Alex, > > So the number of individual pages read would equal the number pages in > the database, which means that the number of dirty pages writes should be > less than or equal the number db pages. > > Sean, take into account that for each deleted record version which changes > (deletes) index key we should modify an index block. If records are not > sorted in database (I know that in Karol's case they seem to be sorted, but > now I talk about generic case) for each deleted record version we need > another index block. We must find it in a cache and if it's missing (which is > quite possible if indexes of current table have size 4 times bigger than > cache), > we must read it from disk. But if we have only dirty pages in cache (which is > also very likely in such a case), some page should be written. With indexes > size N times bigger than cache only 1/N of deleted record versions will not > cause page write in case of random records distribution in database.
Ok, now I see the problem... but it seems that the current approach is less than optimal for a sweep process. To be clear, you are saying that if a row as an index on Field A which has 4 record versions (3 which can be dropped), and the value of the index for those versions where 1, 2, 3 and 4. When the sweep encounters the row, when it "scrubs" version 1, if reads the index and tries to find the version 1 entry and clean it at the same time. And so on. Correct? Sean ------------------------------------------------------------------------------ Master Java SE, Java EE, Eclipse, Spring, Hibernate, JavaScript, jQuery and much more. Keep your Java skills current with LearnJavaNow - 200+ hours of step-by-step video tutorials by Java experts. SALE $49.99 this month only -- learn more at: http://p.sf.net/sfu/learnmore_122612 Firebird-Devel mailing list, web interface at https://lists.sourceforge.net/lists/listinfo/firebird-devel
