Waiting line : M/M/3 with $\lambda = 24/u$ and $\mu = 10/u$ vs M/M/1 with $\lambda = 24/ u$ and $\mu = 30/u$

I'm doing some exercises on waiting line theory, now I've got a question to which I did not note an answer during class. Will the waiting line increase ?

The question is : a company can do 24 jobs / hour at an average waiting time of 6 minutes / job on a cpu of type A. Now we have a new cpu B that can do the job three times faster than a type A cpu. Now they want if they change their 3 cpu's to one, if there is going to be an increase in the waiting line ?

I think changing will have a negative effect :

I did a run in ORSTAT which gave me an average waiting time for 1.30 with the $u = 30$.

However I don't know how to do a simulation with MM3 in orstat...

I think that the average waiting will be lower than 1.30, since there are 3 work units, you will have less impact of changing a job than when there is only one cpu.

Can anyone confirm this ?

  • $\begingroup$ Did I get the right link for Orstat? $\endgroup$
    – whuber
    Aug 16, 2011 at 21:35
  • $\begingroup$ Yes I believe so :p $\endgroup$ Aug 16, 2011 at 22:01

1 Answer 1


The formula for the expected length of the queue in an $M/M/c$ queuing system is:

$$L = {\rho^{c+1} \over (c-1)!(c-\rho)^2}P_0$$

where $\rho = \lambda / \mu$ and $P_0$ is the steady-state probability that the system is empty:

$$P_0 = \left[\sum_{i=0}^{c-1}{\rho^i \over i!} + {\rho^c \over c!(1-{\rho\over c})}\right]^{-1}$$

The derivation is tedious, but you can find these formulae in a variety of sources, e.g., Kleinrock, or online in https://nptel.ac.in/courses/110106046/Module%209/Lecture%204.pdf.

Setting $\lambda = 24$ and $\mu = 30$ in the $M/M/1$ system and $\lambda = 24$ and $\mu = 10$ in the $M/M/3$ system allows us to calculate $L$ explicitly:

$$L_1 = 3.2$$ $$L_3 = 2.59$$

... showing that the expected queue length is less in the $M/M/3$ system, as you suspected.


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