使用以下语句：
SELECT size_for_estimate, buffers_for_estimate, estd_physical_read_factor, estd_physical_reads
FROM V$DB_CACHE_ADVICE
WHERE name = ’DEFAULT’
AND block_size = (SELECT value FROM V$PARAMETER WHERE name = ’db_block_size’)
AND advice_status = ’ON’;
显示结果如下：
Estd Phys Estd Phys
Cache Size (MB) Buffers Read Factor Reads
---------------- ------------ ----------- ------------
30 3,802 18.70 192,317,943
10% of Current Size
60 7,604 12.83 131,949,536
Configuring and Using the Buffer Cache
91 11,406 7.38 75,865,861
121 15,208 4.97 51,111,658
152 19,010 3.64 37,460,786
182 22,812 2.50 25,668,196
212 26,614 1.74 17,850,847
243 30,416 1.33 13,720,149
273 34,218 1.13 11,583,180
304 38,020 1.00 10,282,475
Current Size
334 41,822 .93 9,515,878
364 45,624 .87 8,909,026
395 49,426 .83 8,495,039
424 53,228 .79 8,116,496
456 57,030 .76 7,824,764
486 60,832 .74 7,563,180
517 64,634 .71 7,311,729
547 68,436 .69 7,104,280
577 72,238 .67 6,895,122
608 76,040 .66 6,739,731
200% of Current Size
为什么可以从以上得出以下结论：
if the cache was 212 MB, rather than the current
size of 304 MB, the estimated additional number of physical reads would be over 17
million (17,850,847). Increasing the cache size beyond its current size would not
provide a significant benefit.
请赐教
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