rate vs increase functions#
In an ideal world (where your samples’ timestamps are exactly on the second and your rule evaluation happens exactly on the second) rate(counter[1s]) would return exactly your ICH value and rate(counter[5s]) would return the average of that ICH and the previous 4. Except the ICH at second 1 is 0, not 1, because no one knows when your counter was zero: maybe it incremented right there, maybe it got incremented yesterday, and stayed at 1 since then. (This is the reason why you won’t see an increase the first time a counter appears with a value of 1 – because your code just created and incremented it.)
increase(counter[5s]) is exactly rate(counter[5s]) * 5 (and increase(counter[2s]) is exactly rate(counter[2s]) * 2).
Now what happens in the real world is that your samples are not collected exactly every second on the second and rule evaluation doesn’t happen exactly on the second either. So if you have a bunch of samples that are (more or less) 1 second apart and you use Prometheus’ rate(counter[1s]), you’ll get no output. That’s because what Prometheus does is it takes all the samples in the 1 second range [now() - 1s, now()] (which would be a single sample in the vast majority of cases), tries to compute a rate and fails.
If you query rate(counter[5s]) OTOH, Prometheus will pick all the samples in the range [now() - 5s, now] (5 samples, covering approximately 4 seconds on average, say [t1, v1], [t2, v2], [t3, v3], [t4, v4], [t5, v5]) and (assuming your counter doesn’t reset within the interval) will return (v5 - v1) / (t5 - t1). I.e. it actually computes the rate of increase over ~4s rather than 5s.
increase(counter[5s]) will return (v5 - v1) / (t5 - t1) * 5, so the rate of increase over ~4 seconds, extrapolated to 5 seconds.
Due to the samples not being exactly spaced, both rate and increase will often return floating point values for integer counters (which makes obvious sense for rate, but not so much for increase).