please help with EIGRP

Unanswered Question
Mar 11th, 2007

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

I have this problem too.
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flashsplash Mon, 03/12/2007 - 04:26

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...


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