EIGRP Advertised Distance, when there is multiple "upstream" routers adverting routes

Lion66sixLion66six Member Posts: 13 ■□□□□□□□□□
Hello All,

I have created a small topology while trying to understand Advertised Distance (Or Reported Distance in some documentation!), and Feasible Distance. It was all fine, until I started to understand what Advertised Distance is reported from R2 to R3 in the attached example.

The complication I have that I don't understand how to deal with is that R2 has 2 possible routes to get back to the network on R1. The first route is directly to R1 which gives it an AD of 1, an FD of 11. But it also has another route via R4, with an AD of 4 and an FD of 9. So in this case, what AD would it provide to R3? Would it give R3 an AD of 1 or 4?

I would like to guess that it would pick the route with the lowest AD (makes sense right?), but I can't find anything that clearly states this is the case.


  • EdTheLadEdTheLad Member Posts: 2,111 ■■■■□□□□□□
    R2 has 3 neighbors and four possible paths to

    R2 -> R1 10 + 1 = 11
    R2 -> R4 -> R1 = 5 + 3 + 1 = 9
    R2 -> R3 -> R4 ->R1 = 5 +5 +3 +1 = 14
    R2 -> R3 ->R4 -> R2 -> R1 = 5 + 5 + 5 +10 + 1 = 26

    AD is the cost the neighbor will advertise to it's neighbors i.e. it's lowest cost to the destination for that neighbor.

    R1 AD is 1
    R4 AD is 4
    R3 AD is 9

    The feasible condition states
    The reported distance musts be lower than the feasible distance of the successor.

    From the above 4 paths, R2->R4->R1 is the lowest cost 9 i.e. FD is 9; this is the successor

    The logic is, if my best cost is 9, as long as my neighbor advertises a cost lower than 9 to me, i know it wont loop back to me.

    Looking back at the AD calculated previously, i can see R1 and R4 are lower than 9 good, R3 is 9 bad.

    R4 is the successor, R1 is the feasible successor.
    Networking, sometimes i love it, mostly i hate it.Its all about the $$$$
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