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8641 Views 9 Replies Latest reply: Dec 4, 2010 1:42 PM by Warren Sullivan - CCNP RSS

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EIGRP Feasibility Condition, oh and RD and FD

Dec 4, 2010 6:17 AM

Warren Sullivan - CCNP 934 posts since
Jun 4, 2010

Hey guys,

 

Looking for a laymans definition of Feasibility Condition, i have this explanation, which i understand to an extent, but my question is why?

 

"If, for a destination, a neighbor router advertises a distance that is strictly lower than our feasible distance, then this neighbor lies on a loop free route to this destination"

 

So, for the above statement, i understand that if an EIGRP router advertises to me a composite metric that is lower than what i know to a particular destination network, then it guarantee's it lie's on a loop free route to the destination.

 

How and why does it guarantee this?

 

Thanks guys

 

 

Oh and can someone confirm im on the right track with these two brief statements regarding RD and FD, maybe if someone could expand on it a little for me?

 

-RD is the total metric along a path to a destination network, advertised from an upstream router.

 

-FD is the lowest known distance known from a router to a particular destination. This is the RD + the cost to the router that reported it......

 

I know this is a big question, but, could someone give me the low down on how this works?

 

Thanks again!

Warren

  • shanekillian 230 posts since
    Aug 18, 2008

    Hi Warren,

     

    In order for a route to be considered a Feasible Successor and be  INSTALLED in the Topology Table, the route's Advertised Distance MUST be  LESS than the CURRENT Feasible Distance.

     

    You may find the following useful in understanding the advertised values and the WHY rather than just having to accept that a router will choose the best value.

    http://www.devilwah.com/2010/10/ccnp-route-part-6-eigrp-terminology-in-diagrams/

     

     

    Shane

  • Paul Stewart  -  CCIE Security, CCSI 6,993 posts since
    Jul 18, 2008

    Yep.  The RD is the distance the neighbor told you.  The FD is when you also consider the link to get to the neighbor.  The FD for RTRx should be > the RD received on a link by RTRx.  Otherwise, you are going away from the destination.  As another note, the FD for RTRx is the AD (advertised distance) sent to other routers further downstream.  This AD is their RD and they calculate the full path for their FD.

     

    Now, I live in Kentucky.  I can drive from here to Orlando in about 10 hours.  I could drive to Cincinnati first, but I hear that it is like 13 hours to Orlando from there.  Therefore, that is not a feasible route.

  • Sey 1,388 posts since
    May 4, 2010

    Hi Warren,

     

    So, for the above statement, i understand that if an EIGRP router advertises to me a composite metric that is lower than what i know to a particular destination network, then it guarantee's it lie's on a loop free route to the destination.

     

    How and why does it guarantee this?

     

    Here is how good folks explained it to me, so I'm eager to pass the word on

     

    Consider the following topology.

    feas_cond.png

    R1 is our router which has a route to R5. Suppose the path through R2 is better (the lowest FD), so R2 is becoming our Successor for that route. Now R1 has to decide if R3 will become a Feasible Successor or not. Let's split this into 2 cases.

     

    1) Suppose R1's distance to R5 (which is FD) is 100, whereas R3's distance to R5 through R4 is 120 (which is RD). Let's violate the feasibility condition and choose R3 as a Feasible Successor.

     

    What happens when we configure unequal load balancing? R1 sends some IP packets to R5 through R3. When a packet comes to the next hop, R3 thinks "I need to send it towards R5, so let's consult my routing table". And guess what? The routing table will tell R3 to send the packet back to R1 because R1 is the Successor for that route, in R3's point of view (remember that distance on the left is 100+X, and 120 on the right; X is distance between R1 and R3 which may easily be less than 20). So we have a loop, as the packet can bounce between R1 and R3 never going to other routers.

     

    2) Suppose R1's distance to R5 (which is FD) is 100, whereas R3's distance to R5 through R4 is 95 (which is RD). Now whatever distance X is between R1 and R3, 100+X always greater than 95, so R3 will never route through R1. This is why it's a loop free path.

  • Mohamed Sobair 340 posts since
    Oct 21, 2008
    Currently Being Moderated
    4. Dec 4, 2010 8:48 AM (in response to shanekillian)
    Re: EIGRP Feasibility Condition, oh and RD and FD

    Warren,

     

    Here is the concept of Eigrp FD and RD along with the Feasibility condition:-

     

    1- The Reported Distance Or Advertised distance , is the metric to reported metric from the upstream router to the Destination. (In your example router D recieves two reported distance to reach router A, the first reported distance from router B which advertises the metric from router B to router A. The Second reported Distance from Router C which advertises the metric from Router C to Router A).  So , the reported distance is the metric from the upstream router to the destination.

     

    2- The Feasible Distance is simply , the total calculated metric to the destination. Looking at your example, the FD is the total metric from Router D to router A , So router D has two FD, one through router B and the other through Router C.

     

    3- Now, your first question was if they are both the sme distanc e away? I would say no , because the AD is different and the FD of both paths are different , So they are not in the same distance away.

     

    4- Your last question was, why wouldnt route through router C be a feasible successor? It would indeed meat the feasibility condition and its actually a "Feasible Successor".

     

    Note to Remember:

     

    The Feasibility Condition stays the following:

    " In order for a route to become a feasible Successor, Its advertised distance (RD) must be lesser than the Feasible Distance (FD) of the Successor".

     

    If you look at your example, you will find the path through C meets the feasinbility condition and becomes a "feasible successor" because its:

    RD < FD of path B.

     

     

    I hope I have clarified this for you,

     

     

     

    Regards,

    Mohamed

  • Paul Stewart  -  CCIE Security, CCSI 6,993 posts since
    Jul 18, 2008

    A route is not a feasible successor if it does not meet the feasibility condition. This compares the RD of the FD of the local router. Your last statement says 'greater than the FD of the successor'. This is sort of circularly logic. By successor, we must be talking about a neighbor router. The FD of a neighbor router is it's ad or advertised distance. So that advertised distance becomes the local router's RD for that particular route/neighbor. It is this RD that is used in the feasibility calculation against the local FD. The FD would be the the best calculated local route and includes the link to get to the neighbor(s). It is this bandwidth/delay to get to each neighbor that is not included in the RD value. This can take some time to wrap your head around

  • Mohamed Sobair 340 posts since
    Oct 21, 2008

    Warren,

     

    Your understanding is correct, and regarding your first question, its true the router wont pickup the path through router C because it doesnt meet the feasibility condition.

     

    If you examine the Topolgy indicated on the link above , you will find the RD of path C is greater than the FD of path B which is the successor, thats why it wasnt choosen ad a feasible successor.

     

     

    Regards,

    Mohamed

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