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Given a VM with 1 CPU. I run two similar CPU intensive tasks in two terminals:

/usr/bin/nice -n -20 perl -e 'while(1){$a=1+1;}'
/usr/bin/nice -n 19 perl -e 'while(1){$a=1+1;}'

I would expect that when I check with top, 1 process would take all the CPU and the other none (because there is no idle time left). But...

  PID USER      PR  NI  VIRT  RES  SHR S  %CPU %MEM    TIME+  COMMAND
18175 root       1 -20 20376 1460 1132 R  49.9  0.1   0:18.36 perl
18176 root      39  19 20376 1460 1132 R  49.9  0.1   0:15.16 perl

Why is that? I want my process to humbly only use idle CPU cycles :)

Tested on Linux 3.5.0-17-generic #28-Ubuntu SMP Tue Oct 9 19:31:23 UTC 2012 x86_64 x86_64 x86_64 GNU/Linux

Willem
  • 3,042

1 Answers1

3

There is an "art" used in task scheduling. The standard Linux/Unix scheduling algorithm allows all processes at some point to get CPU. This is done by initially, assigning a process two items: (1) lowest priority in its catagory (as determined from its nice) and (2) a time increment again based on priority cateory. As time goes on the process priority gets incremented, the highest priority process gets CPU time. Also, the process getting CPU is continually redetermined from tick to tick based on highest priority at that time after the recalculation. Once a process gets CPU time, its priority returns again to the lowest priority in its catagory.

You could get what you wanted by using "realtime" priorities, where highest priority class processes gets the CPU and gets to use it until all processes in the priority class are satisified, then processes in the next lower class gets the available time, until either a higher priority process requires the CPU, and so on until the total CPU scheduling increment is used up. A process can relinquish part of the CPU increment, thus causing a reschedule.

Thus you are seeing on using the standard scheduler, your lower priority process still getting CPU time but at a far lower rate.

mdpc
  • 11,914