On Sat, Nov 14, 2020 at 12:23:56PM +0100, Otto Moerbeek wrote:
> It could be I'm mistaken an cache is part of tot.
Indeed both act and cache are part of tot.
-Otto
>
> Mihai Popescu <[email protected]> schreef op 14 november 2020 11:49:35 CET:
> >On Sat, Nov 14, 2020 at 12:30 PM Otto Moerbeek <[email protected]> wrote:
> >
> >> On Sat, Nov 14, 2020 at 12:21:30PM +0200, Mihai Popescu wrote:
> >>
> >> [ .. ]
> >> > CPU0: 22.4% user, 0.0% nice, 3.8% sys, 0.6% spin, 0.6% intr,
> >72.7%
> >> idle
> >> > CPU1: 21.2% user, 0.0% nice, 3.0% sys, 0.2% spin, 0.0% intr,
> >75.6%
> >> idle
> >> > Memory: Real: 1235M/2914M act/tot Free: 4505M Cache: 1054M Swap:
> >0K/7913M
> >>
> >> Cutting some corners, but the basics go like this:
> >>
> >> Currently, no swap is used.
> >>
> >> You see 4505G is free. Roughly you can use that amount more. If free
> >> becomes low, cache will be reduced and/or pages swapped out, making
> >> more pages free so they can become used and part of tot.
> >>
> >> tot + free + cache should add up to available RAM (which is a bit
> >less
> >> than what you have in the machine, since the kernel also needs to fit
> >> somewhere and uses memory of its own).
> >>
> >> -Otto
> >>
> >>
> >Here is my confusion.
> >tot + free + cache would be:
> >2914M + 4505M + 1054M = 8473M
> >
> >dmesg shows me this:
> >real mem = 8029429760 (7657MB)
> >avail mem = 7770787840 (7410MB)
> >here should be the memory for the integrated video card:
> >7657MB - 7410MB = 247MB <- plausible
> >
> >Am I correct, or am I hit by the bit/byte units conversion.
> >
> >Thank you.
>
> --
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