The square-root staffing rule

Up to now we saw that an increase of scale leads to advantages with respect to productivity
and/or service level. These advantages can always be quantified using the Erlang formula.
To obtain a general understanding we formulate a rule of thumb that relates, for a fixed
service level, call volume and the number of agents. In a formula this relation can be
formulated as follows:
overcapacity in % × ps = constant.
The constant in the formula is related to the service level, the formula therefore relates
only overcapacity and the number of agents. The percentage overcapacity in the formula
is given by 100 × (1 − a/s). From the rule of thumb we obtain results such as: if the call
center becomes four times as big, then the overcapacity becomes roughly halve as big. How
we obtain this type type of results is illustrated by the following example.
A call center with 4 agents and _ = 1 and _ = 2 minutes has an average waiting time of a
little over 10 seconds. For this call center the associated constant is 100×(1−2/4)×
p4 =50 × 2 = 100. If we multiply s by 4, than ps doubles. Thus to keep the same service
level (the same constant), we halve the overcapacity to 25%. Thus the productivity becomes
75%, and thus with s = 4×4 = 16 this gives a = 12 and _ = 6. If we verify these numbers
with the Erlang formula, then we find an average waiting time of a little over 6 seconds.
Closest to 10 seconds is s = 15, with approximately 12 seconds waiting time. If we multiply
s again with 4, then the overcapacity can be reduced to 12.5%. This means _ = 28, with
3.2 seconds waiting time. Closest to 6 seconds is s = 62, from which we see that the rule
of thumb works reasonably well.
From the example we see how simply we can get an impression of the allowable call
volume if we change the occupation level. More often we prefer to calculate the number of
agents needed under an increase in call volume. The calculation for this is more complex. If
we denote with c the constant related to the service level divided by 100, then the formula
for s is:
s =c +p
c2 + 4a
2
!2
.
As in the previous example we start with 4 agents, _ = 1 and _ = 2 minutes. The number
c is the constant divided by 100, thus c = 1. Filling in a = _ × _ = 2 and c = 1, then we
find indeed s = 4. Assume that _ doubles. Then a = 4, and with c = 1 we find s _ 6.6.

This is a good approximation: s = 7 gives a waiting time under the 10 seconds, s = 6
above. If _ = 10, the we find s = 25 as approximation. An agent less would give a waiting
time of 9 seconds. If _ doubles again, then we get 47 as approximation, with 45 as best
value according to the Erlang formula. We see that for big values of _ doubling leads to
doubling s.
If c is small with respect to a then we see that s is proportional to a. This means that
the economies of scale become less for very big call centers, because it is already almost at
the highest possible level. What “big” is in this context depends on the service level.
Using this rule of thumb should be done with care. It is only useful to relate _ and s.
Next to that, one should realize that it is only an approximation, the results need to be
checked with the Erlang formule before use in practice. This point was illustrated in the
example.