Laz,
I was re-reading few times your example as well as Rene’s topic . So correct If I am wrong : FD from R2 is 18 and FD from R1 12? I think before any calculation is very to draw the traffic flow and the will give a clue as you will encounter where is split horizon happening right?
Best regards
Yes, you are correct. But remember, we’re talking about the reported distance. This is the distance to the Destination that R3 reports to R2 and the distance to the Destination that R1 reports to R2. We’re looking at the network from the point of view of R2, and we’re analyzing the info it is receiving from the other routers.
At this point, we’re not talking about any user traffic flow. We must first determine the metric of each path, a calculation that will determine what path traffic will take, and that is the process that is being described.
R1 actually has two possible paths to the Destination:
R1-R3-R4-Destination with a metric of 12
R1-R2-R3-R4-Destination with a metric of 25
However, R1 will not report this second path to R2 simply because it learned about it from R2 (split horizon). So R2 will never receive a report from R1 of the second available path to the Destination, so it only learns about R1-R3-R4-Destination with an advertised distance of 12.
Many thanks for your reply. Very clear!!
Maybe I did not explain well my idea. The flow is to found out which interface will not report the second path . In any case it is tricky.
The EIGRP topology table contains all learned paths, whether successors, feasible successors as well as nonfeasible successors. The following command will display only successors and feasible successors:
show ip eigrp topology
However, if you want to display the nonfeasible successors as well, you must issue the following command:
show ip eigrp topology all-links
More info on this command can be found at this Cisco documentation. The topology table does not contain any paths that cause loops since the DUAL algorithm does not allow this.
Hi sir,
Fd is total distance from source to destination
AD is distance from neighbor to destination.
i have doubt sir
lease fd value is selected as best which is successor route
Then what about bout next lease fd valu is not feasible successor route?
Feasible successor route is selected either based on fd value or Ad values of others?
When a router determines the best path, called the SUCCESSOR, it will always use the FD. It does not use the advertised distance (AD) to determine this.
However, when a router determines the backup path, called the FEASIBLE SUCCESSOR, only then will it take into account the AD of the neighbor. This is called the feasibility condition.
In order for a path to become the FEASIBLE SUCCESSOR, the advertised distance of that FEASIBLE SUCCESSOR must be lower than the FEASIBLE DISTANCE of the SUCCESSOR. Now that sounds confusing, so let’s try to clarify. Take a look at this diagram:
Let’s say the router on the left wants to determine the SUCCESSOR and the FEASIBLE SUCCESSOR to the destination. The path via R2 would be the successor since it has the lowest metric. Now how about the FEASIBLE SUCCESSOR? Well, at first glance, that would be via R1 since that’s the next lowest metric. But we must check the feasibility condition. In this case:
The AD to the destination is 10. In other words, R1 is advertising to our router on the left, that it has a metric of 10 to get to the destination.
This AD of 10 must be lower than the total metric from our router to the destination, which is also 10.
Since the AD of 10 is not less than the FD from our router to the destination, this path cannot be a FEASIBLE SUCCESSOR. This is further described in the lesson as well.
I think you mean if the SUCCESSOR is up and the feasible successor fails, will EIGRP put the route into the active state, correct? (FD is feasible distance…)
The active state for EIGRP takes place whenever the finite state machine (FSM) is in the process of calculating either the successor, the feasible successor or when determining the feasibility condition of a potential feasible successor. If you have a stable topology with a successor and a FS to a particular route, and you lose that FS, the FSM will enter an active state.
However, keep in mind that the successor route that is already in the routing table remains there. So in such a scenario, there will be no routing downtime.
I went in and labbed this up to confirm this, and that is indeed what I found. Doing some EIGRP debugging, the FSM does indeed process the change in the network topology, and thus tue FS route does go into an active state, but the successor route remains within the routing table for the duration of this process, so there is no disruption of routing.
R1 will learn the destination network through R3 and R2. R3 will advertise a distance of 5+4 = 9 to R1. R1 would re-advertise the distance to [quote=“ReneMolenaar, post:1, topic:881, full:true”]
This topic is to discuss the following lesson:
Thanks alot and Its realy very nicely explained which is 100% clear with all concepts to understand .
One confirmation just to make sure I am understanding correctly :
If we are looking at the EIGRP topology table of R2 and we want to reach the destination behind R4: R2
R1 FD:19, AD:12
R3 FD:18, AD:9
Successor → R3
Feasible Successor → R1
From the point of view of R2, the numbers you have stated are correct. One additional detail is the feasibility condition that must be met in order for a particular route to be considered a Feasible Successor. The condition states:
The AD of the FS must be lower than the FD of the successor.
That’s a bit difficult to get your head around. But here it is using your numbers:
the AD of the FS is 12
the FD of the successor is 18
12 < 18
So the feasibility condition of the FS is met, so the FS is indeed valid and is placed in the topology table.
The statement says “R2 is sending its feasible distance towards R1 which is 15.” From the point of view of R2, it is indeed the feasible distance of 15 that is being sent to R1, so the statement is correct. Once that feasible distance value reaches R1, from the point of view of R1, the advertised distance is 15.
So it ultimately depends on which router is the point of reference for the context of the statement.