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Mohamed El-Mallah <mie_at_sddpc.sannet.gov> wrote:
>Assumptions :
>DB_FILE_MULTIBLOCK_READ_COUNT =16
>DB_BLOCK_SIZE=4K
>Table 192M total size
>Case 1: 1 Extent 192M
> Number of reads required to perform full scan =
>ceiling(192*1024/(16*4))=3072 Reads
>Case 2: 8 Extents 24M each
> Number of reads =ceiling(24*1024/(16*4)) * 8=3072 Reads
>Case 3: 2048 Extent 96K each
> Number of reads = ceiling(96/(16*4))*2048=4096 Reads
> It needs 2 reads (64K each) to read one extent (96K each)
> There is extra 4096-3072=1024 reads (each 64K)
> I do not think that this is micro seconds, we are talking
>relatively now
> in this case there is 1024/3072%=33% overhead reads
>Best Regards
Ok, I'll bite. Excellent math demonstrating the 33% overhead in LOGICAL database reads. But I'm a little confused. The number of PHYSICAL O/S blocks required to contain the 192M table appears to remain constant in your example. Those logical reads really are micro seconds once the physical plocks have been obtained and you haven't demonstrated any increase in physical reads. (Or am I missing something?)
Mary
Mary Travis
BellSouth Telecommunications
email: Mary.B.Travis_at_bridge.bellsouth.com
Received on Thu Sep 11 1997 - 00:00:00 CDT