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]
image

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
image

2、增大本次计算到的IAT的概率值。

image

  • 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:

image

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)

image

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)

image

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;
}

image

posted @ 2023-02-16 17:31  靑い空゛  阅读(349)  评论(0编辑  收藏  举报