It’s called the CPU Costing model because among other things, it includes the time associated with performing CPU operations.
The CPU Costing model formula once again:
(sum of all the single block I/Os x average wait time for a single block I/O +
sum of all the multiblock I/Os x average wait time for a multiblock I/O +
sum of all the required CPU cycles / CPU cycles per second)
/
average wait time for a single block I/O
So the portion detailing the sum of all required CPU cycles divided by the CPU cycles per second can potentially contribute a significant proportion of the overall costs.
When I previously discussed the costs associated with the CPU model between using an index and a FTS, the FTS example I used had an overall cost of 70 but I calculated that the I/O component costs were only 67. Therefore the costs directly related to CPU operations with the FTS example was 3.
However, these CPU specific costs in this example may vary from database to database, although the FTS might be identical as might the required CPU cycles. However, a variable in all this is the CPU cycles per second system statistic (CPUSPEED) associated with a particular database environment.
Obviously, the faster the CPUs, the quicker it can perform the necessary CPU operations associated with the FTS (or any operation for that matter). Conversely, the slower the CPUs, the longer it will take to complete the necessary CPU related operations. The CPU costing model formula automatically takes all this into consideration.
In the previous example, the CPUSPEED system statistic was 1745.
Let’s now run the identical FTS but this time with a faster and a slower CPU and see how this might adjust the overall related costs when using the CPU costing model.
One can simulate a “faster” CPU by simply adjusting the CPUSPEED system statistic. Let’s make the CPUs appear 10 times faster:
SQL> exec dbms_stats.set_system_stats(pname=>’cpuspeed’, pvalue=>17450);
PL/SQL procedure successfully completed.
OK, let’s now see how this impacts the cost of the FTS:
SQL> SELECT * FROM bowie_stuff WHERE id = 420;
1000 rows selected.
Execution Plan
———————————————————-
Plan hash value: 910563088
——————————————————————-
|Id|Operation |Name |Rows|Bytes|Cost (%CPU)|Time |
——————————————————————-
| 0|SELECT STATEMENT | |1000|18000| 67 (0)|00:00:01|
|*1| TABLE ACCESS FULL|BOWIE_STUFF|1000|18000| 67 (0)|00:00:01|
——————————————————————-
We notice that the overall cost has reduced from 70 down to 67. The cost of 3 that was previously attributed to just the CPU related costs have all disappeared and the costs are now just the 67 in relation to the I/O component.
The CPU is now so fast that it can effectively perform all the necessary operations in a negligible amount of time. An even faster CPU will not further improve the costs associated with this FTS as the costs now only include the I/O related components.
The (%CPU) value of (0) gives us this information if you didn’t follow how I derived the I/O cost of 67 in my previous post.
If we go the other way and now make the CPU about 1/10 the speed of the original example:
SQL> exec dbms_stats.set_system_stats(pname=>’cpuspeed’, pvalue=>175);
PL/SQL procedure successfully completed.
SQL> SELECT * FROM bowie_stuff WHERE id = 420;
1000 rows selected.
Execution Plan
———————————————————-
Plan hash value: 910563088
——————————————————————-
|Id|Operation |Name |Rows|Bytes|Cost (%CPU)|Time |
——————————————————————-
| 0|SELECT STATEMENT | |1000|18000| 93 (28)|00:00:01|
|*1| TABLE ACCESS FULL|BOWIE_STUFF|1000|18000| 93 (28)|00:00:01|
——————————————————————-
We now notice the overall costs have jumped up considerably up from 70 up 93.
The costs associated directly with CPU activities have now increased up from 3 to 26. The CPU component is in the ballpark of 10 times as expensive/significant when you take into account rounding errors (the original 3 value was rounded accordingly). Remember also that these figures are times expressed in units of time it takes to perform a single block I/O.
The CPUs are now so slow that it takes a considerable amount of time to complete all the required CPU operations.
Note that the (%CPU) value is now a significant (28%) of the overall costs as derived from the following formula:
round(cpu related cost/total cost) x 100 = round(26/93 x 100) = 28.
So having a faster (or possibly slower) CPU when performing a hardware upgrade/change can result in potentially different execution plan costings (and as such different execution plans) when using the CPU CBO costing model.
It’s called the CPU costing model for a reason and as one would indeed hope, the speed of said CPU(s) can directly impact the associated costs and decisions made by the CBO.
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作者：czmmiao 文章出处：http://czmmiao.iteye.com/blog/2086252