Path: news.mathworks.com!not-for-mail From: <HIDDEN> Newsgroups: comp.soft-sys.matlab Subject: Re: linear extrapolation Date: Fri, 16 Apr 2010 22:17:20 +0000 (UTC) Organization: The MathWorks, Inc. Lines: 12 Message-ID: <hqanlg$1un$1@fred.mathworks.com> References: <hqa8uu$g95$1@fred.mathworks.com> <hqai74$qb$1@fred.mathworks.com> <hqaipc$alq$1@fred.mathworks.com> <hqampe$hme$1@fred.mathworks.com> Reply-To: <HIDDEN> NNTP-Posting-Host: webapp-03-blr.mathworks.com Content-Type: text/plain; charset=UTF-8; format=flowed Content-Transfer-Encoding: 8bit X-Trace: fred.mathworks.com 1271456240 2007 172.30.248.38 (16 Apr 2010 22:17:20 GMT) X-Complaints-To: news@mathworks.com NNTP-Posting-Date: Fri, 16 Apr 2010 22:17:20 +0000 (UTC) X-Newsreader: MATLAB Central Newsreader 1187260 Xref: news.mathworks.com comp.soft-sys.matlab:627359 "Roger Stafford" <ellieandrogerxyzzy@mindspring.com.invalid> wrote in message <hqampe$hme$1@fred.mathworks.com>... > Now that you've told us this value is the time required in a bubblesort operation, that changes the whole picture. It is known that optimum sorting requires a number of steps of the order of n*log(n) for sorting n numbers. If we assume your bubblesort operation is anywhere near optimum, then the best assumption you could probably make would be to find a best fit - that is, pick the best value for a - for > > t = a*n*log(n) > > over the range from n = 1 to 10000, and use that for n = 1000000. It still wouldn't be reliable but that's about the best you can do under the circumstances. > > Roger Stafford Let me amend that last bit of advice a bit. You should allow for some kind of fixed overhead on your calls to a bubblesort routine by getting the best a and b for t = a*n*log(n) + b, or else take only values of n from, say, n = 1000 to n = 10000, something like that. The low values of n will probably be somewhat misleading for predicting extrapolated values. Roger Stafford