离散事件模拟在windows和linux中生成不同的结果

Discrete Event simulation generates different results in windows and linux

本文关键字:结果 linux 事件 模拟 windows      更新时间:2023-10-16

我正在进行m/m/2/2离散事件模拟。结果与理论值相同,这意味着程序是正确的(在Ubuntu中,g++)。然而,当我在windows(visualstudio)中再次测试代码时,只有当试用次数少于10000时,结果才是正确的。我调试了很长时间,最后发现我必须使用(double) rand() / ( RAND_MAX + 1 )而不仅仅是(double) rand() / RAND_MAX

为什么?这两个编译器是否以不同的方式生成随机?

#include <iostream>
#include <stdlib.h>
#include <fstream>
#include <cmath>
#include "iomanip"
#include <queue>
using namespace std;
#define TRIAL_NUM 10000000
//structure that represent a customer
struct cust
{
int cust_id;
double arrival_time;
double depart_time;
};
//uniform random generator
double uniRand()
{
    return (double) rand() / RAND_MAX;
}
//exponential random generator
double expRand(double lam)
{
double u,x;
u=uniRand();
x=(-1/lam)*log(1-u);
return(x);
}
int main()
{
    //seed the random generator
srand (time(NULL));                   
//queue that hold all customers
std::queue<cust*> customers;
double lam = 0, mu = 0;
bool showStatus = false;
cout<<"M/M/1/1 Discrete Event Simulation with "<<TRIAL_NUM<<" customersn"<<" Please spacify the value of lambda:";
cin>>lam;
while(lam<= 0)
{
    cout<<"lambda value must be a positive number, try again:";
    cin>>lam;
}
cout<<"Please specify the value of mu:";
cin>>mu;
while(mu<= 0)
{
    cout<<"lambda value must be a positive number, try again:";
    cin>>mu;
}
cout<<"Show status for each customer? (y/n)";
char show;
cin>>show;
while( tolower(show) != 'y' && tolower(show) != 'n')
{
    cout<<"Invalid input, try again:";
    cin>>show;
}
if ( tolower(show) =='y' )
{
    showStatus = true;
}
//Generating all arrival time, service time for all customers
double temp_time = 1;
cust* temp;
for (int i = 1; i <= TRIAL_NUM; ++i)
{
    temp= new cust;
    temp -> arrival_time = temp_time + expRand(lam);
    temp_time = temp->arrival_time;
    temp->depart_time = temp->arrival_time + expRand(mu);
    temp->cust_id = i;
    customers.push(temp);
}
//Blocking customer count
double block = 0;
bool blockFlag = false;
temp = customers.front();
double server1 = 0, server2 = 0; 
//perform simulation
while( !customers.empty())
{
    if ( server1 < temp->arrival_time)
    {
        server1 = temp->depart_time;
    }
    else if ( server2 < temp->arrival_time)
    {
        server2 = temp->depart_time;
    }
    else
    {
        block++;
        blockFlag = true;
    }
    if (showStatus)
    {
        cout<<"Customer "<<temp->cust_id<<"tarrived at:"<<temp->arrival_time<<"tservice time:"<<temp->depart_time<<"tstatus:";
        if (blockFlag)
        {   
            cout<<"Blocked"<<endl;
        }
        else{
            cout<<"Served"<<endl;
        }
    }
        customers.pop();
        if (!customers.empty())
        {
            temp = customers.front();
        }
        blockFlag = false;
}
cout<<"Blocking probability is: "<<block/TRIAL_NUM<<endl;
return 0;
}

(double) rand() / (RAND_MAX + 1)是一个危险的表达式。

如果经常发生RAND_MAX等于INT_MAX,则RAND_MAX + 1是溢出和未定义的行为(它通常导致INT_MIN,但任何规范都不能保证这一点)。

在Visual C++中,RAND_MAX是32767,您没有这个问题(但32767是一个小值,它会很快"开始循环")。

无论如何,我认为真正的问题在这里:

u = uniRand();
x = (-1 / lam) * log(1 - u);

uniRand()被定义为(double) rand() / RAND_MAX时,它会产生范围内的值:

0.0 <= uniRand() <= 1.0

当该值等于1.0时,您的CCD_ 11中会出现极点误差。在Visual C++中,由于RAND_MAX的值较小,这种情况会更频繁地发生。