Lesser number of qubits in distributed circuit with four GPUs using cuquantum and cusvaer than on single CPU with qiskit #88
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I am using a VM with allocation of four A40 GPUs and installed cuquantum appliance. I ran docker with all GPUs and didn't give any memory argument. I ran ghz_cusvaer.py which shows max qubits of 31 for single GPU and when I ran with mpirun giving -n 4 it max generate circuit with 33 qubits and max memory usage was not more than 4.7G even all GPUs were 100 percent utilized. Same circuit when ran out of the docker, easily generated ghz state with qubits of around 10k till I have tested and didn't stuck at all. Stats of docker while running circuit to max level And output of docker stats command showed My question is why cuquantum cusvaer is not handling more than 33 qubits which I think for this circuit is quite less since its not using much of entanglement. I need to run circuit with almost 1000 qubits by varying its entanglement, however this much output from cuquantum appliance is not looking encouraging. Should I allocate more resources to docker manually? |
Replies: 2 comments
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The cuQuantum Appliance with the Below, I detail how the cuQuantum Appliance ( With complex-valued single precision, an For double precision, the expression becomes
A system with 4x A40 will have I'll make two observations:
(2) may confuse the |
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@intelligi123 If this wasn't your question, let us know. |
Hi @intelligi123
The cuQuantum Appliance with the
cusvaerbackend implements a numerically dense representation of the state. It does not provide or implement entanglement-based compression, and it doesn't use system memory (RAM). Can you share more about your target workload / circuits?Below, I detail how the cuQuantum Appliance (
cusvaer) determines system size constraints based on available GPU memory.With complex-valued single precision, an$n$ qubit state vector will require $2^{n+3} \mathrm{B}$ . Converting to $1024^{2} = (2^{10})^{2} = 2^{20}$ -- this yields a clean expression for required memory in Mebibytes: $2^{n - 17} \mathrm{MiB}$ .
MiBis done by dividing the result byFor dou…