GPU610/TeamKCM

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Team KCM

Team Members

1. Byungho Kim, Some responsibility
2. Taeyang Chung, Some responsibility
3. SeungYeon Moon, Some responsibility

Email All

Progress

Assignment 1

Byungho's Findings

Introduction

To enhance image quality, most of S/W use convolution algorithm with sharpen kernel. The convolution algorithm is based on pixel window and kernel multiplication and the kernel which consists of 3 x 3 matrix or 5 x 5 matrix. Therefore, as we learned at the GPU610 class, I can speed up the program using CUDA library

Original Image	Enhanced Image

These two images show the result of the S/W. Left picture is the original image and right picture is obtained by the S/W. 3x3 matrix sharpen kernel is used.

Convolution

6 x 4 Pixel Small Image

IP11	IP12	IP13	IP14	IP15	IP16
IP21	IP22	IP23	IP24	IP25	IP26
IP31	IP32	IP33	IP34	IP35	IP36
IP41	IP42	IP43	IP44	IP45	IP46

Kernel

K0	K1	K2
K3	K4	k5
k6	k7	k8

The 33th output pixel will be given by

OP₃₃ = IP₂₂ x K₀ + IP₂₃ x K₁ + IP₂₄ x K₂ + IP₃₂ x K₃ + IP₃₃ x K₄ + IP₃₄ x K₅ + IP₄₂ x K₆ + IP₄₃ x K₇ + IP₄₄ x K₈

Analysis

The profile shows that filter function occupies the most of processing time.

  %   cumulative   self              self     total          
 time   seconds   seconds    calls  ns/call  ns/call  name   
 93.59      1.46     1.46                             filter(image_t*, double*, int, double, double)
  6.41      1.56     0.10  5038848    19.85    19.85  put_pixel_unsafe(image_t*, unsigned int, unsigned int, unsigned char, unsigned char, unsigned char)
  0.00      1.56     0.00        2     0.00     0.00  alloc_img(unsigned int, unsigned int)
  0.00      1.56     0.00        1     0.00     0.00  _GLOBAL__sub_I__Z7get_ppmP8_IO_FILE
  0.00      1.56     0.00        1     0.00     0.00  get_ppm(_IO_FILE*)

In the filter function, 3 x 3 mask walk through all RGB pixels and calculate convolution of them.

image filter(image im, double *K, int Ks, double divisor, double offset)
{  
    image oi;
    unsigned int ix, iy, l;
    int kx, ky;
    double cp[3];
    oi = alloc_img(im->width, im->height);
    if ( oi != NULL ) {
        for(ix=0; ix < im->width; ix++) {
            for(iy=0; iy < im->height; iy++) {
                cp[0] = cp[1] = cp[2] = 0.0;
                for(kx=-Ks; kx <= Ks; kx++) {
                    for(ky=-Ks; ky <= Ks; ky++) {
                        for(l=0; l<3; l++)
                            cp[l] += (K[(kx+Ks) +
                                    (ky+Ks)*(2*Ks+1)]/divisor) *
                                ((double)GET_PIXEL_CHECK(im, ix+kx, iy+ky, l)) + offset;
                    }
                }
                for(l=0; l<3; l++)
                    cp[l] = (cp[l]>255.0) ? 255.0 : ((cp[l]<0.0) ? 0.0 : cp[l]) ;
                put_pixel_unsafe(oi, ix, iy,
                        (color_component)cp[1],
                        (color_component)cp[2],
                        (color_component)cp[0]);
            }
        }
        return oi;
    }
   return NULL;
}

Taeyang's Findings

My program is to solve the heat transfer problem using explicit method. I found this source code at http://www.pengfeidu.net/Media/pdf/1dUnsteadyHeatTransferCode.pdf

The 'Solve()' function takes 3 arguments and as t-value decreases the processing time increases as shown below

The 'Solve()' function:

      void solve(){
               int j=0;
               while(j<tTotal/dt){
                       solutionNew[0]=10;
                       solutionNew[xGridNum]=120;
                       for(int k=1;k<xGridNum;k++){
                            solutionNew[k] = solutionLast[k]+dt*(K/dx/dx*
                                      (solutionLast[k-1]+solutionLast[k+1]-2.*solutionLast[k]
                                      +source(x[k],j*dt)));
                       }
                       j++;
                       solutionLast=solutionNew;
               }
       }

By Implementing parallel solution, I believe I can shorten the processing time.

SeungYeon's Findings

There are many types of image processing or operations can be done. Some of examples are rotating,re-sizing,blurring, etc... For Assignment 1, I decided to work with one of the operation that can be done with an image in c++, which deals with brightness and contrast of an image. I was able to find a open source code from "openCV" website it stands for Open Source Computer Vision, is open source libraries developed by Intel for image and video processing. The libraries can be installed in any platform, and i was able to install it in my windows platform, and all test runs all done using visual studio 2013, Test file is compiled and profiled using performance wizard vs13. Following website openCV.org is an official documentation website of openCV libraries, with code examples for many image and video type operations

The open source code for the program requires 3 user inputs which are path to an image and alpha,beta values which will be multiplied and added to each pixels' in the image. To make sure test run is accurate, i had to change some code in the program and hard code in the values for alpha and beta.

Test run with an image size of 500KB

Total Time takes is 12 seconds

Test run with an image size of 2MB

Total Time takes is 14 seconds

SeungYeons' observation for As_1

Changing contrast and brightness of an image in the function is implemented with a scale * matrix operation. M_A(x,y) = ALPHA * M_B(x,y) + BETA;

Assignment 2

For our Assignment 2 we decided to work with heat transfer problem.

To compare the speed difference between the original code and parallelized code, we first profiled the original code in matrix environment an parallelized code in school computer. (using gporf and nvvp)

Profile results

Each Program was tested with 5 different input data.(500,1000,2500,5000,10000)

Profile #1 (Original code)

Sample Data = 500

Sample Data = 2500

Sample Data = 10000

Profile #2 (GPU implementation version)

Sample Data = 500

Sample Data = 2500

Sample Data = 10000

Summary of profiling

We tested CPU and GPU implementation version with same data sizes and showed significant speedup in result after using kernel.

Summary of profiling in Chart

Charts with some of main operation(function) calls for CPU version and GPU(kernel)

Original

GPU (kernel)

Screen shot of solve function in CPU and GPU(kernel) versions

CPU

GPU(Solve) and Kernel

Assignment 3

For assignment 3, we were checking everything we can do to improve the performance and found 2 thing we can do. Firstly, in the while loop, there were 6 times of memory copy functions called, and we found that we can reduce 6 times to 1 time by using device address pointer switching. Furthermore, we found that if the sample number n is less that 1024, we can use shared memory in the kernel.

Reducing Memory Copy Calls

Assignment 2 Code

void solve(){
    int j=0;
    int d;

    cudaDeviceProp prop;
    cudaGetDevice(&d);
    cudaGetDeviceProperties(&prop, d);
    unsigned ntpb = prop.maxThreadsDim[0];
    unsigned ntpg = ntpb * prop.maxGridSize[0];
    if (xGridNum > ntpg) {
        xGridNum = ntpg;
        std::cout << "n reduced to " << xGridNum << std::endl;
    }

    double* d_x;
    double* d_solutionLast;
    double* d_solutionNew;

    cudaMalloc((void**)&d_x, (xGridNum + 1) * sizeof(float));       
    cudaMalloc((void**)&d_solutionLast, xGridNum * sizeof(double));
    cudaMalloc((void**)&d_solutionNew, xGridNum * sizeof(double));

    while(j<tTotal/dt){
        solutionNew[0]=10;
        solutionNew[xGridNum]=120;

        cudaMemcpy(d_x, x, (xGridNum + 1) * sizeof(float), cudaMemcpyHostToDevice);
        cudaMemcpy(d_solutionLast, solutionLast, xGridNum * sizeof(double), cudaMemcpyHostToDevice);
        cudaMemcpy(d_solutionNew, solutionNew, xGridNum * sizeof(double), cudaMemcpyHostToDevice);

       

        kernel<<<(xGridNum + ntpb - 1) / ntpb, ntpb>>>(d_solutionNew, d_solutionLast, d_x, xGridNum, dt, dx, K, j);
        j++;

        cudaMemcpy(x, d_x, (xGridNum + 1) * sizeof(float), cudaMemcpyDeviceToHost);
        cudaMemcpy(solutionLast, d_solutionLast, xGridNum * sizeof(double), cudaMemcpyDeviceToHost);
        cudaMemcpy(solutionNew, d_solutionNew, xGridNum * sizeof(double), cudaMemcpyDeviceToHost);

        solutionLast=solutionNew;
    }

    for(j = 0; j <= xGridNum; j++){
      std::cout << solutionLast[j] << endl;
    }

    cudaFree(d_x);
    cudaFree(d_solutionLast);
    cudaFree(d_solutionNew);
}

Improved Code

void solve(){
    int j=0;
    int d;
  
    cudaDeviceProp prop;
    cudaGetDevice(&d);
    cudaGetDeviceProperties(&prop, d);
    unsigned ntpb = prop.maxThreadsDim[0];
    unsigned ntpg = ntpb * prop.maxGridSize[0];
    if (xGridNum > ntpg) {
        xGridNum = ntpg;
        std::cout << "n reduced to " << xGridNum << std::endl;
    }

    double* d_x;
    double* d_solutionA;
    double* d_solutionB;

    cudaMalloc((void**)&d_x, (xGridNum + 1) * sizeof(float));       
    cudaMalloc((void**)&d_solutionA, xGridNum * sizeof(double));
    cudaMalloc((void**)&d_solutionB, xGridNum * sizeof(double));
   
    double* d_solutionTemp;

    cudaMemcpy(d_x, x, (xGridNum + 1) * sizeof(float), cudaMemcpyHostToDevice);
    cudaMemcpy(d_solutionA, solutionNew, xGridNum * sizeof(double), cudaMemcpyHostToDevice);
    cudaMemcpy(d_solutionB, solutionLast, xGridNum * sizeof(double), cudaMemcpyHostToDevice);

    std::ofstream myfile;
    myfile.open ("output.txt");                   
  
    // A = new, B = last
    while( j < tTotal / dt){                    
        kernel<<<(xGridNum + ntpb - 1) / ntpb, ntpb>>>(d_solutionA, d_solutionB, d_x, xGridNum, dt, dx, K, j);
        j++;       
        cudaMemcpy(solutionNew, d_solutionA, xGridNum * sizeof(double), cudaMemcpyDeviceToHost); 
        /*
        myfile << "Time" << tTotal/dt << std::endl;
        for(int i = 0; i <= xGridNum; i++){                                   
          myfile << solutionNew[i] << ":";
        }
        */
        d_solutionTemp = d_solutionA;
        d_solutionA = d_solutionB;
        d_solutionB = d_solutionTemp;
    }

    myfile.close();
   
    cudaFree(d_x);
    cudaFree(d_solutionA);
    cudaFree(d_solutionB);
 
    cudaDeviceReset(); // Important
}

Using Shared Memory

Assignment 2 Code

__global__ void kernel(double* solutionNew, double* solutionLast, double* x, int n, float dt, float dx, float K, int j){
        int k = blockIdx.x * blockDim.x + threadIdx.x;

        if( k < n){
                k += 1;
                solutionNew[k] = solutionLast[k]+dt*(K/dx/dx*
                                (solutionLast[k-1]+solutionLast[k+1]-2.*solutionLast[k]
                                 +(x[k]*j*dt*1000))
                                );
        }
}

Improved Code

__global__ void kernel(double* solutionNew, double* solutionLast, double* x, int n, float dt, float dx, float K, int j){
        int k = blockIdx.x * blockDim.x + threadIdx.x;
       
        double *t_c;
        __shared__ double s_c[1024];
   
        if ( n < 1024 && k < (n + 1))
            s_c[k] = solutionLast[k];
        __syncthreads();
               
        if( n < 1024) t_c = s_c;
        else t_c = solutionLast;

                       
        if( k == 0){
            solutionNew[k] = 10;
        } else if( k == n){
            solutionNew[k] = 120;
        }
                       
        if( k < n){
                k += 1;
                solutionNew[k] = t_c[k]+dt*(K/dx/dx*
                                (t_c[k-1]+t_c[k+1]-2.*t_c[k]
                                 +(x[k]*j*dt*1000))
                                );
        }               
}