It could be I'm mistaken an cache is part of tot.
Mihai Popescu <mihscu@gmail.com> schreef op 14 november 2020 11:49:35 CET:
>On Sat, Nov 14, 2020 at 12:30 PM Otto Moerbeek <otto@drijf.net> 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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