In article , GS wrote: :This whole STP/Bridge implementtaion does really care about hardware :interfaces (physical ports)? Or is it everything is in software only, :for example, if the port's state is changed (if the port is got :disconnected)?. Thanks.
I'm still not sure what you are trying to do.
STP is a layer 2 protocol, not a routed protocol. If you have a WAN port which is transmitting BPDU's to you, then the STP information so carried will be taken into account in the construction of the tree.
If you had a situation where you had multiple WAN connections that were linked together at layer 2 on the remote end and so were determined by STP to be part of the same tree, then because WAN links are often much slower than LAN links, those WAN links would tend to be deemed to have a high cost, you could end up with all the WAN traffic directed to the WAN link that happened to have the lowest MAC address (or based on one of the other STP resolution criteria.)
It would, however, be quite unusual for multiple WAN links to be joined together remotely in a common tree -- rare enough that
*probably* it would only occur if you were deliberately bridging BPDU from a remote site to implement an extended LAN.
If the WAN links sent you BPDU at all (which is not particularily likely) then they would very likely appear to be distinct leafs, to be detected by STP as not having any topology loops. There might be a layer 3 topology loop involved, but STP is not designed to detect layer 3 topology loops.
If the STP has determined that a particular WAN link is a leaf, then when that WAN connection goes down, if you are using classic STP then the spanning tree will be recalculated -- and will be determined to be exactly the same as it was before. If you are are using RSTP (Rapid Spanning Tree Protocol) then in the same circumstances, the switches would have figured out that that WAN link was not a potential backup link, and so would know that the topology would not change, and would supress notification of the event to the other switches.