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c++ - openMP 缺乏 yield 递减与更高的线程数

转载 作者:太空宇宙 更新时间:2023-11-04 13:13:24 26 4
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我的代码现在有一个循环,它调用蒙特卡洛函数来计算多个样本的简单积分(y=x,从 0 到 1),并将总时间和积分值写入文本文件。然后循环增加线程数并继续前进。现在大约有 8 个线程,时间峰值约为 2.6 秒。循环迭代超过 64 个线程,我发现速度没有超过 0.2 秒,甚至有时会加速。

对于循环调用蒙特卡洛方法,增加线程数:

//this loop will iterate the main loop for a number of threads from 1 to 16
for (int j = 1; j <= 17; j++)
{
//tell user how many threads are running monte-carlo currently
cout << "Program is running " << number_threads << " thread(s) currently." << endl;

//reset values for new run
num_of_samples = 1;
integration_result = 0;

//this for loop will run throughout number of circulations running through monte-carlo
//and entering the data into the text folder
for (int i = 1; i <= iteration_num; i++)
{
//call monte carlo function to perform integration and write values to text
monteCarlo(num_of_samples, starting_x, end_x, number_threads);

//increase num of samples for next test round
num_of_samples = 2 * num_of_samples;
} //end of second for loop

//iterate num_threads
if (number_threads == 1)
number_threads = 2;
else if (number_threads >= 32)
number_threads += 8;
else if (number_threads >= 16)
number_threads += 4;
else
number_threads += 2;
} //end of for loop

蒙特卡洛的并行部分:

int num_threads;
double x, u, error_difference, fs = 0, integration_result = 0; //fs is a placeholder to hold added values of f(x)
vector< vector<double>> dataHolder(number_threads, vector<double>(1)); //this vector will hold temp values of each thread

//get start time for parallel block of code
double start_time = omp_get_wtime();

omp_set_dynamic(0); // Explicitly disable dynamic teams
omp_set_num_threads(number_threads); // Use 4 threads for all consecutive parallel regions

#pragma omp parallel default(none) private(x, u) shared(std::cout, end_x, starting_x, num_of_samples, fs, number_threads, num_threads, dataHolder)
{
int i, id, nthrds;
double temp = fs;

//define thread id and num of threads
id = omp_get_thread_num();
nthrds = omp_get_num_threads();

//initilialize random seed
srand(id * time(NULL) * 1000);

//if there is only one thread
if(id == 0)
num_threads = nthrds;

//this for loop will calculate a temp value for fs for each thread
for (int i = id; i < num_of_samples; i = i + nthrds)
{
//assign random number under integration from 0 to 1
u = fRand(0, 1); //random number between 0 and 1
x = starting_x + (end_x - starting_x) * u;

//this line of code is from Monte_Carlo Method by Alex Godunov (February 2007)
//calculuate y for reciporical value of x and add it to thread's local fs
temp += function(x);
}

//place temp inside vector dataHolder
dataHolder[id][0] = temp;

//no thread will go beyond this barrier until task is complete
#pragma omp barrier

//one thread will do this task
#pragma omp single
{
//add summations to calc fs
for(i = 0, fs = 0.0; i < num_threads; i ++)
fs += dataHolder[i][0];
} //implicit barrier here, wait for all tasks to be done
}//end of parallel block of code

最佳答案

在使用光散射对简单的蒙特卡洛游走实现相同类型的并行化后,我能够相当多地了解 yield 递减的情况。我认为这里没有 yield 递减,因为积分计算非常简单,线程本身几乎没有单独做的事情,因此它们的开销相对较小。如果其他人有任何其他信息可以证明对这个问题有用,请随时发布。否则我会接受这个作为我的答案。

关于c++ - openMP 缺乏 yield 递减与更高的线程数,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/38647322/

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