I am a begineer of GPGPU. For performance comparison of CPU and GPU I did a little experiment using CUDA Toolkit.<\/p>\n\n\n\n
GPU: GeForce GTX 1060 6GB *
CPU: AMD Ryzen 7 3700X 8-Core Processor
OS: CentOS7.7
CUDA: CUDA11.2
Programing Language: C language
* I bought this GPU according to [1]<\/p>\n\n\n\n
(1) Install CUDA11.2<\/p>\n\n\n\n
wget https:\/\/developer.download.nvidia.com\/compute\/cuda\/11.2.0\/local_installers\/cuda-repo-rhel7-11-2-local-11.2.0_460.27.04-1.x86_64.rpm\nsudo rpm -i cuda-repo-rhel7-11-2-local-11.2.0_460.27.04-1.x86_64.rpm\nsudo yum clean all\nsudo yum -y install nvidia-driver-latest-dkms cuda\nsudo yum -y install cuda-drivers\n<\/code><\/pre>\n\n\n\n(2) Check CUDA is installed or not<\/p>\n\n\n\n
nvcc --version<\/code><\/pre>\n\n\n\n(3) Check GPU is active or not<\/p>\n\n\n\n
nvidia-smi<\/code><\/pre>\n\n\n\n4. Build and Run program<\/h2>\n\n\n\n4.1 Program using CPU<\/h2>\n\n\n\n
(1) Create the following source “calculate_cpu.c”. This program transpose two vectors to one vectors generating an element from two elements corresponding to source vectors in each element of vector.<\/p>\n\n\n\n
#include <stdio.h>\n#include <stdlib.h>\n#include <time.h>\n#include <math.h>\n\nvoid vec_func(float *out, float *a, float *b, long long n) {\n for(int i = 0; i < n; i++){\n out[i] = log1p(expf((-1) * sqrt((log1p(abs(10 * sin(a[i]) * cos(a[i])))) + (log1p(abs(10 * sin(b[i]) * cos(b[i])))))));\n }\n}\nint main(int argc, char *argv[]){\n float *a, *b, *out;\n clock_t start, end;\n long long N = atoll(argv[1]);\n\n \/\/ Allocate memory\n a = (float*)malloc(sizeof(float) * N);\n b = (float*)malloc(sizeof(float) * N);\n out = (float*)malloc(sizeof(float) * N);\n \n \/\/ Initialize array\n for(int i = 0; i < N; i++){\n a[i] = (rand() \/ (float)RAND_MAX); b[i] = (rand() \/ (float)RAND_MAX); out[i] = 0.0f;\n }\n \n \/\/ start\n start = clock();\n\n \/\/ Executing kernel\n vec_func(out, a, b, N);\n\n \/\/ end\n end = clock();\n printf(\"%f\\n\", ((double)(end - start)) \/ CLOCKS_PER_SEC );\n \n if(argc > 2){\n for(int i = 0; i < N; i++){\n printf(\"a:%f v:%f out:%f\\n\", a[i], b[i], out[i]);\n }\n }\n}<\/code><\/pre>\n\n\n\n(2) Build and Run. In my setup, the elapsed time during vec_func function execution is 51.52sec<\/p>\n\n\n\n
gcc -std=c99 -lm .\/calculate_cpu.c -o .\/calculate_cpu\n.\/calculate_cpu 500000000<\/code><\/pre>\n\n\n\n4.2 Program using GPU<\/h2>\n\n\n\n
(1) Create the following source “calculate_gpu.cu”<\/p>\n\n\n\n
#include <stdio.h>\n#include <stdlib.h>\n#include <cuda.h>\n#include <time.h>\n#include <math.h>\n#include <cuda_runtime.h>\n\n__global__ void vec_func(float *out, float *a, float *b, long long n) {\n int i = (blockIdx.x * blockDim.x) + threadIdx.x;\n out[i] = log1p(expf((-1) * sqrt((log1p(abs(10 * sin(a[i]) * cos(a[i])))) + (log1p(abs(10 * sin(b[i]) * cos(b[i])))))));\n}\n\nint main(int argc, char *argv[]){\n float *a, *b, *out;\n float *d_a, *d_b, *d_out;\n cudaEvent_t start, stop;\n cudaEventCreate(&start);\n cudaEventCreate(&stop);\n long long N = atoll(argv[1]);\n\n \/\/ Allocate memory\n a = (float*)malloc(sizeof(float) * N);\n b = (float*)malloc(sizeof(float) * N);\n out = (float*)malloc(sizeof(float) * N);\n \n \/\/ Initialize array\n for(int i = 0; i < N; i++){\n a[i] = (rand() \/ (float)RAND_MAX); b[i] = (rand() \/ (float)RAND_MAX); out[i] = 0.0f;\n }\n \n \/\/ Allocate device memory\n cudaMalloc((void**)&d_a, sizeof(float) * N);\n cudaMalloc((void**)&d_b, sizeof(float) * N);\n cudaMalloc((void**)&d_out, sizeof(float) * N);\n \n \/\/ Transfer data from host to device memory\n cudaMemcpy(d_a, a, sizeof(float) * N, cudaMemcpyHostToDevice);\n cudaMemcpy(d_b, b, sizeof(float) * N, cudaMemcpyHostToDevice);\n\n \/\/ start\n cudaEventRecord(start);\n \n \/\/ Calculation in GPU\n \/\/ Launch (N \/ 1024) blocks with 1024 threads per block with kernel<<<N,M>>>(\u2026);\n vec_func<<<N \/ 1024, 1024>>>(d_out, d_a, d_b, N);\n \n \/\/ end\n cudaEventRecord(stop);\n cudaEventSynchronize(stop);\n \n \/\/ elapsed time\n float milliseconds = 0;\n cudaEventElapsedTime(&milliseconds, start, stop);\n\n \/\/ post process\n cudaEventDestroy(start);\n cudaEventDestroy(stop);\n\n printf(\"vec_func: %f milli sec\\n\", milliseconds );\n \n \/\/ Transfer data back to host memory\n cudaMemcpy(out, d_out, sizeof(float) * N, cudaMemcpyDeviceToHost);\n if(argc > 2){\n for(int i = 0; i < N; i++){\n printf(\"a:%f v:%f out:%f\\n\", a[i], b[i], out[i]);\n }\n }\n \n \/\/ Deallocate device memory\n cudaFree(d_a);\n cudaFree(d_b);\n cudaFree(d_out);\n \n \/\/ Deallocate host memory\n free(a);\n free(b);\n free(out);\n}<\/code><\/pre>\n\n\n\n(2) Build and Run. In my setup, the elapsed time during vec_func function execution is 0.049sec<\/p>\n\n\n\n
nvcc .\/calculate_gpu.cu -o .\/calculate_gpu\n.\/calculate_gpu 500000000<\/code><\/pre>\n\n\n\n5. References<\/h2>\n\n\n\n
[1] CUDA-Enabled GeForce and TITAN Products
https:\/\/developer.nvidia.com\/cuda-gpus<\/a><\/p>\n\n\n\n[2] NVIDIA CUDA Installation Guide for Linux
https:\/\/docs.nvidia.com\/cuda\/cuda-installation-guide-linux\/index.html#axzz4VX488ASJ<\/a><\/p>\n\n\n\n