• RateLimiter的 SmoothBursty(非warmup预热)及SmoothWarmingUp(预热,冷启动)


    SmoothBursty

    主要思想

    记录 1秒内的微秒数/permitsPerSencond = 时间间隔interval,每一个interval可获得一个令牌
    根据允许使用多少秒内的令牌参数,计算出maxPermits
    setRate时初始化下次interval时间,及storedPermits

    acquire时,计算当前nowMicros,如果大于下次interval时间时间,则更新storedPermits和下次interval时间,计算storedPermits能否满足此次acquire,如果能,则需要等待的时间为0,如果不能,则计算还需要多少微秒等待,并在非同步块外执行sleep操作

    如果其他线程已经刷新了nextFreeTicketMicros,会如下情况acquire是无timeout的

    Thread 1: acquire 11 -> storedPermits不能满足要求 -> waitTime = (acquire - stored) * stableIntervalMicros -> nextFreeTicketMicros += waitMicros ----->  out lock sleep
    Thread 2: acquire 2 -> nowMicros < nextFreeTicketMicros , stored = 0,被线程1消耗完了 -> freshPermits = requiredPermits - storedPermitsToSpend 即 = requiredPermits -> waitTime = freshPermits * stableIntervalMicros
    -> nextFreeTicketMicros += waitTime,此时的nextFreeTicketMicros包含了Thread1需要等待的时间 -------> out lock sleep a longer time
    

    tryAquire(num,timeout)逻辑

    timeoutMicros = timeout.toMicros
    lock()
    nowMicros = ...
    canAcquire = nextFreeTicketMicros <= nowMicros + timeoutMicros
    if(!canAcquire){
     return false;
    }
    else{
      microsToWait = ...
    } 
    unlock()
    sleep(microsToWait)
    return true;
    

    SmoothWarmingUp

    主要思想和SmoothBursty相似,由于带预热过程,刚开始由于availablePermitsAboveThreshold>0.0,速率会较慢,如果持续获取令牌,则会使availablePermitsAboveThreshold=0,速率变快

    • 从0->thresholdPermits,生成一个令牌的时间:stableIntervalMicros

    • 从thresholdPermits-> maxPermits ,生成一个令牌的时间:stableIntervalMicros + permits * slope;

      @Override
      final long reserveEarliestAvailable(int requiredPermits, long nowMicros) {
      resync(nowMicros);
      long returnValue = nextFreeTicketMicros;
      //当前需要且尽最大可能消费的
      double storedPermitsToSpend = min(requiredPermits, this.storedPermits);
      //新鲜permits个数,这些个数是一定会产生等待的,除了0
      double freshPermits = requiredPermits - storedPermitsToSpend;
      //计算需要wait的总时间
      long waitMicros =
      //非busty类型的storedPermitsToWaitTime直接返回0
      storedPermitsToWaitTime(this.storedPermits, storedPermitsToSpend)
      + (long) (freshPermits * stableIntervalMicros);
      //下次有票时间
      this.nextFreeTicketMicros = LongMath.saturatedAdd(nextFreeTicketMicros, waitMicros);
      this.storedPermits -= storedPermitsToSpend;
      return returnValue;
      }

       //已知permitsToTake <= storedPermits
       @Override
       long storedPermitsToWaitTime(double storedPermits, double permitsToTake) {
         //减去预热需要保留的permits,剩下的可消耗的数量
         double availablePermitsAboveThreshold = storedPermits - thresholdPermits;
         long micros = 0;
         // measuring the integral on the right part of the function (the climbing line)
         //如果有剩余可用的令牌
         if (availablePermitsAboveThreshold > 0.0) {
          //剩余可用的和需要获取的个数取小值
       	double permitsAboveThresholdToTake = min(availablePermitsAboveThreshold, permitsToTake);
       	// TODO(cpovirk): Figure out a good name for this variable.
       	//用可消耗的数量 + (可消耗的数量 - 实际消耗的数量)permitsToTime
       	//在预热阶段从thresholdPermits到maxPermits的耗时并非是stableIntervalMicros * n
       	//会耗费更多的时间,其计算规则不同,所以才需要把permitsAboveThresholdToTake从permitsToTake减去
       	//length 可能作为一个经验值,相当于补充permitsAboveThresholdToTake个令牌需要的平均时间值*2
                       //剩余可用的-实际需要且最大能消耗的令牌,得到最终剩余的令牌个数,可能是0
       	double length = permitsToTime(availablePermitsAboveThreshold)
       	    + permitsToTime(availablePermitsAboveThreshold - permitsAboveThresholdToTake);
       	//这里确实不好理解,从语义环境来说,它是从 thresholdPermits 到 maxPertmis 过程中
       	//生成 permitsAboveThresholdToTake 个令牌需要耗费的时间
       	//并且带coldFactor的构造函数不是public,SmoothWarmingUp也是private-package的
       	micros = (long) (permitsAboveThresholdToTake * length / 2.0);
       	//从permitsToTake中减去保留预热需留下个数后最终消耗的个数,这部分个数由于是提前存在的、富余的
       	//因此不需要计算到wait时间
       	permitsToTake -= permitsAboveThresholdToTake;
         }
         // measuring the integral on the left part of the function (the horizontal line)
         //如果没有剩余可用令牌,走的是stableIntervalMicros * n
         micros += (stableIntervalMicros * permitsToTake);
         return micros;
       }	
      

    length/2可以理解为下图

    	//permits值越小,需要的时间就越少,值越大,需要的时间就越大
    	private double permitsToTime(double permits) {
    	  //double coldIntervalMicros = stableIntervalMicros * coldFactor;
    	  // thresholdPermits = 0.5 * warmupPeriodMicros / stableIntervalMicros;
          //maxPermits =
          thresholdPermits + 2.0 * warmupPeriodMicros / (stableIntervalMicros + coldIntervalMicros);
          //slope带比率的时间,可以理解为增长因子
    	  //slope =  (coldIntervalMicros - stableIntervalMicros) / (maxPermits - thresholdPermits)
    	  //return表示成这样更易于理解 stableIntervalMicros + (coldIntervalMicros - stableIntervalMicros) * (permits/(maxPermits - thresholdPermits))
    	  return stableIntervalMicros + permits * slope;
    	}
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  • 原文地址:https://www.cnblogs.com/windliu/p/11088320.html
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