Re: longops vs sql

> Wouldn't that also be based upon how many physical reads were actually
> required? Say if you had a huge buffer cache and rarely had to go to
> disk for your data, that value could be quite smaller then, correct?

If the data was cached, then it is a logical read. No physical read required.

> I have a few snap shots from our system, so I looked at our Load
> Profile, it lists:
>
> Load Profile
> ~~~~~~~~~~~~ Per Second Per
> Transaction
> ---------------
> ---------------
> Redo size: 153,299.18
> 3,927.45
> Logical reads: 16,661.37
> 426.86
> Block changes: 540.73
> 13.85
> Physical reads: 1,752.49
> 44.90
> Physical writes: 152.31
> 3.90
>
> So from this I should probably assume that at that point in time we
> were acheiving 1,752 I/O / sec? Would I add the writes into that to see
> the total I/O / sec?

Your Oracle database, during this snapshot interval, was performing 1752.49+152.31= 1,904.80 I/O requests per second, on average. During that interval, you could have had more I/O requests in one second, or less. This is just an average per second on that snapshot interval. Total I/O operations is the sum of physical reads and writes.

The unit of I/O is the database block. Your OS often has a different block size so it could have been performing even more I/O operations per second. For instance, if your database block is 8KB and your OS's filesystem has a block size of 2KB, then one database I/O operation gets translated to 4 2KB I/O operations.

HTH,
Brian

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Brian Peasland
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http://www.peasland.net

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