Measured GFLOPS is greater then theoretical GFLOPS

Measured GFLOPS is greater then theoretical GFLOPS

By : user2957277
Date : November 23 2020, 01:01 AM
I hope this helps you . I have written a script to measure the GFLOPS that I can expect for an element-wise matrix multiplication in Octave. My CPU is a i7-2670QM @ 2.2GHz. Looking at the spec the theoretical GFLOPS is 70.4. Running the script below which uses just one of the four cores of my system I measured 185 GFLOPS. , You are measuring M(ega)FLOPS (1e6), not G(iga)FLOPS (1e9)
code :

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Counting FLOPS/GFLOPS in program - CUDA

Counting FLOPS/GFLOPS in program - CUDA

By : Sudipto Banerjee
Date : March 29 2020, 07:55 AM
it fixes the issue That's not just your opinion; it's simple fact that the number of operations in the case of a sparse matrix is data-dependent, and so you can't get a reasonable answer without knowing something about the data. That makes it impossible to have a one-number-fits-all-data estimate.
This is probably one of the sorts of situations where you could think hard about it for many hours (and do lots of research) to make a maybe-accurate estimate, or you could spend a few minutes writing a variant of your existing implementation that increments a counter each time it does an operation. Sure, that's going to take quite a while to run (especially if you don't do it in a CUDA-enabled form), but probably a lot less time than it would take to do the thinking, and when the answer comes out, you don't have to do a lot of work to convince yourself that it's right.
How to measure the gflops of a matrix multiplication kernel?

How to measure the gflops of a matrix multiplication kernel?

By : user3356763
Date : March 29 2020, 07:55 AM
like below fixes the issue You can measure the GFLOPs by running the algorithm with a large input and measuring the execution time. Then put the execution time and matrix size into that formula. For matrix sizes big enough to keep the entire machine busy, the FLOPs is only weakly dependent on matrix size.
The GPU matrix multiplication algorithm performs the same number of floating-point operations as the naive algorithm.
code :
for (i = 0; i < MatrixSize; i++)
  for (j = 0; j < MatrixSize; j++)
    for (k = 0; k < MatrixSize; k++)
      C[j][i] += A[j][k] * B[k][i];
Calculation of gflops for double precision

Calculation of gflops for double precision

By : Roberto Miguez
Date : March 29 2020, 07:55 AM
this will help No. 1 double-precision floating-point operation is still one floating-point operation.
Most GPUs process double-precision data slower than single-precision, so there should be two specifications of peak GFLOPS. One peak single-precision GFLOPS spec, and one peak double-precision GFLOPS spec. Sometimes it is broken done further, so that (for example) peak division performance is listed separately from peak addition performance.
C++ calculate GFlops

C++ calculate GFlops

By : apicht
Date : March 29 2020, 07:55 AM
I wish this help you There are many issues with this code. First, you are using a float variable (j) to maintain the counter of the loop with strict termination condition j<999999999. This is probably the reason why the loop may run forever. The type of j should be an integral type such as int.
Second, the number of flops in the loop depends on the compiler you are using, the compiler options you are passing to the compiler and the target architecture. The best way to figure this out is to see the generated assembly code.
code :
flop=999999;   // Actually is 999999999, but integer overflow in expression
flop/=secs;    // To get the Flops in second, multiply with time elapsed
cout<<"\n\n\n Floating-points Operations Per Second\n\n
How to calculate Gflops of a kernel

How to calculate Gflops of a kernel

By : Mengnan Wang
Date : March 29 2020, 07:55 AM
I wish this helpful for you First some general remarks:
In general, what you are doing is mostly an exercise in futility and is the reverse of how most people would probably go about performance analysis.
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