I'm currently preparing my BSCI certification exam, I need some explication about variance concept of EIGRP routing protocol.
In my lab I was set variance value to 80 on all routers, just to make sure my redundant path meet requirement. But the redundant path does'nt appear in the routing table and still only appears as feasable successor only. I was think I had correctly understanding variance concept in my reading, I just setup this lab to view result in routing table and how routing table work to load balance trafic base on there appropriate bandwith.
In my reading, I read thos rule for a reduntant path become also a successor :
The local best metric (the current FD) must be greater than the best metric (the AD) learned from the next router. In other words, the next router in the path must be closer to the destination than the current router. This prevents routing loops.
The variance multiplied by the local best metric (the current FD) must be greater than the metric through the next router (the alternative FD). This condition is true if the metric of the alternate path is within the variance.
(here is a copy paste from my CCNP documentation)
For 3600 point of view, I would like to understand why "192.168.1.0/24 via 192.168.3.1 (857600/281600), Serial0/0" is not appear in routing table.... is only appear in ip eigrp topology as a feasable successor.... But I set variance value to 80, the best path as a feasable distance of 307200 if I check variance rule 857600 < 80 * 307200, this path should become a feasable successor... Why is not ??
and you and Xine is the same person ?
>> if it's not a successor and not only a feasable successor how should I call this one ?
you can call it less used next-hop to net 192.168.2.0/24 if you like.
There is not an exact term for this case as you noted
Hope to help