webrtc QOS笔记一 Neteq直方图算法浅读

webrtc QOS笔记一 Neteq直方图算法浅读

想起博客园帐号了，回来填点webrtc qos的坑，本文分析个很好用的直方图算法，不仅可以在音频里面计算抖动延迟，我发现用来统计丢包率也很好用.

Histogram Algorithm

DelayManager::Update()->Histogram::Add() 会根据计算的iat_packet(inter arrival times, =实际包间间隔 / 打包时长)，将该iat_packet插入IATVector直方图对应数组下标内。并更新该直方图的数据下标下概率参数。[M88 SRC]

一共有四步操作：

1、用遗忘因子，对历史数据的出现概率进行遗忘, 并统计概率合

2、增大本次计算到的IAT的概率值。

例：

假如历史bucket 数据为:

buckets_ = {0，0，1，0}

遗忘因子为 0.9:

forget_factor = 0.9

新来的抖动延迟数据为66ms， 桶间为20ms一个单位, 那插入位置为 66 / 20 = 3，则更新后

buckets = {0，0，0.9，0.1}

假若使用%95分位的值作为目标延迟， 则更新后的目标延迟为 60ms.

3、调整本次计算到的IAT的概率，使整个IAT的概率分布之和近似为1。调整方式为假设当前概率分布之和为tempSum，则：

4、更新forget_factor_, 使遗忘因子forget_factor_逼近base_forget_factor_

a.使用start_forget_weight_更新(默认初始值start_forget_weight_ = 2，base_forget_factor_=0.9993)

获取目标延迟

依据probability获取此百分位的值作为目标延迟(初始值0.97)

int Histogram::Quantile(int probability) {

  // Find the bucket for which the probability of observing an

  // inter-arrival time larger than or equal to |index| is larger than or

  // equal to |probability|. The sought probability is estimated using

  // the histogram as the reverse cumulant PDF, i.e., the sum of elements from

  // the end up until |index|. Now, since the sum of all elements is 1

  // (in Q30) by definition, and since the solution is often a low value for

  // |iat_index|, it is more efficient to start with |sum| = 1 and subtract

  // elements from the start of the histogram.

  int inverse_probability = (1 << 30) - probability;

  size_t index = 0;        // Start from the beginning of |buckets_|.

  int sum = 1 << 30;       // Assign to 1 in Q30.

  sum -= buckets_[index];

  while ((sum > inverse_probability) && (index < buckets_.size() - 1)) {

    // Subtract the probabilities one by one until the sum is no longer greater

    // than |inverse_probability|.

    ++index;

    sum -= buckets_[index];

  }

  return static_cast<int>(index);

}

遗忘因子曲线

测试曲线,调整遗忘因子提高抖动估计灵敏度:

#include <iostream>

#include <cstdint>

#include <vector>

uint32_t packet_loss_rate_ = 0;

int main()

{

  std::vector<int> input;

  std::vector<float> buckets;

  float forget_factor = 0.9993;

  float val = 0;

  for (size_t k = 0; k < 1000; k ++) {

    val = val * forget_factor + (1-forget_factor);

    buckets.push_back(val);

  }

  for (int i = 0; i < 1000; ++i) {

    std::cout << buckets[i]<< " ";

  }

  return 0;

}