STP Bandwidth Rules

[Snifferson]

Hi All,

Can anyone tell me:

I have 2 (a and b) redundancies. (a) goes through 1x 100baseT switch and (b) through 2x 1000baseT and 1x10baseT switches.

Will STP choose route a or b?

Effectively, I am asking, will STP choose the lowest average or does it do it a different way?

Thanks,

Joe

[Michel Hegeraat]

If you want to know how STP or RSTP works then google and wikipedia are your friends

Check out this page

http://en.wikipedia.org/wiki/Spanning_Tree_Protocol

Determine the least cost paths to the root bridge.

The computed  spanning tree has the property that messages from any connected device  to the root bridge traverse a least cost path, i.e., a path from the  device to the root that has minimum cost among all paths from the device  to the root. The cost of traversing a path is the sum of the costs of  the segments on the path.

The table below shows the default cost of an interface for a given data rate.

Data rate	STP Cost (802.1D-1998)	RSTP Cost (802.1W-2001)
4 Mbit/s	250	5,000,000
10 Mbit/s	100	2,000,000
16 Mbit/s	62	1,250,000
100 Mbit/s	19	200,000
1 Gbit/s	4	20,000
2 Gbit/s	3	10,000
10 Gbit/s	2	2,000

So in your case cost of a is: 19 and cost of b is: 2 x 4  + 100

Cheers,

Michel

[Marvin Rhoads]

...what Michel said (assuming the root bridge is reachable using BPDUs up both paths).