It could be I'm mistaken an cache is part of tot. 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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