Can someone clarify what is the correct feasibility condition?
A.The neighbor's advertised distance(reported distance) must be less or equal to the current successor's feasible distance
B. The neighbor's advertised distance(reported distance) must be less than the feasible distance of the current successor.
Whatever the answer is ,does the same thing apply to the variance condition ie: Does the Successor(FD)* variance has to be less or equal Or the Feasible successor' advertised distance has to be strictly less than the FD*variance
Option A is correct. Advertised distance must be less than the feasible distance. With variance, it can be less than or equal.
Darlington,
I racked my brain over this one as well when I was studying for route....answer is...
If you are talking about a router creating an EIGRP feasible successor, then answer B is correct..
If you are talking about variance to establish criteria for load balancing then answer A is correct.
HTH,
Wes
Wouldnt the awnser be B. regardless
isnt the varince command used on anything already in the topolgy table?
Maning that the feasabillity condition would have aready been run and only feasable successors are eligable for loadbalancing.
There was a big discussion on this a while back, please see the following:
Flexin,
The feasible successor decision with regards to variance is strictly dependant upon the variance variable configured. If the variance variable is 2, meaning that any route less than or "equal" to twice the successor value, then that route would qualify as a feasible successor.
Hi Wesley,
If the variance variable is 2, meaning that any route less than or "equal" to twice the successor value, then that route would qualify as a feasible successor.
to be a FS you must meet the FC which is : AD < FD of successor
the variance is for unequal cost load balancing: each FS whose FD is <= FD of successor * variance is installed in the routing table and based on composite metric the load sharing on each path is automatically calculated.
Alain.
this is the way i though it would work.
so the feasabillity condition is met then the variance is run on any fesable successors in the topology table.
Have i got that right ?
Variance essentially takes the feasible distance into consideration. Suppose we set the variance to 2, any paths through routers that have their FD < = 2 * FD (of successor) will also be added into the routing table and packets would be load-balanced across these paths.
Of course, the advertised distance MUST be less than the feasible distance of the successor even before it can be considered for load-balancing. If the AD is not less than the FD, then even if (because of the variance command) the FD is less than 2 * FD (of successor) that route will not be considered for load balancing.
I would explain this with a diagram but I am at work right now. Maybe once I get back I can. I'm sure someone else would've explained this better than me by then
Alain,
I agree, I believe we said the same thing in just a different way.
Wes
Thanks,
this has cleared it up for me and is what i originally thought.
so in essence the varience is only ever calculated of feasable successors.
Totaly agree with Wesley.
sorry guys.
was just reading over the whole thread and isn't this what i said in the first place ?
If the variance variable is 2, meaning that any route less than or "equal" to twice the successor value, then that route would qualify as a feasible successor.
above is what confused me. a variance of 2 does not automatically make a route a FS. The feasabillity condition determinse what routes are valid as FS then the variance is run on those FS to decide on which uneaqual cost load balancing paths will be used.
Hi wesley,
you said:
The feasible successor decision with regards to variance is strictly dependant upon the variance variable configured
I don't agree because even if we didn't use variance the FC would still be the same to elect a FS but this FS would not be considered a successor until the successor failed.
Alain
Hey Flexin,
I just described this in the CCNA Study Group, in detail; but I will post it here for you Flexin.
The variance command will not affect routes that do not meet the feasability condition: Lets say you have 4 equal metric routes for a destination using EIGRP, and the destination being 10.0.0.1. This is without using variance at all; all 4 routes just happen to all be equal. They ALL meet the feasibility condition because their reported distance (in red) is less than the feasible distance 323072.
All these routes are in the routing table by default, load balancing as they are equal metric routes. Are there any routes that don't meet the feasability condition? Well let's see here. Lets issue the show ip eigrp topology all-links command and see
P 10.0.0.1/24, 4 successors, FD is 323072, tag is 13979
via 192.168.41.93 (323072/322816), GigabitEthernet1/13
via 192.168.41.97 (323072/322816), GigabitEthernet1/14
via 192.168.41.221 (323072/322816), GigabitEthernet1/15
via 192.168.41.225 (323072/322816), GigabitEthernet1/16
show ip eigrp topology all-links
#Lines omitted
P 10.0.0.1, 4 successors, FD is 323072, tag is 13979, serno 345876
via 192.168.41.93 (323072/322816), GigabitEthernet1/13
via 192.168.41.97 (323072/322816), GigabitEthernet1/14
via 192.168.41.221 (323072/322816), GigabitEthernet1/15
via 192.168.41.225 (323072/322816), GigabitEthernet1/16
via 192 168.38.130 (323328/323072), Port-channel17
Look, the above route in red does not meet the feasability condition as its reported distance is NOT LESS than the feasable distance. In this case if you wanted to add this route to the routing table as an equal metric route with the variance command, you would first need to modify the metric so that it meets the feasibility condition; either by changing bandwidth or delay, or by using an offset list.
Message was edited by: Joshua, CCNA - Oh and Offset lists aren't part of the CCNA track.
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