0
$\begingroup$

In Spanning Tree Protocol, a root switch is selected at first, and then somehow, the shortest path from each other switch to the root is obtained. Thus we established a tree network.

My questions are:

  • After the tree network is established, does the root switch act as a center in the network? In other words, all the traffic originated from other switch would first flow into the root switch along the shortest path and then routed to its destination by the root switch, also along the shortest path.
  • If it's not the case, why does the protocol calculate a tree such that the path between each non-root switches and the root is minimized? As far as I can see, a minimum spanning tree would work better, which is not difficult to calculate.
$\endgroup$
1
$\begingroup$

I can find no requirement that all traffic is routed through the root switch. For example, in A<-->B<-->C, with C the root switch, traffic fom A to B does not pass through the root switch.

The spanning tree protocol first selects a root switch. In Cisco, this is the switch with the lowest bridge ID. So the selection of the root is arbitrary. Now a tree is built that connects all switches, without creating cycles. Even if initially a minimum spanning tree is created because links with maximum bandwidth (minimum cost) are selected, at some future point this may no longer be when a link fails and a redundant link or path is enabled. This new link may result in the total tree not being a minimum spanning tree anymore. I can imagine that reconfiguring the whole tree to be minimum again can cause unwanted overhead.

I found this information on Wikipedia and Cisco. I am not a network guru so my answer may not be complete. I found no information that the initial tree created is minimum though it seems to meet the requirements and seems to follow e.g. Kruksal's algorithm.

$\endgroup$
  • $\begingroup$ I think the Spanning Tree protocol builds a shortest-path tree for the root switch other than a minimum spanning tree. $\endgroup$ – Mengfan Ma Dec 22 '18 at 7:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.