Re: If AD = FD in EIGRP

From: Ram Shummoogum (rshummoo@ca.ibm.com)
Date: Wed Jun 13 2007 - 10:19:03 ART

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It has to be less to actually meet the feasibility condition. "tested"

RAM 12304

                                                                           
             "Naresh Myaka"
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                                       If AD = FD in EIGRP
             06/13/2007 03:33
             AM
                                                                           
                                                                           
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              "Naresh Myaka"
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Hi GS,

We all are aware that in EIGRP, FC is met if AD is less then FD. I'm not
sure whether if neighbor's AD = FD will it still meet the FC and also will
it be in the topology table.

Regards,
Naresh M.

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