Prometheus 如何求普罗米修斯速率的分位数

我正在看这个

我不明白为什么

histogram_quantile(0.9, 
    rate(prometheus_http_request_duration_seconds_bucket[5m])
)

不使用单位观察事件/秒给出速率的分位数，而是使用单位秒/观察事件给出请求持续时间的分位数

应给出5分钟内特定桶/秒平均观察事件数

我可以想象，分位数会给你速率分位数

我一定是理解错误了

此处的速率函数用于指定分位数计算的时间窗口，如中所示。翻译过来就是在过去的5分钟内，我的90%用户所经历的最大http响应时间是多少

直方图_分位数函数通过假设桶内的线性分布来插值分位数值，le给出最大观察时间。桶是一个计数器，用于测量自过程开始以来发生的观察次数。rate通过计算平均每秒发生的观测次数来建立链接，从中可以插值整个时间窗口内的平均响应时间

你是对的，这不是一个100%准确的测量，因为平均值，但函数正在做很多假设，桶的选择已经引入了偏差

我想你可以用irate来计算瞬时分位数，但很可能它会更嘈杂。

这是普罗米修斯的历史分位数代码

举个例子,

assumed the original bucket is :
[50][100][150][200][200] with corresponding upperbound 5s,10s,15s,20s,+Inf.

then the rate(xx[5m]) returned a bucket like this:
[20/5*60][40/5*60][60/5*60][80/5*60][80/5*60]

histogram_quantile will delegate the returned bucket to another function bucketQuantile.
It used the rough following logic to compute the percentile: 

1) get the total rank of the percentile 
such as 90ile is 0.9 * total counts = 0.9 * (80/5*60)
2) compute the value of 90ile
last upperbound before the total rank position is 15 secs;
current upperbound of the total rank is 20 secs;
the count in the bucket that 90ile position belongs is (80/5*60)-(60/5*60);
the internal rank in that single bucket of 90ile position is (0.9 * 80/5*60)-(60/5*60);
finally, the value of 90ile is: 15 sec + (internal rank / that bucket count) * (20sec-15sec) = 15 + 3 * ( (0.9 * 80/5*60)-(60/5*60) / (80/5*60)-(60/5*60) ) = 
15 + 3 * ( (0.9*80 - 60)/(80-60) ) = 15 + 3 * ( 12/20) = 15+3*0.6= 16.8 sec

就是这样，你可以看到分母5*60在计算中实际上没有影响。因此，rate func只是用来指定5分钟的时间窗口

rate(prometheus_http_request_duration_seconds_bucket[5m]

assumed the original bucket is :
[50][100][150][200][200] with corresponding upperbound 5s,10s,15s,20s,+Inf.

then the rate(xx[5m]) returned a bucket like this:
[20/5*60][40/5*60][60/5*60][80/5*60][80/5*60]

histogram_quantile will delegate the returned bucket to another function bucketQuantile.
It used the rough following logic to compute the percentile: 

1) get the total rank of the percentile 
such as 90ile is 0.9 * total counts = 0.9 * (80/5*60)
2) compute the value of 90ile
last upperbound before the total rank position is 15 secs;
current upperbound of the total rank is 20 secs;
the count in the bucket that 90ile position belongs is (80/5*60)-(60/5*60);
the internal rank in that single bucket of 90ile position is (0.9 * 80/5*60)-(60/5*60);
finally, the value of 90ile is: 15 sec + (internal rank / that bucket count) * (20sec-15sec) = 15 + 3 * ( (0.9 * 80/5*60)-(60/5*60) / (80/5*60)-(60/5*60) ) = 
15 + 3 * ( (0.9*80 - 60)/(80-60) ) = 15 + 3 * ( 12/20) = 15+3*0.6= 16.8 sec