zoukankan      html  css  js  c++  java
  • Spark Application的调度算法

    要想明白spark application调度机制,需要回答一下几个问题:

    1.谁来调度?

    2.为谁调度?

    3.调度什么?

    3.何时调度?

    4.调度算法

    前四个问题可以用如下一句话里来回答:每当集群资源发生变化时,active master 进程所有已注册的并且没有调度完毕的application调度Worker节点上的Executor进程。

       "active master" , spark集群可能有多个master,但是只有一个active master 参与调度,standby master不参与调度。

       集群资源发生变化是什么意思呢?这里的集群资源指的主要是cores的变化,注册/移除Executor进程使得集群的freeCores变多/变少,添加/移除Worker节点使得集群的freeCores变多/变少... ...,所有导致集群资源发生变化的操作,都会调用schedule()重新为application和driver进行资源调度。

    spark提供了两种资源调度算法:spreadOut和非spreadOut。spreadOut算法会尽可能的将一个application 所需要的Executor进程分布在多个worker几点上,从而提高并行度,非spreadOut与之相反,他会把一个worker节点的freeCores都耗尽了才会去下一个worker节点分配。

    为了详细说明这两种算法,我们先来以一个具体的例子来介绍,最后再介绍源码。

    基本概念

      每一个application至少包含以下基本属性:

      coresPerExecutor:每一个Executor进程的core个数

      memoryPerExecutor:每一个Executor进程的memory大小

      maxCores: 这个application最多需要的core个数。

      每一个worker至少包含以下基本属性:

      freeCores:worker 节点当前可用的core个数

      memoryFree:worker节点当前可用的memory大小。

    假设一个待注册的application如下:

      coresPerExecutor:2

      memoryPerExecutor:512M

      maxCores: 12

    这表示这个application 最多需要12个core,每一个Executor进行都要2个core,512M内存。

    假设某一时刻spark集群有如下几个worker节点,他们按照coresFree降序排列:

    Worker1:coresFree=10  memoryFree=10G

    Worker2:coresFree=7   memoryFree=1G

    Worker3:coresFree=3   memoryFree=2G

    Worker4:coresFree=2   memoryFree=215M

    Worker5:coresFree=1   memoryFree=1G

     

    其中worker5不满足application的要求:worker5.coresFree < application.coresPerExecutor

    worker4也不满足application的要求:worker4.memoryFree < application.memoryPerExecutor

     

    因此最终满足调度要求的worker节点只有前三个,我们将这三个节点记作usableWorkers。

     

    spreadOut算法

     先介绍spreadOut算法吧。上面已经说了,满足条件的worker只有前三个:

    Worker1:coresFree=10  memoryFree=10G

    Worker2:coresFree=7   memoryFree=1G

    Worker3:coresFree=3   memoryFree=2G

     

    第一次调度之后,worker列表如下:

    Worker1:coresFree=8  memoryFree=9.5G  assignedExecutors=1  assignedCores=2

    Worker2:coresFree=7   memoryFree=1G    assignedExecutors=0  assignedCores=0

    Worker3:coresFree=3   memoryFree=2G    assignedExecutors=0  assignedCores=0

    totalExecutors:1,totalCores=2

     

    可以发现,worker1的coresFree和memoryFree都变小了而worker2,worker3并没有发生改变,这是因为我们在worker1上面分配了一个Executor进程(这个Executor进程占用两个2core,512M memory)而没有在workre2和worker3上分配。

     

    接下来,开始去worker2上分配:

    Worker1:coresFree=8  memoryFree=9.5G      assignedExecutors=1  assignedCores=2

    Worker2:coresFree=5   memoryFree=512M    assignedExecutors=1  assignedCores=2

    Worker3:coresFree=3   memoryFree=2G        assignedExecutors=0  assignedCores=0

    totalExecutors:2,totalCores=4

    此时已经分配了2个Executor进程,4个core。

     

    接下来去worker3上分配:

    Worker1:coresFree=8  memoryFree=9.5G      assignedExecutors=1  assignedCores=2

    Worker2:coresFree=5   memoryFree=512M    assignedExecutors=1  assignedCores=2

    Worker3:coresFree=1   memoryFree=1.5G     assignedExecutors=1  assignedCores=2

    totalExecutors:3,totalCores=6

     

    接下来再去worker1分配,然后worker2...  ...以round-robin方式分配,由于worker3.coresFree<application.coresPerExecutor,不会在他上面分配资源了:

    Worker1:coresFree=6  memoryFree=9.0G      assignedExecutors=2  assignedCores=4

    Worker2:coresFree=5   memoryFree=512M    assignedExecutors=1  assignedCores=2

    Worker3:coresFree=1   memoryFree=1.5G     assignedExecutors=1  assignedCores=2

    totalExecutors:4,totalCores=8

     

    Worker1:coresFree=6  memoryFree=9.0G      assignedExecutors=2  assignedCores=4

    Worker2:coresFree=3   memoryFree=0M       assignedExecutors=2  assignedCores=4

    Worker3:coresFree=1   memoryFree=1.5G     assignedExecutors=1  assignedCores=2

    totalExecutors:5,totalCores=10

    此时worker2也不满足要求了:worker2.memoryFree<application.memoryPerExecutor

    因此,下一次分配就去worker1上了:

    Worker1:coresFree=4  memoryFree=8.5G      assignedExecutors=3  assignedCores=6

    Worker2:coresFree=3   memoryFree=0M        assignedExecutors=2  assignedCores=4

    Worker3:coresFree=1   memoryFree=1.5G     assignedExecutors=1  assignedCores=2

    totalExecutors:6,totalCores=12

    ok,由于已经分配了12个core,达到了application的要求,所以不在为这个application调度了。

    非spreadOUt算法

    那么非spraadOut算法呢?他是逮到一个worker如果不把他的资源耗尽了是不会放手的:

    Worker1:coresFree=8  memoryFree=9.5G  assignedExecutors=1  assignedCores=2

    Worker2:coresFree=7   memoryFree=1G    assignedExecutors=0  assignedCores=0

    Worker3:coresFree=3   memoryFree=2G    assignedExecutors=0  assignedCores=0

    totalExecutors:1,totalCores=2

    Worker1:coresFree=6  memoryFree=9.0G  assignedExecutors=2  assignedCores=4

    Worker2:coresFree=7   memoryFree=1G    assignedExecutors=0  assignedCores=0

    Worker3:coresFree=3   memoryFree=2G    assignedExecutors=0  assignedCores=0

    totalExecutors:2,totalCores=4

     

    Worker1:coresFree=4  memoryFree=8.5    assignedExecutors=3  assignedCores=6

    Worker2:coresFree=7   memoryFree=1G    assignedExecutors=0  assignedCores=0

    Worker3:coresFree=3   memoryFree=2G    assignedExecutors=0  assignedCores=0

    totalExecutors:3,totalCores=6

    Worker1:coresFree=2  memoryFree=8.0G  assignedExecutors=4  assignedCores=8

    Worker2:coresFree=7   memoryFree=1G    assignedExecutors=0  assignedCores=0

    Worker3:coresFree=3   memoryFree=2G    assignedExecutors=0  assignedCores=0

    totalExecutors:4,totalCores=8

     

    Worker1:coresFree=0  memoryFree=7.5G  assignedExecutors=5  assignedCores=10

    Worker2:coresFree=7   memoryFree=1G    assignedExecutors=0  assignedCores=0

    Worker3:coresFree=3   memoryFree=2G    assignedExecutors=0  assignedCores=0

    totalExecutors:5,totalCores=10

     可以发现,worker1的coresfree已经耗尽了,好可怜。由于application需要12个core,而这里才分配了10个,所以还要继续往下分配:

    Worker1:coresFree=0  memoryFree=7.5G      assignedExecutors=5  assignedCores=10

    Worker2:coresFree=5   memoryFree=512G    assignedExecutors=1  assignedCores=2

    Worker3:coresFree=3   memoryFree=2G        assignedExecutors=0  assignedCores=0

    totalExecutors:6,totalCores=12

     

    ok,最终分配来12个core,满足了application的要求。

     

    对比:

    spreadOut算法中,是以round-robin方式,轮询的在worker节点上分配Executor进程,即以如下序列分配:worker1,worker2... ... worker n,worker1... ....

    非spreadOut算法中,逮者一个worker就不放手,直到满足一下条件之一:

      worker.freeCores<application.coresPerExecutor 或者  worker.memoryFree<application.memoryPerExecutor 。

    在上面两个例子中,虽然最终都分配了6个Executor进程和12个core,但是spreadOut方式下,6个Executor进程分散在不同的worker节点上,充分利用了spark集群的worker节点,而非spreadOut方式下,只在worker1和worker2上分配了Executor进程,并没有充分利用spark worker节点。

     

    小插曲,spreadOut + oneExecutorPerWorker 算法

    spark还有一个叫做”oneExecutorPerWorker“机制,即一个worker上启动一个Executor进程,下面只是简单的说一下得了:

    Worker1:coresFree=8  memoryFree=9.5G  assignedExecutors=1  assignedCores=2

    Worker2:coresFree=7   memoryFree=1G    assignedExecutors=0  assignedCores=0

    Worker3:coresFree=3   memoryFree=2G    assignedExecutors=0  assignedCores=0

    totalExecutors:1,totalCores=2

     

    Worker1:coresFree=8  memoryFree=9.5G      assignedExecutors=1  assignedCores=2

    Worker2:coresFree=5   memoryFree=512M    assignedExecutors=1  assignedCores=2

    Worker3:coresFree=3   memoryFree=2G        assignedExecutors=0  assignedCores=0

    totalExecutors:2,totalCores=4

     

     

    Worker1:coresFree=8  memoryFree=9.5G      assignedExecutors=1  assignedCores=2

    Worker2:coresFree=5   memoryFree=512M    assignedExecutors=1  assignedCores=2

    Worker3:coresFree=1   memoryFree=1.5G     assignedExecutors=1  assignedCores=2

    totalExecutors:3,totalCores=6

    Worker1:coresFree=6  memoryFree=9.0G      assignedExecutors=1  assignedCores=4

    Worker2:coresFree=3   memoryFree=512M     assignedExecutors=1  assignedCores=2

    Worker3:coresFree=1   memoryFree=1.5G     assignedExecutors=1  assignedCores=2

    totalExecutors:3,totalCores=8

     

    Worker1:coresFree=6  memoryFree=9.0G      assignedExecutors=1  assignedCores=4

    Worker2:coresFree=2   memoryFree=0   M     assignedExecutors=1  assignedCores=4

    Worker3:coresFree=1   memoryFree=1.5G     assignedExecutors=1  assignedCores=2

    totalExecutors:3,totalCores=10

     

    Worker1:coresFree=4  memoryFree=9.5G      assignedExecutors=1  assignedCores=6

    Worker2:coresFree=2   memoryFree=0   M     assignedExecutors=1  assignedCores=4

    Worker3:coresFree=1   memoryFree=1.5G     assignedExecutors=1  assignedCores=2

    totalExecutors:3,totalCores=12

     

    和spreadOut+非oneExecutorPerWorker对比发现,唯一的不同就是Executor进程的数量,一个是6,一个是3。

    (

       这里在额外扩展一下,假设application的maxCores=14,而不是12,那么接着上面那个worker列表来:

        Worker1:coresFree=4  memoryFree=9.5G      assignedExecutors=1  assignedCores=6

        Worker2:coresFree=0   memoryFree=0   M     assignedExecutors=1  assignedCores=6

        Worker3:coresFree=1   memoryFree=1.5G     assignedExecutors=1  assignedCores=2

        totalExecutors:3,totalCores=12

        

       虽然worker2.memoryFree=0,但是仍然可以继续在他上面分配core,因为onExecutorPerWorker机制不检查内存的限制。

    )

     

     

     接下来看看源码是怎么实现的:

    了解了上面写的,在阅读源码就很轻易了,这里简单说一下。

    org.apache.spark.deploy.master.Master收到application发送的RegisterApplication(description, driver)消息后,开始执行注册逻辑:

        case RegisterApplication(description, driver) => {
          // TODO Prevent repeated registrations from some driver
    //standby master不调度
    if (state == RecoveryState.STANDBY) { // ignore, don't send response } else { logInfo("Registering app " + description.name) val app = createApplication(description, driver)
    //注册app,即将其加入到waitingApps中 registerApplication(app) logInfo(
    "Registered app " + description.name + " with ID " + app.id)
    //将app加入持久化引擎,主要是为了故障恢复 persistenceEngine.addApplication(app)
    //向driver发送RegisteredApplication消息表明master已经注册了这个app driver.send(RegisteredApplication(app.id, self))
    //为waitingApps中的app调度资源 schedule() } }

    上面的注释已经写的很清楚了... ...

      /**
       * Schedule the currently available resources among waiting apps. This method will be called
       * every time a new app joins or resource availability changes.
       */
      private def schedule(): Unit = {
        if (state != RecoveryState.ALIVE) { return }
        // Drivers take strict precedence over executors
    //为了避免每次schedule,总是在相同的worker上分配资源,所有这里打乱worker顺序。
    val shuffledWorkers = Random.shuffle(workers) // Randomization helps balance drivers
    //下面这个for循环是为driver调度资源,因为这里只将application的调度,所以driver的调度不说了。
    for (worker <- shuffledWorkers if worker.state == WorkerState.ALIVE) { for (driver <- waitingDrivers) { if (worker.memoryFree >= driver.desc.mem && worker.coresFree >= driver.desc.cores) { launchDriver(worker, driver) waitingDrivers -= driver } } }

    //为application调度资源 startExecutorsOnWorkers() }
      /**
       * Schedule and launch executors on workers
       */
      private def startExecutorsOnWorkers(): Unit = {
        // Right now this is a very simple FIFO scheduler. We keep trying to fit in the first app
        // in the queue, then the second app, etc.
    // 为waitingApps中的app调度资源,app.coresLeft是app还有多少core没有分配
    for (app <- waitingApps if app.coresLeft > 0) { val coresPerExecutor: Option[Int] = app.desc.coresPerExecutor // Filter out workers that don't have enough resources to launch an executor
    // 筛选出状态为ALIVE并且这个worker剩余内存,剩余core都大于等于app的要求,然后按照coresFree降序排列
    val usableWorkers = workers.toArray.filter(_.state == WorkerState.ALIVE) .filter(worker => worker.memoryFree >= app.desc.memoryPerExecutorMB && worker.coresFree >= coresPerExecutor.getOrElse(1)) .sortBy(_.coresFree).reverse
    //在usableWorkers上为app分配Executor val assignedCores
    = scheduleExecutorsOnWorkers(app, usableWorkers, spreadOutApps) // Now that we've decided how many cores to allocate on each worker, let's allocate them
    // 在worker上启动Executor进程
    for (pos <- 0 until usableWorkers.length if assignedCores(pos) > 0) { allocateWorkerResourceToExecutors( app, assignedCores(pos), coresPerExecutor, usableWorkers(pos)) } } }

      这个方法做了如下事情:

     1.筛选出可用的worker,即usableWorkers,如果一个worker满足以下所有条件,那么这个worker就被添加到usableWorkers中:

       Alive

       worker.memoryFree >= app.desc.memoryPerExecutorMB

       worker.coresFree >= coresPerExecutor

      2.assignedCores是一个数组,assignedCores[i]里面存储了需要在usableWorkers[i]上分配的core个数,譬如如果assingedCores[1]=2,那么就需要在usableWorkers[1]上分配2个core。

      /**
       * Schedule executors to be launched on the workers.
       * Returns an array containing number of cores assigned to each worker.
       *
       * There are two modes of launching executors. The first attempts to spread out an application's
       * executors on as many workers as possible, while the second does the opposite (i.e. launch them
       * on as few workers as possible). The former is usually better for data locality purposes and is
       * the default.
       *
       * The number of cores assigned to each executor is configurable. When this is explicitly set,
       * multiple executors from the same application may be launched on the same worker if the worker
       * has enough cores and memory. Otherwise, each executor grabs all the cores available on the
       * worker by default, in which case only one executor may be launched on each worker.
       *
       * It is important to allocate coresPerExecutor on each worker at a time (instead of 1 core
       * at a time). Consider the following example: cluster has 4 workers with 16 cores each.
       * User requests 3 executors (spark.cores.max = 48, spark.executor.cores = 16). If 1 core is
       * allocated at a time, 12 cores from each worker would be assigned to each executor.
       * Since 12 < 16, no executors would launch [SPARK-8881].
       */
      private def scheduleExecutorsOnWorkers(
          app: ApplicationInfo,
          usableWorkers: Array[WorkerInfo],
          spreadOutApps: Boolean): Array[Int] = {
        val coresPerExecutor = app.desc.coresPerExecutor
        val minCoresPerExecutor = coresPerExecutor.getOrElse(1)
        val oneExecutorPerWorker = coresPerExecutor.isEmpty
        val memoryPerExecutor = app.desc.memoryPerExecutorMB
        val numUsable = usableWorkers.length
        val assignedCores = new Array[Int](numUsable) // Number of cores to give to each worker
        val assignedExecutors = new Array[Int](numUsable) // Number of new executors on each worker
        var coresToAssign = math.min(app.coresLeft, usableWorkers.map(_.coresFree).sum)
    
        /** Return whether the specified worker can launch an executor for this app. */
    //是否可以在一个worker上分配Executor def canLaunchExecutor(pos: Int): Boolean = { val keepScheduling = coresToAssign >= minCoresPerExecutor val enoughCores = usableWorkers(pos).coresFree - assignedCores(pos) >= minCoresPerExecutor // If we allow multiple executors per worker, then we can always launch new executors. // Otherwise, if there is already an executor on this worker, just give it more cores. val launchingNewExecutor = !oneExecutorPerWorker || assignedExecutors(pos) == 0 if (launchingNewExecutor) {
    //在不里,需要检查worker的空闲core和内存是否够用 val assignedMemory
    = assignedExecutors(pos) * memoryPerExecutor val enoughMemory = usableWorkers(pos).memoryFree - assignedMemory >= memoryPerExecutor val underLimit = assignedExecutors.sum + app.executors.size < app.executorLimit keepScheduling && enoughCores && enoughMemory && underLimit } else { // We're adding cores to an existing executor, so no need // to check memory and executor limits
    //尤其需要注意的是,oneExecutorPerWorker机制下,不检测内存限制,很重要。 keepScheduling && enoughCores } } // Keep launching executors until no more workers can accommodate any // more executors, or if we have reached this application's limits var freeWorkers = (0 until numUsable).filter(canLaunchExecutor) while (freeWorkers.nonEmpty) { freeWorkers.foreach { pos => var keepScheduling = true while (keepScheduling && canLaunchExecutor(pos)) {
    //要分配的cores coresToAssign
    -= minCoresPerExecutor
    //已分配的cores assignedCores(pos)
    += minCoresPerExecutor // If we are launching one executor per worker, then every iteration assigns 1 core // to the executor. Otherwise, every iteration assigns cores to a new executor.
    //一个worker只启动一个Executor if (oneExecutorPerWorker) { assignedExecutors(pos) = 1 } else { assignedExecutors(pos) += 1 } // Spreading out an application means spreading out its executors across as // many workers as possible. If we are not spreading out, then we should keep // scheduling executors on this worker until we use all of its resources. // Otherwise, just move on to the next worker.
    //如果没有开启spreadOUt算法,就一直在一个worker上分配,直到不能再分配为止。
    if (spreadOutApps) { keepScheduling = false } } } freeWorkers = freeWorkers.filter(canLaunchExecutor) } assignedCores }

     

     

     

      /**
       * Allocate a worker's resources to one or more executors.
       * @param app the info of the application which the executors belong to
       * @param assignedCores number of cores on this worker for this application
       * @param coresPerExecutor number of cores per executor
       * @param worker the worker info
       */
      private def allocateWorkerResourceToExecutors(
          app: ApplicationInfo,
          assignedCores: Int,
          coresPerExecutor: Option[Int],
          worker: WorkerInfo): Unit = {
        // If the number of cores per executor is specified, we divide the cores assigned
        // to this worker evenly among the executors with no remainder.
        // Otherwise, we launch a single executor that grabs all the assignedCores on this worker.
    //计算要创建多少个Executor进程,默认值是1.

    val numExecutors = coresPerExecutor.map { assignedCores / _ }.getOrElse(1)
        val coresToAssign = coresPerExecutor.getOrElse(assignedCores)
        for (i <- 1 to numExecutors) {
          val exec = app.addExecutor(worker, coresToAssign)
    //真正的启动Executor进程了。 launchExecutor(worker, exec) app.state
    = ApplicationState.RUNNING } }

    由于本人接触spark时间不长,如有错误或者任何意见可以在留言或者发送邮件到franciswbs@163.com,让我们一起交流。

     

    作者:FrancisWang 

    邮箱:franciswbs@163.com
    出处:http://www.cnblogs.com/francisYoung/
    本文地址:http://www.cnblogs.com/francisYoung/
    本文版权归作者和博客园共有,欢迎转载,但未经作者同意必须保留此段声明,且在文章页面明显位置给出原文连接,否则保留追究法律责任的权利。

  • 相关阅读:
    java注解说明
    paypal
    eclispe查看jdk源码后特别卡顿导致未响应解决
    ubuntu+tomcat,多环境、自动化部署脚本,git+maven+tomcat+ubuntu
    ubuntu+let's encrypt生成永久免费https证书 ubuntu+tomcat+nginx+let's encrypt
    MySQL创建数据库与创建用户以及授权
    linux查找并杀死进程shell
    redmine安装笔记
    C#动态获取本机可用串口的两种方式
    C# 控件缩写规范
  • 原文地址:https://www.cnblogs.com/francisYoung/p/5205420.html
Copyright © 2011-2022 走看看