Tugas Pemograman Parallel Irsyad.docx
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Tugas Pemograman Parallel Benchmark program Pi, Perkalian Matrix dan fibonacci
OLEH:
MUH.IRSYAD ASHARI
(D42116305)
DEPARTEMEN TEKNIK INFORMATIKA FAKULTAS TEKNIK UNIVERSITAS HASANUDDIN 2019
1. Program Pi #include <stdio.h> #include double sum; int checkParallel = 0; void piP(long num_steps, int red){ sum = 0.0; double runTime, step = 1.0/(double) num_steps, start = omp_get_wtime(); if(red){ #pragma omp for reduction(+:sum) for (int i=1;i<= num_steps; i++){ double x = (i-0.5)*step; sum += 4.0/(1.0+x*x); } }else{ double sumP = 0.0; #pragma omp for for (int i=1;i<= num_steps; i++){ double x = (i-0.5)*step; sumP += 4.0/(1.0+x*x); } #pragma omp atomic sum += sumP; } double pi = step * sum; runTime = omp_get_wtime() - start; #pragma omp atomic checkParallel++; #pragma omp single { printf("step = %d, pi = %.10lf, %f s",num_steps,pi,runTime); if(checkParallel == 1) printf("(serial + overhead)"); else printf("(parallel)"); if(red) printf("(with red)\n"); else printf("(non-red)\n"); checkParallel = 0; } } void pi(long num_steps){ double start, runTime, step = 1.0/(double) num_steps; start = omp_get_wtime(); sum = 0.0; for (int i=1;i<= num_steps; i++){
double x = (i-0.5)*step; sum += 4.0/(1.0+x*x); } double pi = step * sum; runTime = omp_get_wtime() - start; printf("step = %d, pi = %.10lf, %f detik(serial)\n",num_steps,pi,runTime); } int main () { long steps = 500000000; for(int i = 0; i < 1; i++){ #pragma omp parallel num_threads(1) piP(steps, 1); pi(steps) #pragma omp parallel piP(steps,1); printf("\n"); } }
Data :
serial+overhead serial overhead parallel speed up 1.586 1.58 0.006 0.39 405.12821 Program efektif karena efisienso waktu speed up hingga 405.128.
2. Matrix #include<stdio.h> #include<stdlib.h> #include #include void MMulT(int n); void MMul(int n); int main(){ int nList[] = {500,1000,2000,4000}; for(int i = 0; i < 4; i++){ MMul(nList[i]); } for(int i = 0; i < 4; i++){ MMulT(nList[i]); } } void MMul(int n){ clock_t start, finish; start = clock(); int checkParallel = 0; double *arrayA = malloc(sizeof(double) * n * n); double *arrayB = malloc(sizeof(double) * n * n); double *result = malloc(sizeof(double) * n * n); double *arrayTemp = malloc(sizeof(double) * n * n); #pragma omp parallel { #pragma omp for for(int i = 0; i < n; i++){ for(int j = 0; j < n; j++){ int sum = 0; for(int k = 0; k < n; k++){ sum += arrayA[(n*i)+k] * arrayB[(n*k)+j]; } result[(n*i)+j] = sum; } } #pragma omp atomic checkParallel++; } finish = clock(); printf("processing time B not transposed, n = %d", n); if(checkParallel == 1) printf("(serial) : %.7lf\n", ((double)(finish-start) / CLOCKS_PER_SEC)); else printf("(parallel) : %.7lf\n", ((double)(finish-start) / CLOCKS_PER_SEC)); } void MMulT(int n){
clock_t start, finish; start = clock(); int checkParallel = 0; double *arrayA = malloc(sizeof(double) * n * n); double *arrayB = malloc(sizeof(double) * n * n); double *result = malloc(sizeof(double) * n * n); double *arrayTemp = malloc(sizeof(double) * n * n); #pragma omp parallel { //transpose matrix B #pragma omp for for(int i = 0; i < n; i++){ for(int j = 0; j < n; j++){ arrayTemp[(n*i)+j] = arrayB[(n*j)+i]; } } #pragma omp single { free(arrayB); arrayB = arrayTemp; } #pragma omp for for(int i = 0; i < n; i++){ for(int j = 0; j < n; j++){ int sum = 0; for(int k = 0; k < n; k++){ sum += arrayA[(n*i)+k] * arrayB[(n*j)+k]; } result[(n*i)+j] = sum; } #pragma omp atomic checkParallel++; } finish = clock(); printf("processing time B transposed, n = %d", n); if(checkParallel == 1) printf("(serial) : %.7lf\n", ((double)(finish-start) / CLOCKS_PER_SEC)); else printf("(parallel) : %.7lf\n", ((double)(finish-start) / CLOCKS_PER_SEC)); }
DATA YANG MUNCUL :
KESIMPULAN : Dapat dilihat dari data hasil percobaan diatas. Processing time yang dibutuhkan jauh lebih cepat apabila program di-run secara parallel.
3.Fibonacci #include<stdio.h> #include #include long fib(int n){ long x,y; if(n < 2) return n; else{ x = fib(n - 1);
y = fib(n - 2); return x + y; } } long fibP(int n, int cut){ long x,y; if(n < 2) return n; else{ #pragma omp task shared(x) { if(cut == 1) x = fib(n - 1); else x = fibP(n - 1, cut-1); } #pragma omp task shared(y) { if(cut == 1) y = fib(n - 2); else y = fibP(n - 2, cut-1); } #pragma omp taskwait return x + y; } } void runner(int n, int s,int proc, int cut){ double start, finish; long result; start = omp_get_wtime(); if(s){ result = fib(n); }else{ #pragma omp parallel num_threads(proc) #pragma omp single result = fibP(n, cut); } finish = omp_get_wtime(); printf("fib(%d) = %ld waktu eksekusi = %.5lf", n, result, (finish start)); if(s) printf("(serial)\n"); else if(proc == 1) printf("(serial + overhead) task num = %d\n",(1<<cut)); else printf("(parallel(%d)) task num = %d\n",proc, (1<<cut));
} int main(){ int nList[] = {42,43,44,45}; for(int j = 0; j < 1; j++){ for(int i = 0; i < 4; i++){ runner(nList[i], 1, 0, 0); } runner(45, 0, 1, 18); runner(45, 0, 1, 19); runner(45, 0, 1, 20); runner(45, 0, 1, 22); runner(45, 0, 1, 24); runner(45, 0, 1, 28); runner(45, 0, 4, 2); } }
OLEH:
MUH.IRSYAD ASHARI
(D42116305)
DEPARTEMEN TEKNIK INFORMATIKA FAKULTAS TEKNIK UNIVERSITAS HASANUDDIN 2019
1. Program Pi #include <stdio.h> #include
double x = (i-0.5)*step; sum += 4.0/(1.0+x*x); } double pi = step * sum; runTime = omp_get_wtime() - start; printf("step = %d, pi = %.10lf, %f detik(serial)\n",num_steps,pi,runTime); } int main () { long steps = 500000000; for(int i = 0; i < 1; i++){ #pragma omp parallel num_threads(1) piP(steps, 1); pi(steps) #pragma omp parallel piP(steps,1); printf("\n"); } }
Data :
serial+overhead serial overhead parallel speed up 1.586 1.58 0.006 0.39 405.12821 Program efektif karena efisienso waktu speed up hingga 405.128.
2. Matrix #include<stdio.h> #include<stdlib.h> #include
clock_t start, finish; start = clock(); int checkParallel = 0; double *arrayA = malloc(sizeof(double) * n * n); double *arrayB = malloc(sizeof(double) * n * n); double *result = malloc(sizeof(double) * n * n); double *arrayTemp = malloc(sizeof(double) * n * n); #pragma omp parallel { //transpose matrix B #pragma omp for for(int i = 0; i < n; i++){ for(int j = 0; j < n; j++){ arrayTemp[(n*i)+j] = arrayB[(n*j)+i]; } } #pragma omp single { free(arrayB); arrayB = arrayTemp; } #pragma omp for for(int i = 0; i < n; i++){ for(int j = 0; j < n; j++){ int sum = 0; for(int k = 0; k < n; k++){ sum += arrayA[(n*i)+k] * arrayB[(n*j)+k]; } result[(n*i)+j] = sum; } #pragma omp atomic checkParallel++; } finish = clock(); printf("processing time B transposed, n = %d", n); if(checkParallel == 1) printf("(serial) : %.7lf\n", ((double)(finish-start) / CLOCKS_PER_SEC)); else printf("(parallel) : %.7lf\n", ((double)(finish-start) / CLOCKS_PER_SEC)); }
DATA YANG MUNCUL :
KESIMPULAN : Dapat dilihat dari data hasil percobaan diatas. Processing time yang dibutuhkan jauh lebih cepat apabila program di-run secara parallel.
3.Fibonacci #include<stdio.h> #include
y = fib(n - 2); return x + y; } } long fibP(int n, int cut){ long x,y; if(n < 2) return n; else{ #pragma omp task shared(x) { if(cut == 1) x = fib(n - 1); else x = fibP(n - 1, cut-1); } #pragma omp task shared(y) { if(cut == 1) y = fib(n - 2); else y = fibP(n - 2, cut-1); } #pragma omp taskwait return x + y; } } void runner(int n, int s,int proc, int cut){ double start, finish; long result; start = omp_get_wtime(); if(s){ result = fib(n); }else{ #pragma omp parallel num_threads(proc) #pragma omp single result = fibP(n, cut); } finish = omp_get_wtime(); printf("fib(%d) = %ld waktu eksekusi = %.5lf", n, result, (finish start)); if(s) printf("(serial)\n"); else if(proc == 1) printf("(serial + overhead) task num = %d\n",(1<<cut)); else printf("(parallel(%d)) task num = %d\n",proc, (1<<cut));
} int main(){ int nList[] = {42,43,44,45}; for(int j = 0; j < 1; j++){ for(int i = 0; i < 4; i++){ runner(nList[i], 1, 0, 0); } runner(45, 0, 1, 18); runner(45, 0, 1, 19); runner(45, 0, 1, 20); runner(45, 0, 1, 22); runner(45, 0, 1, 24); runner(45, 0, 1, 28); runner(45, 0, 4, 2); } }
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