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Re: Simple vm question
Refer the below link to understand /proc/meminfo
http://www.redhat.com/advice/tips/meminfo.html
--- "Crawford, Andrew (IT)"
<Andrew.Crawford@xxxxxxxxxxxxxxxxx> wrote:
> Dear all,
>
> I'm trying to understand the data in /proc/meminfo.
>
> MemTotal: 8258560 kB
> MemFree: 17116 kB
> MemShared: 0 kB
> Buffers: 27976 kB
> Cached: 7739640 kB
> SwapCached: 0 kB
> Active: 6474448 kB
> ActiveAnon: 270376 kB
> ActiveCache: 6204072 kB
> Inact_dirty: 1240656 kB
> Inact_laundry: 241704 kB
> Inact_clean: 131008 kB
>
> Am I wrong in believing that
>
> MemFree = MemTotal - Buffers - SwapCached - Active -
> Inact_dirty - Inact_laundry
> - Inact_clean - (some space for the
> kernel itself)
>
> In other words, the things which use up memory are
>
> * Buffers
> * Stuff swapped out but still in RAM
> * Active pages
> * Inactive pages
> * Some kernel space
>
> What am I missing? Because if we try this with the
> numbers above we get:
>
> 17116 = 8258560 - 27976 - 0 - 6474448 - 1240656 -
> 241704 - 131008 - X
>
> 17116 = 142768 - X
>
> X = 125662
>
> This would suggest that "space for the kernel
> itself" on this machine is 122MB. That seems like
> quite a lot considering I run the same version at
> home on a machine with 16MB total RAM!
>
> What's going on here? Anyone?
>
> Supplemental question: If "Cached" is meant to be
> the whole size of the page cache (minus Buffers for
> some reason), how come all the active pages + all
> the inactive pages - Buffers adds up to way more
> than Cached?
>
> Cheers,
>
> Andrew
>
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