如何合理地估算線程池大?。?/h1>
來源:蔣小強 ,
ifeve.com/how-to-calculate-threadpool-size/
如何合理地估算線程池大???
這個問題雖然看起來很小,卻并不那么容易回答。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(shè)要求一個系統(tǒng)的TPS(Transaction Per Second或者Task Per Second)至少為20,然后假設(shè)每個Transaction由一個線程完成,繼續(xù)假設(shè)平均每個線程處理一個Transaction的時間為4s。那么問題轉(zhuǎn)化為:
如何設(shè)計線程池大小,使得可以在1s內(nèi)處理完20個Transaction?
計算過程很簡單,每個線程的處理能力為0.25TPS,那么要達到20TPS,顯然需要20/0.25=80個線程。
很顯然這個估算方法很天真,因為它沒有考慮到CPU數(shù)目。一般服務(wù)器的CPU核數(shù)為16或者32,如果有80個線程,那么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為CPU總核數(shù)):
如果是CPU密集型應(yīng)用,則線程池大小設(shè)置為N+1
如果是IO密集型應(yīng)用,則線程池大小設(shè)置為2N+1
如果一臺服務(wù)器上只部署這一個應(yīng)用并且只有這一個線程池,那么這種估算或許合理,具體還需自行測試驗證。
接下來在這個文檔:服務(wù)器性能IO優(yōu)化 中發(fā)現(xiàn)一個估算公式:
最佳線程數(shù)目 = ((線程等待時間+線程CPU時間)/線程CPU時間 )* CPU數(shù)目
比如平均每個線程CPU運行時間為0.5s,而線程等待時間(非CPU運行時間,比如IO)為1.5s,CPU核心數(shù)為8,那么根據(jù)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(zhuǎn)化為:
最佳線程數(shù)目 = (線程等待時間與線程CPU時間之比 + 1)* CPU數(shù)目
可以得出一個結(jié)論:
線程等待時間所占比例越高,需要越多線程。線程CPU時間所占比例越高,需要越少線程。
上一種估算方法也和這個結(jié)論相合。
一個系統(tǒng)最快的部分是CPU,所以決定一個系統(tǒng)吞吐量上限的是CPU。增強CPU處理能力,可以提高系統(tǒng)吞吐量上限。但根據(jù)短板效應(yīng),真實的系統(tǒng)吞吐量并不能單純根據(jù)CPU來計算。那要提高系統(tǒng)吞吐量,就需要從“系統(tǒng)短板”(比如網(wǎng)絡(luò)延遲、IO)著手:
盡量提高短板操作的并行化比率,比如多線程下載技術(shù)
增強短板能力,比如用NIO替代IO
第一條可以聯(lián)系到Amdahl定律,這條定律定義了串行系統(tǒng)并行化后的加速比計算公式:
加速比=優(yōu)化前系統(tǒng)耗時 / 優(yōu)化后系統(tǒng)耗時
加速比越大,表明系統(tǒng)并行化的優(yōu)化效果越好。Addahl定律還給出了系統(tǒng)并行度、CPU數(shù)目和加速比的關(guān)系,加速比為Speedup,系統(tǒng)串行化比率(指串行執(zhí)行代碼所占比率)為F,CPU數(shù)目為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。
寫到這里,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,部分原因在于:
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(zhì)的原因在于:Redis基本都是內(nèi)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(wǎng)絡(luò)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(jié)合系統(tǒng)真實情況(比如是IO密集型或者是CPU密集型或者是純內(nèi)存操作)和硬件環(huán)境(CPU、內(nèi)存、硬盤讀寫速度、網(wǎng)絡(luò)狀況等)來不斷嘗試達到一個符合實際的合理估算值。
最后來一個“Dark Magic”估算方法(因為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();
}
然后自己繼承這個抽象類并實現(xiàn)它的三個抽象方法,比如下面是我寫的一個示例(任務(wù)是請求網(wǎng)絡(luò)數(shù)據(jù)),其中我指定期望CPU利用率為1.0(即100%),任務(wù)隊列總大小不超過100,000字節(jié):
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(wù)
* @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): 100000
createTask() produced pool_size_calculate.AsyncIOTask which took 40 bytes in a queue
Formula: 100000 / 40
* Recommended queue capacity (bytes): 2500
Number of CPU: 4
Target utilization: 1
Elapsed time (nanos): 3000000000
Compute time (nanos): 47181000
Wait time (nanos): 2952819000
Formula: 4 * 1 * (1 + 2952819000 / 47181000)
* Optimal thread count: 256
推薦的任務(wù)隊列大小為2500,線程數(shù)為256,有點出乎意料之外。我可以如下構(gòu)造一個線程池:
ThreadPoolExecutor pool =
new ThreadPoolExecutor(256, 256, 0L, TimeUnit.MILLISECONDS, new LinkedBlockingQueue(2500));
-更多文章-
大數(shù)據(jù)就業(yè)前景,分析的太到位了!
當CPU飆高時,它在做什么
阿里分布式事務(wù)框架GTS開源了!
SpringCloud隨筆:記錄在使用OAuth2遇到的巨坑
-關(guān)注我-
https://mp.weixin.qq.com/s?__biz=MzAxNjk4ODE4OQ==&mid=2247484967&idx=1&sn=4417f52860f457b9c0a971ab4f38d477&chksm=9bed2755ac9aae4336876361f99066f3a7b363bdb377df372f14c35e744707e56f83132c846b&scene=21#wechat_redirect
任務(wù)調(diào)度 網(wǎng)絡(luò)
版權(quán)聲明:本文內(nèi)容由網(wǎng)絡(luò)用戶投稿,版權(quán)歸原作者所有,本站不擁有其著作權(quán),亦不承擔相應(yīng)法律責任。如果您發(fā)現(xiàn)本站中有涉嫌抄襲或描述失實的內(nèi)容,請聯(lián)系我們jiasou666@gmail.com 處理,核實后本網(wǎng)站將在24小時內(nèi)刪除侵權(quán)內(nèi)容。
版權(quán)聲明:本文內(nèi)容由網(wǎng)絡(luò)用戶投稿,版權(quán)歸原作者所有,本站不擁有其著作權(quán),亦不承擔相應(yīng)法律責任。如果您發(fā)現(xiàn)本站中有涉嫌抄襲或描述失實的內(nèi)容,請聯(lián)系我們jiasou666@gmail.com 處理,核實后本網(wǎng)站將在24小時內(nèi)刪除侵權(quán)內(nèi)容。