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Unformatted text preview: Comments NOTE: 1. The answers to Question 4.6 in both standard answers are not totally correct. Here is my answer to this question. If there is any question, please tell me: I. There are two kinds of answers: (ts + m * tw)(p 1) or (ts + m * tw * p /2) * log(p); II. Step 1: all-to-one accumulate: ts * log(p) + m * tw * (p 1) (from the view of receiver) Step 2: one-to-all broadcast: (ts + m' * tw) * log(p) (Here size of m and m' may be different, since they are different content, and you can also treat them as the same thing for simplifying.) III. One-to-all broadcast, however, pass the array of count numbers: Since the size of message will not be changed, if we use the hypercube way, then we can achieve the lower bound of ts and tw at the same time: ts * log(p) + m' * tw * (p 1) Here the factor of tw is m'*(p -1) rather than m'*log(p) due to the reason that there still are contentions in the network; we using m' rather than m is because it is the size of counts array,...
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- Winter '11