I have attached a pdf drawing just to clarify my question. More specifically, an n -vertex graph has a tree of diameter n 1 as a subgraph if, and only if, it has a Hamiltonian path (the tree is the Hamiltonian path). I am also trying to understand why the 6 seconds for the BPDU propagation delays since the worst case scenario is when we loose the first root switch and only BPDU must only go through 5 switches to reach the last switch, BPDU propagation delay should correspond to 5 switches. This (probably) can't be done efficiently since a maximum-diameter spanning tree could be a Hamiltonian path. My question is how can used the value 6 as Message_age_overestimate with 7 bridge when the age between the root bridge and the first bridge is 0, the max age when reaching the last switch should be set to 5 not 6. Sling Style SEAT HEIGHT: 20 PACKABLE: Yes FASTENERS: 1-2 Nylon Strap with Slide Buckle SAFETY: Fall Arrest System MINIMUM TREE SIZE: 9 Diameter. = ((lost_msg + 1) x hello) + ((BPDU_Delay x (dia – 1)) = End-to-end_BPDU_propa_delay + essage_age_overestimate = (dia – 1) x overestimate_per_bridge (1 as per Cisco standard) When this protocol mentions that the maximum diameter is 7, does that means that a maximum of 7 or 8 bridges allowed in a spanning-tree domain. How can we find a tree of maximum diameter within an undirected unweighted graph Note that the tree does not have to be a spanning tree. If we consider that the Message_age_overestimate is The diameter of a graph is the largest of all shortest-path distances in it. When this protocol mentions that the maximum diameter is 7, does that means that a maximum of 7 or 8 bridges allowed in a spanning-tree domain. Time Complexity: O (N), where N is the number of nodes in the binary tree. I am having some difficulties to understand the spanning-tree protocol.
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