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rhofing
Starting Member
15 Posts |
 Posted - 2004-03-18 : 17:07:26
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| At the risk of sounding like a mathematical moron I have a question about DBCC SHRINKDATABASE. How is the actual free space calculated based on the percentage free value specified in the command? I take an example from BOL:mydb data file contains 6 MB of data, if you specify a target_percent of 25 for shrinking mydb. SQL Server calculates the target size for this file to be 8 MB (6 MB of data plus 2 MB of free space). I understand that 2MB is 25% of 8MB. But how did they arrive at the 8MB value in the first place? Another way to ask this question is: 6MB plus how many MB's would represent 25% of the resulting value?I ask this question because I am taking the 70-228 practice tests and there are a number of similar questions.I am pulling out what little hair I have left :)Thank you.Ric |
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tkizer
Almighty SQL Goddess
38200 Posts |
Posted - 2004-03-18 : 17:17:56
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| It just depends on your environment. In most environments, you leave 10% of free space in the MDF file. So that would be 6.6MB. I guess for their example, they wanted 20% free. There isn't one answer for this because it just depends on what your environment's needs are.Tara |
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MuadDBA
628 Posts |
Posted - 2004-03-19 : 13:19:52
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| Algebra. You need to phrase the question in this way:6MB is 75% of what value?or you could phrase it this way:What value minus 25% equals 6MB.Either way, once you write it algebraicly, you are going to get an equation looking something like this:X - .25X = 6.75X = 6X = 6/.75X = 6 * 4 /3 (to divide by a fraction [.75 = 3/4] you multiply by its inverse, or 4/3)X = 24/3X = 8Is this the answer you were looking for? |
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DBCC SHRINKDATABASE question