Re: Simple vm question

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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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