I'm hoping here that this post isn't out of line with the scope of the NANOG list, of which I've been a long time lurker. If so, please just ignore me. We're trying to calculate Jitter of a variable (non-limited) size data set. One Jitter formula that we see cited occasionally (and is in RFC 1889 - I believe iPerf uses this formula for it's Jitter #'s) looks something like this: J = J+(|D(i-1,i)|-J)/16 The problem with this formula is that it works best on small sample sets, and it also favors more recent samples. As the sample size grows, the jitter of early samples seem to get factored down to basic "noise", and then aren't really well represented in the overall Jitter number. We're trying to find a viable formula for showing a general Jitter "average" over a period of time. One possibility here is just to iterate all samples like this: Jsum = Jsum+|D(i-1,i)| and then calculating the jitter like this: J = Jsum / (sample count - 1) The sample count could be anywhere from 2 to 1 million (or more). This formula does seem to represent early sample in the "Jitter" number just as strongly as later samples, but seems like it might be a bit simplistic. Does anyone have any feedback on this alternate way of calculating Jitter, or any better ways to do this? Thanks in advance for any input. Jeff Murri Nessoft, LLC jeff@nessoft.com www.nessoft.com
On Fri, 10 Jun 2005 03:53:09 -0600 Jeff Murri <jeff@nessoft.com> wrote:
I'm hoping here that this post isn't out of line with the scope of the NANOG list, of which I've been a long time lurker. If so, please just ignore me.
Hello Jeff; These are both moving averages, the question is the memory of the moving average. The RTP version has a specific finite memory, the one you describe has an infinite memory. The statistical trouble with infinite memory moving average estimates is that they eventually converge to a fixed value (after one million samples, say, even a large change in jitter will take a long time to produce a small change in the average), and, if the underlying process is not stationary, then they need not converge to anything like the correct current value. The RTCP protocol has a finite memory estimator that has have enough memory to smooth out statistical fluctuations somewhat, but which responds to real changes in the underlying jitter fairly rapidly, and which is comparable across implementations. Yes, this means that older data is ignored (that's the finite memory part), but its intended use is to try and estimate what's happening in the network now, not last week. (I have had RTCP sessions up for months; even a solid day of high jitter would hardly budge a total average over months.) If what you want is the jitter averaged over some long period of time (say so you can say that the average jitter on your network was X msec in 2005), then what you want is indeed Jsum = Jsum+|D(i-1,i)| J = Jsum / (sample count - 1) (assuming that the sample count is the number of delay measurements, not the number of delay differences). Note that that is the same as J[i] = J[i-1] * (i-1 /i) + |D(i-1,i)| * (1/i) assuming that the first sample is i = 0 and J[0] is finite; this shows clearly how new data gets down-weighted as time goes on and i increases. Regards Marshall
We're trying to calculate Jitter of a variable (non-limited) size data set. One Jitter formula that we see cited occasionally (and is in RFC 1889 - I believe iPerf uses this formula for it's Jitter #'s) looks something like this:
J = J+(|D(i-1,i)|-J)/16
The problem with this formula is that it works best on small sample sets, and it also favors more recent samples. As the sample size grows, the jitter of early samples seem to get factored down to basic "noise", and then aren't really well represented in the overall Jitter number.
We're trying to find a viable formula for showing a general Jitter "average" over a period of time. One possibility here is just to iterate all samples like this:
Jsum = Jsum+|D(i-1,i)|
and then calculating the jitter like this:
J = Jsum / (sample count - 1)
The sample count could be anywhere from 2 to 1 million (or more). This formula does seem to represent early sample in the "Jitter" number just as strongly as later samples, but seems like it might be a bit simplistic.
Does anyone have any feedback on this alternate way of calculating Jitter, or any better ways to do this?
Thanks in advance for any input.
Jeff Murri Nessoft, LLC jeff@nessoft.com www.nessoft.com
you saw marshall's comment. If you're interested in a moving average, he's pretty close. If I understood your question, though, you simply wanted to quantify the jitter in a set of samples. I should think there are two obvious definitions there. A statistician would look, I should think, at the variance of the set. Reaching for my CRC book of standard math formulae and tables, it defines the variance as the square of the standard deviation of the set, which is to say sum of ((x(i) - xmean)^2) ------------------------ n - 1 where the n values x(i) are the members of the set, xmean is the mean of those values, and n is the number of x(i). A sample set with a larger standard deviation or variance than another set has contains more jitter. In this context, the other thought that comes to mind is the variation from nominal. If the speed-of-light delay between here and there is M, the jitter might be defined as the root-mean-square difference from M, which would be something like sum of ((x(i) - xmin)^2) ----------------------- n - 1 with the same variables except that xmin is the least value in the set.
At 09:56 AM 6/10/2005, Fred Baker wrote:
you saw marshall's comment. If you're interested in a moving average, he's pretty close.
If I understood your question, though, you simply wanted to quantify the jitter in a set of samples. I should think there are two obvious definitions there.
A statistician would look, I should think, at the variance of the set. Reaching for my CRC book of standard math formulae and tables, it defines the variance as the square of the standard deviation of the set, which is to say
That is one thing I have never understood, if you can pretty much just look at a standard dev and see it is high, and yeah that means your numbers are flopping all over the place, then what good is the square of it? Does it just make graphing better in some way? Thanks, Eric
On Fri, 10 Jun 2005 10:07:06 -0700 Eric Frazier <eric@dmcontact.com> wrote:
At 09:56 AM 6/10/2005, Fred Baker wrote:
you saw marshall's comment. If you're interested in a moving average, he's pretty close.
If I understood your question, though, you simply wanted to quantify the jitter in a set of samples. I should think there are two obvious definitions there.
A statistician would look, I should think, at the variance of the set. Reaching for my CRC book of standard math formulae and tables, it defines the variance as the square of the standard deviation of the set, which is to say
That is one thing I have never understood, if you can pretty much just look at a standard dev and see it is high, and yeah that means your numbers are flopping all over the place, then what good is the square of it? Does it just make graphing better in some way?
Hello Eric; <statistics details> Because (under some broad assumptions, primarily that the underlying process is stationairy) estimates of the variance are distributed as a CHI**2 distribution. More exactly, summation( (x[i] - mean(x))^2) / true_variance is distributed as CHI**2(N), which means that as i increases, then summation( (x[i] - mean(x))^2) / true_variance is distributed as a normal distribution with a mean of N and a variance (of the variance estimate) of 2N, so that V = summation( (x[i] - mean(x))^2) / N is an efficient estimate of the true variance, with a sigma (of the variance estimate) of sqrt (2 / N) * V (Since you have to estimate the mean from the same data, you can show that the estimator is less biased if you use 1 / N - 1 rather than 1 / N in actual calculations.) Basically, if you want to perform the true rites in the Church of Linear Statistics, you worry about variances and CHI**2 distributions. If (like most of us in the real world) you are dealing with non-stationary processes and unknown distributions, you can ignore this, just calculate the standard deviation, see whether or not things differ by more than 3 standard deviations, and be done with it. </statistics details> Note (from the days when spacecraft had 4 kilobytes of memory) that if you estimate s[i] = s[i-1] + x[i] v[i] = v[i-1] + x[i]^2 then the mean estimate at any time is m[i] = s[i] / i the total variance is V[i] = v[i] / i and the standard deviation for i > 1 is sigma[i] = sqrt[(v[i] - (i * m[i]^2))/(i-1)] so, you can do this on the fly without storing all of the data. Regards Marshall
Thanks,
Eric
On June 10, 2005 10:07 am, Eric Frazier wrote:
At 09:56 AM 6/10/2005, Fred Baker wrote:
you saw marshall's comment. If you're interested in a moving average, he's pretty close.
If I understood your question, though, you simply wanted to quantify the jitter in a set of samples. I should think there are two obvious definitions there.
A statistician would look, I should think, at the variance of the set. Reaching for my CRC book of standard math formulae and tables, it defines the variance as the square of the standard deviation of the set, which is to say
That is one thing I have never understood, if you can pretty much just look at a standard dev and see it is high, and yeah that means your numbers are flopping all over the place, then what good is the square of it? Does it just make graphing better in some way?
Pedantic mode on. Jitter != Variance Variance is how spread out a set is (hence the squaring). [it has further subclasses at sample variance and population variance] To try to explain it a little more simply than CHI-square distributions and formulas.... (Mark Twain said, why use a 25 cent word when a nickel one will do... :-) The squaring lets you get an indication of the range with added bias to wildly differing samples so they won't be drowned out by the large number of samples close to the mean. Sample variance is the sum of the squares of the differences from the mean divided by number of samples. It is a numeric value. But... Jitter is the measure of deviation from a predetermined constant clock rate. It can also be the phase deviation of a signal. You can think of it as clock skew or drift or the uncertainty level in timing. Jitter is normally expressed as a plus minus range in either peak, mean, or rms. They are similar but not the same thing. Pedantic mode off... cheers, --dr -- World Security Pros. Cutting Edge Training, Tools, and Techniques Tokyo, Japan November 14/15 2005 http://pacsec.jp/core05 pgpkey http://dragos.com/ kyxpgp
participants (5)
-
Dragos Ruiu -
Eric Frazier -
Fred Baker -
Jeff Murri -
Marshall Eubanks