EIGRP Question.

Discussion in 'Cisco' started by cozzmo1@hotmail.com, May 7, 2006.

  1. Guest

    Looking at this, Could someone please look at my notes in parentheses
    and tell me if I got this correct?

    If the AD (advertised distance) is less than the FD (feasible Distance)
    of the best route, then it is a feasible successor.
    The successor is the best route and is put into the IP routing table.

    Langley#show ip eigrp topology
    P 10.1.2.0/24, 1 successors, FD is 768
    via 10.1.3.1 (768/256), Serial0 (This is the successor,
    currently used route)
    via 10.1.5.2 (1280/256), Serial1 (256 < 768, this is a FS).

    P 172.16.90.0 255.255.255.0, 2 successors, FD is 0 (am I correct in
    assuming that the FD is really 46251776?)
    via 172.16.80.28 (46251776/46226176), Ethernet0 (This is the
    successor)
    via 172.16.81.28 (46251776/46226176), Ethernet1 (This is the
    successor)
    via 172.16.80.31 (46277376/46251776), Serial0 (not a FS, as
    the AD = FD)


    Thanks
    crzzy1
    , May 7, 2006
    #1
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  2. thrill5 Guest

    You are correct. Not sure why the FD is reported as 0 for the 172.16.90.0
    route. Could be a bug in your IOS version.

    Scott
    <> wrote in message
    news:...
    > Looking at this, Could someone please look at my notes in parentheses
    > and tell me if I got this correct?
    >
    > If the AD (advertised distance) is less than the FD (feasible Distance)
    > of the best route, then it is a feasible successor.
    > The successor is the best route and is put into the IP routing table.
    >
    > Langley#show ip eigrp topology
    > P 10.1.2.0/24, 1 successors, FD is 768
    > via 10.1.3.1 (768/256), Serial0 (This is the successor,
    > currently used route)
    > via 10.1.5.2 (1280/256), Serial1 (256 < 768, this is a FS).
    >
    > P 172.16.90.0 255.255.255.0, 2 successors, FD is 0 (am I correct in
    > assuming that the FD is really 46251776?)
    > via 172.16.80.28 (46251776/46226176), Ethernet0 (This is the
    > successor)
    > via 172.16.81.28 (46251776/46226176), Ethernet1 (This is the
    > successor)
    > via 172.16.80.31 (46277376/46251776), Serial0 (not a FS, as
    > the AD = FD)
    >
    >
    > Thanks
    > crzzy1
    >
    thrill5, May 8, 2006
    #2
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  3. Guest

    I think that perhaps since this has 2 load balancing successors which
    both have equal cost AD and FD, the route just has an incorrect zero
    for the FD.
    I wonder if you had unequal cost successors (using the variance
    command) if they too would have an FD of zero then.

    P 172.16.90.0 255.255.255.0, 2 successors, FD is 0 (am I correct in
    assuming that the FD is really 46251776?)
    via 172.16.80.28 (46251776/46226176), Ethernet0 (This is the
    successor)
    via 172.16.81.28 (46251776/46226176), Ethernet1 (This is the
    successor)
    , May 8, 2006
    #3
  4. Guest

    One other question,

    Why does the last entry show up (via 172.16.80.31)?
    Wouldn't you need "#show ip eigrp topology all-links" in order to see
    that which is not a feasible successor? and yet, there it is.

    According to my text,
    routes that are not successors or feasible successors are not shown in
    #show ip eigrp topology
    yet these routes can be found in
    #show ip eigrp topology all-links


    thanks,
    crzzy1
    , May 10, 2006
    #4
  5. mos33

    Joined:
    Aug 15, 2008
    Messages:
    8
    I wants to ask :
    what happen if the feasible distance equal the advertised distance ??
    in this case the neighbor is considered as feasible successor??
    mos33, Sep 7, 2008
    #5
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