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Although secondary storage (usually magnetic disks) is significantly larger than main memory, it s also significantly slower. A disk I/O involves either moving a data block from disk to memory (a disk read) or writing a data block to disk from memory (a disk write). Typically, it takes about 10 40 milliseconds (0.01 0.04 seconds) to perform a single disk I/O. Suppose your update transaction involves 25 I/Os you could spend up to 1 second just waiting to read or write data. In that same second, your CPUs could have performed millions of instructions the update takes a negligible amount of time compared to the disk reads and disk writes. If you already have the necessary data in Oracle s memory, the retrieval time would be much faster, as memory read/writes take only a few nanoseconds. This is why avoiding or minimizing disk I/Os plays such a big role in providing high performance in Oracle databases.

Upon COMMIT, Oracle will process each of these lists of 20 block pointers, and if the block is still available, it will perform a fast cleanout So, as long as the number of blocks we modify does not exceed 10 percent of the number of blocks in the cache and our blocks are still in the cache and available to us, Oracle will clean them out upon COMMIT Otherwise, it just skips them (ie, does not clean them out) With this understanding, we can set up artificial conditions to see how the cleanout works I set my DB_CACHE_SIZE to a low value of 16MB, which is sufficient to hold 2,048 8KB blocks (my blocksize is 8KB) Next I create a table such that a row fits on exactly one block I ll never have two rows per block.

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logs are sized appropriately, indexes won t be slowing things down, and I m running on a machine with two hyperthreaded Xeon CPUs this should run fast. What I discovered afterward, however, is the following: Snap Id Snap Time Sessions Curs/Sess Comment --------- ------------------ -------- --------- ------------------Begin Snap: 793 29-Apr-05 13:45:36 15 3.9 End Snap: 794 29-Apr-05 13:46:34 15 5.7 Elapsed: 0.97 (mins) Top 5 Timed Events ~~~~~~~~~~~~~~~~~~ % Total Event Waits Time (s) Call Time -------------------------------------------- ------------ ----------- --------CPU time 165 53.19 buffer busy waits 368,698 119 38.43 log file parallel write 1,323 21 6.86 latch: cache buffers chains 355 2 .67 enq: HW - contention 2,828 1 .24 I collectively waited 119 seconds, or about 24 seconds per session, on buffer busy waits. These waits are caused entirely by the fact that there are not enough freelists configured on my table for the type of concurrent activity that is taking place. I can eliminate most of that wait time easily, just by creating the table with multiple freelists: ops$tkyte@ORA10GR1> create table t ( x int, y char(50) ) 2 storage( freelists 5 ) tablespace MSSM; Table created. or by altering the object: ops$tkyteORA10GR1> alter table t storage ( FREELISTS 5 ); Table altered. You will find that the buffer busy waits goes way down, and the amount of CPU needed (since you are doing less work here; competing for a latched data structure can really burn CPU) also goes down along with the elapsed time: Snap Id Snap Time Sessions Curs/Sess Comment --------- ------------------ -------- --------- ------------------Begin Snap: 809 29-Apr-05 14:04:07 15 4.0 End Snap: 810 29-Apr-05 14:04:41 14 6.0 Elapsed: 0.57 (mins) Top 5 Timed Events ~~~~~~~~~~~~~~~~~~ % Total Event Waits Time (s) Call Time -------------------------------------------- ------------ ----------- --------CPU time 122 74.66

All computers use memory, which actually consists of a hierarchy of different levels of memory. The heart of this hierarchy is main memory, which contains all the instruction executions and data manipulations. All main memories are random access memory (RAM), which means that you can read any byte in memory in the same amount of time. Typically, you can access main memory data in the 10 100 nanosecond range. An important part of the information Oracle stores in the RAM allocated to it is the program code that is executing currently or that has been executed recently. If a new user process needs to use the same code, it s available in memory in a compiled form, making the processing time a whole lot

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