# The Response Time Stretch Factor

Posted in Math, Scalability on Oct 30, 2016

Computer systems, and for that matter all types of systems that receive requests and process them, have a response time that includes some time waiting in queue if the server is busy when a request arrives. The wait time increases sharply as the server gets busier. For simple M/M/m systems there is a simple equation that describes this exactly, but for more complicated systems this equation is only approximate. This has rattled around in my brain for a long time, and rather than keeping my notes private I’m sharing them here (although since I’m still trying to learn this stuff I may just be putting my ignorance on full display).

I’ll skip all the derivations and go through only the basics to get to the results.

The equation describing the curve above is derived as follows. When a request arrives at the server, it will have to wait time \(S\), the service time, to be completed. But if the server is busy, which happens with some probability, then it will be delayed some additional wait time \(W\) for the request in process, and potentially for other requests are already waiting, before it enters service. The total time is the residence time, \(R = W + S\).

The wait time \(W\) is some fraction of the total residence time, which can be at most 100%, and if this is denoted by \(\rho\) then \(W = \rho R\). Thus,

$$ R = W + S = \rho R + S $$

Which, when rearranged and divided by service time to make it a relative “stretch factor,” becomes

$$ R = \frac{1}{1- \rho} $$

None of this is original; I’ve basically cribbed this from Neil Gunther’s book
*Analyzing Computer Systems Performance with Perl::PDQ*. Later in that same
book, Gunther shows the derivation of a related formula for the stretch factor
in a multiserver queue with \(m\) servers,

$$ R \approx \frac{1}{1-\rho^m} $$

Where \(\rho\) denotes server utilization. This heuristic approximation does arise analytically, but is not exact when there are more than 2 servers. It underestimates wait time when utilization is high, especially with large numbers of servers (say, 64). The exact solution is given by

$$ R = \frac{C(m, \rho) S}{m(1-\rho)} + S $$

Where the first term is just \(W\), and the function \(C(m,\rho)\) is the Erlang C function. If we put it all together and rearrange into “stretch factor” form, the Erlang equation for stretch factor is

$$ R(m, \rho) = 1 + \frac{ \frac{(m \rho)^m}{m!} }{ (1-\rho) \sum_{n=0}^{m-1} \frac{(m \rho)^n}{n!} + \frac{(m \rho)^m}{m!} } \frac{1}{m(1-\rho)} $$

If you’d like to see how this looks, check out this Desmos calculator.

Now here’s the part my brain has been trying to connect, almost idly in shower-thought time, for a while. If intuition led from one direction to the heuristic approximation \(R=\frac{1}{1-\rho^m}\), and an exact derivation leads to a formula that in the single- and dual-server case is the same as the heuristic, then what is the heuristic missing to extend to an exact multiserver queueing system? Can it be extended, or is it just a coincidence that it’s an exact solution for \(m=1, m=2\)?

A few trains of thought have sprung into my mind.

First, I observed that the Erlang C formula includes \(m!\), and it happens to be the case that 1! = 1, and 2! = 2. A clue, or a distraction? What if I add a term to the heuristic, multiplied by \(m/m!\) or similar, which would just reduce to the same thing for the single- and dual-server cases? What form would that term take?

Secondly, what if I work backwards from the Erlang formula and see if it reduces to a different form of the heuristic?

Thirdly, what if the heuristic’s exactness for the base cases is just a coincidence after all? In that case, perhaps a new, simpler approximation to the Erlang formula is waiting to be invented. Approximations are highly useful to me in tools such as spreadsheets and the like, or even in using intuition, which is workable with the heuristic form. Approximations are easy to think about and a lot easier to type and troubleshoot. They’re also faster to compute, should performance matter.

I’ll take each of these cases in turn.

If the heuristic is exact for 1- and 2-server cases because of the coincidence that the factorial function for these values is the value itself, then what is the missing term? What shape might it have?

To gain some intuition about this, I wrote a simple Desmos
calculator to show
the *value* of the hypothetical “missing term” in the context of the heuristic’s
value. In other words, the heuristic’s *error function*. Here’s a graph of
that for several values of \(m\). Red is 1 server, green is 8, purple is 16, orange
is 64.

Note that I showed utilization extending out beyond the value 1 because the shape of the function is interesting, but that value is impossible—a server can never be more than 100% utilized. What’s interesting, to me anyway, is that the error function looks kind of like the PDF of a Gamma distribution. I’ll leave that thought there.

I’m not sure what form a term including, say, \(m/m!\) would take. This is
something I haven’t explored a lot yet. *TODO*.

What does the Erlang formula reduce to in the \(m=1\) case? The Erlang C formula itself reduces to \(\rho\), and when decorated with the additional stuff to get it into stretch-factor form, it reduces to

$$ R(1, \rho) = 1 + \frac{\rho}{1-\rho} $$

Which is just a rearrangement of the heuristic function. (Interestingly, Wolfram Alpha will simplify it to include the Gamma function and list the heuristic as an approximation. Gamma again, though Gamma function and Gamma distribution are different. The Gamma function is closely related to the factorial function.)

In the \(m=2\) case, if I’m doing my algebra right, the Erlang C function simplifies to

$$ C(2, \rho) = \frac{2\rho^2}{ (1-\rho) + (1-\rho)2\rho + 2\rho^2} $$

Which further simplifies to \(\frac{2\rho^2}{\rho+1}\), which when rewritten into stretch-factor form, becomes

$$ R(2, \rho)=1+\frac{2\rho^2}{\rho+1} \frac{1}{2(1-\rho)} $$

Which is exactly \(\frac{1}{1-\rho^2}\), the heuristic form.

At \(m=3\) and above, the heuristic is only approximate. What does the Erlang
form reduce to for the first of those cases? Does it result in the missing term
that will extend to 4 and beyond too? (*Note: I wrote about this in a followup
post*).

Approximations to complicated equations are often really useful. You use them all the time, although you may not know it. For example, a computer uses approximations to calculate functions such as sine, square root, and logarithm. The field of numeric methods is dedicated to studying such algorithms and their errors.

In 1977, Sakasegawa discovered an approximation to the length of a queue, which is more accurate than the heuristic function, but wasn’t derived analytically. You can find the paper here. In a nutshell, his method was to begin with a well-known queueing theory formula derived by Pollaczek and Khinchine, which is valid (exact) for M/G/1 systems, that is, single-server queueing systems in some specific cases. Sakasegawa observed the error curve at various parameters and essentially did a least-squares sum of errors regression to arrive at an approximation for the queue length, which in the simplest types of queueing systems reduces to:

$$ L_q \approx \frac{ \rho^{\sqrt{2(m+1)}} }{ 1-\rho} $$

In “stretch factor” form, this becomes

$$ R(m, \rho) \approx 1+\frac{\rho^{\sqrt{2(m+1)}}}{\rho m(1-\rho)} $$

This is such an accurate approximation that it’s more than good enough for real-life applications, and I use it all the time. (It’s hard to put Erlang’s formula into a spreadsheet, because it has iterative computations.)

Given the usefulness and simplicity of approximations, perhaps an approximation to the stretch factor can be derived from the heuristic form \(1/(1-\rho^m)\). This idea appeals to me because it might lead to insights about what’s disappeared or simplified out of the exact Erlang formula in the base case.

One possible way to do this is to approximate the error function of the heuristic. I already mentioned that it might be possible to do this with the Gamma distribution’s PDF. Finding an approximation to that, and then adding or multiplying by it, might result in a usable approximation.

Another idea is a sigmoid such as the classic logistic function. In fact, if I wanted to approximate the error in the range (0,1) this isn’t too bad an approximation for \(m=16\), for example:

$$ \frac{.5}{1+e^{-10(x-1)}} $$

Finally, instead of adding a term or multiplying the heuristic by a term, perhaps it’s the \(\rho^m\) portion in the denominator that needs to be tweaked. A little analysis led me to the following conclusions.

The error at \(m>3\) could be explained by the exponent \(m\) being too large. If so, then the correct value for the exponent could be a function of \(\rho\), and an adjustment to it would need to be of the form

$$ A_{exp} = \frac{log\left( \frac{E(\rho)-1}{E(\rho)} \right)}{log(\rho)} $$

Where \(E(\rho)\) is the Erlang formula for the stretch factor. I arrived at this by solving the error function for \(\rho\). For convenience, this can be divided by \(m\) to normalize it relative to the number of servers. I’ve made a Desmos calculator illustrating the shape of this adjustment term for 1, 4, 8, and 16 servers:

One of the challenges with this is that due to the limitations of floating-point math in computers, the heuristic function appears to have no error at low utilization. I think it does, but it’s just a small value. That’s why the shape of that curve has a discontinuity at low utilization.

An interesting observation: If you zoom out and remove the limits on the range
of values plotted, you get something that looks like a part of the Gamma
function. Is this a coincidence? Could the Gamma function be used as an
approximation to this error? Or is the missing term a Gamma function, which
would result in an exact solution? *TODO*.

Another way to nudge the heuristic to approximate the Erlang residence time stretch factor would be to examine whether the base, \(\rho\), is too large or too small. Following a similar train of thought as before and solving the error function for the number of servers, I found that the error would need to be of the form

$$ A_{base} = \left( \frac{E(\rho)-1}{E(\rho)}\right)^{1/m} $$

Normalizing this relative to \(\rho\) by dividing, I got a function that has similar discontinuities as before, and is of the following shape. It looks like it might be possible to approximate with something like a quadratic from 0 to 1, but if you zoom out further, it looks more like… wait for it… part of the Gamma function. You can see this on Desmos.

I experimented with this in a different way, by trying to approximate \(A_{base}\) directly. I just guessed at its shape and came up with the following, which isn’t too far off for \(2<m<5\):

$$ A_{base} \approx \rho - \frac{2}{15} \sqrt{m} (\rho-1)\rho $$

You can see this shape, compared with the actual \(A_{base}\), at this Desmos.

And this one shows what this looks like when included as a term in the heuristic stretch factor.

Red is Erlang, blue is my heuristic, and black dashed is Gunther’s. Don’t be fooled; mine may look better, but if you examine high utilizations you’ll see it’s much worse. Small errors in the approximation to the error function make big differences in the result. (This is true if the error is defined as a function of number of servers, too.)

There’s a lot of outright superstition in this blog post, but hopefully some of the intuition I’m trying to develop comes through as well. Guessing at approximations (and seeing Gamma-things everywhere, like shapes in the clouds) is probably more time-wasting than productive, but I find that by playing with things I learn a lot. There’s something still unfinished in my mind, something unsatisfactory about coming towards an exact solution from two directions, finding two different forms, and finding that one of them runs aground at a certain point, even though it’s just a simplification of the other under some circumstances. You might replace “even though” with “because” in the previous sentence, and be satisfied, but as for me I feel there’s still something to be learned here. And potentially a useful result might come out of it, even if it’s only a fast approximation like Sakasegawa’s.

Besides, I just enjoy exploring math and its shapes; this is a great form of relaxation or stress relief for me when I want to concentrate on something so as to put other things out of my mind. Hopefully there’s a kindred soul out there who finds this interesting too. If so, hello, and enjoy!