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hadoop2.9.0之前的版本yarn RM fairScheduler调度性能优化

时间:2019-11-12 18:39:02      阅读:90      评论:0      收藏:0      [点我收藏+]

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对一般小公司来说 可能yarn调度能力足够了 但是对于大规模集群1000 or 2000+的话  yarn的调度性能捉襟见肘

恰好网上看到一篇很好的文章https://tech.meituan.com/2019/08/01/hadoop-yarn-scheduling-performance-optimization-practice.html

参考了YARN-5969 发现hadoop2.9.0已经修正了该issue 实测提高了调度性能 

FairScheduler 调度方式有两种 

心跳调度:Yarn的NodeManager会通过心跳的方式定期向ResourceManager汇报自身状态 伴随着这次rpc请求 会触发Resourcemanager 触发nodeUpdate()方法 为这个节点进行一次资源调度

持续调度:有一个固定守护线程每隔很短的时间调度 实时的资源分配,与NodeManager的心跳出发的调度相互异步并行进行

心跳调度作为一个线程 每次运行

每次nodeUpdate 走的都是相同的逻辑

    // If the node is decommissioning, send an update to have the total
    // resource equal to the used resource, so no available resource to
    // schedule.
    if (nm.getState() == NodeState.DECOMMISSIONING) {
      this.rmContext
          .getDispatcher()
          .getEventHandler()
          .handle(
              new RMNodeResourceUpdateEvent(nm.getNodeID(), ResourceOption
                  .newInstance(getSchedulerNode(nm.getNodeID())
                      .getUsedResource(), 0)));
    }

    if (continuousSchedulingEnabled) {
      if (!completedContainers.isEmpty()) {  //心跳调度
        attemptScheduling(node);
      }
    } else {
      attemptScheduling(node);  //持续调度
    }

    // Updating node resource utilization
    node.setAggregatedContainersUtilization(
        nm.getAggregatedContainersUtilization());
    node.setNodeUtilization(nm.getNodeUtilization());

 

 

 

continuousSchedulingAttempt

/**
* Thread which attempts scheduling resources continuously,
* asynchronous to the node heartbeats.
*/
private class ContinuousSchedulingThread extends Thread {

@Override
public void run() {
while (!Thread.currentThread().isInterrupted()) {
try {
continuousSchedulingAttempt();
Thread.sleep(getContinuousSchedulingSleepMs());
} catch (InterruptedException e) {
LOG.warn("Continuous scheduling thread interrupted. Exiting.", e);
return;
}
}
}
}

之后进行一次node节点 根据资源宽松情况的排序

void continuousSchedulingAttempt() throws InterruptedException {
    long start = getClock().getTime();
    List<NodeId> nodeIdList = new ArrayList<NodeId>(nodes.keySet());
    // Sort the nodes by space available on them, so that we offer
    // containers on emptier nodes first, facilitating an even spread. This
    // requires holding the scheduler lock, so that the space available on a
    // node doesn‘t change during the sort.
    synchronized (this) {
      Collections.sort(nodeIdList, nodeAvailableResourceComparator);
    }

    // iterate all nodes
    for (NodeId nodeId : nodeIdList) {
      FSSchedulerNode node = getFSSchedulerNode(nodeId);
      try {
        if (node != null && Resources.fitsIn(minimumAllocation,
            node.getAvailableResource())) {
          attemptScheduling(node);
        }
      } catch (Throwable ex) {
        LOG.error("Error while attempting scheduling for node " + node +
            ": " + ex.toString(), ex);
        if ((ex instanceof YarnRuntimeException) &&
            (ex.getCause() instanceof InterruptedException)) {
          // AsyncDispatcher translates InterruptedException to
          // YarnRuntimeException with cause InterruptedException.
          // Need to throw InterruptedException to stop schedulingThread.
          throw (InterruptedException)ex.getCause();
        }
      }
    }

依次对node遍历分配Container 

queueMgr.getRootQueue().assignContainer(node) 从root遍历树 对抽象的应用资源遍历
    boolean validReservation = false;
    FSAppAttempt reservedAppSchedulable = node.getReservedAppSchedulable();
    if (reservedAppSchedulable != null) {
      validReservation = reservedAppSchedulable.assignReservedContainer(node);
    }
    if (!validReservation) {
      // No reservation, schedule at queue which is farthest below fair share
      int assignedContainers = 0;
      Resource assignedResource = Resources.clone(Resources.none());
      Resource maxResourcesToAssign =
          Resources.multiply(node.getAvailableResource(), 0.5f);
      while (node.getReservedContainer() == null) {
        boolean assignedContainer = false;
        Resource assignment = queueMgr.getRootQueue().assignContainer(node);
        if (!assignment.equals(Resources.none())) {
          assignedContainers++;
          assignedContainer = true;
          Resources.addTo(assignedResource, assignment);
        }
        if (!assignedContainer) { break; }
        if (!shouldContinueAssigning(assignedContainers,
            maxResourcesToAssign, assignedResource)) {
          break;
        }
      }
接下来在assignContainer 方法中对子队列使用特定的比较器排序这里是fairSchduler
  @Override
  public Resource assignContainer(FSSchedulerNode node) { 对于每一个服务器,对资源树进行一次递归搜索
    Resource assigned = Resources.none();

    // If this queue is over its limit, reject
    if (!assignContainerPreCheck(node)) {
      return assigned;
    }

    // Hold the write lock when sorting childQueues
    writeLock.lock();
    try {
      Collections.sort(childQueues, policy.getComparator());
    } finally {
      writeLock.unlock();
    }

对队列下的app排序

/*
     * We are releasing the lock between the sort and iteration of the
     * "sorted" list. There could be changes to the list here:
     * 1. Add a child queue to the end of the list, this doesn‘t affect
     * container assignment.
     * 2. Remove a child queue, this is probably good to take care of so we
     * don‘t assign to a queue that is going to be removed shortly.
     */
    readLock.lock();
    try {
      for (FSQueue child : childQueues) {
        assigned = child.assignContainer(node);
        if (!Resources.equals(assigned, Resources.none())) {
          break;
        }
      }
    } finally {
      readLock.unlock();
    }
    return assigned;
assignContainer 可能传入的是app 可能传入的是一个队列 是队列的话 进行递归 直到找到app为止(root(FSParentQueue)节点递归调用assignContainer(),最终将到达最终叶子节点的assignContainer()方法,才真正开始进行分配)

技术图片

 

 我们在这里 关注的就是排序

hadoop2.8.4 排序类 FairSharePolicy中的 根据权重 需求的资源大小 和内存占比 进行排序 多次获取

getResourceUsage() 产生了大量重复计算 这个方法是一个动态获取的过程(耗时)
  @Override
public int compare(Schedulable s1, Schedulable s2) {
double minShareRatio1, minShareRatio2;
double useToWeightRatio1, useToWeightRatio2;
Resource minShare1 = Resources.min(RESOURCE_CALCULATOR, null,
s1.getMinShare(), s1.getDemand());
Resource minShare2 = Resources.min(RESOURCE_CALCULATOR, null,
s2.getMinShare(), s2.getDemand());
boolean s1Needy = Resources.lessThan(RESOURCE_CALCULATOR, null,
s1.getResourceUsage(), minShare1);
boolean s2Needy = Resources.lessThan(RESOURCE_CALCULATOR, null,
s2.getResourceUsage(), minShare2);
minShareRatio1 = (double) s1.getResourceUsage().getMemorySize()
/ Resources.max(RESOURCE_CALCULATOR, null, minShare1, ONE).getMemorySize();
minShareRatio2 = (double) s2.getResourceUsage().getMemorySize()
/ Resources.max(RESOURCE_CALCULATOR, null, minShare2, ONE).getMemorySize();
useToWeightRatio1 = s1.getResourceUsage().getMemorySize() /
s1.getWeights().getWeight(ResourceType.MEMORY);
useToWeightRatio2 = s2.getResourceUsage().getMemorySize() /
s2.getWeights().getWeight(ResourceType.MEMORY);
int res = 0;
if (s1Needy && !s2Needy)
res = -1;
else if (s2Needy && !s1Needy)
res = 1;
else if (s1Needy && s2Needy)
res = (int) Math.signum(minShareRatio1 - minShareRatio2);
else
// Neither schedulable is needy
res = (int) Math.signum(useToWeightRatio1 - useToWeightRatio2);
if (res == 0) {
// Apps are tied in fairness ratio. Break the tie by submit time and job
// name to get a deterministic ordering, which is useful for unit tests.
res = (int) Math.signum(s1.getStartTime() - s2.getStartTime());
if (res == 0)
res = s1.getName().compareTo(s2.getName());
}
return res;
}
}

新版优化后如下

@Override
    public int compare(Schedulable s1, Schedulable s2) {
      int res = compareDemand(s1, s2);

      // Pre-compute resource usages to avoid duplicate calculation
      Resource resourceUsage1 = s1.getResourceUsage();
      Resource resourceUsage2 = s2.getResourceUsage();

      if (res == 0) {
        res = compareMinShareUsage(s1, s2, resourceUsage1, resourceUsage2);
      }

      if (res == 0) {
        res = compareFairShareUsage(s1, s2, resourceUsage1, resourceUsage2);
      }

      // Break the tie by submit time
      if (res == 0) {
        res = (int) Math.signum(s1.getStartTime() - s2.getStartTime());
      }

      // Break the tie by job name
      if (res == 0) {
        res = s1.getName().compareTo(s2.getName());
      }

      return res;
    }


    private int compareDemand(Schedulable s1, Schedulable s2) {
      int res = 0;
      Resource demand1 = s1.getDemand();
      Resource demand2 = s2.getDemand();
      if (demand1.equals(Resources.none()) && Resources.greaterThan(
              RESOURCE_CALCULATOR, null, demand2, Resources.none())) {
        res = 1;
      } else if (demand2.equals(Resources.none()) && Resources.greaterThan(
              RESOURCE_CALCULATOR, null, demand1, Resources.none())) {
        res = -1;
      }
      return res;
    }


    private int compareMinShareUsage(Schedulable s1, Schedulable s2,
                                     Resource resourceUsage1, Resource resourceUsage2) {
      int res;
      Resource minShare1 = Resources.min(RESOURCE_CALCULATOR, null,
              s1.getMinShare(), s1.getDemand());
      Resource minShare2 = Resources.min(RESOURCE_CALCULATOR, null,
              s2.getMinShare(), s2.getDemand());
      boolean s1Needy = Resources.lessThan(RESOURCE_CALCULATOR, null,
              resourceUsage1, minShare1);
      boolean s2Needy = Resources.lessThan(RESOURCE_CALCULATOR, null,
              resourceUsage2, minShare2);

      if (s1Needy && !s2Needy) {
        res = -1;
      } else if (s2Needy && !s1Needy) {
        res = 1;
      } else if (s1Needy && s2Needy) {
        double minShareRatio1 = (double) resourceUsage1.getMemorySize() /
                Resources.max(RESOURCE_CALCULATOR, null, minShare1, ONE)
                        .getMemorySize();
        double minShareRatio2 = (double) resourceUsage2.getMemorySize() /
                Resources.max(RESOURCE_CALCULATOR, null, minShare2, ONE)
                        .getMemorySize();
        res = (int) Math.signum(minShareRatio1 - minShareRatio2);
      } else {
        res = 0;
      }

      return res;
    }


    /**
     * To simplify computation, use weights instead of fair shares to calculate
     * fair share usage.
     */
    private int compareFairShareUsage(Schedulable s1, Schedulable s2,
                                      Resource resourceUsage1, Resource resourceUsage2) {
      double weight1 = s1.getWeights().getWeight(ResourceType.MEMORY);
      double weight2 = s2.getWeights().getWeight(ResourceType.MEMORY);
      double useToWeightRatio1;
      double useToWeightRatio2;
      if (weight1 > 0.0 && weight2 > 0.0) {
        useToWeightRatio1 = resourceUsage1.getMemorySize() / weight1;
        useToWeightRatio2 = resourceUsage2.getMemorySize() / weight2;
      } else { // Either weight1 or weight2 equals to 0
        if (weight1 == weight2) {
          // If they have same weight, just compare usage
          useToWeightRatio1 = resourceUsage1.getMemorySize();
          useToWeightRatio2 = resourceUsage2.getMemorySize();
        } else {
          // By setting useToWeightRatios to negative weights, we give the
          // zero-weight one less priority, so the non-zero weight one will
          // be given slots.
          useToWeightRatio1 = -weight1;
          useToWeightRatio2 = -weight2;
        }
      }

      return (int) Math.signum(useToWeightRatio1 - useToWeightRatio2);
    }

  }

用了测试环境集群 比较了修改前后两次队列排序耗时

技术图片

 

 上面红框里为 新版本 下面红框为老版本 虽然没有进行压测 但是在同样的调度任务前提下 是有说服力的 在大集群上每秒调度上千万乃至上亿次该方法时  调度优化变的明显

hadoop2.9.0之前的版本yarn RM fairScheduler调度性能优化

标签:新版本   html   ddt   type   tno   递归   rtl   mic   clock   

原文地址:https://www.cnblogs.com/songchaolin/p/11844217.html

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