Thanks Alot

It’s a good feeling while I’m looking at these questions and able to answer them with no problem.

Today pass CCNA 640-802 Score: 986/1000.
Thanks to 9tut.

I took the exam twice too, same exact labs, slight modifications. VTP EIGRP and Access lists.

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If you want to use a feasible successor for “un-equal-cost load balancing”, you must first get that route into the routing table. Remember that FS(Feasible Successor) only lives in the topology table while the successor exists in the routing table. So the condition that has to be met in order to get the FS to the routing table is this:

eigrp cost of the successor route(Feasible Distance) * variance >= ( feasible distance of the feasible successor)

And yes, if a variance of ’3′ satisfies the above formula for a Feasible Successor route, it means that FS route is 3 times worse like you said.

Basically this means: “sure we can use both routes to send packets to the same destination, and we can use the two routes to do unequal cost load balancing.. but packets going through the FS route is most likely to reach the destination slower than the successor route(hence the name unequal cost).

Whereas with equal cost load balancing, all equal cost routes are equally good to send packets to the destination..”

In this case the variance of 3 says- for the feasible successor route with the FD 3 times worse than the successor’s Cost is Ok to become the successor as well (no need to be exactly equal, be 3 times worse only – and you are the successor as well).

people, what do you think?

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chanaka, I think you got the variance concept totally wrong. *Please read EIGRP chapter on Wendell Odom’s, ICND2 Official Cert Guide, 3rd edition (pg 436 – 437 to be exact)*

The above question lists a variance that is set to some number. As soon as the variance set to a value above 1, we can be confident that it’s DEFINITELY going to be un-equal cost load balancing.

The output also shown that the maximum paths is 4. So that’s how we get the answer “maximum 4 un-equal cost load balancing paths are possible with this configuration”

i’ll try to briefly explain how the ‘variance’ works. You should actually refer to the text for proper explanation.

Remember you can load balance only with the routes that are in your routing table. And you also know that the router only puts the EIGRP successor route in the routing table. For multiple successor routes to exist in the routing table, their eigrp cost must be exactly the same! Then only the router can use these paths for “equal-cost load balancing” up-to the number of paths defined by the max paths setting (default to 4 max, can go upto 16 as i can remember)

if you want to use a feasible successor for “un-equal-cost load balancing”, you must first get that route into the routing table. Remember that FS(Feasible Successor) only lives in the topology table while the successor exists in the routing table. So the condition that has to be met in order to get the FS to the routing table is this:

eigrp cost of the successor route(Feasible Distance) <= ( variance * feasible distance of the feasible successor)

Then only it will be added to the routing table for un-equal cost load balancing. but remember this choice is only available for Feasible Successors. The condition for feasible successors is that:
The Advertised Distance(aka Reported Distance) of the Feasible Successor < The feasible distance of the successor

hth someone. it's not difficult guys.. just needs to take sometime to get your head around the concept and it's pretty clear after that.

We can use varience for load balance as follow. if there are 4 path and put varience 3
so there are 4 unequal path having. if we put 2, have 2 same paths per each.

(confg- touter)#varience 3

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Thanks

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Everywhere i read is the cost multiplixer. But for example if you have 2 path to one destination and use the variance 2 comman. it will multiples to by to 2 and still have the same result for both the path. I Can´t understand.

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