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Simple vm question

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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