Re: question about rollback and redo

(assuming no indexes)
When you delete a row the row plus about 80 bytes of overhead gets put into the undo. A few bytes of change are made to the data block (deleted flag, ITL entry).

The redo then contains a record consisting of two change vectors - each with an overhead of about 80 bytes. One change vector describes the change to the data block, and the other change vector describes the change to the redo block, which include the detail of the deleted row.

Delete 1MB as 10,000 rows of 100 bytes, and you have 10,000 undo records - which is the 10,000 rows with 10,000 overheads - and
10,000 redo records which is 10,000 copies of the row plus about 20,000 lots of overhead.

--
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Jonathan Lewis
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"Ryan Gaffuri" <rgaffuri_at_cox.net> wrote in message
news:1efdad5b.0309191139.5eab5f6d_at_posting.google.com...


> Documentation isnt too clear on this(or Im just not following it).

>

> When you do a delete the entire row is copied to the rollback

segment.


> Now if Im deleting 1MB worth of data in the datafile(lets assume

there


> is no pctfree or pct used, just data), will 1MB of data be written

to


> rollback? Im assuming there will be less due to block header,

etc...,


> but how much less?

>

> Now when the rollback is written to redo, how much data does that

> entail? What goes in a 'change vector'? Does the entire row get

copied


> twice? Once to the rollback and once to the redo logs?