CPU Performance

Here is some more interesting script.

Process CPU distribution

This one shows how one process run on different cpus.

global one, two
global time
function div:string (a:long, b:long)
{
    total = a + b;
    mod = (100*a)%total;
    return sprintf("%d.%d%%", (100*a)/total, mod*100/total);
}
probe timer.profile
{
    if (cpu() == 0)
        one[execname(), tid()]++
    if (cpu() == 1)
        two[execname(), tid()]++
}
probe timer.s(5)
{
    exit();
}
probe begin
{
    time = gettimeofday_s()
}
probe end
{
    printf("%20s %5s %19s %19s\n", "process", "TID", "CPU0", "CPU1");
    foreach([p, t] in two){
        printf("%20s %5d %19s %19s\n", p, t,
            div(one[p, t], two[p, t]), div(two[p, t], one[p,t]));
        delete one[p,t];
    }
    foreach([p, t] in one){
        printf("%20s %5d %18d%% %18d%%\n", p, t, 100, 0);
    }
    time = gettimeofday_s() - time;
    printf("\nTotal............................: %5d secs.\n", time);
}

Next one would be more suitable for more cpus, while a little better for less process.

#Argument on line
# ($1) is the number of cpus - 1 in the system
# ($2) is the process name regex
global sample
global time
probe timer.profile
{
    process = execname();
    if ($# == 2 && process =~ @2)
        sample[process, tid()] <<< cpu();
}
probe timer.s(5)
{
    exit();
}
probe begin
{
    time = gettimeofday_s()
}
probe end
{
    foreach([p, t] in sample){
        printf("process: %-20s TID: %5d\ncpu\tsamles\n", p, t)
        print (@hist_linear(sample[p,t], 0, $1, 1))
    }
    time = gettimeofday_s() - time
    printf("\nTotal............................: %5d secs.\n", time);
}