Nacos12# 随机权重负载均衡算法

瓜农老梁

发布于 2021-09-02 15:16:03

7030

发布于 2021-09-02 15:16:03

文章被收录于专栏：瓜农老梁瓜农老梁

引言

Nacos在Client选择节点时提供了一种基于权重的随机算法，通过源码分析掌握其实现原理，方便实战中加以运用。

一、内容提要

下面以图示的方式贯穿下随机权重负载均衡算法的流程：

节点列表

假设注册了5个节点，每个节点的权重如下。

组织递增数组

目的在于形成weights数组，该数组元素取值[0~1]范围，元素逐个递增，计算过程如下图示。另外注意非健康节点或者权重小于等于0的不会被选择。

随机算法

通过生成[0~1]范围的随机数，通过二分法查找递增数组weights[]接近的index，再从注册节点列表中返回节点。

二、源码分析

随机权重负载均衡算法是在NacosNamingService#selectOneHealthyInstance提供，一起走查下。

@Override
public Instance selectOneHealthyInstance(String serviceName, String groupName, boolean subscribe)
  throws NacosException {
  return selectOneHealthyInstance(serviceName, groupName, new ArrayList<String>(), subscribe);
}

@Override
public Instance selectOneHealthyInstance(String serviceName, String groupName, List<String> clusters,
                                         boolean subscribe) throws NacosException {
  String clusterString = StringUtils.join(clusters, ",");
  // 注解@1
  if (subscribe) {
    ServiceInfo serviceInfo = serviceInfoHolder.getServiceInfo(serviceName, groupName, clusterString);
    if (null == serviceInfo) {
      serviceInfo = clientProxy.subscribe(serviceName, groupName, clusterString);
    }
    return Balancer.RandomByWeight.selectHost(serviceInfo);
  } else {
    // 注解@2
    ServiceInfo serviceInfo = clientProxy
      .queryInstancesOfService(serviceName, groupName, clusterString, 0, false);
    return Balancer.RandomByWeight.selectHost(serviceInfo);
  }
}

注解@1 已订阅「从缓存获取注册节点列表」，默认subscribe为true。

注解@2 从「从服务器获取注册节点列表」

protected static Instance getHostByRandomWeight(List<Instance> hosts) {
  NAMING_LOGGER.debug("entry randomWithWeight");
  if (hosts == null || hosts.size() == 0) {
    NAMING_LOGGER.debug("hosts == null || hosts.size() == 0");
    return null;
  }
  NAMING_LOGGER.debug("new Chooser");
  List<Pair<Instance>> hostsWithWeight = new ArrayList<Pair<Instance>>();
  for (Instance host : hosts) {
    if (host.isHealthy()) {  // 注解@3
      hostsWithWeight.add(new Pair<Instance>(host, host.getWeight()));
    }
  }
  NAMING_LOGGER.debug("for (Host host : hosts)");
  Chooser<String, Instance> vipChooser = new Chooser<String, Instance>("www.taobao.com");
  // 注解@4
  vipChooser.refresh(hostsWithWeight);
  NAMING_LOGGER.debug("vipChooser.refresh");
  // 注解@5
  return vipChooser.randomWithWeight();
}

注解@3 非健康节点不会被选中，组装Pair的列表，包含健康节点的权重和Host信息

注解@4 刷新需要的数据，具体包括三部分：所有健康节点权重求和、计算每个健康节点权重占比、组织递增数组。

public void refresh() {
    Double originWeightSum = (double) 0;
    // 注解@4.1
    for (Pair<T> item : itemsWithWeight) {

        double weight = item.weight();
        // ignore item which weight is zero.see test_randomWithWeight_weight0 in ChooserTest
        // weight小于等于 0的将会剔除
        if (weight <= 0) {
            continue;
        }

        items.add(item.item());

        // 值如果无穷大
        if (Double.isInfinite(weight)) {
            weight = 10000.0D;
        }

        // 值如果为非数字值
        if (Double.isNaN(weight)) {
            weight = 1.0D;
        }

        // 累加权重总和
        originWeightSum += weight;
    }

    // 注解@4.2
    double[] exactWeights = new double[items.size()];
    int index = 0;
    for (Pair<T> item : itemsWithWeight) {
        double singleWeight = item.weight();
        //ignore item which weight is zero.see test_randomWithWeight_weight0 in ChooserTest
        if (singleWeight <= 0) {
            continue;
        }
        // 每个节点权重的占比
        exactWeights[index++] = singleWeight / originWeightSum;
    }

    // 注解@4.3
    weights = new double[items.size()];
    double randomRange = 0D;
    for (int i = 0; i < index; i++) {
        weights[i] = randomRange + exactWeights[i];
        randomRange += exactWeights[i];
    }
  
    double doublePrecisionDelta = 0.0001;

    if (index == 0 || (Math.abs(weights[index - 1] - 1) < doublePrecisionDelta)) {
        return;
    }
    throw new IllegalStateException(
            "Cumulative Weight caculate wrong , the sum of probabilities does not equals 1.");
}

注解@4.1 所有健康节点权重求和originWeightSum

注解@4.2 计算每个健康节点权重占比exactWeights数组

注解@4.3 组织递增数组weights，每个元素值为数组前面元素之和

以一个例子来表示这个过程，假设有5个节点：

1.2.3.4 100
1.2.3.5 100
1.2.3.6	100
1.2.3.7	80
1.2.3.8	60

步骤一计算节点权重求和

originWeightSum = 100 + 100 + 100 + 80 + 60 = 440

步骤二计算每个节点权重占比

exactWeights[0] = 0.2272

exactWeights[1] = 0.2272

exactWeights[2] = 0.2272

exactWeights[3] = 0.1818

exactWeights[4] = 0.1363

步骤三组织递增数组weights

weights[0] = 0.2272

weights[1] = 0.4544

weights[2] = 0.6816

weights[3] = 0.8634

weights[4] = 1

注解@5 随机选取一个，逻辑如下：

public T randomWithWeight() {
    Ref<T> ref = this.ref;
    // 注解@5.1
    double random = ThreadLocalRandom.current().nextDouble(0, 1);
    // 注解@5.2
    int index = Arrays.binarySearch(ref.weights, random);
    // 注解@5.3
    if (index < 0) {
        index = -index - 1;
    } else {
        // 注解@5.4
        return ref.items.get(index);
    }

    // 返回选中的元素
    if (index >= 0 && index < ref.weights.length) {
        if (random < ref.weights[index]) {
            return ref.items.get(index);
        }
    }

    /* This should never happen, but it ensures we will return a correct
     * object in case there is some floating point inequality problem
     * wrt the cumulative probabilities. */
    return ref.items.get(ref.items.size() - 1);
}

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原始发表：2021-08-04，如有侵权请联系 cloudcommunity@tencent.com 删除

负载均衡