The first one is correct. The route from E to A via B will be a feasible successor and with a variance of 2 it will load balance 2:1 across both path.
The second example i'm not so sure about. Where are you applying the variance command ?
If your'e doing it on Router A then there will only be two paths to F. I don't believe you would see the A-D-G-H-F path. Even if you did when the packet got to Router G that router would always forward the packet straight to F rather than via H. The path taken is relevant locally to each router.
This one is a bit more complicated. Bear in mind also that the examples you are using are based purely on link costs which in the real world is not how EIGRP does things but there's nothing wrong with this approach. In fact as you may know Jeff Doyle uses the same cost examples in his CCIE book for simplicity. So based on using costs alone.
A very important point to note and one which i overlooked in your previous example (apologies) is that you are not measuring FD to a router but to a network. So if we assume that the network we are working on is the network between F -> H which we will refer to as Net1.
The feasible distance from A -> Net 1 with the path A -> C -> F -> Net1 = 10 + 10 + 5. Note that you must include the cost of the Net1 link.
As you can probably see all your paths have an FD of 25 to get to Net1 so they would all be seen as equal cost paths to Net1.
Notice also that in contrast to what i said before Router G will also have 2 equal cost paths to Net1 via Router F and Router H.
Hi everyone, I would like to thank you in advance for any help you can provide a newcomer like myself!
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