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 datadependent, and so you can't get a reasonable answer without knowing something about the data. That makes it impossible to have a onenumberfitsalldata 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 maybeaccurate 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 CUDAenabled 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?
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 floatingpoint 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
By : Roberto Miguez
Date : March 29 2020, 07:55 AM
this will help No. 1 doubleprecision floatingpoint operation is still one floatingpoint operation. Most GPUs process doubleprecision data slower than singleprecision, so there should be two specifications of peak GFLOPS. One peak singleprecision GFLOPS spec, and one peak doubleprecision 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
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
if(flop/1000000000>=1flop/1000000000<1){
if(flop/1000000000>=1){
flop/=secs; // To get the Flops in second, multiply with time elapsed
cout<<"\n\n\n Floatingpoints Operations Per Second\n\n

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.

