Hello,
I am trying to understand the below paragraph.
“Sia Status (Stuck in Active): This is a bad one…it means that EIGRP has not received a reply to a query packet from one of the neighbors within the allowed time (about 3 minutes). When this happens, EIGRP will drop the neighbor adjacency, and it will be stuck in active. More on this later!”
Now my question is, If a network is unreachable in EIGRP then it will send a query packet to its neighbors. If the router does not receive any reply for that query then why would it drop its neighbor adjacency ???
The thing I struggled with is I assumed that this was the distance for each router to the destination. It is not about that. What this lesson is saying is that each of these routers is advertising a route with a certain feasible distance to R3 while following the split horizon rule. After all of that R3 looks at all of the routes for the destination that it has received from all of the routers and looks at the lowest feasible distance while following the feasibility condition.
one of the questions I had for a while was if split horizon blocks the interface on R1 from advertising wouldn’t it block it if we had 3 routers like this R1 → R2 → R3. It would not because split horizon just stops us from advertising a route out of the same interface it doesn’t stop us from receiving a route and adding our interface metric to it.
just wrote this just in case it helps anyone else.
The af-interface command is used to define user defaults to apply to EIGRP interfaces that belong to a particular address-family. The command can only be used with EIGRP’s named mode.
The command can be issued from the address-family configuration mode within the EIGRP router configuration. You can enter the configuration mode of all EIGRP interfaces on the router and issue commands for all of them, or you can choose individual interfaces to configure. For example, the following configuration sets all of the EIGRP interfaces to passive mode. Also, notice the options offered for the af-interface configuration mode command:
The result of the default keyword is that all EIGRP interfaces associated with that address family have been configured to be passive. The af-interface command also allows you to specify specific interfaces for which to apply EIGRP configuration parameters. The following commands change the EIGRP hold time for the GigabitEthernet0/1 interface:
You can achieve the same thing by going to each EIGRP interface and applying whatever EIGRP command you like, however, this makes it easier to do so directly from the configuration mode of the address family.
You can find out more information bout this command at the following Cisco command reference:
Thanks so much for your kind words! We work hard to make networking concepts as crystal clear as possible. I hope you find the rest of the lessons and the forum useful!
When it comes to the Feasible and Advertised distances. These are not directly advertised by the routers, correct?
To be more specific, routers calculate both the FD and the RD based off the metric parameters the neighbor sent to them, correct? The neighbor doesn’t directly say “my FD is xxxx”, it instead provides the parameters which it used to calculate it. The receiving router then uses these parameters to locally calculate both the RD and its FD.
And one extra thing. So in order to consider a route a feasible successor, its RD must be less than the FD of the current successor. This is to prevent loops, correct? Is there an example of where these loops could occur or is it just the potential for loops to happen in the first place?
Feasible distance and advertised distances are all examined from the point of view of the local router. These comparisons are performed internally and their results are not advertised to other routers. They are used internally only to determine the successor and feasible successor. However, one thing to note is that the advertised distance by definition is actually advertised . It’s the value that a neighboring EIGRP router has reported as its distance to the destination in question. Now just to be clear, if you look at an EIGRP update in Wireshark, you will not see the actual metric value, but you will see the various components that comprise it, such as delay, bandwidth, reliability, load, and MTU. It is those values that are included in the update. When that is received by the local router, it will use the EIGRP formula using configured K values to calculate a specific number as the metric. That metric will then be used to compare with the feasible distance to determine if a path becomes a feasible successor or not.
Here is an example of the contents of an EIGRP update message that includes information about a particular prefix:
Note that you can see, under the heading “Legacy Metric” the delay, bandwidth, MTU, reliability and load that comprise the EIGRP metric.
Another way to say the feasibility condition is to say that the “candidate feasible successor router should be closer to the destination than the local router.” Yes, this is indeed used to prevent loops. A potential scenario where a loop could occur might be if a router incorrectly advertises a lower-cost path to a network that actually routes through the receiving router. This could cause the receiving router to start routing packets to that network through the advertising router, creating a loop.
This is probably very random but I was wondering if there is a way to support you guys and get like one of the shirts in this video by purchasing it or something similar. I don’t know if you guys do anything like that if it’s only for employees.
“R4 will advertise the destination network to R3.
R3 will advertise the network to R1 and R2.
R1 will advertise the network to R2.
R2 will advertise the network to R1.
R1 will advertise this network back to R3.
R2 will advertise this network back to R3.”
is there a certain rule in here? I mean for example why firstly R2 advertise back to R3 before R1?
but it’s not so , Why for example R2 advertised distance is not 18 but 19 for example, why it’s choosing to advertised distance from a specific route path?
Here’s the list of steps again, but with numbers as shown in the lesson:
R4 will advertise the destination network to R3.
R3 will advertise the network to R1 and R2.
R1 will advertise the network to R2.
R2 will advertise the network to R1.
R1 will advertise this network back to R3.
R2 will advertise this network back to R3.
Steps 3 and 4 typically take place at the same time independently from one another. Similarly steps 5 and 6 are also simultaneous and independent. So the list could have been:
R4 will advertise the destination network to R3.
R3 will advertise the network to R1 and R2.
R1 will advertise the network to R2 AND R2 will advertise the network to R1.
R1 will advertise this network back to R3 AND R2 will advertise this network back to R3.
This is probably a more correct listing of the steps taken. Does that make sense?
Concerning your second question, you must keep in mind that the values of AD and FD are calculated from the point of view of R3. Secondly, you have to remember the split horizon rule. A router will not advertise a network to a neighbor from which it received that route. So When R1 sends its AD to R3, it sends this path:
Thank you for explanation. I understand the sequence of operations But still didn’t get the split horizon rule. I mean I understand what split horizon rule is but am not able to grasp in that example.
R3 is advertising the path via R4 to destination with 9 metric. But R1 can’t advertise the route via R3 to R3, So you can’t advertise a network to a router that you received the network update? But this is valid in here because we have only one destination network. If we would have 2 destination network then the one that’s not advertised through R3 to R1 can be advertised to R3 via R1? Right?
R4 will advertise the destination network to R3. - This should be 5 right? so… 4 + 5 = 9
R3 will advertise the network to R1 and R2 as AD = 9.
R1 will advertise the network to R2 as AD = 3+4+5 == 12
R2 will advertise the network to R1 as AD = 9+4+5 == 18
I cant seem to find the numbers you calculate. Where am i going wrong?
Something doesn’t make sense to me here. R3 should never see the AD from R1 and R2 for two specific reasons:
These routes should not even be advertised from R1 and R2 to R3. Why? Because of Split Horizon. R1 and R2 are using R3 as the successor here thus they should not advertise this network back to it
These routes should be poison reversed upon their reception
Thanks for your response. I think you’ve answered my question in a roundabout way (I also asked my question without reading a little bit further. sorry!). Split horizon is a new concept for me but makes complete sense. Rene does go on to breakdown how split horizon plays into this, however your confusion is relatable. I think the exercise is there to trick you a little and dig deeper into the further technicalities of it all. That is ultimately how we learn when having to think critically! Cheers!
The split horizon rule is applied to each destination network individually so we are only talking about one network at any time. So for the specific destination network connected to R4, R3 has advertised this network to R1. R1 now has that network in its routing table, and it knows it has learned it via R3. Therefore it will never advertise the route it learned from R3 back to R3. In the same way, it will never tell R3 the advertised distance to the destination network using the path via R3. It will only announce its advertised distance via any other route it may have, and that is the route via R2, that’s why it uses the R1-R2-R3-R4 route as the advertised distance. Does that make sense?
Remember, we are dealing with two different measurements here: Advertised Distance (AD) and Feasible Distance (FD), and we are looking at them from the point of view of R3. Having this in mind, I will attempt to respond to each of your statements above:
R4 sends its FD to R3. Once it reaches R3, it is considered the AD, which is the distance that R4 reports that it has to the destination. So R4 informs R3 with a message that says: “I can get to the destination with a metric of 5”. So the AD that R4 reports is 5. But R3 will thus have a resulting FD of 9 because, from the point of view of R3, the FD includes the total cost to the destination.
Yes, R3 will send its own FD to R1 and to R2. When R1 and R2 receive it, it is considered the AD that R3 reports to the destination, essentially sending a message saying “I can get to the destination with a metric of 9”.
You would be correct if the split horizon rule was not used. The path 3+4+5=12 that you state uses the path R1-R3-R4. However, because R1 learned about this route from R3, the split horizon rule says that it will never re-advertise that route back to the router from which it was learned. So, R1 will only send its AD to R3 if it has an alternate route to the destination. And it does, via R1-R2-R3-R4, which has a total distance of 7+9+4+5=25. So the AD that reaches R3 from R1 will have a metric of 25.
Again, this uses the path R2-R3-R4 which violates the split horizon rule. So R2 will only send an AD via an alternate path, if it has one, and it does, which is via R2-R1-R3-R4 which has a metric of 19.
. It’s the value that a neighboring EIGRP router has reported as its distance to the destination in question. Now just to be clear, if you look at an EIGRP update in Wireshark, you will not see the actual metric value, but you will see the various components that comprise it, such as delay, bandwidth, reliability, load, and MTU. It is those values that are included in the update. When that is received by the local router, it will use the 
