consisten gets question

Hello Experts,

According to the thread in ask tom web site, http://asktom.oracle.com/pls/asktom/f?p=100:11:0:NO:P11_QUESTION_ID:880343948514 , it was said that the total value of the compatible get equals (number of lines) / Fetch arraysize (default is 15 for SQL * MORE) + number of blocks. However, I do not understand the thing that why we add the total number of blocks added to the calculation?

Because, from my point of view, shouldn't the blocks not already in memory? Otherwise, it should be physical reads, is not it?


Thanks in advance.

NightWing wrote:

Used blocks 150/40 = 4 blocks

First extraction - 15 lu, 25 left (block 1) - 1 becomes consistent

Second Fetch - 15 lu, 10 left (block 1)-2 becomes consistent

Third Fetch - 10 read, 0 (block 1) left-3 becomes consistent

Fourth Fetch - 5 read: Left35 (block 2) 35 - 4 in line gets

Fifth Fetch - 10 lu, 25 left (block 2) - 5 becomes consistent

SixthFetch - 15 lu, 10 left (block 2) - 6 becomes consistent

SeventhFetch - 10 read, 0 left (block 2) - 7 becomes consistent

EightFetch - 5 read, 35 left (block 3) - 8 becomes consistent

and so on...

First of all - don't forget that the formula is an approximation,

Second - your example has errors: from where you said the 3rd fetch reads 10 lines, is not she reads 15 ranks and takes two compatible lets you read - 10 rows of block 1 and block 2 5 lines.  Continue from there and the list of all 10 checkouts, without forgetting to let compatible 2 gets when an extraction cross block limits. It is probably easier if you draw an image.

Simple guideline for the simplest cases: each time that you perform an extraction you do conform; but some extractions will cross a border in a second block, resulting in two gets for extraction. As an approximation, Tom has assumed that none of the extractions ends exactly at the end of a block, and he also has an additional extraction (if you have N blocks, you have borderline N-1 to consider).

Concerning

Jonathan Lewis

Tags: Database

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