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I am trying to compile a kernel ( 3.14, arm64 arch ) for a low memory system ( 16MB ram ). I have managed to turn off all the features that I dont need, basically no devices, no fs, no modules, no ZONE_DMA, a very very basic kernel.

However, when I boot the kernel I see this:

Memory: 860K/16384K available (789K kernel code, 67K rwdata, 56K rodata, 64K init, 38K bss, 15524K reserved)

  • What is this reserved memory?

  • How can I reduce this? The reserve eats up a lot of my RAM, leaving only 860K available

Thanks in advance!

Jonathan
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eldukae
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2 Answers2

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It is the minimum amount of memory that should always be there to satisfy critical memory allocations. Setting it too low might lead to a broken system and setting to high might instantly OOM your system. You can modify this value by writing to /proc/sys/vm/min_free_kbytes.

To read it:

$ cat /proc/sys/vm/min_free_kbytes 
67584

To set it to 1024KB (1MB):
$ echo 1024 > /proc/sys/vm/min_free_kbytes

Karim Manaouil
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Memory: 860K/16384K available (789K kernel code, 67K rwdata, 56K rodata, 64K init, 38K bss, 15524K reserved)

This is the structure of process in memory: text/code segment, data segment(bss,rodata etc), heap and stack. Kernel always resides in RAM and reserve some memory space for its usages. But when the RAM is small it is shared by kernel and user space processes.

To know more on embedded linux read this post.

Community
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skaushal
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    This still does not explain what the 'Reserved' section of memory is used for and how can it be tuned – eldukae Mar 02 '15 at 22:10