EIGRP Feasible Successor Election

Question:

I have a quick question Catalyst Switches Price regarding Feasible Successor election process.

I've read that Feasibility Condition is AD<FD

So for example we have the ff feasible distances on R1

        (Cost+AD)

R2     10 + 11     = 21 (Successor)

R3     4  +  20     = 24

R4     6 +  19     = 25

Which would be chosen as the Feasible Successor?

     In this case, we have 2 entries which satisfies the condition: 20<21 ;  19<21

     Will it be both R3 & R4?

If yes, will R3 be more prioritized as the backup route when R2 goes down since it has the lesser metric?

Answer:

I've read that Feasibility Condition is AD<FD

Correct.

So for example we have the ff feasible distances on R1

Be careful here. There are two very strong misunderstandings about the feasible distance:

Some sources say that the feasible distance is the current lowest distance to the destination. This is wrong. Such a distance is simply the current distance.

Some other sources say that the feasible distance is the distance to the destination via a particular neighbor. This is also wrong. Such a distance is called computed distance or total distance.

In reality, the feasible distance is the historical (i.e. not necessarily current) lowest distance to the destination. The history starts anew when the route goes from Active to Passive state. In other words, the feasible distance is a record of the lowest distance since the last time the route entered the Passive state. It is not necessarily equal to the current lowest distance, and most certainly, there is no per-neighbor feasible distance.

In this sense, the feasibility condition says: "If a router is closer to the destination than I have ever been (now or in the past), it can not be on a routing loop."

R2     10 + 11     = 21 (Successor)

R3     4  +  20     = 24

R4     6  +  19     = 25

Which would be chosen as the Feasible Successor?

Assuming that there was no shorter path since the last time the route entered the Passive state, the feasible distance is 21. Every neighbor whose reported distance is less than 21 is a feasible successor. So in this case, both R3 and R4 would be feasible successors.

If yes, will R3 be more prioritized as the backup route when R2 goes down since it has the lesser metric?

Yes, R3 would be the next choice after R2 fails because the total distance through R3 is the next best.

Note a different thing: assume that the costs are as follows:

R2     10 + 11     = 21 (Successor)

R3     4  +  20     = 24 (feasible successor)

R4     6  +  19     = 25 (feasible successor)

R5     1  + 22     = 23

R3 and R4 are feasible successors. R5 is not a feasible successor because it does not meet the feasibility condition, yet, in case R2 fails, R5 provides the next least-cost path.

Most textbooks about EIGRP say that if the successor fails, we'll start using the feasible successor that provides the next shortest path. In this case, it would be R3 - however, the total distance through R3 is 24 while R5, it would be just 23. Staying with R3 would actually cause EIGRP to hang on a worse path than which is currently available.

So what really happens is that if the successor fails, the router will first look up the neighbor that appears to provide the next least-cost path - in this case, R5 - and only then it verifies whether it meets the feasibility condition. If it does, we start using it right away. If it does not, as in this case, we will not start using it, nor will we start using R3. Instead, we will enter the Active state and start sending queries, trying to find out if the R5 is really an appropriate next hop and can be promoted to the successor role. If it can, then after receiving all replies, the feasible distance can be reset and set to the new minimum which is now 23, thereby allowing R5 to pass the feasibility distance and to become the next successor.

Not quite an easy thing to digest, I know, but this is really what happens inside EIGRP.

You're welcome to Catalyst Switches ask further!