View source for Question: Why does the effective service rate f(x) go to zero when x = 0 in the example on queuing systems?

{{Question
|Chapter=System Modeling
|Author(s)=Richard Murray
|Date=2009/10/10
}}
Q: The formula for <math>f(x)</math> which scales <math>\mu_\text{max}</math> is <math>x / (x + 1)</math> . This is zero when <math>x = 0</math> (no queue) and 1 when <math>x</math> goes to infinity. Why is this the right model?

A: The model used by Agnew (1976)  is that the rate at which jobs are processed is linear in the queue length when the length is small, and saturates and the maximum service rate . At the extreme where there are no jobs on the queue, there is no need to process incoming requests and Agnew's assumption was the more processing would be applied as the queue got longer, until it saturates for  very large.

Notice that the term  in the denominator means that the service rate very rapidly reaches it maximum as the queue increases. If there is 1 job on the queue then the service rate is <math>0.5 \mu_\text{max}</math>, 3 jobs gives a rate of <math>0.75 \mu_\text{max}</math>, and 9 jobs gives a rate of <math>0.9 \mu_\text{max}</math>.