How is kernel oom score calculated?

Looked on The Google, and couldn't find anything that explained how the score in proc//oom_score is calculated. Why use this score instead of just using the total memory used?

Answer

See Goldwyn Rodrigues's 2009 article for the implementation at that time, Jonathan Corbet's 2010 article for what I believe is the current behavior, and Jonathan Corbet's 2013 article for ideas about future changes.

From the 2010 article:

In David's patch set, the old badness() heuristics are almost entirely
gone. Instead, the calculation turns into a simple question of what
percentage of the available memory is being used by the process. If
the system as a whole is short of memory, then "available memory" is
the sum of all RAM and swap space available to the system. If,
instead, the OOM situation is caused by exhausting the memory allowed
to a given cpuset/control group, then "available memory" is the total

amount allocated to that control group. A similar calculation is made
if limits imposed by a memory policy have been exceeded. In each case,
the memory use of the process is deemed to be the sum of its resident
set (the number of RAM pages it is using) and its swap usage.

This calculation produces a percent-times-ten number as a result; a
process which is using every byte of the memory available to it will
have a score of 1000, while a process using no memory at all will get
a score of zero. There are very few heuristic tweaks to this score,
but the code does still subtract a small amount (30) from the score of

root-owned processes on the notion that they are slightly more
valuable than user-owned processes.

One other tweak which is applied is to add the value stored in each
process's oom_score_adj variable, which can be adjusted via /proc.
This knob allows the adjustment of each process's attractiveness to
the OOM killer in user space; setting it to -1000 will disable OOM
kills entirely, while setting to +1000 is the equivalent of painting a
large target on the associated process.