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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 <>，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 |
Wait Events
s - second
cs - centisecond - 100th of a second
ms - millisecond - 1000th of a second
us - microsecond - 1000000th of a second
ordered by wait time desc, waits desc (idle events last)
| 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
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