RTX 2080 Ti & GTX 1080 Ti low PPD?

[Foldingwolf]

Hey guys,
im new to folding and testing my GPUs.
I read on this forum that a RTX 2080Ti can do like 2-2.5M PPD, im getting like only 1,4M with 63% power limit, i tried it with 70,80,90,100% power limit but the highest I got was 1,6M with 100%.
Am I doing something wrong? Are there any other options?

With my 1080Ti, im getting 1M PPD at 55% power limit.

Im trying to keep the power limit as low as I can because the electricity cost is here a bit high.

Can you give me some help here?

Thanks in advance

[Joe_H]

Welcome to the folding support forum.

There are two main possibilities for the low PPD reported. First, do you have a passkey entered into the folding client?

Second, for about the last month some projects from specific servers will report points based on just the base points, no Quick Return Bonus included. This is due to an ongoing updating of the server code to a new version. You can spot this in the logs, the estimated points will be the same as the base points from the Project Summary page for that project. At some point in the future there will be a manual crediting of the bonus points.

Basically for the second case the client acts as if you did not enter a passkey.

[kiore]

Hi I also have a RTX 2080ti and a GTX 1080ti those numbers in windows don't look so bad for some projects, as they are both very powerful cards there can be a very large difference in PPD between units. I know people always report their best ever score as if it their average. I expect you will get units that go well over 2million ppd on the RTX card, just not all of them do this.

[Foldingwolf]

Welcome to the folding support forum.

There are two main possibilities for the low PPD reported. First, do you have a passkey entered into the folding client?

Second, for about the last month some projects from specific servers will report points based on just the base points, no Quick Return Bonus included. This is due to an ongoing updating of the server code to a new version. You can spot this in the logs, the estimated points will be the same as the base points from the Project Summary page for that project. At some point in the future there will be a manual crediting of the bonus points.

Basically for the second case the client acts as if you did not enter a passkey.

Yes I have a passkey entered
Well, alright. I will keep on going then !

Hi I also have a RTX 2080ti and a GTX 1080ti those numbers in windows don't look so bad for some projects, as they are both very powerful cards there can be a very large difference in PPD between units. I know people always report their best ever score as if it their average. I expect you will get units that go well over 2million ppd on the RTX card, just not all of them do this.

It explains a lot... Thanks for this!

[foldy]

Do you use Windows or Linux? What is the pcie slot speed of your GPUs?

[Foldingwolf]

Do you use Windows or Linux? What is the pcie slot speed of your GPUs?

Im using Windows, where can I see the pcie slot speed?

[foldy]

Using GPU-Z tool.

[Foldingwolf]

Using GPU-Z tool.

[JimboPalmer]

Both cards have 16 lanes of PCIe 3.0, but only 8 lanes are used. When the card needs to talk to the main PC, it will be slower.

Common communication events that happen in each WU are: Load the data onto the card, return the result from the card, and periodic check points; saving data from the card to the main PC's Hard disk. The first two happen once per WU so are not much of a slowdown, Checkpoints should happen multiple times per WU.

The actual Protein work happens on the card, so is not slowed down by this.

[toTOW]

PCIe 3.0 8x is still enough for most cards ...

What projects are you running ? What GPUZ shows in Sensors tab ?

[Foldingwolf]

PCIe 3.0 8x is still enough for most cards ...

What projects are you running ? What GPUZ shows in Sensors tab ?

Im currently running project 11726 on the GTX 1080 Ti & project 14179 on the RTX 2080 Ti.

[foldy]

Looks good. Performance varies between different work units. So you may want to try rtx 2080ti with higher power limit and different work units to get more PPD. In general Linux is faster than Windows.

[QuintLeo]

63% power limit is WHY you are getting low performance.
Folding wants your cards to run as much power as they can to get max PPD, due to the way Quick Return Bonus is calculated even a SMALL amount of higher clock rate makes a BIG difference.

[MeeLee]

Power limit shouldn't matter much to bonus points.
The RTX2080 Ti does need 180W minimum, to get full points (combined with an overclock).
Settings allow you to go as low as 125W, and in some cases even lower.
PPD will be very low that way.
I don't know what the GTX 1080 needs, but my RTX 2060 needs 125-130W, 2070 needs 133-136W, 2080 needs 136-149W.
The 2080 Ti needs 177-182W. This is under linux with overclocking enabled.

Since the 1080 Ti is about as fast as an RTX, but at older technology, I'd presume it will need at least ~140-160W of power to get it's optimal points.

[bruce]

Power limit shouldn't matter much to bonus points.

Bonus points depend non-linearly on total processing time. Every WU spends some time waiting for the shaders to complete the calculations of forces for each step, some time Integrating those forces, some time preparing for the next step (updating the coordinates of every atom) and some time transferring data to/from main RAM. The proportions of time spent doing each necessary step/process depend on how much of that process can be parallelized to fit within the number of shaders as well as their GHz (effectively proportional to GFLOPS) although other processing steps cannot be parallelized so everything is NOT directly proportional to the number of shaders or their GHz. With a small number of atoms, having more shaders doesn't help because they can't all be kept busy.

Forces are calculated for each pair of nearby atoms and those forces become negligible for atoms that are not nearby. Those forces are added to forces produced by differences in the charge associated with distant atoms. Motions of each atom depends on the integral of those total forces. The classical F=M*A starts out by becoming d²x - ʃ ʃ (F/M) dt² + other factors

(Sorry, I've used "steps" in two different ways in this explanation, but I don't know a pair of words that would work better.)

More details here: http://docs.openmm.org/latest/userguide/index.html