java模拟具有固定工作量的持续性任务

Question

我正在模拟一个CPU受限的任务。每个CPU受限的任务计算800的阶乘。每个CPU绑定的任务都是线程执行的可运行对象。当我增加线程数（每个线程运行一个可运行的tak）时，我发现一些线程运行得非常快，以至于服务时间趋于零。我无法理解这种行为。代码如下

import java.math.BigInteger;    
public class CpuBoundJob  implements Runnable {     
    public void run() {    
         BigInteger factValue = BigInteger.ONE;
            long t1=System.nanoTime();      
            for ( int i = 2; i <= 800; i++){
              factValue = factValue.multiply(BigInteger.valueOf(i));
            }
        long t2=System.nanoTime();              
        System.out.println("Service Time(ms)="+((double)(t2-t1)/1000000));
    }    
}

---

public class TaskRunner extends Thread {
    CpuBoundJob job=new CpuBoundJob();
    public void run(){

    job.run();  
    }
}

---

public class Test {
int numberOfThreads=5;
public Test(){
    for(int i=1;i<=numberOfThreads;i++){
        TaskRunner t=new TaskRunner();
        t.start();
        }
}
public static void main(String[] args) {
    new Test(); 
    }
}

如果有5个线程，输出如下

Service Time(ns)=28.765821
Service Time(ns)=33.489663
Service Time(ns)=29.19727
Service Time(ns)=34.259404
Service Time(ns)=37.347448

如果有10个线程，输出如下

Service Time(ns)=45.647232
Service Time(ns)=3.972654
Service Time(ns)=23.494475
Service Time(ns)=12.210069
Service Time(ns)=19.382478
Service Time(ns)=15.34706
Service Time(ns)=54.769652
Service Time(ns)=20.646827
Service Time(ns)=3.28936
Service Time(ns)=29.809905
Service Time(ns)=60.798897
Service Time(ns)=50.718839
Service Time(ns)=2.727253
Service Time(ns)=2.882779
Service Time(ns)=63.864835
Service Time(ns)=42.601425
Service Time(ns)=4.029496
Service Time(ns)=4.339761
Service Time(ns)=79.396239
Service Time(ns)=2.923832
Service Time(ns)=5.773848
Service Time(ns)=3.064359
Service Time(ns)=2.446592
Service Time(ns)=2.205802
Service Time(ns)=2.212513
Service Time(ns)=2.265408
Service Time(ns)=82.51073
Service Time(ns)=2.200276
Service Time(ns)=2.289487
Service Time(ns)=2.322645
Service Time(ns)=2.201459
Service Time(ns)=2.217644
Service Time(ns)=2.197908
Service Time(ns)=2.252381
Service Time(ns)=13.564814
Service Time(ns)=2.238171
Service Time(ns)=2.199486
Service Time(ns)=2.179355
Service Time(ns)=2.237381
Service Time(ns)=2.593041
Service Time(ns)=2.444225
Service Time(ns)=2.42054
Service Time(ns)=38.745219
Service Time(ns)=81.232565
Service Time(ns)=19.612216
Service Time(ns)=22.31381
Service Time(ns)=59.521916
Service Time(ns)=59.511258
Service Time(ns)=54.439255
Service Time(ns)=11.582434

我无法理解2.4等的服务时间，有时服务时间降至0.8。为什么运行固定工作量的线程执行得如此之快

Answer 1

你是怎么开始这些测试的？如果每次都冷启动它们，我怀疑JVM是“warming up”和Just In Time compiling代码。如果是这种情况，那么带有几个线程的测试将以解释代码的形式运行，随后的运行将被编译，甚至以后的运行也会根据以前的运行进行优化The JVM is magic that way

降低优化影响的想法：

• 使用随机数作为工作参数（而不是循环索引）
• 将时间用作工作参数（同上）
• 加入一些字符串连接
• 根据工作参数改变回路的长度

Answer 2

在某个时刻，JIT可能会启动并优化代码。尝试通过先“预热”系统来运行实验，即进行N次迭代（你需要试验N的最佳值），在不进行测量的情况下运行系统，然后进行测量，例如：

public class Test {

private int numberOfThreads;
private int warmUpIterations;

public Test(int numberOfThreads, int warmUpIterations) {
    this.numberOfThreads = numberOfThreads;
    this.warmUpIterations = warmUpIterations;
}

public void runTests() {
    for (int i = 0; i < warmUpIterations; i++) {
        test(); // don't collect timing here
    }

    test(); // collect timing here
}

private void test() {
    for (int i = 0; i < numberOfThreads; i++) {
        TaskRunner t = new TaskRunner();
        t.start();
    }
}

public static void main(String[] args) {
    new Test(10, 10000).runTests();
}

}

还打印从{{CD1>}内的统计数据有问题，考虑将它们积聚在一些集合中，并在测试完成后打印它们。

Answer 3

很可能是计时功能太不准确，无法测量如此快速的操作。试试100000的阶乘或类似的东西

nanotime方法的文件称，它不能保证提供纳秒精度：

This method provides nanosecond precision, but not necessarily nanosecond resolution (that is, how frequently the value changes) - no guarantees are made except that the resolution is at least as good as that of currentTimeMillis().