# The Erlang Response Time Stretch Factor For 3 And 4 Servers

Posted in Math, Scalability on Dec 11, 2016

In a previous post I explored a few variations of equations that express the M/M/m queueing theory response time “stretch factor,” and tried to indicate some areas where I wanted to dig into the relationships between these formulas a bit more. In this post I discuss the divergence between the official Erlang C formula and Neil Gunther’s heuristic approximation to it. I introduced this before thusly:

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?

Here’s the Erlang form again:

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

And the equivalent heuristic by Gunther, which is exact for 1 and 2 service channels:

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

Note that this is a stretch factor relative to the service time, i.e. to find the average residence time, you must multiply by the service time.

When \(m\) is 1 or 2, the Erlang formula simplifies exactly to the heuristic. When \(m=3\), the Erlang formula simplifies to:

$$ R(3, \rho) = 1 + \frac{3\rho^3}{(1-\rho)(3\rho^2+4\rho+2)} $$

It’s not immediately obvious to me how that might be related to the heuristic form of the response time stretch factor, \(\frac{1}{1-\rho^3}\).

At \(m=4\), the Erlang formula simplifies to

$$ R(4, \rho) = 1 + \frac{8\rho^4}{(1-\rho)(8\rho^3+24\rho^2-15\rho+15)} $$

The relationship between this and \(\frac{1}{1-\rho^4}\) is is no more obvious to me. Importantly, I think, at this point the Erlang formula and the heuristic start behaving very differently. With 3 service channels, the formulas have essentially the same shapes (same poles, etc), but there’s a slight ripple in one that’s not in the other.

But with 4 service channels they’re fundamentally different over portions of their range (outside the physically meaningful performance region of \(0<=\rho<=1\), that is).

What is the error in the heuristic for 3 and 4 service channels? I touched on this briefly in the previous post. By subtracting the heuristic from the Erlang formula and simplifying, I found that when there are 3 service channels, the difference is

$$ \frac{\rho^3}{3\rho^4+7\rho^3+9\rho^2+6\rho+2} $$

But with 4, the difference between the functions isn’t expressible neatly, and is sometimes infinity, reinforcing the impression that the coincidence between Erlang and heuristic is just that.

I still feel that something useful is hiding just around the corner, because the heuristic arises analytically but doesn’t scale correctly to more than 2 service channels, and Erlang also arises analytically but is correct. The fact that they are exactly the same for 1 and 2 service channels just doesn’t feel accidental. But intuition has led me astray many times.

You can graph and visualize all of the above with a Desmos calculator that I made for you.