why feasible successor must less than successor

hello all,

i just need to know why in eigrp

“To qualify as a feasible successor, a router must have an AD less than the FD of the current successor route“

what if it was higher ??

* my opinion here as if fasiable successor (ad) was higher than successor (fd) , so if successor fails so it will not enter again as successor when he back to work

correct me and guide me

Find more posts tagged with

Comments

fredrikjj

It's a loop prevention mechanism. If a neighbor's metric for a route is lower than the metric I'm using, they didn't learn that route from me and sent it back to me. If it's loop free, it's safe to quickly change to that route.

bermovick

It's simply to guarantee a loop-free path.

If a neighbor is advertising a route with a cost that's higher than another neighbor's cost for the same route but less than the routers own cost for that better route, then it's mathematically impossible for the higher-cost route to have gone through the router once and looped around to have reached it again.

EdTheLad

While i'm sure you probably know the answer bermovick, that explanation is terribly confusing.
As Fred put it, if my neighbor has a higher cost to the destination that me, he might have learnt the route from me.

JHON CENA

Thanks all for your feedback but ... it still confused ..i didn`t understand why there is a loop ??

While i'm sure you probably know the answer bermovick, that explanation is terribly confusing.
As Fred put it, if my neighbor has a higher cost to the destination that me, he might have learnt the route from me.

still confused , if there is a pic or example it will be perfect

fredrikjj

EIGRP Feasible Successor Routes - PacketLife.net

Though, split horizon will prevent this scenario typically.

gorebrush

How would split horizon help? That stops you sending out information on the same port as you received it in?

EdTheLad

A)
If my cost to reach network X is 100, and my connected neighbor has a cost 50 to reach network X, i know 100% that my neighbor did not learn the route from me. Why? because if they did, their cost would be 100 ( my cost ) + whatever additional cost they have. So if their cost is less than mine, i can be sure they are not looped through me.

If my cost to reach network X is 100, and my connected neighbor has a cost of 200, there is a possibility that they learnt this route from me, 100 ( my cost ) + 100, that additional 100 could be their cost to me.

So with scenario A, i can be sure there is no loop. With Scenario B, there could be a loop, maybe the neighbor just has a high cost to the destination network, but if i follow the feasible rule that my neighbor must have a lower cost, i can be absolutely sure that i'm not causing a loop.

fredrikjj

gorebrush wrote: »

How would split horizon help? That stops you sending out information on the same port as you received it in?

R3 will not receive R5's network back from R2 or R1.

PS.
Disabling split horizon on R1/R2 results in R3 having multiple routes to R5 (via R1/R2), but it still won't cause issues beyond unnecessary entries in the topology table because they'll never pass the feasibility condition.

EdTheLad

Yes, R2's next hop is R3, so it wont send the update to R3.

JHON CENA

thanks all i got it

Danielh22185

I might also add to be careful with topics like this. It can become easily confusing and makes for good exam questions.

For the longest time I thought a route could be used as a feasible successor that normally would not due to having a higher AD than the FD of the current successor IF the variance was set to a higher value. EIGRP can be tricky and I personally think this is the most easily confused subjects surrounding EIGRP. Study up!