loops in ospf - TCP-IP

This is a discussion on loops in ospf - TCP-IP ; Hi I'm taking a course in computer networks and right now we're studying about Link State protocols (i.e. - OSPF) My question is regarding loops. It's my understanding that when a broken connection goes back up, the two routers on ...

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  1. loops in ospf

    Hi

    I'm taking a course in computer networks and right now we're studying
    about Link State protocols (i.e. - OSPF)
    My question is regarding loops.

    It's my understanding that when a broken connection goes back up, the
    two routers on both sides have to exchange their information and
    rebuild their internal maps in order to avoid loops. But how can a
    loop occur? let's assume that a broken line goes back up and the two
    routers don't exchange information, how can a loop occur then. Can
    someone please give an example or an illustration?


  2. Re: loops in ospf

    On Apr 28, 11:12*am, hyperboogie wrote:
    > Hi
    >
    > I'm taking a course in computer networks and right now we're studying
    > about Link State protocols (i.e. - OSPF)
    > My question is regarding loops.
    >
    > It's my understanding that when a broken connection goes back up, the
    > two routers on both sides have to exchange their information and
    > rebuild their internal maps in order to avoid loops. But how can a
    > loop occur? let's assume that a broken line goes back up and the two
    > routers don't exchange information, how can a loop occur then. Can
    > someone please give an example or an illustration?


    Take a simple case:

    In this mesh, routers A and B each have a link to C. There is also a
    slow, i.e. costly, standby link between A and B.

    A-------C
    |
    B---+

    Now the A-C link breaks, so the mesh reconfigures with the spare link,
    to:

    A------B------C

    Okay, so now all packets from A to C must go through B. B knows it has
    to forward all packets from A to C, and from C to A.

    Now, the A-C link is restored, B figures this out, but C does not.
    What happens when C wants to send a packet back to A?

    Well, C sends the packet to B, thinking that's still the only way to
    get to A. But B has placed that slow link in standby again, so its
    route to A is through C.

    So when B receives a packet addressed to A, and sends it back to C,
    along its computed best route. And so on.

    This packet from C to A will bounce back and forth between C and B,
    until C finally updates its routing tables, and finds that direct link
    to A.

    Bert

  3. Re: loops in ospf

    On Apr 28, 9:56 pm, Albert Manfredi wrote:
    > On Apr 28, 11:12 am, hyperboogie wrote:
    >
    > > Hi

    >
    > > I'm taking a course in computer networks and right now we're studying
    > > about Link State protocols (i.e. - OSPF)
    > > My question is regarding loops.

    >
    > > It's my understanding that when a broken connection goes back up, the
    > > two routers on both sides have to exchange their information and
    > > rebuild their internal maps in order to avoid loops. But how can a
    > > loop occur? let's assume that a broken line goes back up and the two
    > > routers don't exchange information, how can a loop occur then. Can
    > > someone please give an example or an illustration?

    >
    > Take a simple case:
    >
    > In this mesh, routers A and B each have a link to C. There is also a
    > slow, i.e. costly, standby link between A and B.
    >
    > A-------C
    > |
    > B---+
    >
    > Now the A-C link breaks, so the mesh reconfigures with the spare link,
    > to:
    >
    > A------B------C
    >
    > Okay, so now all packets from A to C must go through B. B knows it has
    > to forward all packets from A to C, and from C to A.
    >
    > Now, the A-C link is restored, B figures this out, but C does not.
    > What happens when C wants to send a packet back to A?
    >
    > Well, C sends the packet to B, thinking that's still the only way to
    > get to A. But B has placed that slow link in standby again, so its
    > route to A is through C.
    >
    > So when B receives a packet addressed to A, and sends it back to C,
    > along its computed best route. And so on.
    >
    > This packet from C to A will bounce back and forth between C and B,
    > until C finally updates its routing tables, and finds that direct link
    > to A.
    >
    > Bert


    Thank you so very much :-)
    You've been very helpful
    This is precisely what I needed to understand

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