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Thomas Barrett


Last seen: 4 meses hace |  Con actividad desde 2021

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Any suggestions for upgrading desktop GPU for doing CUDA computing?
Hi, We are currently running Matlab on a regular office desktop workstation which has a Nvidia Quadro P4000, that we regularly ...

alrededor de 1 año hace | 0 respuestas | 0

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Can this loop containing different indices be vectorized using implicit expansion (or otherwise)?
I have a code which is doing some processing over every point in a 3D matrix. Until recently, I was doing this by reshaping the ...

más de 2 años hace | 1 respuesta | 0

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Is it possible to correctly perform a multi-dimensional FFT on a 1D linearised version of a 3D array?
I have a 3D array indexed by A_3D(ind_y,ind_x,ind_z). This has then been converted to a single 1D linear vector B_1D(ind) such t...

más de 2 años hace | 2 respuestas | 0

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Speeding up code which integrates a 2D gpuArray matrix using Simpson's Rule
I have a (real) 2D gpuArray, which I am using as part of a larger code, and now am trying to also integrate the array using the ...

alrededor de 3 años hace | 1 respuesta | 1

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Help to optimize the execution time of this function, modifying the entries of a matrix (GPU? Parallel CPU?).
Hello, I am looking for some support with how to increase the speed of a piece of code, and would be really grateful for some ...

alrededor de 3 años hace | 1 respuesta | 0

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Errors when using gpuArray: "CUSPARSE_INTERNAL_ERROR" ,followed by "CUDA-capable devices are busy or unavailable"
This was fixed when upgrading to a more recent version of Matlab (R2020b).

alrededor de 3 años hace | 0

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Errors when using gpuArray: "CUSPARSE_INTERNAL_ERROR" ,followed by "CUDA-capable devices are busy or unavailable"
I am creating a 400x400 sparse complex double matrix, and then trying to convert it to a gpuArray. However, the conversion resul...

alrededor de 3 años hace | 1 respuesta | 0

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Possible to speed up this gpuArray calculation with arrayfun() (or otherwise)?
I have a complex matrix A, and would like to modify it Nt times according to A = exp( -1i*(A + abs(A).^2) ). The size of A is ty...

alrededor de 3 años hace | 1 respuesta | 0

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