Posted - 05/29/2013 : 03:25:20
| Completed subject:
Any solution for better performance about a cube with 3 distinct counts on the same source DB table on SQL Server 2008 R2?
I have a big table about 180G for size and 4 billion rows on SQL server 2008R2 64bit so far, and growing much very day.
So far, we partitioned source DB table by ColumnA, and we have 3 distinct count on ColumnA, ColumnB and ColumnC, then we have 3 measure groups(as some sqlserver&doc required), for example, MG_A, MG_B and MG_C for each distinct count, and we want to partition MG_A, MG_B and MG_C. The final goal is that we can build each measure group partitions incrementally.
Better performance for building cube with these 3 distinct counts on the same source DB table.
For MG_A, ColumnA is OK, source DB table and MG_A are all partitioned by ColumnA, but how about MG_B and MG_C?
If source DB table is partitioned by ColumnA, and MG_B is partitioned by ColumnB(this is recommended by some docs), then these two partitions are not aligned one by one(Alignment is required by some doc), then any impact on performance?
Any idea for better performance of cube building?
Since 3 distinct count is based on the same source DB table, any chance to improve the performance of cube building? otherwise, 3 times work to do just as 3 distinct counts on 3 DB tables.
If we partition measure group NOT by distinct column, for example, partition MG_B by ColumnA, instead of ColumnB, is this a correct solution? That means, is it mandatory that partition MG_B by ColumnB?
If we partition measure group by distinct column, any chance for only incremental processing? For example, just build the latest partition? Otherwise, we have to re-build all MGs from the scratch. It's too slow.
Because I saw some doc said for distinct count aggregatonfunction, you have to re-build all partitions.
I know too much questions here, but one depends on another one.
All in all, any suggestions about distinct count and incremental cube building?