Setup
Let's assume, you are working in a software company with a large number of clients. To help them with technical or usability issues your company runs a small help desk group which is able to take 50% of the incoming calls. You want to achieve 95% by adding more people to the group. Service level is appropriate and you can easily train new people, you just need enough hands to pick up calls. How many people do you have to hire?Exponential relationship
With zero resources you can't take any calls so you performance is zero. On the other hand even with a very large number of seats it may happen that there is some funny error in your app which causes problems exactly on Friday the 13th, 2011 at 11:30 and you are suddenly flooded with calls. With other words you would need a huge amount of people (comparable to the number of users) to take all the calls anything might happen. This is typical to an exponential relationship: p = 1-exp(-ar). Where p is the ratio of calls taken compared to all calls, r is the resources (number of help desk people), a is a constant and exp is the natural logarithm. The relationship looks like this.
Model 1
Using this assumption our current situation is: ,5 = 1-exp(-ar). If we rearrange this equtation:-0,5 = -exp(-ar)
0,5 = exp(-ar)
ln(,5) = -ar // using the natural logarithm
0,6931 = ar
r = 0,6931/a
in general
r = -ln(1-p)/a
for the desired 95%:
r = 2,9957/a
So if you want to reach 95% so yo have to increase the number of seats by 2,9957/0,6931 = 4,3! So if you have two people you will need 9!
Refined model
The situation is not so bleak because callers are willing to wait a while which soothes a bit the fluctuations. That I dare to neglect because people don't wait longer than 1 or 2 minutes which is not significant. However they most probably call back later, which changes the situation. So making a new assumption we say that in average a customer who couldn't be served will call once again before writing an angry e-mail, ask a friend, look for solution on the Internet or simple give up.
Let say that we have two - imaginary - groups of people; the first group takes only first calls the second group only the repeated calls. If the ratio of calls taken is p, then the first group takes p portion of the first time calls. The situation with the second group is more complicated; 1-p portion is already lost but the second group - which takes the same ratio of calls as the first - will answer p part of the remaining 1-p repeated calls, that is p(1-p). The second group is our penalty for not answering all calls in time. So this is extra resource. This increases our "ideal" resource by (p + (1-p)p)/p = 2-p, this is the ratio of the size of both group versus the first group. This is actually decreases the performance of taking first time calls. Putting this ratio into the original formula p = 1-exp(-ar) will be:
p = 1-exp(-ar/(2-p))
ln (1-p) = -ar/(2-p)
r = -ln(1-p) * (2-p)/ a
r2/r1= (ln(1-p2) * (2-p2)) / ((ln-p1) * (2-p1))
The calculation for our orignal problem:
(2,9957 * 1,05) / ( 0,6931 * 1,5) = 3,032
That is to move from 50% to 95% of calls taken we have to triple the workforce.
That is to move from 50% to 95% of calls taken we have to triple the workforce.
I don't know - yet - if this model works at all. For those who want to know more there are several good resources on the net. E.g.:
http://en.wikipedia.org/wiki/Erlang_%28unit%29
http://ganymedes.lib.unideb.hu:8080/dea/bitstream/2437/87268/1/Call%20Centerek%20Matematikai%20Modellez%C3%A9se.pdf
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