Is this correct interpretation of EIGRP load balancing?

[hfakoor222]

From what I read in the 250 CCNP ENCOR book it says this page 157:

EIGRP supports unequal-cost load balancing, which allows
installation of both successor routes and feasible successors
into the EIGRP RIB. EIGRP supports unequal-cost load
balancing by changing EIGRP’s variance multiplier. The
EIGRP variance value is the feasible distance (FD) for a route
multiplied by the EIGRP variance multiplier. Any feasible
successor’s FD with a metric below the EIGRP variance value is
installed into the RIB. EIGRP installs multiple routes where the
FD for the routes is less than the EIGRP multiplier value up to
the maximum number of ECMP routes, as discussed earlier.

Dividing the feasible successor metric by the successor route
metric provides the variance multiplier. The variance multiplier
is a whole number, so any remainders should always round up.
Using Figure 7-2 as the example topology and output from the
EIGRP topology table in Figure 7-3, the minimum EIGRP
variance multiplier can be calculated so that the direct path
from R1 to R4 can be installed into the RIB. The FD for the
successor route is 3328, and the FD for the feasible successor is
5376. The formula provides a value of about 1.6 and is always
rounded up to the nearest whole number to provide an EIGRP
variance multiplier of 2. Figure 7-12 displays the calculation.

My Note:  Note above for the successor route it is 2816 +256+256 = 3328

The path from R1 -> R4 is 5376 ......    

5376/2238 = 1.6  as the book drawing below lays out

My Note: Now if we calculate R1  ->  R2  -> R4 path we see 7936 metric   (2560+2560 + 2816)

7936/3328   is 2.3846,  rounded up it is 3.

3*  3328 = 9984  which is the variance value 

As mentioned previously :

The
EIGRP variance value is the feasible distance (FD) for a route multiplied by the EIGRP variance multiplier.

The FD in this case is 7936 which is < 9984

So this would get installed into the RIB except for the fact that it doesn't meet the feasability condition. The reported distance from R2 (2560+2816= 5276) is higher than the feasible distance calculated locally at R1 so it does not get installed into RIB.

as the table below mentions:

Is the logic above how backup routes are decided in EIGRP?

Given the other calculations given in the book (1 other example) and that I've tried I believe this is to be correct.

Any thoughts/opinions?

[MHM Cisco World]

@David Ruess Is the best to answer you, 
please David check his request and answer him. 
good luck friends

[David Ruess]

Hello,

In your example (I opened the exact same book you have with that example) it does NOT meet the feasibility condition if the variance is 3. In the book it only gives the example of a variance of 2.

So in your example the FD for the route 10.4.4.0/24 route at R1 is 3328 for the R1->R3->R4 path. If you multiply that by 2 (variance) then you get 6656. Any metric for a route to 10.4.4.0/24 (in this case) that is less than 6656 AND meets the feasibility condition will be installed into the RIB and be used for Unequal Cost Load Balancing. In your second example of the FD being 7936. It will NOT be installed in the routing table even if the variance is set to 3. As you pointed out 3328*3 = 9984. 7936 is below that number but it does not meet the FC, however as previously mentioned the book gives a variance of 2, which in that case the 7936 route will not be installed in the routing table.

The reason the book presents it the way that you have with dividing the routes and getting 1.6 to round up and such is to make it easier to calculate. When you get into Named EIGRP mode and wide metrics these numbers can be in the billions. So instead of trying to multiply the FD of a route which could be something like 2358456958 x2, ok that didn't meet my needs, maybe maybe not, now try that same number x3 to see if that metric coves more paths, its easier just to divide what's given, get a small number and round up. That's your variance.

EDITED: @Jon Marshall thank you for catching my mistake. I've tried to edit out incorrect information.

Hope that helps. If you need more clarification on anything more specific please let me know.

-David

[Jon Marshall]

Hi David

Apologies I responded as you were responding but we seem to be saying different things. 

If I understand you correctly you are saying if you set the variance to 3 then it would add the route R1 -> R2 -> R4 but as far as I understood it that wouldn't happen as R1 -> R2 -> R4 is not a feasbile successor and so can't be used ie. variance can only be applied to feasible successor routes. 

Jon

[David Ruess]

Thanks for the catch Jon. I guess I was more focused on the variance aspect vs the feasibility condition that's affected by it and got a little lost in the sauce with my explanation. I guess that's what happens when you stop studying EIGRP for a few months without a refresher.

[Jon Marshall]

No problem, I know what it's like. 

I always need to refresh my memory when it comes to EIGRP and the way it chooses routes etc. 

Jon

[Jon Marshall · Accepted solution]

Basically yes to what you say. 

If a route does not meet the feasibility condition then it cannot be used regardless of the variance or to put it another way you can only load balance across unequal cost paths if those paths are feasible successors. 

Jon

[hfakoor222]

Thank you

[hfakoor222]

This article helped me to be able to calculate the EIGRP metric off-hand especially if you calculate all 4 routers EIGRP at the end it stuck with me how to do it

https://www.practicalnetworking.net/stand-alone/eigrp-metric/