經典面試題：如何合理地估算線程池大小？

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作者：蔣小強

來源：ifeve.com/how-to-calculate-threadpool-size/

如何合理地估算線程池大小？

這個問題雖然看起來很小，卻並不那麼容易回答。大家如果有更好的方法歡迎賜教，先來一個天真的估算方法：假設要求一個系統的TPS（Transaction Per Second或者Task Per Second）至少爲20，然後假設每個Transaction由一個線程完成，繼續假設平均每個線程處理一個Transaction的時間爲4s。那麼問題轉化爲：

如何設計線程池大小，使得可以在1s內處理完20個Transaction？

計算過程很簡單，每個線程的處理能力爲0.25TPS，那麼要達到20TPS，顯然需要20/0.25=80個線程。

很顯然這個估算方法很天真，因爲它沒有考慮到CPU數目。一般服務器的CPU核數爲16或者32，如果有80個線程，那麼肯定會帶來太多不必要的線程上下文切換開銷。

再來第二種簡單的但不知是否可行的方法（N爲CPU總核數）：

如果是CPU密集型應用，則線程池大小設置爲N+1
如果是IO密集型應用，則線程池大小設置爲2N+1

如果一臺服務器上只部署這一個應用並且只有這一個線程池，那麼這種估算或許合理，具體還需自行測試驗證。

接下來在這個文檔：服務器性能IO優化中發現一個估算公式：

最佳線程數目 = （（線程等待時間+線程CPU時間）/線程CPU時間）* CPU數目

比如平均每個線程CPU運行時間爲0.5s，而線程等待時間（非CPU運行時間，比如IO）爲1.5s，CPU核心數爲8，那麼根據上面這個公式估算得到：((0.5+1.5)/0.5)*8=32。這個公式進一步轉化爲：

最佳線程數目 = （線程等待時間與線程CPU時間之比 + 1）* CPU數目

可以得出一個結論：

線程等待時間所佔比例越高，需要越多線程。線程CPU時間所佔比例越高，需要越少線程。

上一種估算方法也和這個結論相合。

一個系統最快的部分是CPU，所以決定一個系統吞吐量上限的是CPU。增強CPU處理能力，可以提高系統吞吐量上限。但根據短板效應，真實的系統吞吐量並不能單純根據CPU來計算。那要提高系統吞吐量，就需要從“系統短板”（比如網絡延遲、IO）着手：

儘量提高短板操作的並行化比率，比如多線程下載技術
增強短板能力，比如用NIO替代IO

第一條可以聯繫到Amdahl定律，這條定律定義了串行系統並行化後的加速比計算公式：

加速比=優化前系統耗時 / 優化後系統耗時

加速比越大，表明系統並行化的優化效果越好。Addahl定律還給出了系統並行度、CPU數目和加速比的關係，加速比爲Speedup，系統串行化比率（指串行執行代碼所佔比率）爲F，CPU數目爲N：

Speedup <= 1 / (F + (1-F)/N)

當N足夠大時，串行化比率F越小，加速比Speedup越大。

寫到這裏，我突然冒出一個問題。

是否使用線程池就一定比使用單線程高效呢？

答案是否定的，比如Redis就是單線程的，但它卻非常高效，基本操作都能達到十萬量級/s。從線程這個角度來看，部分原因在於：

多線程帶來線程上下文切換開銷，單線程就沒有這種開銷
鎖

當然“Redis很快”更本質的原因在於：Redis基本都是內存操作，這種情況下單線程可以很高效地利用CPU。而多線程適用場景一般是：存在相當比例的IO和網絡操作。

所以即使有上面的簡單估算方法，也許看似合理，但實際上也未必合理，都需要結合系統真實情況（比如是IO密集型或者是CPU密集型或者是純內存操作）和硬件環境（CPU、內存、硬盤讀寫速度、網絡狀況等）來不斷嘗試達到一個符合實際的合理估算值。

最後來一個“Dark Magic”估算方法（因爲我暫時還沒有搞懂它的原理），使用下面的類：

package pool_size_calculate;
import java.math.BigDecimal;import java.math.RoundingMode;import java.util.Timer;import java.util.TimerTask;import java.util.concurrent.BlockingQueue;
/** * A class that calculates the optimal thread pool boundaries. It takes the * desired target utilization and the desired work queue memory consumption as * input and retuns thread count and work queue capacity. * * @author Niklas Schlimm * */public abstract class PoolSizeCalculator {
  /**   * The sample queue size to calculate the size of a single {@link Runnable}   * element.   */  private final int SAMPLE_QUEUE_SIZE = 1000;
  /**   * Accuracy of test run. It must finish within 20ms of the testTime   * otherwise we retry the test. This could be configurable.   */  private final int EPSYLON = 20;
  /**   * Control variable for the CPU time investigation.   */  private volatile boolean expired;
  /**   * Time (millis) of the test run in the CPU time calculation.   */  private final long testtime = 3000;
  /**   * Calculates the boundaries of a thread pool for a given {@link Runnable}.   *   * @param targetUtilization   *            the desired utilization of the CPUs (0 <= targetUtilization <=    *            1)    * @param targetQueueSizeBytes    *            the desired maximum work queue size of the thread pool (bytes)    */   protected void calculateBoundaries(BigDecimal targetUtilization,       BigDecimal targetQueueSizeBytes) {     calculateOptimalCapacity(targetQueueSizeBytes);     Runnable task = creatTask();     start(task);     start(task); // warm up phase     long cputime = getCurrentThreadCPUTime();     start(task); // test intervall     cputime = getCurrentThreadCPUTime() - cputime;     long waittime = (testtime * 1000000) - cputime;     calculateOptimalThreadCount(cputime, waittime, targetUtilization);   }   private void calculateOptimalCapacity(BigDecimal targetQueueSizeBytes) {     long mem = calculateMemoryUsage();     BigDecimal queueCapacity = targetQueueSizeBytes.divide(new BigDecimal(         mem), RoundingMode.HALF_UP);     System.out.println("Target queue memory usage (bytes): "         + targetQueueSizeBytes);     System.out.println("createTask() produced "         + creatTask().getClass().getName() + " which took " + mem         + " bytes in a queue");     System.out.println("Formula: " + targetQueueSizeBytes + " / " + mem);     System.out.println("* Recommended queue capacity (bytes): "         + queueCapacity);   }   /**    * Brian Goetz' optimal thread count formula, see 'Java Concurrency in    * Practice' (chapter 8.2)    *     * @param cpu    *            cpu time consumed by considered task    * @param wait    *            wait time of considered task    * @param targetUtilization    *            target utilization of the system    */   private void calculateOptimalThreadCount(long cpu, long wait,       BigDecimal targetUtilization) {     BigDecimal waitTime = new BigDecimal(wait);     BigDecimal computeTime = new BigDecimal(cpu);     BigDecimal numberOfCPU = new BigDecimal(Runtime.getRuntime()         .availableProcessors());     BigDecimal optimalthreadcount = numberOfCPU.multiply(targetUtilization)         .multiply(             new BigDecimal(1).add(waitTime.divide(computeTime,                 RoundingMode.HALF_UP)));     System.out.println("Number of CPU: " + numberOfCPU);     System.out.println("Target utilization: " + targetUtilization);     System.out.println("Elapsed time (nanos): " + (testtime * 1000000));     System.out.println("Compute time (nanos): " + cpu);     System.out.println("Wait time (nanos): " + wait);     System.out.println("Formula: " + numberOfCPU + " * "         + targetUtilization + " * (1 + " + waitTime + " / "         + computeTime + ")");     System.out.println("* Optimal thread count: " + optimalthreadcount);   }   /**    * Runs the {@link Runnable} over a period defined in {@link #testtime}.    * Based on Heinz Kabbutz' ideas    * (http://www.javaspecialists.eu/archive/Issue124.html).    *     * @param task    *            the runnable under investigation    */   public void start(Runnable task) {     long start = 0;     int runs = 0;     do {       if (++runs > 5) {        throw new IllegalStateException("Test not accurate");      }      expired = false;      start = System.currentTimeMillis();      Timer timer = new Timer();      timer.schedule(new TimerTask() {        public void run() {          expired = true;        }      }, testtime);      while (!expired) {        task.run();      }      start = System.currentTimeMillis() - start;      timer.cancel();    } while (Math.abs(start - testtime) > EPSYLON);    collectGarbage(3);  }
  private void collectGarbage(int times) {    for (int i = 0; i < times; i++) {      System.gc();      try {        Thread.sleep(10);      } catch (InterruptedException e) {        Thread.currentThread().interrupt();        break;      }    }  }
  /**   * Calculates the memory usage of a single element in a work queue. Based on   * Heinz Kabbutz' ideas   * (http://www.javaspecialists.eu/archive/Issue029.html).   *   * @return memory usage of a single {@link Runnable} element in the thread   *         pools work queue   */  public long calculateMemoryUsage() {    BlockingQueue queue = createWorkQueue();    for (int i = 0; i < SAMPLE_QUEUE_SIZE; i++) {      queue.add(creatTask());    }    long mem0 = Runtime.getRuntime().totalMemory()        - Runtime.getRuntime().freeMemory();    long mem1 = Runtime.getRuntime().totalMemory()        - Runtime.getRuntime().freeMemory();    queue = null;    collectGarbage(15);    mem0 = Runtime.getRuntime().totalMemory()        - Runtime.getRuntime().freeMemory();    queue = createWorkQueue();    for (int i = 0; i < SAMPLE_QUEUE_SIZE; i++) {      queue.add(creatTask());    }    collectGarbage(15);    mem1 = Runtime.getRuntime().totalMemory()        - Runtime.getRuntime().freeMemory();    return (mem1 - mem0) / SAMPLE_QUEUE_SIZE;  }
  /**   * Create your runnable task here.   *   * @return an instance of your runnable task under investigation   */  protected abstract Runnable creatTask();
  /**   * Return an instance of the queue used in the thread pool.   *   * @return queue instance   */  protected abstract BlockingQueue createWorkQueue();
  /**   * Calculate current cpu time. Various frameworks may be used here,   * depending on the operating system in use. (e.g.   * http://www.hyperic.com/products/sigar). The more accurate the CPU time   * measurement, the more accurate the results for thread count boundaries.   *   * @return current cpu time of current thread   */  protected abstract long getCurrentThreadCPUTime();
}

然後自己繼承這個抽象類並實現它的三個抽象方法，比如下面是我寫的一個示例（任務是請求網絡數據），其中我指定期望CPU利用率爲1.0（即100%），任務隊列總大小不超過100,000字節：

package pool_size_calculate;
import java.io.BufferedReader;import java.io.IOException;import java.io.InputStreamReader;import java.lang.management.ManagementFactory;import java.math.BigDecimal;import java.net.HttpURLConnection;import java.net.URL;import java.util.concurrent.BlockingQueue;import java.util.concurrent.LinkedBlockingQueue;
public class SimplePoolSizeCaculatorImpl extends PoolSizeCalculator {
  @Override  protected Runnable creatTask() {    return new AsyncIOTask();  }
  @Override  protected BlockingQueue createWorkQueue() {    return new LinkedBlockingQueue(1000);  }
  @Override  protected long getCurrentThreadCPUTime() {    return ManagementFactory.getThreadMXBean().getCurrentThreadCpuTime();  }
  public static void main(String[] args) {    PoolSizeCalculator poolSizeCalculator = new SimplePoolSizeCaculatorImpl();    poolSizeCalculator.calculateBoundaries(new BigDecimal(1.0), new BigDecimal(100000));  }
}
/** * 自定義的異步IO任務 * @author Will * */class AsyncIOTask implements Runnable {
  @Override  public void run() {    HttpURLConnection connection = null;    BufferedReader reader = null;    try {      String getURL = "http://baidu.com";      URL getUrl = new URL(getURL);
      connection = (HttpURLConnection) getUrl.openConnection();      connection.connect();      reader = new BufferedReader(new InputStreamReader(          connection.getInputStream()));
      String line;      while ((line = reader.readLine()) != null) {        // empty loop      }    }
    catch (IOException e) {
    } finally {      if(reader != null) {        try {          reader.close();        }        catch(Exception e) {
        }      }      connection.disconnect();    }
  }
}

得到的輸出如下：

Target queue memory usage (bytes): 100000createTask() produced pool_size_calculate.AsyncIOTask which took 40 bytes in a queueFormula: 100000 / 40* Recommended queue capacity (bytes): 2500Number of CPU: 4Target utilization: 1Elapsed time (nanos): 3000000000Compute time (nanos): 47181000Wait time (nanos): 2952819000Formula: 4 * 1 * (1 + 2952819000 / 47181000)* Optimal thread count: 256

推薦的任務隊列大小爲2500，線程數爲256，有點出乎意料之外。我可以如下構造一個線程池：

ThreadPoolExecutor pool = new ThreadPoolExecutor(256, 256, 0L, TimeUnit.MILLISECONDS, new LinkedBlockingQueue(2500));

---END---

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經典面試題：如何合理地估算線程池大小？

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