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!
