Parallell multiplication matrix and inverse iteratively julia9

New to Julia
1

I’m new julia programmer. I want to write a program that can calculate multiple and inverse iteratively. But I can’t return result. It’s my code.

addprocs(4);
@everywhere using Distributed , 
   DistributedArrays
C=dzeros(1000000);
@time @distributed for i in 1:100 
A=rand(10000,200); 
B=rand(10000,1);
P=localpart(C);
P[1+(i-1)%Int(100/nworkers())]= 
    Vec((A×A')\B);
end

When I wanna see c as result. C is `0.
How do I fix my code?

2

Can you describe these two statements in a bit more detail?

Maybe you could also provide a serial code which does what you want, and then we can help you parallelise it.

3

I wanna calculate inv(A)×B that both of them matrices. This action must do several times on different data. I wanna use 4 process for doing it. When I use one process, everything is OK. I wanna speed up with multi processes. Result is zero. It is my problem.

4

First of all, you should be using B\A which will be faster and more accurate.

5

Also, BLAS is muti-threaded already, so you probably won’t get much of a speedup.

1 Like
6

everywhere using Distributed
C=;
@time @distributed for i in 1:100
V= (A×A’)\B;
append!(C,V);
end

This code is doing correctly. It is done on one process. I wanna do this action for 4 process.

7

Yes I use it.

8

Parallelising linear system solves is highly nontrivial. I strongly recommend you do this using a software package rather than writing your own code.

10

The expression will error. @distributed gives back a Task object, which will collect exceptions from the worker processes. Check this for the error.

One reason why your expression won’t work, is that you try to assign a vector A*A'\B to a float P[1+(i-1)...]. The latter is a single entry of the array P. Make sure to assign the former to a chunk of P of proper length.

Also, it’s vec, not Vec. But that shouldn’t be needed anyway.

By the way, which Julia version are you on? I tried the code on 1.5.3 and the \times symbol for multiplication is no longer valid. Same on 1.6.

11

You can get 10X speedup using your GPU.

using CUDA, BenchmarkTools 

A = rand(10000,200)
B = rand(10000,1)

gpuA = cu(A)
gpuB = cu(B)

@btime ($A*$A')\$B 
  5.324 s (8 allocations: 1.49 GiB)                                                

@btime ($gpuA*$gpuA')\$gpuB 
  555.748 ms (76 allocations: 2.80 KiB)
12

Thanks.