Hi Rene,
This is all I have to say, THANK YOU! 
Complicated concepts, made very objective.
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Hi Rene In your lesson you said R1 will adveretise a Distance of 5+4+3 = 12 to R2, however you said that R1 will not advertise 5+4+3=12 to R3 Because of Split Horizon Correct ? I’m just making sure because you said
(“R1 will learn the destination network through R3 and R2. R3 will advertise a distance of 5+4 = 9 to R1. So why didn’t I place “9” in the advertised distance field in the table? Good question! Remember split-horizon? Don’t advertise to your neighbor whatever you learned from them….R1 is not sending information about this network back to R3. To be more specific: whatever you learn on an interface you don’t advertise back out of the same interface”). Basically saying when R3 recieves an AD from R1, it will not be 12 but instead 25. Is that correct ?
Hello Dennis
Yes, what you’re saying is correct.
- R1 will advertise a distance of 25 to the destination to R3 because it learned this distance from R2.
- Similarly, R1 will advertise a distance of 12 to the destination to R2 because it learned this from R3.
So, R1 will always give the distance to the destination to each router that it did not learn from that particular router.
I hope this has been helpful!
Laz
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I Labbed it up and according to what you said, R3 should have the route for the destination network through R1 and R2 in the topology table if you use the command show ip eigrp topology all-links. the only way to see the routes with that command is when you turn off split-horizon on the interface. i do not know if it can happen without disabling split horizon. thanks
I think i figured it out. i forced R3 to not to advertise it to R1 and now R1 is advertising it to R3 and i could see the two routes in the topology table with the command show ip eigrp topology all-links. but now what is the point of not accepting these routes since split horizon is there to prevent loops from happening. thanks
Hello Hany
Can you let us know which topology you labbed up so that we can help you out? Rene has three different tolologies in this lesson.
Thanks!
Laz
HI laz,
this one https://networklessons.com/wp-content/uploads/2013/02/eigrp-successor-rule-example.png
Hello again Hany
So, in the original setup, with split horizon functioning, R3 will advertise the destination network to R1 and R2, however, R1 and R2 will not re-advertise the destination back to R3 because of the split horizon rule. This is why you can’t see these routes in the topology table. Now I noticed that @ReneMolenaar had the following list of advertisements that would occur:
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.
The last two should not occur, since the split horizon rule is in effect. I will let @ReneMolenaar know about this issue.
If you forced R3 not to advertise to R1, then R1 will learn of the destination from R2 and R2 from R3. So, R2 can advertise to R3 without violating the split horizon rule.
Under normal circumstances, you shouldn’t need to manually change what R3 is advertising to R1, split horizon will take care of avoiding loops.
I hope this has been helpful!
Laz
Hello Laz,
I think you mean R1 can advertise to R3 without violating split horizon rule. There is no way R2 will advertise the network to R3 as it has originally learned the route through it and since split horizon is on
Hello Hany
Yes you are correct, the phrase should read “So, R1 can advertise to R3 without violating the split horizon rule”
Thanks!
Laz
Hello,
I still have doubt to get Advertised distance and Feasible distance as per topology. As mentioned that Advertised distance: How far the destination is away for your neighbor and Feasible distance: The total distance to the destination.
I followed the given steps:
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.
But i am not getting same value as mentioned in result.
As per my understanding the result is mentioned below:
- R4 will advertise the destination Network to R3 which is 5 so AD for R3= 5
- Now R3 will advertise to R1 and R2 which is 9 so as of now AD for R1 and R2 will be 9 (5+4)
- R1 will advertise to R2 what it has learned from R3 so AD for R2= 12 (5+4+3)
- R2 will advertise to R1 what it has learned from R3 so AD fro R1=18 (5+4+9)
- R1 will advertise back to R3 what it has learned from R2 so AD for R3=25 (5+4+9+7)
- R2 will advertise back to R3 what it has learned from R1 so AD for R3=19 (5+4+3+7)
We have two AD values for each router here so now which Advertised distance (AD) would be selected.
And how R4 Feasible distance would be 9 ? as it is directly connected to destination.
R3 must have AD=5 and FD=9 but in case R4 i am confused !
Please help me out to know this concept !
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Hello Abhishek
I understand your confusion, and it is complicated further because of the names that are used (advertised and feasible distance etc.) I will attempt to clear it up.
In this example, we are looking at the network from the point of view of R3. So imagine yourself physically at that location and attempt to view the network from that perspective.
A few definitions:
Advertised Distance: the distance that OTHER routers are at from the destination. The OTHER routers advertise this information to R3 and R3 populates its topology accordingly.
Feasible Distance: the distance to the destination via each possible route as recorded in the topology of R3.
Remember, we are looking at the network from R3’s perspective.
So now for Advertised Distances
- Advertised distance from R4 to the destination is 5. Why? Because R4 informs R3 that to reach the destination, it has a distance of 5. Straightforward.
- Advertised distance from R1 to the destination is 25. Here’s the tricky part. R1 has two paths to the destination: R1–>R3–>R4–>Destination and R1–>R2–>R3–>R4–>Destination The first has an advertised distance of 3+4+5=12 and the second has an advertised distance of 7+9+4+5=25. We use 25 not 12. Why? Because, to quote Rene directly: “Remember split-horizon? Don’t advertise to your neighbor whatever you learned from them….R1 is not sending information about this network back to R3. To be more specific: whatever you learn on an interface you don’t advertise back out of the same interface.” So R1 does not advertise the R1–>R3–>R4–>Destination administrative distance because of split-horizon. This is why 25 remains in the table.
- Advertised distance from R2 to the destination is 19. Again, like before, R2 has two routes to the destination: R2–>R3–>R4–>Destination and R2–>R1–>R3–>R4–>Destination with distances at 9+4+5=18 and 1+3+4+5=19. Again, because of split horizon, the first is not used, so the advertised distance is 19.
For Feasible Distances from R3:
- For the route via R4, the Feasible distance is 4+5=9
- For the route via R1, the Feasible distance is 3+7+9+4+5=28
- For the route via R2 the Feasible distance is 9+7+3+4+5 = 28
So to answer your questions directly:
Yes, for R1 and R2 we have two Advertised Distances. But the only ones that will reach R3 are those that have been learned from a different interface. Remember split horizon. So only one is chosen.
Remember we are looking at the network from R3s perspective. So the Feasible Distance of 9 indicates that the route to the destination FROM R3 via R4 will have a total distance of 4+5=9.
I hope this has been helpful!
Laz
Hello Laz,
Thanks for clarification. Now i am getting your point.
Please make me correct if i am wrong. Now i am taking AD and FD value from R2’s perspective.
- Advertised distance from R3 is 9 (Path R3->R4) directly
- Advertise distance from R1 is 25 (Path R1->R2->R3->R4->Destnation). Second path would not be selected as per split horizon rule which is (R1->R3->R4->Destination).
Feasible distance
-
FD from R3 is = 9+4+5=18
-
FD from R1 is = 25+7= 32
Thanks,
Abhishek Saini
Hello again Abhishek
You’ve almost got it. I think you understand it, but you just misread some costs.
From R2’s perspective:
- Advertised distance of R3 to destination is 4+5=9 (Path R3–>R4–>Dest). You are correct here.
- Advertised distance of R1 to destination is 3+4+5=12 (Path R1–>R3–>R4–> Dest). The second path that will not be included would have been R1–>R2–>R3–>R4–Dest. which has a distance of 7+9+4+5=25. This will not be used as per split horizon rule. Remember, we are looking at the network from the point of view of R2 now.
Feasible Distance
- FD of R2 VIA R3 is 9+4+5=18 You are correct here.
- FD of R2 VIA R1 is 7+3+4+5=19
I hope this has been helpful!
Laz
Hello Laz,
I believe, Now i got the point.
Thank you so much for your help.
Just for my practice i found out AD and FD values from R1 perspective.
Advertised distance from R3 is 9 (Path R3->R4) directly
Advertise distance from R1 is 18 (Path R2->R3->R4->Destnation). Second path would not be selected as per split horizon rule which is (R2->R1->R3->R4->Destination).
Feasible distance
FD from R3 is = 9+3=12
FD from R2 is = 18+7=25
Thanks,
Abhishek
Hello Abhishek
Yes you’ve got it! I assume however that you R2 and not R1 in the following statement:
Good work!
Laz
Hi Rene
In this topology
I don’t understand where all of this value come from it see so complicate
Do u have better way help me to understand this ?
Hi Heng,
These values are metric values that I made up. On real routers running EIGRP, you will see very large numbers for the metrics which makes it harder to explain.
The advertised distance is the metric that a router has received from its neighbor. The feasible distance is the total distance to get to the destination. In other words, the advertised distance + the metric on your own interface.
To understand the values of the table in your screenshot, it’s best to fill in the topology tables of R1, R2, and R3 yourself first:
R3’s topology table:
| R3 | Advertised distance | Feasible distance |
|---|---|---|
| R1 | ||
| R2 | ||
| R4 |
R1’s topology table:
| R1 | Advertised distance | Feasible distance |
|---|---|---|
| R2 | ||
| R3 |
R2’s topology table:
| R2 | Advertised distance | Feasible distance |
|---|---|---|
| R1 | ||
| R3 |
For example, here’s the first entry for R3:
| R3 | Advertised distance | Feasible distance |
|---|---|---|
| R1 | ||
| R3 | ||
| R4 | 5 | 9 |
In the topology table of R3, you see an entry for the advertised distance of 5. That’s what R4 advertises to R3. R3 adds its own metric on the interface that connects to R4 (4) to the advertised distance. 4+5 = 9 so that becomes the feasible distance.
R3 advertises this entry to R1 and R2.
R1’s topology table looks like this:
| R1 | Advertised distance | Feasible distance |
|---|---|---|
| R2 | ||
| R3 | 9 | 12 |
R3 advertises a metric of 9 so that’s what we add in the advertised distance. It adds its own interface metric (3) to this value, so the feasible distance becomes 12.
Now see if you can enter the remaining values yourself. Keep in mind that that the route is also advertised from R3 > R1 > R2 and back to R3. It’s also advertised from R3 > R2 > R1 > R3.
hi rene i have question regarding the below example in the picture:
when R3 received 25AD from R1 and 19AD from R2 will R3 send those values back to R4?
Hello Ziad
R3 will share its own advertised distance to the destiation with R4 and not the advertised distance reported by R1 and R2. So R3 will share an AD of 9 to the destination with R4.
In general, in EIGRP a router will share its own AD with other routers and the the AD reported by other routers.
I hope this has been helpful!
Laz
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