In previous posts I’ve talked about queuing theory, and the application of Little’s Law in particular, to Internet Service Provider (ISP) customer support call centre wait times. We can define Little’s Law, as it applies to a call centre, as:

The long-term average number of support staff (N) in a stable system is equal to the total number of customers in the queue (Q) multiplied by the average time support staff spend resolving a customers’ technical problems (T), divided by the total time waited in the queue (W); or expressed algebraically: N=QT/W.

Thinking things through a bit more, the total number of customers in the queue (Q) at a point in time in a stable system should be equal to the rate at which people joined the queue (* λ*), minus the rate at which the support desk dealt with technical problems (i.e. N/T) over the period of observation. Obviously Q>=0.

So N=QT/W and Q=* λ*-N/T which all comes out in the wash as:

**N=λT/(W+1)**

I thought might be a bit of fun to see if this could be applied to the customer support call centre waiting statistics published by one of Australia’s largest ISPs, *iiNet*.

iiNet make some support centre data available via their customer toolbox page. Below is a screenshot of call activity and wait times graphed each hour by iiNet on 10 January 2012. The green line (in conjunction with the scale on the left hand side of the graph) represents the average time (in minutes) it took to speak to a customer service representative (CSR), including call-backs. The grey bars (in conjunction with the right hand scale) represents the total number of incoming phone calls to iiNet’s support desk.

It may be possible to use the formula derived above to estimate how many CSRs iiNet had on the support desk handling calls that day. For example, during the observed peak period of 8am to 1pm on Tuesday, 10 January 2012, the iiNet support desk was getting around 732 calls per hour on average. The expected wait time in the queue over the same period was around 11 minutes.

If we assume that the average time taken for a CSR to resolve a technical problem is, let’s say, 12.5 minutes, then we can estimate that the number of CSRs answering calls in a typical peak-hour between 8am to 1pm on 10 January 2012 as:

732*(12.5/60) / (11/60 + 1)

= 129 CSRs actively handling calls.

Sounds sort of a reasonable for a customer service-focussed ISP the size of iiNet. But if iiNet wanted to bring the average time in the queue down even more – to a more reasonable 3 minutes, for example – they’d need 145 CSRs (all else remaining equal) during a typical peak-hour answering calls.

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Filed under: ISP statistics, research, statistical concepts, statistics | Tagged: queuing theory |

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