GPGPU-Assignment_1 Solved

Assignment #1: The Big Dot 

The dot product of two vectors 𝑎 = (𝑎$,             𝑎’,                     …, 𝑎()’ ) and 𝑏 = (𝑏$,               𝑏’,          …, 𝑏()’ ), written 𝑎 ∙ 𝑏, is simply the sum of the component-by-component products:

             𝑎 ∙ 𝑏 = ∑(-/)$’ 𝑎- × 𝑏-

Dot products are used extensively in computing and have a wide range of applications. For instance, in 3D graphics (n = 3), we often make use of the fact that 𝑎 ∙ 𝑏 = |𝑎||𝑏|𝑐𝑜𝑠𝜃, where | | denotes vector length and 𝜃 is the angle between the two vectors.  In this assignment, you are expected to:

1.     Write CUDA code to compute in parallel the dot product of two (possibly large N = 100,000, or N = 1024*1024) random single precision floating point vectors;

2.     Write two functions to compute the results on the CPU and GPU, and compare the two results to check for correctness (1.0e-6);

•      float *CPU_big_dot(float *A, float *B, int N);

•      float *GPU_big_dot(float *A, float *B, int N);

3.     Print performance statistics with timer function;

•      CPU: Tcpu = Total computation time for CPU_big_dot();

•      GPU: Tgpu = Total computation time for GPU_big_dot();

•      Memory allocation and data transfer from CPU to GPU time

•      Kernel execution time

•      Data transfer from GPU to CPU time

•      Speedup = CPU/GPU

4.     Analyze the performance results in a few sentences.

•      Which one runs faster?

•      What’s the reason for that? Problem size, overhead, etc.  

 

Timer functions #include <sys/time.h> long long start_timer() { struct timeval tv; gettimeofday(&tv, NULL); return tv.tv_sec * 1000000 + tv.tv_usec;  

}

 

long long stop_timer(long long start_time, char *name) { struct timeval tv; gettimeofday(&tv, NULL);

long long end_time = tv.tv_sec * 1000000 + tv.tv_usec;

Printf(“%s: %.5f sec\n”, name, ((float) (end_time – start_time)) /

(1000 * 1000));  return end_time – start_time;

}