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Back to Lecture Thumbnails Mayank

Question Does higher bandwidth solve the bus contention problem? For example, what happens if nodes 0 and 7 and nodes 1 and 6, communicate small amounts of data at a very high rate? Here it seems having 2 separate low bandwidth links between the roots might avoid collisions. papamix

Good observation. The picture in the slide actually abstracts away implementation details. Fat trees typically employ a larger number of routers and links to implement the upper layers that appear to have "fatter" links in the picture. E.g. the the 8-node network in the slide could look something like this: http://www.design-reuse.com/news_img/20040402_brescia2.gif

On a side note: if you notice in the last picture all routers have 4 links, except for the ones in the top row, which only have 2. To build fat trees where all routers have the same radix (same number of links) the top row routers are often "folded". To get an idea of how this looks, go here http://users.ece.cmu.edu/~mpapamic/connect/ and select "Fat Tree" as the topology to see an example. poop

Question: Is the cost in this case also O(lgN)? kayvonf

In this lecture, we've discussed cost in terms of the number of resources to build the network. Is this was not a "fat tree" and all links were one unit, then the cost of the network would be O(N).

If the width of a link is proportional to the number of subtree nodes it services, the cost would be O(NlgN). I'll let you puzzle over why. aakashr

Because a tree with n vertices has (n-1) edges, the cost for all links being one unit is O(n). The second one, you can get from the recurrence C(n) = 2C(n/2) + O(n) because this is a balanced tree.