extract data points from moving average

I'm wondering if anyone can help me write a formula that will extract individual data points from a rolling average.

I have data which represents a 30-day rolling average that I can get to every day. I'd like to figure out what _one_day's_ value is, if I can. It seems that if I know the rolling average every day I ought to be able to figure out the data that went into it?

Example: My ISP gives me a daily usage report that represents my 30 day rolling average of downloaded megabites usage. I have years of data. Can I look at one 30-day period and the adjacent one and by looking at what falls off one end and adds on to the other figure out what the daily usage was for that day?


Thanks for any help you can provide.

if you know at some point all 30 values then it is easy
last one went out and new first one went in so the new one was
new first = 30*(new moving average - old moving average) + old last

otherwise - I think not.

I do think mathematics also says "NO":

a1 = (x1+x2+...+x29+x30)/30
a2 = (x2+x3+...+x30+x31)/30
a3 = (x3+x4+...+x31+x32)/30
...
you have more and more equations, more and more known values (ak) but with each one you also get one more unknown value

Sorry, but in my opinion - no way.

As you say, RheaLocci, if there is an answer to the question, we need to work it out mathematically before we can even begin to talk about putting it into Excel.

In the spirit of brainstorming ideas (Kaper, comment on this possibility): what if we start combining averages:

a1-a2=(x1+...x30)/30-(x2+...+x31)/30=x1/30-x31/30 one equation in two unkowns still no unique solution. But it seems that, if we have enough averages, the number of combinations will increase quickly, and we may end up with enough unique equations to solve for some of the data points.

If I did this right, with 4 averages, we get a system of 6 equations in 6 unkowns, which we should be able to solve. Extend that to enough combinations of moving averages, and we may be able to eventually have enough information to solve for each data point. Did I do that right?

Obviously, It is a kind of challenge. So let's play a bit longer.
a1-a2=(x1+...x30)/30-(x2+...+x31)/30=x1/30-x31/30
a2-a3=(x2+...+x31)/30-(x3+...+x32)/30=x2/30-x32/30
2 equations 4 unknown
one could say that we can have 3 equations and 4 unknown if we take:
a1-a3=(x1+...x30)/30-(x3+...+x32)/30=x1/30+x2/30-x31/30-x32/30
Promissing? Not really, because it is just linear combination of the two above, so it will shorten back to 2 equations, 4 unknown

Of course, there is a chance that I'm simply to sceptical.
If it shows true, I can handle, so if somebody knows how-to, I will welcome the news.

Promissing? Not really, because it is just linear combination of the two above, so it will shorten back to 2 equations, 4 unknown

You're correct. I knew I had to be missing something -- it seemed too easy.