Efficiently calculate Erlang C for large numbers

Report · Apr 21, 2011

I need to calculate the Erlang C function for some contact center software. A picture of the function and a brief discussion is here.

I cobbled together a solution based on the approach normally used in Excel udsing the poisson distribution:

private function poissonA():Number
{
    return (Math.pow(trafficIntensity(), _numberAgents) *
            Math.exp(-1 * trafficIntensity())) /
            com.subhumanmedia.Main.factorial(_numberAgents);
}
private function poissonB():Number
{
    var result:Number = 0;
    for (var i:int = -1; i < _numberAgents; i++)
    {
        result += (Math.pow(trafficIntensity(), i) *
                    Math.exp( -1 * trafficIntensity()) /
                    com.subhumanmedia.Main.factorial(i));
    }
    return (result);
}
public function erlangCFunction():Number
{
    var pA:Number = poissonA();
    return(pA / (pA + ((1 - agentOccupancy()) * poissonB())));
}

This works and it's ok for the few people who are currently using it. But it has limits I need to solve. It will only work for up to roughly 170 agents because of the maximum 64-bit floating point number. Also, on the face of it it seems pretty inefficient. I have tried to overcome the number size limitations by using a BigInteger implementation but that is stalled because Math.exp( -1 * trafficIntensity()) is never going to be an integer and I don't know how to multiply a BigIntger by a float. For BigIntegers I'm using As3Crypto .

I'm not much of a coder and even less of a mathematician. Can anyone give some suggestions for how to approach this? I've spent literally hours reading articles related to this but didn't find any practical help (at least that I could reccognize and knew how to apply).

I'm looking for either (1) a way to implement the current approach using only BigIntegers or preferably (2) a way to recast the original formula that is inherently more efficient to calculate.

Thanks in advance.

Ron

Report · Sep 27, 2017

I had the same question, and got some help at probability - Erlang C for Large Numbers - Mathematics Stack Exchange

In VBA

Function ErlangC(u As Double, m As Long) As Double

' Returns the probability of waiting when dialing into a call center

' u = call intensity = calls/s * duration

' m = number of agents on hand

' m > u or wait times will grow without limit

Dim LnFactM As Double

Dim LnU As Double

Dim k As Long

Dim d As Double

With WorksheetFunction

LnFactM = .GammaLn(m + 1)

LnU = Log(u)

For k = 0 To m - 1

d = d + Exp(LnFactM - .GammaLn(k + 1) + (k - m) * LnU)

Next k

End With

ErlangC = 1 / (1 + (1 - u / m) * d)

End Function

For example, =ErlangC(1000, 1030) returns 24.9%

Report · Sep 28, 2017

With further credit to user Dap at math.stackexchange, a version that requires no iteration:

Function DapErlangC(u As Double, m As Long) As Double

' shg 2017

' Credit to user Dap at

' https://math.stackexchange.com/questions/2446752/erlang-c-for-large-numbers