Hi,
Could you explain the solution to the bonus exercise of lab 1, where you calculate the latency as being 2d * single link latency? Thanks
Best,
Stephane
Hello Stephane,
In the tree structure, all of the nodes will be leaves of the tree (the non-leaf nodes will all be routers).
So if you select a node from one half of the tree, it needs to go up to the root (depth d) and then all the way back down to leaves (depth d) in order to communicate with other half of the nodes.
Thus, the latency is 2 * d * single link latency.
Best,
Edin
In the tree structure, all of the nodes will be leaves of the tree (the non-leaf nodes will all be routers).
So if you select a node from one half of the tree, it needs to go up to the root (depth d) and then all the way back down to leaves (depth d) in order to communicate with other half of the nodes.
Thus, the latency is 2 * d * single link latency.
Best,
Edin
Perfect thanks Edin
