Oracle数据库中平均事务响应时间的计算公式
注： 该计算公式取自：白鳝 <<oracle 优化日记>>，p316页
Top 5 Timed Events
| Event | Waits | Time(s) | Avg Wait(ms) | % Total Call Time | Wait Class |
|---|---|---|---|---|---|
| direct path read | 327,284 | 15,555 | 48 | 86.4 | User I/O |
| CPU time | 1,093 | 6.1 | |||
| db file sequential read | 283,101 | 509 | 2 | 2.8 | User I/O |
| db file scattered read | 99,520 | 180 | 2 | 1.0 | User I/O |
| enq: TX - row lock contention | 274 | 133 | 485 | .7 | Application |
| Event | Waits | %Time -outs | Total Wait Time (s) | Avg wait (ms) | Waits /txn |
|---|---|---|---|---|---|
| direct path read | 327,284 | 0.00 | 15,555 | 48 | 32.66 |
| db file sequential read | 283,101 | 0.00 | 509 | 2 | 28.25 |
| db file scattered read | 99,520 | 0.00 | 180 | 2 | 9.93 |
| enq: TX - row lock contention | 274 | 98.91 | 133 | 485 | 0.03 |
| log file sync | 6,791 | 0.00 | 93 | 14 | 0.68 |
| control file sequential read | 16,168 | 0.00 | 91 | 6 | 1.61 |
| log file parallel write | 7,816 | 0.00 | 82 | 10 | 0.78 |
| name-service call wait | 1,199 | 0.42 | 68 | 57 | 0.12 |
以"direct path read"这个等待事件为例子来计算： A---"direct path read" 的 Total Wait Time（以ms来计算）为15555*1000=15555000ms B---"direct path read" 的 Waits 为327284 C---"direct path read" 的 Waits /txn 为32.66 D---"direct path read" 的 % Total Call Time 为 86.4%=0.864 平均事务响应时间=A/B*C/D=15555000/327284*32.66/0.864=1796.584626 ms
原文地址：http://blog.csdn.net/msdnchina/article/details/42366381
