EIGRP Variance not working

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I am trying to get a pair of routers to load balance EIGRP with the variance command. There two sites each with two routers connected by identical T1 links (speed, delay etc). Eigrp is configured and sees it's own connection as the best link, and the neighbor router as a successor with a slightly higher FD due to the extra LAN segment. Output looks like this:

#sh ip eigrp all-l

P 192.168.64.0/24, 1 successors, FD is 2172416, serno 4

via 192.168.16.2 (2172416/28160), Serial0/1/0:0

via 192.168.1.1 (2174976/2172416), FastEthernet0/0


So the best path (FD 2,172,416) is this routers s0/1/0:0 link (a T1 link to the site) and the other route is the successor (FD 2,174,976) all of that looks ok.


Now I want to load balance the links, so I configure "variance 2" in that n*minimum metric or 2*2,172,416 = 4,344,832 which is greater than the FD of the second successor route. So I should now see both routes in the routing table, but I still only see the one route:

D 192.168.64.0/24 [90/2172416] via 192.168.16.2, 01:01:58, Serial0/1/0:0


I have tried clearing the neighbors, setting variance to 128, and even in the lab rebooting the entire router and I cant seem to get both routes installed into the routing table.


I am observing this behavior in the lab as well as on the physical routers I am working on.


Any help?



System image file is "flash:c2800nm-advipservicesk9-mz.124-9.T7.bin"



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Giuseppe Larosa Thu, 08/28/2008 - 09:35
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Hello Bill,

what you see is correct EIGRP behaviour.

Variance does not allow to override the feasibility condition = it applies only to successor and feasible successors if any.


So if neighborx's Advertised distance > Current Best Distance for the prefix neighbor x is not considered a feasible successor and isn't considered by the variance even if the calculated metric via neighor x is less then variance * best metric.


This is required because otherwise the DUAL protocol wouldn't be loop free in all possible cases.

So variance applies to scenarios where at least a feasible successor exists for the prefix.


so to be used by variance a path for prefix A via neighbor X must be:


AD_neighX (Prefix_A) < FD (Prefix_A)


AND

Feas_Distance_via_neighx(Prefix_A) < Variance * FD (Prefix_A)


Hope to help

Giuseppe

Maybe I am slightly off track here ... this question should clarify things for me. Is there only ONE FD for any given network that is used to check for a valid route, or does the router check every AD against it's FD thus as long as the FD > AD of any route learned from a neighbor it is potentially valid; or must the AD of all neighbor routers be less than the *best* FD to be valid


On the router I am looking at I have one successor route (it's own T1 link) and one feasible successor (neighbor router on the LAN). The FS is a viable route since it passes the test of Advertised Distance < Feasible Distance so it is not a 'loop'.

#sh ip eigrp all-l

P 192.168.64.0/24, 1 successors, FD is 2172416, serno 4

via 192.168.16.2 (2172416/28160), Serial0/1/0:0

via 192.168.1.1 (2174976/2172416), FastEthernet0/0


This shows I have two routes,

1 via serial with FD of 2,172,416 > AD 28160

2 via fa0/0 with FD of 2,174,976 > AD 2,172,416


Both routes pass the FD > AD test, so they are not creating a network loop.


To prove my point, on the other side of the link where I have a switch that is configured to participate in EIGRP it sees both routers and installs both as Successors automatically since it sees both as equal cost links:

P 192.168.1.0/24, 2 successors, FD is 2172672, serno 634

via 192.168.64.251 (2172672/2172416), Vlan1

via 192.168.64.253 (2172672/2172416), Vlan1


Again the setup I am working with is 2 routers each with a T1 link to the remote networks which are also connected on a local LAN via the fa0/0 interface.


R1-------R2

R3-------R4


Where R1 and R3 are on the same LAN and R2 & R4 are on the same LAN; ie the redundant path for R2 is through R4, and vice versa.


By default from R2's perspective, s0/0 to R1 is the best link to use, however R4 via fa0/0 should be a feasible successor, and with the variance command I should be able to get it installed in the routing table.


Giuseppe Larosa Thu, 08/28/2008 - 10:45
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Hello Bill,


let's take this part of show


P 192.168.64.0/24, 1 successors, FD is 2172416, serno 4

via 192.168.16.2 (2172416/28160), Serial0/1/0:0

via 192.168.1.1 (2174976/2172416), FastEthernet0/0


the AD via 192.168.1.1 is 2172416 that happens to be = to FD = 2172416


the feasibility condition says:

AD < FD

not AD <= FD


so the second path via 192.168.1.1 does not pass the feasibility condition as I had suggested in previous post.




Hope to help

Giuseppe

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