I have a script converting video files and I run it at server on test data and measure its time by time. In result I saw:

real    2m48.326s
user    6m57.498s
sys     0m3.120s

Why real time is that much lower than user time? Does this have any connection with multithreading? Or what else?

Edit: And I think that script was running circa 2m48s


4 Answers 4


The output you show is a bit odd, since real time would usually be bigger than the other two.

  • Real time is wall clock time. (what we could measure with a stopwatch)
  • User time is the amount of time spend in user-mode within the process
  • Sys is the CPU time spend in the kernel within the process.

So I suppose if the work was done by several processors concurrently, the CPU time would be higher than the elapsed wall clock time.

Was this a concurrent/multi-threaded/parallel type of application?

Just as an example, this is what I get on my Linux system when I issue the time find . command. As expected the elapsed real time is much larger than the others on this single user/single core process.

real    0m5.231s
user    0m0.072s
sys     0m0.088s

The rule of thumb is:

  • real < user: The process is CPU bound and takes advantage of parallel execution on multiple cores/CPUs.
  • real ≈ user: The process is CPU bound and takes no advantage of parallel exeuction.
  • real > user: The process is I/O bound. Execution on multiple cores would be of little to no advantage.
  • I don't know if avconv is multi-threaded. May it be. avconv is the new generation of ffmpeg. I was converting 7 short flv files (about 20 seconds each).
    – kobylecki
    Commented Jun 13, 2012 at 15:59
  • real time would usually be bigger than the other two - but I ask about other situation
    – kobylecki
    Commented Jun 13, 2012 at 16:17
  • 4
    This explanation is correct. It looks like this process was run on 4 cores. See also my explanation of hyperthreading for more on how real/sys/user time is calculated. It doesn't relate exactly, but the concepts are the same.
    – bahamat
    Commented Jun 13, 2012 at 16:32
  • @kobylecki real time is less than the others because it looks like avconv is run on multiple cores. Since I don't know that software, nor how it was run, I don't want to make a 100% claim, but this is what it looks like based on the available information (3 lines of time measurements, and knowledge :-)
    – Levon
    Commented Jun 13, 2012 at 17:36
  • In the find example usr value is much lower because most of the time has been spent during interrupts, even if find would have been multithreaded it would have stayed low (sorry if I don't master english tenses).
    – Emmanuel
    Commented Oct 7, 2013 at 12:07

Just to illustrate what has been said, with a two threaded processes doing some calculation.

    #include <pthread.h>
    static void  * dosomething () {
        unsigned long a,b=1;
        for (a=1000000000; a>0; a--) b*=3;
        return NULL;
    main () {
        pthread_t one, two;
        pthread_create(&one,NULL, dosomething, NULL);
        pthread_create(&two,NULL, dosomething, NULL);
        pthread_join (one, NULL);
        pthread_join (two, NULL);
/* end of a.c */


gcc a.c -lpthread

(This is just to illustrate, in real life I should have added the -D_REENTRANT flag)

$ time ./a.out

real    0m7.415s
user    0m13.105s
sys     0m0.032s

(Times are on an Intel Atom that has two slow cores :) )

  • Hi I comment on this old entry, with no optimization the compiler don't detect that the function is not providing any result. When compiled with the -O2 option the calculation are skipped.
    – Emmanuel
    Commented May 23, 2023 at 8:52

It doesn’t need to be multi-threaded program(s); you can get the same effect with multiple processes.  I wrote a C program to do some trivial number crunching:

$ time ./runner

real    1m12.746s
user    1m12.493s
sys     0m0.015s

Then I wrote a script to run two copies of the process:

$ cat run_script
./runner &
./runner &

$ time ./run_script

real    1m31.231s
user    3m0.538s
sys     0m0.215s

As Levon said:

I have a i7 (8 processors), so if run in bash:

$ time for ((i=0;i<1000000;i++)); do echo $i; done | wc -l

real    0m4,878s
user    0m4,368s
sys     0m3,083s

Then if restrict bash to one processos with taskset -cp 0 $BASHPID

$ taskset -cp 0 $BASHPID                                                                                         
pid 30551's current affinity list: 0-7
pid 30551's new affinity list: 0

now the result is

$ time for ((i=0;i<1000000;i++)); do echo $i; done | wc -l

real    0m7,120s
user    0m4,282s
sys     0m2,824s



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