并行执行比Serial更多的时间？

Question

我正在研究TBB中的任务实现,并运行了Fibonacci系列的并行和连续计算代码.

守则是:

#include <iostream>
#include <list>
#include <tbb/task.h>
#include <tbb/task_group.h>
#include <stdlib.h>
#include "tbb/compat/thread"
#include "tbb/task_scheduler_init.h"
using namespace std;
using namespace tbb;

#define CutOff 2

long serialFib( long n ) {
if( n<2 )
return n;
else
return serialFib(n-1) + serialFib(n-2);
}


class FibTask: public task 
{
    public:
    const long n;
    long* const sum;

    FibTask( long n_, long* sum_ ) : n(n_), sum(sum_) {}

    task* execute() 
    {
        // cout<<"task id of thread is \t"<<this_thread::get_id()<<"FibTask(n)="<<n<<endl;  // Overrides virtual function task::execute    
                // cout<<"Task Stolen is"<<is_stolen_task()<<endl;
        if( n<CutOff ) 
        {
            *sum = serialFib(n);
        }
         else
         {
            long x, y;
            FibTask& a = *new( allocate_child() ) FibTask(n-1,&x);
            FibTask& b = *new( allocate_child() ) FibTask(n-2,&y);
            set_ref_count(3); // 3 = 2 children + 1 for wait // ref_countis used to keep track of the number of tasks spawned at                            the current level of the task graph
            spawn( b );
                      // cout<<"child id of thread is \t"<<this_thread::get_id()<<"calculating n ="<<n<<endl;
            spawn_and_wait_for_all( a ); //set tasks for execution and wait for them
            *sum = x+y;
        }
        return NULL;
    }
};


long parallelFib( long n ) 
{
    long sum;
    FibTask& a = *new(task::allocate_root()) FibTask(n,&sum);
    task::spawn_root_and_wait(a);
    return sum;
}


int main()
{     
     long i,j;
     cout<<fixed;

     cout<<"Fibonacci Series parallelly formed is "<<endl;
      tick_count t0=tick_count::now();
     for(i=0;i<50;i++)
     cout<<parallelFib(i)<<"\t";
    // cout<<"parallel execution of Fibonacci series for n=10 \t"<<parallelFib(i)<<endl;

     tick_count t1=tick_count::now();
     double t=(t1-t0).seconds();
     cout<<"Time Elapsed in Parallel Execution is  \t"<<t<<endl;
     cout<<"\n Fibonacci Series Serially formed is "<<endl;
     tick_count t3=tick_count::now();

     for(j=0;j<50;j++)
     cout<<serialFib(j)<<"\t";
     tick_count t4=tick_count::now();
     double t5=(t4-t3).seconds();
     cout<<"Time Elapsed in Serial  Execution is  \t"<<t5<<endl;
     return(0);
}

与串行执行相比,并行执行需要更多时间.在此并行执行需要2500秒,而串行需要大约167秒.任何人都可以解释一下这个原因吗？

Answer 1

高架.

当您的实际任务是轻量级时,协调/通信占主导地位,您不会(自动)从并行执行中获益.这是一个非常常见的问题.

尝试连续计算M Fibonacci数(足够高的成本),然后并行计算它们.你应该看到收益.