 3. Because the AD is smaller than the best-route metric but larger than the FD, no feasible successor is placed in the topology table. Router C has a feasible successor identified as well as router E because the route is loop free and because the AD for the next hop router is less than the FD for the successor."

Please, could anyone explain why the book says that the FD for router D to router A is 2? Shouldn't it be 5?

Thank you

2 REPLIES
New Member

Re: please help with EIGRP

It's seems like a mistake from the book yes. And the diagram doesn't really make sense. Is it like this exactly in the book?

The FD/AD rule is very simple...

IF the AD of the feasible successor is less then de FD of the successor then it kan become a feasible successor. (this is also a criteria for useing the variance option)

Don't let some books/documents/movies distracted you. Just stick by the formule and u will be fine...

Re: please help with EIGRP

Hi, lunarsyatems

Here is a better example from cisco(regarding the FD,AD and feasible successor):

http://www.cisco.com/en/US/tech/tk365/technologies_white_paper09186a0080094cb7.shtml#feasibleandreported

HTH

SSLIN

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