Scaling down Haswell from 4.8GHz to 3.4GHz.

[Idontcare]

Technically, lower clocks are better is real. Running high frequency increases the performance but also destroys much more energy in heat and increased resistance.
Unless we could speak about superconductivity, increased frequencies will cause faster performance but also more energy wasting, which is opposite of the efficiency. Very visibly, this occurs especially in overclocking, when OC CPUs offer 10-15% performance increase yet their TDP is doubled or tripled against stock setting.
The OP is probably trying to find a good P-state level where his CPU offers the fastest performance for lower power draw.
I would advice the OP to try to downgrade under stock settings where the raw performance remains only slightly changed but the power draws and temps go down very quickly, yet the peak efficiency also depends on if he was delidding or not.

52.6W is not fixed in strict sense, it's because the CPU runs on 1.6 GHz idle P-state, if you OC and are not doing anything you still have 1.6 and same power draw because the OC frequency was not yet triggered.

Of course running lower clocks results in lower power, but that isn't what I was talking about.

If the goal is to maximize power efficiency (FPS/W) then you have to account for the fixed power consumption of the entire system, not just the CPU.

And that "big picture" accounting results in a maximum in the performance/W curve.

 

[BallaTheFeared]

Anyways, here are three results using PoV Ray.

2.8GHz (Power Savings Enabled)

Idle 52.6w Load: 75w
Max Core Temp: 38C
PoV 768 .3 Benchmark: 10m59s


4.3GHz (Power Savings Enabled)

Idle: 52.6w Load: 116w
Max Core Temp: 54C
PoV 768 .3 Benchmark: 7m09s


4.8GHz (Power Savings Enabled)

Idle: 52.6w Load: 178w
Max Core Temp: 80C
PoV 768 .3 Benchmark: 6m21s

Well I guess I'll try it myself, derp math mode engaged.



2.8GHz = 75 Joules /s over 659 seconds, so 75 x 659 = 49425 Joules to complete the task


4.3GHz = 116 Joules /s over 429 seconds, so 116 x 429 = 49764 Joules to complete the task


4.8GHz = 178 Joules /s over 381 seconds, so 178 x 381 = 67818 Joules to complete the task

I imagine I derped my way through that, seems simple enough but derp is as derp does.

 

[Idontcare]

Can you figure out the ROI on these clocks for me IDC?

System power consumption with all power savings features enabled and a 7950 installed is 52.6w, it doesn't matter what I do after that for clocks it's always 52.6w, which also means my 7950 uses about 10w more than the HD 4600 on the desktop at idle. This is with C7s enabled, it doesn't matter if it's reading 1.35v idle or .5v idle, power consumption doesn't change. I dunno what it's doing, but according to RealTemp it's constantly in C7 state, so my guess is it's aggressively power gating and perhaps even parking/turning off cores at idle.


We should then assume that 52.6w is platform fixed rate, correct?

My apologies Balla, somehow in skimming through the thread while bringing myself up to speed I managed to miss your post above.

But yes, assuming the fixed offset based on idle power numbers is a good enough approximation. It isn't perfect, but this isn't the basis for sending a rocket to Mars so it will serve our purpose quite nicely here on earth

Well I guess I'll try it myself, derp math mode engaged.



2.8GHz = 75 Joules /s over 659 seconds, so 75 x 659 = 49425 Joules to complete the task


4.3GHz = 116 Joules /s over 429 seconds, so 116 x 429 = 49764 Joules to complete the task


4.8GHz = 178 Joules /s over 381 seconds, so 178 x 381 = 67818 Joules to complete the task

I imagine I derped my way through that, seems simple enough but derp is as derp does.

Those numbers are right by my estimation as well. Your derp-mode is a good derp-mode

Since 2.8 and 4.3 basically yield the same value, look to see where inbetween those two clockspeed you get the peak efficiency.

A spot check at 3.5GHz is all that is needed to determine of you have an efficiency plateau (like I do with my 2600K in the graph I posted above), or if you have more curvature to your efficiency curve.