03-06-2007 10:39 PM - edited 03-03-2019 04:03 PM
I pasted another one including attachment :-)
Hi, can you clarify this?
Imagine my "root" bridge has been identified. Please find attached diagram. Then I have two switches "Switch2" and "Switch3". I understand that at that point a root port for each switch should be determined determined based on the lowest cost to get to root bridge. If I look at "Switch3", what happens if I have two paths with same cost to get to "Root"?
Which path the STP algorithm would take at that point thus making that the "Root Port" and the path with the highest cost the "blocked" port??
03-06-2007 11:00 PM
Generally the root port is elected based on the lowest cost to the root bridge (bandwidth based). In case the cost remains the same the tie breaking criteria is Lowest Sender Port ID.
03-07-2007 05:04 AM
Hi
The root port will be determined by the following
1) Lowest Root path cost. if these are equal
2) Lowest sending Bridge ID. If these are equal
3) Lowest Port ID
HTH
Jon
03-07-2007 05:44 AM
Friend,
STP uses the following order sequence to decide on its parameters
Step 1. Lowest Root BID
Step 2. Lowest Path Cost to Root Bridge
Step 3. Lowest Sender BID
Step 4. Lowest Port ID
In your case the step 1 & 2 would yield the same result and hence the deciding factor would be the 3 point which is the lowest sender bridge ID (which could be either be the root or the swith2 in your case)
If you had both the links terminated on the same root switch directly, then the deciding factor would have been the port ID and the lowest port would be in forwarding state by default.
HTH, rate if it does
Narayan
03-07-2007 05:50 AM
Hi Narayan
I don't believe Step 1 comes into the decision in electing the root port. The original poster said that the root bridge had already been elected so step 1 is not taken into account.
The decision as to which port is the root port is started at step 2.
Jon
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