A mundane reason could be a degradation in the cluster factor of one or both
indexes (insufficient to change the plan, but causing an increase in
The sledge hammer approach to rule this out is to physically re-order the
table in the order of the index being used for range scans.
Another mundane reason would be if some column(s) have recently been
expanded in length sufficiently to give you a lot of row relocation and/or
actual multi-block storage for a single row. The sledge hammer approach will
mitigate relocation but be less effective for multi-block rows.
Please notice the words sledge hammer: If someone leaps from here to a
periodic table rebuild strategy they get what they deserve. Depending on the
size of your table, ruling this out (or in) may be more or less expensive
than further analysis.
As JL pointed out, trying to beat the LRU buffer cache for tables that are
expanding (and especially involving indexed access) is not a likely win.
That strategy is well matched to pretty small lookup tables that are slowly
changing. But even those can be kept in the regular cache with a
"heart-beat" periodic triple full scan via index that also references a
non-indexed column, or if small enough by FTS.
These are not center case solutions, but rather are exceptional methods for
specific edge cases that are not really rare, but are not that frequent
And they are WAY down the list from better plans, but you've documented that
your plans did not change.
From: oracle-l-bounce@xxxxxxxxxxxxx [mailto:oracle-l-bounce@xxxxxxxxxxxxx]
On Behalf Of Jonathan Lewis
Sent: Thursday, March 14, 2019 6:23 PM
Subject: Re: Keep buffer cache question
Did any of the relevant segments appear in "Segments by physical reads" ?
You need to find out whether it's the indexes or the tables.
As a basic guideline you almost certainly WON'T beat the LRU algorithm by
setting up the keep cache.
A change like this can happen simply because objects (and particularly
indexes) get bigger over time as the data sizes grow. You can get into the
position (especially when the number of queries grows) that a query reads a
leaf block into memory but causes another leaf block from the same index to
be flushed, and a few seconds later some other query wants the leaf block
that was flushed.
Consider an index on (customer_id, order_date) - when the data is small
index entries for "the most recent order for customer X" may find two or
three customers in the same leaf block, so one query benefits from the
caching caused by another. As the data gets larger you get to a position
where every customer has several leaf blocks and every query for "the most
recent order for customer X" has to read a different leaf block and queries
don't get any benefit from each other. At this point your only solution is
to increase the buffer cache to ensure that one block for each customer can
stay in memory long enough for its next usage.
If you do try implementing a KEEP pool, don't forget to check for the
effects of read-consistency. Depending how CR blocks are created you may
find them as copies created in the default cache, or the recycle cache (if
you have one), and some (because of operation "copy current to new buffer")
will be in the keep cache. Sizing the keep cache to keep the blocks AND the
CR blocks can be problematic. You'll have to check what actually happens
because the behaviour changes with version of Oracle and I haven't checked
From: oracle-l-bounce@xxxxxxxxxxxxx <oracle-l-bounce@xxxxxxxxxxxxx> on
behalf of dmarc-noreply@xxxxxxxxxxxxx <dmarc-noreply@xxxxxxxxxxxxx>
Sent: 14 March 2019 21:22
Subject: Keep buffer cache question
In doing an AWR report comparison for comparable times one major difference
I saw was that 2 frequently run queries were suddenly doing a lot of
physical i/o. For a comparable 2 hour period they went from 1.5 million to
1.8 million executions but physical reads increased from 0 to 1.2 million. I
sampled a few other random times and this was consistent. The queries are
both doing index access. One is an index range scan and the other a unique
scan against the primary key.