spanning tree - TCP-IP

This is a discussion on spanning tree - TCP-IP ; Hi I have a basic question. Why spanning tree protocol is used only in Layer 2 i.e. for switches and why not in Layer 3 i.e. for routers ? Looping may exist even among routers. How is the loop avoided ...

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  1. spanning tree

    Hi

    I have a basic question. Why spanning tree protocol is used only in
    Layer 2 i.e. for switches and why not in Layer 3 i.e. for routers ?
    Looping may exist even among routers. How is the loop avoided at layer
    3.

    Thanks


  2. Re: spanning tree

    "Ask" writes:
    > I have a basic question. Why spanning tree protocol is used only in
    > Layer 2 i.e. for switches and why not in Layer 3 i.e. for routers ?
    > Looping may exist even among routers. How is the loop avoided at layer
    > 3.


    Because that's what routing protocols are for.

    --
    James Carlson, KISS Network
    Sun Microsystems / 1 Network Drive 71.232W Vox +1 781 442 2084
    MS UBUR02-212 / Burlington MA 01803-2757 42.496N Fax +1 781 442 1677

  3. Re: spanning tree

    In article , James Carlson writes:
    > "Ask" writes:
    >> I have a basic question. Why spanning tree protocol is used only in
    >> Layer 2 i.e. for switches and why not in Layer 3 i.e. for routers ?
    >> Looping may exist even among routers. How is the loop avoided at layer
    >> 3.

    >
    > Because that's what routing protocols are for.


    Of course, routing protocols do not entirely prevent layer 3 routing
    loops.

    That's what TTL is for. You put a TTL field in the IP packet header
    and decrement it with every hop so that packets don't loop through
    the network forever but instead have a finite maximum lifetime.

    That way you can live with the routing loops while the routing protocol
    reconverges after a topology change or while the techies recover from
    a configuration error.

  4. Re: spanning tree

    > >> I have a basic question. Why spanning tree protocol is used only in
    > >> Layer 2 i.e. for switches and why not in Layer 3 i.e. for routers ?
    > >> Looping may exist even among routers. How is the loop avoided at layer
    > >> 3.

    > >
    > > Because that's what routing protocols are for.

    >
    > Of course, routing protocols do not entirely prevent layer 3 routing
    > loops.
    >
    > That's what TTL is for. You put a TTL field in the IP packet header
    > and decrement it with every hop so that packets don't loop through
    > the network forever but instead have a finite maximum lifetime.
    >
    > That way you can live with the routing loops while the routing protocol
    > reconverges after a topology change or while the techies recover from
    > a configuration error.


    Why can't bridges use routing protocols and routers STP ?


  5. Re: spanning tree

    In article <1156266288.653912.37770@74g2000cwt.googlegroups.co m>,
    Ask wrote:

    >Why can't bridges use routing protocols


    Because by definition routing protocols are not relevant to bridges.

    The task of a linked series of bridges is to get a packet from
    source to destination based *only* upon the MAC address (which must
    be left unchanged.)

    There is no relationship between MAC addresses of hosts that are "near
    by" each other, so in order for bridges to share information about the
    best way to get to a destination given the destination MAC address,
    they would effectively have to share all the MAC addresses that
    they learn about: any summarization to reduce the tables would have
    to be based upon chance groupings of MAC prefixes, and most bridges
    would end up having to know about the existance all the other linked
    bridges.

    Routing protocols work fairly efficiently for routers because
    hosts "near by" each other must have IP addresses that fall into
    clear groups. The information about individual IPs is seldom
    exchanged (not very far, anyhow), only the information about groups.
    The groups are constructed in such a way that it is easy to
    find "supernets" of IPs that all have to be routed the same way
    at any particular routing hop. As you get further away,
    the groups get larger and larger, only getting smaller in the case
    where there are multiple ways to get somewhere. Notice that for
    routing, you do not wish to force all traffic to go the same way
    (as would be the case for STP): you want traffic to go the most
    efficient way, or to be split up between the choices in proportion
    to the cost of travel.


    Working through this, the question would seem to evolve into,
    "Why is STP designed so that traffic sometimes has to go through
    'extra' hops to get somewhere, with 'extra' meaning compared to
    the alternative design of being able to notice that more direct
    links exist that could be used without looping if you were selective
    about which traffic you allowed through the links?" I won't try to answer
    that at the moment.

  6. Re: spanning tree

    In article , roberson@hushmail.com (Walter Roberson) writes:
    > In article <1156266288.653912.37770@74g2000cwt.googlegroups.co m>,
    > Ask wrote:
    >
    >>Why can't bridges use routing protocols

    >
    > The task of a linked series of bridges is to get a packet from
    > source to destination based *only* upon the MAC address (which must
    > be left unchanged.)


    [...]

    Mr. Roberson goes on to argue that the large number of MAC addresses
    and an inability to aggregate them usefully makes it hard to come up
    with a good routing protocol.

    Not all switches and bridges have great gobs of memory within which
    to maintain complete routing tables or routing protocol state databases.

    By itself, that pretty well puts the kibosh on routing protocols
    being the solution to "routing" loops on layer 2 switch infrastructure.

    In addition, switches and bridges need to deal with multicast and
    broadcast traffic. That means that ordinary unicast routing protocols
    won't do the job. You need to compute a forwarding tree.

    And if you're going to deploy a multicast routing protocol to compute
    a forwarding tree...

    ....well, that's what STP amounts to.

  7. Re: spanning tree

    "Ask" wrote:
    >
    > I have a basic question. Why spanning tree protocol is used only in
    > Layer 2 i.e. for switches and why not in Layer 3 i.e. for routers ?
    > Looping may exist even among routers. How is the loop avoided at layer
    > 3.


    Good question, and I think I disagree with most of the answers you got.

    STP could be used at Layer 3, between IP subnets. The protocol could
    operate between routers, just as well as it operates now between
    switches. In doing so, it could be used to find the path between any
    pair of IP subnets, without creating routing loops.

    No problem. The fact that IP addresses are hierarchical, and MAC
    addresses are not, is almost inconsequential to this discussion. Because
    any routing protocol has to be capable of routing between IP subnets
    that are completely unrelated to each other. There's no mandatory
    relationship between IP subnet IDs. Two IP subnets that are very far
    apart may be different by just one bit. The value of the Subnet ID
    cannot be assumed to carry any intrinsic information. It's just an
    arbitrary number. So, in this regard, very much the same as bridge IDs
    used in STP.

    I think the goal in Layer 3 routing, though, is to try to find the
    shortest path between a source and a destination IP subnet, rather than
    just avoiding the formation of loops by creating a minimal spanning
    tree. So that's why distance-vector or link-state algorithms are used.
    Presumably, at Layer 2, using the shortest path is not as critical,
    although that may be changing.

    So basically, what these routing algorithms do *is* to create spanning
    trees, but each router behaves as the root of one of these trees. That
    way, the minimal spanning tree rooted at that router becomes the
    shortest path to all other reachable Layer 3 networks from that router.
    And other routers elsewhere in the world are not going to use that same
    spanning tree as their solution. They will work out their own solution.

    Bert


  8. Re: spanning tree

    wrote:
    > Not all switches and bridges have great gobs of memory within which
    > to maintain complete routing tables or routing protocol state databases.


    In the early days of switches (well, bridges, anyway), it used to be
    not uncommon for switches to not have enough memory to hold all the
    MAC addresses they needed in their bridging tables. When the MAC
    address working sets got bigger than the tables, the bridge was forced
    to fall back into forwarding everything.

    I remember one DEC LanBridge-100 we had where this was a chronic
    problem. It really has a bad effect on network performance when that
    happens :-(

    I don't think that happens much anymore, for a couple of reasons.
    One, is that the switches have a lot more memory than they used to.
    Another is that people tend not to build huge flat networks anymore
    because routers have gotten cheap enough and fast enough to put in
    every wiring closet.



  9. Re: spanning tree

    "Albert Manfredi" writes:
    > No problem. The fact that IP addresses are hierarchical, and MAC
    > addresses are not, is almost inconsequential to this
    > discussion. Because any routing protocol has to be capable of routing


    I don't think that's entirely true. The real issue here is that IP
    addresses are assigned and managed by administrators, but that MAC
    addresses are essentially (from a topology point of view) arbitrary
    numbers. Thus, with L3 addresses, it makes a lot of sense to talk
    about (and put algorithmic effort into) prefixes and aggregation, but
    with L2 addresses, you're talking about the moral equivalent of host
    routes.

    > I think the goal in Layer 3 routing, though, is to try to find the
    > shortest path between a source and a destination IP subnet, rather
    > than just avoiding the formation of loops by creating a minimal
    > spanning tree.


    Actually, I'd say that both are important and not completely
    independent goals. If L3 route computation results in a loop, then
    that's toxic to the network (TTL notwithstanding), as it results in a
    traffic multiplier -- and, of course, a service outage. It's also, by
    definition, not the optimal path.

    (Another poster responded saying that TTL protects the routing network
    from loops. While that's roughly true, anyone who has suffered
    through a loop knows that it's fairly poor protection. It's really
    best at keeping _transient_ forwarding loops from becoming a total
    melt-down, not for handling persistent errors or preventing overload.)

    > So that's why distance-vector or link-state algorithms
    > are used. Presumably, at Layer 2, using the shortest path is not as
    > critical, although that may be changing.


    Indeed. And by the same token, routing protocols analogous to those
    used at L3 could potentially be used for L2 as well. I think Radia
    has some thoughts on that -- see:

    http://research.sun.com/minds/2005-0208/

    --
    James Carlson, KISS Network
    Sun Microsystems / 1 Network Drive 71.232W Vox +1 781 442 2084
    MS UBUR02-212 / Burlington MA 01803-2757 42.496N Fax +1 781 442 1677

  10. Re: spanning tree

    "James Carlson" wrote:

    > I don't think that's entirely true. The real issue here is that IP
    > addresses are assigned and managed by administrators, but that MAC
    > addresses are essentially (from a topology point of view) arbitrary
    > numbers. Thus, with L3 addresses, it makes a lot of sense to talk
    > about (and put algorithmic effort into) prefixes and aggregation, but
    > with L2 addresses, you're talking about the moral equivalent of host
    > routes.


    Yes, I understand about prefix aggregation, which is why I said "almost
    inconsequential." But the point is, prefix aggregation is a helpful
    technique to make routing tables more efficient. It is not something
    that is mandatory for these routing algorithms to function. Prefixes
    must also be allowed to be unrelated even if the subnets are
    geographically close to one another, which is exactly the same thing you
    get with switch IDs.

    And not to forget that switch IDs are not just the MAC address. There
    are also 16 configurable bits in there that net admins can play with, to
    make STP behave in some specific way. So even if MAC addresses are
    topologically arbitrary, as you say, switch IDs are not necessarily
    topologically arbitrary. Much like IP subnet IDs, in this regard.

    Bottom line: subnet IDs and switch IDs are not so different. In both
    cases, the algorithms must work when these numbers are topologically
    arbitrary, and in both cases the algorithms may operate more efficiently
    if the numbers are given topological significance.

    >> I think the goal in Layer 3 routing, though, is to try to find the
    >> shortest path between a source and a destination IP subnet, rather
    >> than just avoiding the formation of loops by creating a minimal
    >> spanning tree.

    >
    > Actually, I'd say that both are important and not completely
    > independent goals.


    That's why I used the word "just." Obviously loops must be avoided. In
    addition, you also want to minimize the path length from a source to a
    destination in routing protocols. Which clearly is not a goal for STP.

    > Indeed. And by the same token, routing protocols analogous to those
    > used at L3 could potentially be used for L2 as well. I think Radia
    > has some thoughts on that -- see:


    Yes, I agree. This is also true, and being discussed, in multicast
    routing. As L2 nets become bigger, it will be hard to justify why
    routing and bridging algorithms should be different. And switch IDs, as
    currently defined for STP, can certainly provide any amount of "L2 route
    aggregation" functionality you might require.

    Bert


  11. Re: spanning tree


    "Ask" wrote in message
    news:1156250174.918337.281840@h48g2000cwc.googlegr oups.com...
    > Hi
    >
    > I have a basic question. Why spanning tree protocol is used only in
    > Layer 2 i.e. for switches and why not in Layer 3 i.e. for routers ?
    > Looping may exist even among routers. How is the loop avoided at layer
    > 3.
    >
    > Thanks
    >


    Since you have a basic question, I will try to give a basic answer

    The Spanning-Tree Protocol (STP) uses the SPA (spanning-tree algorithm) to
    resolve and shut down the redundant paths or to avoid loops by creating a
    spanning tree. Switches that use STP send BPDUs (a Spanning-Tree Protocol
    hello packet) out all its ports to let other switches know of its existence.
    This information is used to elect a root bridge for the network, to detect
    loops, and then remove the loops by shutting down selected port. Each port
    on a switch that uses STP exists in one of the following five states:
    Blocking, Listening, Learning, Forwarding, Disabled

    Routers use different protocols (Distance-vector protocols, Link-state
    protocols) and each protocol uses different algorithm ( SPF algorithm - a
    calculation performed on the database that results in the SPF tree,
    Diffusing Update Algorithm (DUAL), etc.).

    In other words, can Chinese communicate with Greek using their own
    languages?

    The Dude



  12. Re: spanning tree

    "Albert Manfredi" writes:
    > And not to forget that switch IDs are not just the MAC address. There
    > are also 16 configurable bits in there that net admins can play with,
    > to make STP behave in some specific way. So even if MAC addresses are
    > topologically arbitrary, as you say, switch IDs are not necessarily
    > topologically arbitrary. Much like IP subnet IDs, in this regard.


    No, not really. The BIDs (Bridge Identifiers) have nothing to do with
    the locations of the individual MAC addresses that need to be
    forwarded. Knowing the topology of BIDs tells you nothing specific
    about how you need to forward a given MAC frame. In contrast, having
    the IP prefix from a routing protocol *does* tell you how to forward
    anything that matches the prefix.

    I can't say I've seen any use of BIDs that has the sort of topological
    significance you're suggesting (do you have any references?).
    Instead, the protocol simply uses the BID to identify a switch in a
    network for path computation and to select the root node. At most,
    the high-order bits (the 16-bit "priority value") could be tweaked to
    force a particular node to become the root -- but that's hardly
    necessary, it's meaningless on the N-1 nodes that aren't selected as
    root, and the protocol works fine without it.

    --
    James Carlson, KISS Network
    Sun Microsystems / 1 Network Drive 71.232W Vox +1 781 442 2084
    MS UBUR02-212 / Burlington MA 01803-2757 42.496N Fax +1 781 442 1677

  13. Re: spanning tree

    "The Dude" Dude@thedu.de> wrote:

    > "Ask" wrote:


    >> I have a basic question. Why spanning tree protocol is used only in
    >> Layer 2 i.e. for switches and why not in Layer 3 i.e. for routers ?
    >> Looping may exist even among routers. How is the loop avoided at
    >> layer
    >> 3.

    >
    > Since you have a basic question, I will try to give a basic answer
    >
    > The Spanning-Tree Protocol (STP) uses the SPA (spanning-tree
    > algorithm) to resolve and shut down the redundant paths or to avoid
    > loops by creating a spanning tree. Switches that use STP send BPDUs (a
    > Spanning-Tree Protocol hello packet) out all its ports to let other
    > switches know of its existence. This information is used to elect a
    > root bridge for the network, to detect loops, and then remove the
    > loops by shutting down selected port. Each port on a switch that uses
    > STP exists in one of the following five states: Blocking, Listening,
    > Learning, Forwarding, Disabled
    >
    > Routers use different protocols (Distance-vector protocols, Link-state
    > protocols) and each protocol uses different algorithm ( SPF
    > algorithm - a calculation performed on the database that results in
    > the SPF tree, Diffusing Update Algorithm (DUAL), etc.).
    >
    > In other words, can Chinese communicate with Greek using their own
    > languages?


    The question was, "Why do the French speak French and the English speak
    English? Can't the French speak English?

    And several of the answers came back, "Because the French speak French,
    and the English speak English."

    IMO, the correct answer is, "Yes, the French *could* speak English, and
    the English *could* speak French. There are no insurmountable obstacles
    to this at all."

    Routing protocols do provide the extra feature of always looking to
    minimize path length (cost) for any pair of hosts, but there's no reason
    at all to believe that routing protocols cannot be modified slightly to
    operate entirely at L2. Or STP at L3.

    Bert


  14. Re: spanning tree

    "James Carlson" wrote:

    > "Albert Manfredi" writes:
    >> And not to forget that switch IDs are not just the MAC address. There
    >> are also 16 configurable bits in there that net admins can play with,
    >> to make STP behave in some specific way. So even if MAC addresses are
    >> topologically arbitrary, as you say, switch IDs are not necessarily
    >> topologically arbitrary. Much like IP subnet IDs, in this regard.

    >
    > No, not really. The BIDs (Bridge Identifiers) have nothing to do with
    > the locations of the individual MAC addresses that need to be
    > forwarded. Knowing the topology of BIDs tells you nothing specific
    > about how you need to forward a given MAC frame. In contrast, having
    > the IP prefix from a routing protocol *does* tell you how to forward
    > anything that matches the prefix.


    Bridge identifiers can be configured to have the spanning tree formed in
    specific ways, rather than just arbitrarily. So they can be used to give
    the Bridge ID topological significance. Topological significance means
    that the bridge ID, or the subnet ID in the case of a routing algorithm,
    are coded to provide some intrinsic information related to the topology
    of the network.

    Certainly, how a spanning tree is formed will affect through which
    switch ports specific end system MAC addresses will be routed. The
    network designer can arrange it so that certain end system pairs, for
    example, preferentially get shorter paths than others through the
    spanning tree. In the context of a single spanning tree, which STP
    provides, I'd say that qualifies as "topologically significant" IDs.

    If STP morphs into a true routing protocol, there is no reason to
    believe that this same Bridge ID format cannot be used to provide true
    route aggregation, just like subnet IDs do now.

    So I think the comment that subnet IDs may be topologically significant,
    while Bridge IDs are not, does not hold up under scrutiny. Even if, in
    the existing STP, topological significance does not provide exactly the
    same effect as it does in RIP or OSPF.

    > I can't say I've seen any use of BIDs that has the sort of topological
    > significance you're suggesting (do you have any references?).
    > Instead, the protocol simply uses the BID to identify a switch in a
    > network for path computation and to select the root node. At most,
    > the high-order bits (the 16-bit "priority value") could be tweaked to
    > force a particular node to become the root -- but that's hardly
    > necessary, it's meaningless on the N-1 nodes that aren't selected as
    > root, and the protocol works fine without it.


    The specific switch to act as the root bridge and the specific switches
    to become the designated bridges in the spanning tree can be
    "encouraged" by intelligently coding the 16 upper bits of the Bridge ID.
    I'd say there's not a lot more "topological significance" one can hope
    for, in the existing STP. And, repeating myself, if STP becomes a
    routing algorithm, exactly the same Bridge ID format can be repurposed
    for use in route aggregation.

    Bert


  15. Re: spanning tree


    "Albert Manfredi" wrote in message
    news:J4ID90.L6w@news.boeing.com...
    > "The Dude" Dude@thedu.de> wrote:
    >
    >> "Ask" wrote:

    >
    >>> I have a basic question. Why spanning tree protocol is used only in
    >>> Layer 2 i.e. for switches and why not in Layer 3 i.e. for routers ?
    >>> Looping may exist even among routers. How is the loop avoided at layer
    >>> 3.

    >>
    >> Since you have a basic question, I will try to give a basic answer
    >>
    >> The Spanning-Tree Protocol (STP) uses the SPA (spanning-tree algorithm)
    >> to resolve and shut down the redundant paths or to avoid loops by
    >> creating a spanning tree. Switches that use STP send BPDUs (a
    >> Spanning-Tree Protocol hello packet) out all its ports to let other
    >> switches know of its existence. This information is used to elect a root
    >> bridge for the network, to detect loops, and then remove the loops by
    >> shutting down selected port. Each port on a switch that uses STP exists
    >> in one of the following five states: Blocking, Listening, Learning,
    >> Forwarding, Disabled
    >>
    >> Routers use different protocols (Distance-vector protocols, Link-state
    >> protocols) and each protocol uses different algorithm ( SPF algorithm - a
    >> calculation performed on the database that results in the SPF tree,
    >> Diffusing Update Algorithm (DUAL), etc.).
    >>
    >> In other words, can Chinese communicate with Greek using their own
    >> languages?

    >
    > The question was, "Why do the French speak French and the English speak
    > English? Can't the French speak English?
    >
    > And several of the answers came back, "Because the French speak French,
    > and the English speak English."
    >
    > IMO, the correct answer is, "Yes, the French *could* speak English, and
    > the English *could* speak French. There are no insurmountable obstacles to
    > this at all."
    >
    > Routing protocols do provide the extra feature of always looking to
    > minimize path length (cost) for any pair of hosts, but there's no reason
    > at all to believe that routing protocols cannot be modified slightly to
    > operate entirely at L2. Or STP at L3.
    >
    > Bert


    What I meant was: can we apply Greek in China and Chinese in Greece?
    Anyway, I think it's all about the way algorithms are written. You have to
    re-write the algorithms in order to apply STP to routers.

    The Dude



  16. Re: spanning tree

    In article ,
    "Albert Manfredi" wrote:

    > The question was, "Why do the French speak French and the English speak
    > English? Can't the French speak English?
    >
    > And several of the answers came back, "Because the French speak French,
    > and the English speak English."
    >
    > IMO, the correct answer is, "Yes, the French *could* speak English, and
    > the English *could* speak French. There are no insurmountable obstacles
    > to this at all."


    I think this is a poor analogy. French and English are just different
    ways of solving the same problem. But routing protocols and spanning
    tree protocols are intended to solve *different* problems. If you want
    an (admittely poor) analogy, it would be "Why are hammers used to push
    nails in, when we could use the handle of a screwdriver?" Yes, you
    *could* do this, but it wouldn't work as well -- it's best to use tools
    that are designed specifically to solve the problem at hand.

    The specific problem that spanning tree protocols are intended to solve
    is forwarding of broadcasts without causing loops. Routing protocols
    are designed to find the best path to a particular destination, which is
    needed for unicast forwarding. As they're solving different problems,
    it's hardly surprising that different mechanisms would be best.

    --
    Barry Margolin, barmar@alum.mit.edu
    Arlington, MA
    *** PLEASE post questions in newsgroups, not directly to me ***
    *** PLEASE don't copy me on replies, I'll read them in the group ***

  17. Re: spanning tree

    "Barry Margolin" wrote:

    > The specific problem that spanning tree protocols are intended to
    > solve
    > is forwarding of broadcasts without causing loops. Routing protocols
    > are designed to find the best path to a particular destination, which
    > is
    > needed for unicast forwarding. As they're solving different problems,
    > it's hardly surprising that different mechanisms would be best.


    Which at the very least says that there should be no problem using a
    routing protocol as a L2 protocol, to replace STP. Since after all,
    routing protocols ALSO must prevent loops, and there's nothing to say
    that the L2 protocol should NOT find the shortest path, just because STP
    is not doing so today.

    The other half of this is that for small L3 networks, where finding the
    shortest path is not so essential, STP would work just fine. I mean,
    build a single spanning tree and use it to route packets at L3.

    Both routing protocols and L2 protocols are used to create connectivity
    in mesh networks, generically speaking. It's misleading to say that the
    two jobs are so fundamentally different that we're talking apples and
    oranges, IMO.

    I think it's instructive to consider that the IEEE is loking into
    replacing GMRP with a different L2 multicast routing protocol, along the
    same lines as how multicast routing is done at L3.

    Bert


  18. Re: spanning tree

    In article ,
    "Albert Manfredi" wrote:

    > "Barry Margolin" wrote:
    >
    > > The specific problem that spanning tree protocols are intended to
    > > solve
    > > is forwarding of broadcasts without causing loops. Routing protocols
    > > are designed to find the best path to a particular destination, which
    > > is
    > > needed for unicast forwarding. As they're solving different problems,
    > > it's hardly surprising that different mechanisms would be best.

    >
    > Which at the very least says that there should be no problem using a
    > routing protocol as a L2 protocol, to replace STP. Since after all,
    > routing protocols ALSO must prevent loops, and there's nothing to say
    > that the L2 protocol should NOT find the shortest path, just because STP
    > is not doing so today.


    The problem with trying to use a routing protocol instead of STP is that
    a routing protocol computes paths for specific destinations. STP is
    used only for BROADCAST packets. It's not used for unicast messages,
    those use the learning feature of the switch -- when it receives a
    packet on an interface, it looks at the source MAC address and uses that
    interface to send packets back to that address.

    Some routing protocols, like OSPF, perform this computation by first
    determining the complete topology of the network. So a similar protocol
    *could* be used in L2 to gather the topology information, and then
    compute a spanning tree from this. But in practice, the standard STP is
    good enough for most switched networks. It's unusual to have a complex
    enough L2 topology that you need a more complex protocol.

    --
    Barry Margolin, barmar@alum.mit.edu
    Arlington, MA
    *** PLEASE post questions in newsgroups, not directly to me ***
    *** PLEASE don't copy me on replies, I'll read them in the group ***

  19. Re: spanning tree


    "Albert Manfredi" wrote in message
    news:J4Iyuu.EyK@news.boeing.com...
    > "Barry Margolin" wrote:
    >
    >> The specific problem that spanning tree protocols are intended to solve
    >> is forwarding of broadcasts without causing loops. Routing protocols
    >> are designed to find the best path to a particular destination, which is
    >> needed for unicast forwarding. As they're solving different problems,
    >> it's hardly surprising that different mechanisms would be best.

    >
    > Which at the very least says that there should be no problem using a
    > routing protocol as a L2 protocol, to replace STP. Since after all,
    > routing protocols ALSO must prevent loops, and there's nothing to say that
    > the L2 protocol should NOT find the shortest path, just because STP is not
    > doing so today.
    >
    > The other half of this is that for small L3 networks, where finding the
    > shortest path is not so essential, STP would work just fine. I mean, build
    > a single spanning tree and use it to route packets at L3.
    >
    > Both routing protocols and L2 protocols are used to create connectivity in
    > mesh networks, generically speaking. It's misleading to say that the two
    > jobs are so fundamentally different that we're talking apples and oranges,
    > IMO.
    >
    > I think it's instructive to consider that the IEEE is loking into
    > replacing GMRP with a different L2 multicast routing protocol, along the
    > same lines as how multicast routing is done at L3.
    >
    > Bert
    >


    Another thing that I would like to add is that switches are faster devices
    than routers. However, many network administrators do not like STP on
    because BPDUs slow the network down. Think about applying STP to routers
    that are slower devices than switches.
    Anyway, the OP can try for himself and I would be happy to hear the results!

    The Dude



  20. Re: spanning tree

    In article ,
    The Dude Dude@thedu.de> wrote:

    >Another thing that I would like to add is that switches are faster devices
    >than routers.


    Please don't repeat marketing professional noise. What does "faster"
    mean? When it comes to most packets (or frames) forwarded per second
    among all connected networks, the biggest routers are clearly fastest.
    If you care about per-packet latency, then yes, cut-through bridges
    are faster than cut-through routers because a bridge can start forwarding
    sooner, but many people think cut-through forwarding is a Very Bad Idea.
    For a given number of dollars, you would expect a bridge to forward
    more packets per second than a router with the same cost (not price),
    because bridging needs fewer computations per packet. However, the
    general proposition is as nonsensical today as marketing professional
    baloney 15 years ago about boxes containing ASICs being fastest.


    Vernon Schryver vjs@rhyolite.com

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