Spark Application的调度算法
来自： http://www.cnblogs.com/francisYoung/p/5205420.html
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 为 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/
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