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.
