my question is about rapid spanning tree operation.
Assume a RSTP topology as : 1-2-3-4-5-6-7-8-9-R
,where R is the root bridge of the switched network
and 1,2,3,4,5,6,7,8,9 are bridges connected through "-", so
that the active switching path towards R for all bridges passes through bridge 9. Let the delay for a bridge to switch a frame from one port to another is 1 sec
Now assuming that bridge 1 had a link to
R which just came up: R-1-2-3-4-5-6-7-8-9-R (R appears twice but actually consider it as the same one bridge)
According to 802.1w standard and cisco on line documentation, R will negotiate with 1 to block its port with 2 before R starts forwarding on its port connected to 1, 1 with 2 and so on(proposal-agreement mechanism).
Let now this new path, is selected only from 1 and 2(let 2 block its port with 3 and 3 will be the designated for the ethernet segment between 2 and 3).
According to 802.1w, 1 will Tx a BPDU with TC flag = on through R (so that all rest bridges notified for the topology change and flush their dynamic CAM table entries) immediately when 1 put its port towards R to Forwarding. No other bridge (neither 2 when blocks its port with 3) will Tx a BPDU with TC flag on.
Assuming that bridge 2 has also a port where host H1 connects to so that H1 Tx a broadcast frame just before bridge 1 negotiates with 2(proposal-agreement)t.e. before bridge 2 blocks its port with bridge 3 and bridge 1 begins forwarding on its port to R, then :this frame that just Txed ,travels towards R through bridge 3.
this frame will be Rxed by bridge 1 and bridge 2 after 8 and 9 seconds respectively, since bridge 1 after this time period has put its port to R in forwarding already.
The BPDU with TC flag set,Txed by bridge 1, will be arriving bridge 3, when the broadcast frame Txed by H1 arrives 1, travelling in oposite direction.
Since 1 and 2 have already flushed their CAM when H1's frame arrives they will record the MAC address of H1 on a wrong port.
Am i missing something here?