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physical space taken by a file on a NetApp



 
 
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  #1  
Old July 10th 07, 07:51 PM posted to comp.arch.storage
[email protected]
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Posts: 9
Default physical space taken by a file on a NetApp

How can I determine the physical amount of space taken by a file on a
NetApp (WAFL)?

  #3  
Old July 28th 07, 10:35 AM posted to comp.arch.storage
Raju Mahala
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Posts: 47
Default physical space taken by a file on a NetApp

On Jul 11, 2:55 am, Faeandar wrote:
On Tue, 10 Jul 2007 11:51:44 -0700, wrote:
How can I determine the physical amount of space taken by a file on a
NetApp (WAFL)?


Divide the 'du' output by 4k (block size)?

Unless you have some weird file size or layout I would think a simple
'du' would work just fine.

Not sure how it would work in windows, sorry.

~F


Why are you dividing by 4. I feel if you give #du -sk Dir It
should give size in KB consumed bye dir. and it shouldn't concern with
WAFL which have 4k block size. du will send request to NetApp which
will provide size to du and du will provide to user. Can you comment
on it and can explain your view ?

Raju

  #4  
Old October 2nd 07, 08:12 PM posted to comp.arch.storage
AnthonyL[_2_]
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Posts: 4
Default physical space taken by a file on a NetApp

I'll take a stab at it...

By physical I'm assuming you mean actual sectors on physical media? So
if this is the case it depends on the following factors that I'm aware of.

1) WAFL overhead
each aggregate needs 10% for WAFL overhead, you could consider this as
using additional physical space.

2) Right-sizing
Drives are right-sized to account for minor deviations in total storage
capacity among hard-drive vendors. I.E. All "146GB" drives are sized to
a certain number to make things clean/safe from a RAID config
perspective. You loose a marginal amount of storage from each drive due
to this. (I'm not sure of the amount)

3) Aggregate/RAID Configuration
Depending on the disk layout differing amounts of physical space will be
consumed. Assume an Aggregate with raidsize set at 9 with RAID-DP with
18 total drives. So the amount of physical space consumed would be an
additional ~22% (2/9) so a 1MB file would consume 1.22MB of physical
space due to RAID overhead.

4) Checksums
If using a V-series block checksums consume an additional *12.5%* from
each physical disk. This is obviously a big hit, zoned checksums
consume only a marginal amount of space, but block is the
recommendation. This consumes additional physical space as well. Note
this is only with a V-series.

5) Snap reserve
Aggregate and Volume snap reserves should probably also be calculated if
you have snapshots turned on just to give a true view of amount of space
required, even though this physical space isn't needed until a snap is
taken, and I overwrite those blocks from the production filesystem.

Also dont forget to convert the marketing number to real usable
gigabytes etc... I.E. a drive advertised as a "146GB" drive really
means there are roughly 146,000,000,000 bytes available.
146,000,000,000/1000/1000/1000 = 146 Gigabytes. However the computer
doesn't use decimal and the division should be by 2^10 (which is 1024)
so a "146GB" drive is really 146,000,000,000/1024/1024/1024 =~ 136GB

As for the 4k block size of WAFL, I dont think that has an impact really
on files that are stored, meaning, if I store a 1k file it doesnt take
4k minimum. However if I increase/decrease the size of my flexvol I
have to do it in 4k increments.

Thats all I can think of! Please add or correct if necessary!

-Anthony

Raju Mahala wrote:
On Jul 11, 2:55 am, Faeandar wrote:
On Tue, 10 Jul 2007 11:51:44 -0700, wrote:
How can I determine the physical amount of space taken by a file on a
NetApp (WAFL)?

Divide the 'du' output by 4k (block size)?

Unless you have some weird file size or layout I would think a simple
'du' would work just fine.

Not sure how it would work in windows, sorry.

~F


Why are you dividing by 4. I feel if you give #du -sk Dir It
should give size in KB consumed bye dir. and it shouldn't concern with
WAFL which have 4k block size. du will send request to NetApp which
will provide size to du and du will provide to user. Can you comment
on it and can explain your view ?

Raju


 




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