Whopps. I made some mistakes in the previous post. Toke Eskildsen [t...@statsbiblioteket.dk]:
> Extrapolating from 1.4M documents and 180 clients, let's say that > there are 1.4M/180/5 unique terms for each sort-field and that their > average length is 10. We thus have > 1.4M*log2(1500*10*8) + 1500*10*8 bit ~= 23MB > per sort field or about 4GB for all the 180 fields. That would be 10 bytes and thus 80 bits. The results were correct though. > So 1 active searcher and 2 warming searchers. Ignoring that one of > the warming searchers is highly likely to finish well ahead of the other > one, that means that your heap must hold 3 times the structures for > a single searcher. This should be taken with a grain of salt as it depends on whether or not there is any re-use of segments. There might be for sorting. Apologies for any confusion, Toke Eskildsen
