I made a 3nm×3nm×3nm supercell mesh of pbs atoms which contains about 2000 atoms and QE gave me a 13256 GB requirement of RAM. What am I doing wrong? Many people have already done this calculation, but surely no one has 13TB of RAM available on a single node for a calculation like this.
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2$\begingroup$ +1 and welcome to our new community! Since you say that "many people have already done this calculation", and Tristan's comments on his answer say that it's "probably not" possible to do this, I wonder if you could show us a reference for your statement that many people have already done this calculation. Was the calculation done in some research paper to which you can provide us a URL? Who did this calculation? Did they use 2000 atoms? Or 100-200 atoms, which is what Tristan says is more typical? $\endgroup$– Nike DattaniDec 19, 2022 at 5:26
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2$\begingroup$ 13TByte does seem a lot. I can't comment on QE but doing a quick back of the envelope calculation for the code I am an author of (CRYSTAL) suggests 0.5 TByte should be enough. It does depend on the chemical composition of the system and the basis set employed, though - if the OP could provide that I might be able to make a better estimate. And we have done full geometry optimisations with CRYSTAL for systems this size and larger, but you'll require a chunky computer - 1000 cores+. $\endgroup$– Ian BushDec 19, 2022 at 6:51
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1$\begingroup$ I suggest you to move to other codes less memory heating like SIESTA, for example. $\endgroup$– Camps ♦Dec 19, 2022 at 17:10
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3$\begingroup$ You could also consider asking a separate question about strategies to address the physics you are interested in with a different approach requiring less memory. $\endgroup$– uhohDec 20, 2022 at 0:43
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$\begingroup$ Can you please tell me how did you know how much RAM is needed by QE? $\endgroup$– CamillaApr 18 at 14:41
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2 Answers
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Memory use to store the KS orbitals in a plane wave DFT calculation is proportional to cell-volume*number-of-states, and in turn proportional to the square of the number of atoms in the large limit. So 13 TB does not surprise me at all. I once computed 4096 Si atoms using CASTEP but that was on a supercomputer, and might easily have used a few TB of RAM.
I respectfully suggest that your claim that "many people have done this calculation" is in error, and that the calculations you may have seen either did not use plane-wave DFT, and/or used a massively parallel supercomputer.
If you still wish to use full KS-DFT for this system, one of the O(N) or linear-scaling local basis set codes would be a more reasonable choice.
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2000 atoms is quite a lot. If you see 3x3x3, often that's the primitive cell repeated by that much. A 3 nm particle will involve a lot of computational power even with approximations. Be sure you are using only the gamma point as well and not using excessive vacuum.
answered Dec 18, 2022 at 17:30
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$\begingroup$ I used the gamma points and not a lot of vacuum. Pbs qdot. Is there any better way $\endgroup$ Dec 18, 2022 at 17:46
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2$\begingroup$ Probably not. Typical calculations are on the order of 100-200 atoms not 2000. You may need to look at a different method / code. $\endgroup$ Dec 18, 2022 at 18:07