How To Boot with "mem=1024m" Argument using GRUB - Ubuntu 10.04
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Published on 2010-06-15T19:35:04Z
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2010/06/15
19:43 UTC
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I am still working on this question. This new one is a different question so I thought it would be good to post a new question. Is this the proper protocol or should I have just edited the other question?
I'm running Ubuntu 10.04 with the kernel 2.6.32-22-generic on a Toshiba Satellite laptop. When I enter the GRUB menu (I have Ubuntu 9.10 installed as well), I can choose which kernel to boot. I use scroll down to the one I want and press "e" and I expect to be able to enter mem=1024m
and force the kernel to use this much memory. But when I run cat /proc/meminfo
or look in the process manager after booting wth this argument I still see all the RAM: ~2 GB.
Am I using this boot argument incorrectly? The boot configuration (before I add anything) looks like this:
insmod ext2
set root=(hd0,1)
search --no-floppy --fs-uuid --set 10270f21-1c42-494b-bd3f-813c23f6d\
518
linux /boot/vmlinuz-2.6.32-22-generic root=UUID=10270f21-1c42-494b-b\
d3f-813c23f6d518 ro quiet splash
initrd /boot/initrd.img-2.6.32-22-generic
The way I did this was that I added the mem=1024m
after the last line and pressed Ctrl+x (Emacs save and boot the kernel) and the system booted.
I tried adding mem=1024m
to the end and the beginning of this list and it appeared to not change the RAM allocation.
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