Re: cpufreq: intel_pstate: map utilization into the pstate range

[Julia Lawall <julia.lawall@inria.fr>]

On Tue, 4 Jan 2022, Rafael J. Wysocki wrote:

> On Tue, Jan 4, 2022 at 4:49 PM Julia Lawall <julia.lawall@inria.fr> wrote:
> >
> > I tried the whole experiment again on an Intel w2155 (one socket, 10
> > physical cores, pstates 12, 33, and 45).
> >
> > For the CPU there is a small jump a between 32 and 33 - less than for the
> > 6130.
> >
> > For the RAM, there is a big jump between 21 and 22.
> >
> > Combining them leaves a big jump between 21 and 22.
>
> These jumps are most likely related to voltage increases.
>
> > It seems that the definition of efficient is that there is no more cost
> > for the computation than the cost of simply having the machine doing any
> > computation at all.  It doesn't take into account the time and energy
> > required to do some actual amount of work.
>
> Well, that's not what I wanted to say.

I was referring to Francisco's comment that the lowest indicated frequency
should be the most efficient one.  Turbostat also reports the lowest
frequency as the most efficient one.  In my graph, there are the pstates 7
and 10, which give exactly the same energy consumption as 12.  7 and 10
are certainly less efficient, because the energy consumption is the same,
but the execution speed is lower.

> Of course, the configuration that requires less energy to be spent to
> do a given amount of work is more energy-efficient.  To measure this,
> the system needs to be given exactly the same amount of work for each
> run and the energy spent by it during each run needs to be compared.

This is bascially my point of view, but there is a question about it.  If
over 10 seconds you consume 10J and by running twice as fast you would
consume only 6J, then how do you account for the nest 5 seconds?  If the
machine is then idle for the next 5 seconds, maybe you would end up
consuming 8J in total over the 10 seconds.  But if you take advantage of
the free 5 seconds to pack in another job, then you end up consuming 12J.

> However, I think that you are interested in answering a different
> question: Given a specific amount of time (say T) to run the workload,
> what frequency to run the CPUs doing the work at in order to get the
> maximum amount of work done per unit of energy spent by the system (as
> a whole)?  Or, given 2 different frequency levels, which of them to
> run the CPUs at to get more work done per energy unit?

This is the approach where you assume that the machine will be idle in any
leftover time.  And it accounts for the energy consumed in that idle time.

> The work / energy ratio can be estimated as
>
> W / E = C * f / P(f)
>
> where C is a constant and P(f) is the power drawn by the whole system
> while the CPUs doing the work are running at frequency f, and of
> course for the system discussed previously it is greater in the 2 GHz
> case.
>
> However P(f) can be divided into two parts, P_1(f) that really depends
> on the frequency and P_0 that does not depend on it.  If P_0 is large
> enough to dominate P(f), which is the case in the 10-20 range of
> P-states on the system in question, it is better to run the CPUs doing
> the work faster (as long as there is always enough work to do for
> them; see below).  This doesn't mean that P(f) is not a convex
> function of f, though.
>
> Moreover, this assumes that there will always be enough work for the
> system to do when running the busy CPUs at 2 GHz, or that it can go
> completely idle when it doesn't do any work, but let's see what
> happens if the amount of work to do is W_1 = C * 1 GHz * T and the
> system cannot go completely idle when the work is done.
>
> Then, nothing changes for the busy CPUs running at 1 GHz, but in the 2
> GHz case we get W = W_1 and E = P(2 GHz) * T/2 + P_0 * T/2, because
> the busy CPUs are only busy 1/2 of the time, but power P_0 is drawn by
> the system regardless.  Hence, in the 2 GHz case (assuming P(2 GHz) =
> 120 W and P_0 = 90 W), we get
>
> W / E = 2 * C * 1 GHz / (P(2 GHz) + P_0) = 0.0095 * C * 1 GHz
>
> which is slightly less than the W / E ratio at 1 GHz approximately
> equal to 0.01 * C * 1 GHz (assuming P(1 GHz) = 100 W), so in these
> conditions it would be better to run the busy CPUs at 1 GHz.

OK, I'll try to measure this.

thanks,
julia