Eigrp

CiderCider Member Posts: 88 ■■□□□□□□□□
Hi Guys,

Im good with EIGRP except I cannot wrap my head around Feasible Succ ---> Admin Distance (reported distance) < Feasible Distance.

Looking at the attached which I got from CBT in their EIGRP unleashed video , I cannot understand why R3 cannot take the route to R5 as a FS. What happens if the link between R3 and R4 go down? The only way it can get to R4, R6, R7 is through this link so logically it should be the backup link.

What am I missing here?

Any help would be appreciated.

Comments

  • mikeybinecmikeybinec CCNA 200-120 Member Posts: 484 ■■■□□□□□□□
    It would be helpful to see a sh ip ei to command but since that is not available I'll take a shot. .Your reported distance is what your neighbor's router is saying it costs to get to the destination. Obviously, it's gonna be less than the feas distance to the destination from your router. Looking at your topology, I would say that you are load balanced to either R4 or R5 unless you modify the bandwidth of the links. R6 and r7 would not be in the topology because they are outside the AS, but I could be wrong. You oughta build this on packet tracer-run sh ip ei to and then sh ip ro to get an idea on on FSs and FDs.
    Cisco NetAcad Cuyamaca College
    A.S. LAN Management 2010 Grossmont College
    B.S. I.T. Management 2013 National University
  • StarwarsStarwars Member Posts: 12 ■□□□□□□□□□
    The feasible condition is an eigrp loop avoidance mechanism, i.e. its a calculation that's performed to guarantee traffic sent to the next-hop will no be looped back to the sender. This does not mean that if the calculation fails a guaranteed loop is going to happen, it just means that if the calculation passes you are 100% sure there is no loop.

    Lets say i have 3 routers A->B-C

    The feasible condition states that, if our neighbors reported distance is lower than our successors feasible distance for a specific prefix ,we are sure its a loop free path.

    Lets say C advertises a prefix 10.0.0.0/24 to B, the cost from B to C is 10, B has only one path to C, this path is the successor and its cost is 10. B will advertise this cost to A, when A receives this cost( B's reported distance) it adds its egress interface cost which is 10 and now A has a cost of 20 to get to 10.0.0.0/24.

    A has only one path to C, i.e. one successor, the cost to C is 20 i.e. the feasible distance. The reported distance i.e. the cost B advertised to A is 10. Looking at the feasible condition. in respect to A

    The reported distance from B(10) is less than the feasible distance from A(20), therefore it's impossible to have a loop.In simple terms if my neighbor has a lower cost to a destination than you, you can be 100% sure you have a loop free path to the destination.


    Let's now look at your topology.

    R3,R4 and R5, for simplicity lets say R4 is advertising a loopback 4.4.4.4/32

    R3->R4 is fastethernet, lets say cost is 10
    R3->R5 is fastethernet, lets say cost is 10
    R4->R5 is serial, lets say cost 64

    R3 will work out the shortest path to 4.4.4.4/32; lets say R5 advertises the loopback with a cost of 1(reported distance) to both R3 and R4. R3 will add it's local fast ethernet egress interface cost 10 to the reported distance from R5(1) which gives a cost 11 for the path R3->R4.

    R5 will receive a reported distance(1) from R4 and add its local serial interface cost(64) to give a local cost of 65 to R4. This cost 65(reported distance) will be advertised to R3, R3 will add it's egress interface fa1/1 cost 10 and have a local cost 75.

    Now looking at R3, we have 2 paths

    R3 -> 4.4.4.4/32 via f1/0 path cost 11, reported distance 1
    R3 -> 4.4.4.4/32 via fa1/1 path cost 75, reported cost 65

    R3 Successor is the path via f1/0 due to lowest path cost 11, this value 11 is set as the FD for prefix 4.4.4.4, this can only be changed to a higher value if a recalculation occurs. If a new path to the destination becomes available with a lower cost this can be updated without a recalculation.

    Now we need to check if the second path via R5 adheres to the feasible condition i.e. can we make this path a backup as in a feasible successor. Is the advertised distance from R5(65) less than the our distance to R5(75), yes its a feasible successor.

    Let's imagine the serial link had a cost 100, in that case R5 advertises a cost 101 to R3, since 101 is greater than R3's successor feasible distance(75) this does not adhere to the FC and the path is not used as a backup.

    Now you are asking what happens if the link between R3 and R4 fails? i have another path, why cant i use it as a backup. The answer is you can, you just need eigrp to perform a recalculation to work out the new topology, the link between R4 and R5 is down so it obviously wont be used in the recalculation. This recalculation involves sending queries/replies to neighbors etc and working out new successors and feasible successors, it takes some time. If you had a feasible successor in the first place this recalculation would not have to be performed and your backup path could kick in straight away.
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