I have an Efergy home energy monitor that sends a packet to the cloud every 10 seconds, from which a graph of electricity use can be plotted. The download (Report) shows usage good to one minute resolution.

It's real cool - but it's also problematic because there's no packet loss checks whatsoever. Packet values are simply added, period. It will not even trigger anything if packets completely stop. It just shows up as zero usage. If you're at the edge of reception, you could be losing a lot of packets and not know it.

The only sure indication of a problem is if no packets whatsoever make it in a given minute (all six are lost), i.e., a zero usage reading. This is because an "empty" minute reading is not ordinarily possible. It reports down to 6 W; most modern houses use dozens of watts even when very quiet.

To make a long story short, I want to say something simple and eye-opening to Efergy. Because I'm sure they're not expecting a stats-based enhancement request (lol). Like "did you know you can consistently lose over half your data before it even begins registering as one 'empty' minute per hour?" (rhetorical example!)

So I guess I am looking for the rate expected to produce six failures in a row out of 360 (60 minutes $\times$ 6 packets/minute), on average - but only ONE set of six in a row, in the 360.

Be sure to think of it in the overall context (the hour). Which is to say, your approach should also be able to easily say what packet loss rate would produce, say, 5 lost minutes in an hour, as well. (They don't need to be contiguous minutes.) Or even, one lost minute per day. And, of course, 100% lost packets should result in 60 lost minutes per hour.

For extra credit you can discuss minute boundaries. The Efergy report states usage by time of day down to the minute. Six fails in a row would not show up on the Efergy as an empty minute if the six failures started in the middle of a minute. Does this mean we need to talk about up to 9(?) fails in a row... or does this sort of cancel out because we're talking about averages anyway? I suspect what we need here is actually 7.5 packets (the average of the 6 to 9 that would be expected to produce one empty minute).

FWIW, I'm sure internet lag can affect this at some level. But let's keep this simple for now - I'm just trying to make a simple point to the Efergy people.

Thanks if you can help!

P.S. My first post! I joined Stack Exchange to ask this.


1 Answer 1


The fundamental problem is that single packet losses cannot be assumed to be independent in a probabilistic model. There is a nontrivial correlation due to the fact that, if a packet is lost, it is more likely that subsequent packets will also be lost. If you assume independence to make the model simple, it is very likely you will drastically underestimate the failure probability.

Consequently, we would need to impose certain distributional assumptions if we were to attempt to fit a time series to this process, which is how a statistician would probably begin to approach your question.

Furthermore, your question is not posed in a way that is mathematically precise. After reading your entire post, I still am not entirely clear on the following:

  1. How is packet loss detected;
  2. The most likely failure modes of packet transmission;
  3. The relationship between packet loss in a given time interval and assignment of "failure";
  4. What specific statistical inference you are wishing to make in regard to the incidence of "failure."
  • $\begingroup$ Thanks for responding, Heropup! 1) Packet loss isn't detected (that's the problem) but you can confirm loss by turning off transmission briefly. Otherwise steady usage will appear reduced by however long it's off. 2) Most likely mode is being near the edge of transmission capability, I guess. 3) Strictly speaking, this is not well known. It also touches on the idea of internet lag. But if they don't even know packets are missing, I doubt they are doing much that's sophisticated. 4) Please see the kind of statement I would like to make, in the OP. (Should I be editing the OP or replying here?) $\endgroup$ Sep 24, 2015 at 17:40

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