In EIGRP does AD < FD or AD <=FD

[mirror512]

I am confused with Cisco EIGRP issue for feasible successor

Does AD < FD or AD <=FD

[JohnTylerPearce]

In EIGRP you have Successors and Feasible Successors.

Successors have the best composite metric to a specific destination and hince the best Feasible Distance which

is that routers best distance to the destination. You will only have one Successor in the EIGRP Route Table at

all times unless you are doing Load Balancing.

A Feasible Successor is a valid loop free path basically but has and Administrative Distance that is less than

the routes Feasible Distance. This makes sense when you think about it, if the Administrative Distance is more

than the Feasible Distance than you can have a loop, which is why it is not even imported into the EIGRP Topology

Table.

Successor -> Best Feasible Distance of the specified route

Feasible Successor -> Loop-free valid path but not the best

Feasible Distance -> Best composite metric to a specific destination based on that local router

Administrative Distance -> The neighboring routers distance to that specific destination.

I hope that helped.

[mirror512]

But mymain doubt is what if AD == FD

In some book i have seen AD can be = FD to become feasible successor.

Is that correct

[JohnTylerPearce]

I wouldn't think a route that has AD == FD would be installed in the Topology Table at all. Since the

Feasbility Condition is any route(s) that has an AD less than the FD of the current successor will make

the Feasbility Condiiton. I haven't technically tested this in a lab but I'm pretty sure it wouldn't be installed.

[Peter Paluch]

Gentlemen,

The inequality is strict, i.e. AD < FD. Unfortunately, I was not able to create a network scenario showing that if AD was allowed to be equal to FD, a loop would form. What I know is that the mathematical proof of the Feasibility Condition is based on contradiction (reductio ad absurdum) and requires the inequality to be strict, otherwise, the contradiction will not form and the correctness of the Feasibility Condition would not be established. This does not help much, I know, but so far it is the only exact explanation I have.

If I manage to concoct a scenario showing a loop with AD==FD, I will certainly be back. Until then, we should simply rely on the correctness of the proof that requires AD

One remark: the AD

Best regards,

Peter

[JohnTylerPearce]

That's what I was thinking Peter. I'm not even sure how you could get AD = FD. That would mean

that when the composite metric of RA(AD in this instance) comes into RB it will add that to the composite

metric of RB. Which I know that the composite metric itself isn't really sent to the other eigrp router but

the values of BW, Delay, Reliability, Load, and MTU (which isn't even used).

[lgijssel]

He guys,

From the context it looks as if you discussing FD versus RD, Feasible Distance vs the Reported Distance in EIGRP.

So I do not quite understand where the AD (Administrative Distance afik) is involved.

Please enlighten me if I missed something here.

Leo

[Peter Paluch]

Leo,

Sorry for the confusion. The AD is an overloaded acronym - it can stand both for Advertised Distance (the Cisco's unfortunate choice for Reported Distance) and Administrative Distance. With respect to EIGRP, it can be particularly confusing. You are right, we should be using the Reported Distance to avoid any confusion. Sadly, Networking Academy and even some other Cisco materials use the term Advertisted Distance and the acronym AD, adding greatly to the confusion.

In this context, however, the AD meant Reported (Advertised) Distance in EIGRP.

Best regards,

Peter

[Peter Paluch · Accepted solution]

Hello John,

I'm not even sure how you could get AD = FD.

I think I got it. Consider the following exhibit.

The R2 and R3 are interested about the distance to R1 (forget about composite metric for now). Both R2 and R3 use the direct link to R1 and their current FD is 100. When R2 and R3 tell each other their distances, they also put 100 into the advertisement. Hence, for R2, the R3's AD 100 is equal to R2's own FD, and vice versa.

Now, assume for a moment that it was allowed that AD == FD, in other words, R2 could consider R3 to be its feasible successor towards R1, and vice versa. That would mean that R2 could, at any time and without any "warning", use R3 to reach R1 because it would be sure that no routing loop can form - this is the basis of unequal-cost load balancing. The same would go for R3 - it could use R2 at any time to reach R1.

Assume now that the unequal-cost load balancing was indeed configured. For a certain statistical fraction of packets, R2 would send the packets to R1 via R3, and R3 could repeat them back to R2! That would form a routing loop - even if perhaps not for all packets, only those for which R2 would decide to send to R3 and R3 at the same time decide to go through R2 - to reach R1. Still, the network is not loop-free anymore.

Hence, the AD==FD could have at least this negative aspect. Another negative aspect would be apparent if, say, R3's direct link to R1 failed. R3 would merely select the R2 as its new successor. Until the update about the changed distance would arrive from R3 to R2, R2 could still consider R3 as its feasible successor to R1 and send packets to it which would guaranteedly be returned back by R3 to R2. Another routing loop.

So indeed, it is necessary for the strict inequality to hold, i.e., AD

MTU (which isn't even used).

MTU is indeed not used in metric computation. However, if there are more routes to the same destination having the same resulting EIGRP metric, the MTU will be used as the tiebreaker: the route with the higher MTU wins.

EDIT (August 20th 2012): MTU is indeed unused in EIGRP path calculation/selection. See my last post.

Best regards,

Peter

[mirror512]

There was one MCQ question where i had to choose three feasible successors and one of them had AD = FD.

so i was confused whether i should choose that as answer or not.

[Peter Paluch · Accepted solution]

Hello,

To avoid confusion, I will be using the acronym RD from now on to denote the reported distance as advertised by a neighbor.

I would certainly not choose the router having RD = FD as the feasible successor. I have tried to show in my previous post that such behavior would lead to routing loops which is something EIGRP ardently tries to avoid.

Best regards,

Peter

[Iluvnetwork]

AD = RD right?

[JohnTylerPearce]

Well, if you have routes to the same destination having the same resulting EIGRP metric wouldn't they

just use equal-cost load balancing by default?

[Peter Paluch]

John,

 Well, if you have routes to the same destination having the same resulting EIGRP metric wouldn't they just use equal-cost load balancing by default? 

Are you asking about the MTU issue? Well, if you had two routes with the same metric to the same destination but they would have different MTUs, only the route with the larger MTU would make it into the routing table.

EDIT (August 20th 2012): MTU is not used in EIGRP path calculation or selection. See my last post.

Best regards,

Peter

[JohnTylerPearce]

Yeah, I was really asking about the MTU Issue. I honestly didn't know if you had two routes with

equal cost composite metrics they would not equal-load balance if there MTUs mistmatched.

Thanks for your input Peter.