为什么多次调用curand时我的代码会失败？

问题描述

当我将numPaths增加到1000000时，除非除非同时减少numSteps，否则下面的蒙特卡洛代码将失败（“进程退出，代码为-2147483645，价格函数的输出未打印）。”如果删除所有对curand的引用并使用固定数字，则代码始终有效，因此我认为问题与生成太多随机数有关。我尝试用每个线程（而不是新的子序列）的新种子调用curand_init，但是它不能解决问题。

此外，在使用“推力”而不是“ curand”生成随机数时，我也遇到了相同的问题。

有人可以帮忙吗？

#include <math.h>
#include <curand_kernel.h>

#include "cuda_runtime.h"
#include <thrust/device_vector.h>
#include <thrust/reduce.h>
#include <thrust/transform_reduce.h>
#include <thrust/transform.h>
#include <thrust/functional.h>
#include <thrust/iterator/transform_iterator.h>
#include <thrust/iterator/counting_iterator.h>
#include <thrust/iterator/zip_iterator.h>
#include <device_launch_parameters.h>
#include <thrust/execution_policy.h>

using State3D = thrust::tuple<double,double,double>;

template<typename State>
struct AbstractModel : public thrust::binary_function<State,int,State>
{
private:
    double seed = 0;
    int numSteps;
public:
    double dt,sqdt;

    AbstractModel(double T,int numSteps)
        : numSteps(numSteps),dt(T / numSteps),sqdt(sqrt(T / numSteps)) {}

    __device__ virtual void nextStep(State& state,curandState &rand_state) const = 0;

    __device__ State operator() (const State &state,const int &i) const
    {
        curandState rand_state;
        curand_init(seed,i,&rand_state);

        State currentState = state;
        for (unsigned i = 0; i < numSteps; i++)
            nextStep(currentState,rand_state);
        return currentState;
    }
};

struct Model : public AbstractModel<State3D>
{
    double lambda,m,w,theta,eta;

    Model(double T,int numSteps,double lambda,double m,double w,double theta,double eta)
        : AbstractModel(T,numSteps),lambda(lambda),m(m),w(w),theta(theta),eta(eta) {}

    __device__ void nextStep(State3D &state,curandState &rand_state) const
    {
        double level = fmax(state.get<2>(),0.0),sqlevel = sqrt(level),vol = lambda * sqlevel;
        state.get<0>() += vol * sqdt * curand_normal_double(&rand_state);
        state.get<1>() += vol * vol * dt;
        state.get<2>() += theta * (1.0 - level) * dt + eta * sqlevel * sqdt * curand_normal_double(&rand_state);
    }
};

struct ModelInitializer : public thrust::unary_function<int,State3D>
{
    __device__ State3D operator() (const int idx) const
    {
        return thrust::make_tuple<double,double>(0.0,0.0,0.0);
    }
};

struct ModelPayoffTest : public thrust::unary_function<thrust::tuple<double,double>,double>
{
    double operator()(const thrust::tuple<double,double> &state) const
    {
        return state.get<2>();
    }
};

template<typename Model,typename Payoff>
double price(unsigned numPaths,Model &model,Payoff &payoff)
{
    thrust::device_vector<State3D> markov_states(numPaths);
    //initialize paths with start value
    thrust::transform(thrust::make_counting_iterator(0),thrust::make_counting_iterator(0) + numPaths,markov_states.begin(),ModelInitializer()/*model*/);

    //diffuse paths
    thrust::transform(markov_states.begin(),markov_states.end(),thrust::make_counting_iterator(0)/*rand_states.begin()*/,model);

    return thrust::transform_reduce(markov_states.begin(),payoff,thrust::plus<double>()) / numPaths;
}

int main()
{
    double lambda = 0.05,m = 20,w = 0.5,theta = 0.05,eta = 0.1;
    double T = 10;
    int numSteps = 10 * 365;

    Model cheyette(T,numSteps,lambda,eta);
    ModelPayoffTest payoff;

    int numPaths = 100000;
    std::cout << price<Model,ModelPayoffTest>(numPaths,cheyette,payoff) << "\n";
}

解决方法

暂无找到可以解决该程序问题的有效方法，小编努力寻找整理中！

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