0
$\begingroup$

Assume there is only one machine (server) at a factory that produces j types of products and $\lambda_j$ and $\mu_j$ are known individually.

For example, the 1st type of product has an arrival rate of $\lambda_1=0.10$ products/min and type 1 can be produced by the server with a service rate, $\mu_1=0.70$ products/min (deterministic). The 2nd type of product has an arrival rate of $\lambda_2=0.30$ products/min and type 2 can be produced by the server with a service rate, $\mu_2=0.40$ products/min (deterministic). By saying deterministic, I mean the service times for each j are constant and there is no variation within each type.

My questions:

In this case, would this queue be an M/G/1 queue?

If it is an M/G/1 queue, can you please explain how I can use the closed form formulation of $W$,

$$ W=\frac{1}{\mu} +\frac{\lambda(\frac{1}{\mu^2}+CV^2)}{2(1-\frac{\lambda}{\mu)}}, where~~CV~is~Coefficient~of~Variation $$

to calculate the expected waiting time in the system?

How can I calculate $\lambda$ based on $\lambda_1$ and $\lambda_2$, similarly $\mu$ based on $\mu_1$ and $\mu_2$. Besides, how is CV calculated in such a situation?

EDIT1:

Well, it seems that from this answer: M/G/1 Queue or not?, this is an M/G/1 queue.

EDIT2:

On the other hand, from this answer: M/G/1 queuing system with two arrivals, it may not be an M/G/1 queue.

EDIT3:

Okay, I also knew that $\lambda=\lambda_1+\lambda_2$ and just learned how to calculate $\mu$ from the M/G/1 queuing system with two arrivals (though they had a little mistake), which is

$\frac{1} {\mu}=(\frac{λ_1}{λ_1+λ_2})\frac{1} {\mu_1}+(\frac{λ_2}{λ_1+λ_2})\frac{1}{\mu_2}$.

Assuming these calculations are true and this is an M/G/1 queue (which still needs a clarification), my question becomes:

How can I find $CV^2$?

EDIT4:

Actually, this post Queueing Delay(W) for M/D/1 queue with different value of service times kind of convinced me that my problem is an M/G/1 queue, furthermore, it has a specific name, called Multiclass M/G/1 queue. Also, I think I got the answer for $CV^2$ from the book "Analysis of Queues Methods and Applications" authored by Natarajan Gautam. Please correct me, if it is wrong.

$CV^2 = \frac1 \lambda \sum_{j=1}^J \lambda_j (\frac1 \mu)^2$

$\endgroup$
0
$\begingroup$

From the wiki page: The coefficient of variation (CV) is defined as the ratio of the standard deviation $\sigma$ to the mean $\mu$. $$CV=\frac {\sigma }{m}$$

CV I believe is for service time distribution, then let the service time be $t$, since $$\mu=\frac { \text{Unit Time} }{t}$$ then $$t = \frac { \text{Unit Time} }{\mu}$$ Unit Time in your example is 1 minute. $$ \begin{align} CV & = \frac {\sigma }{m} \\ \sigma_{j} & = \sqrt{ \frac { \sum_{j=1}^{J}( t_{j}-{\overline {t} )^{2} } }{J-1} } \\ CV_{j} & = \frac { \sigma_{j} }{\overline { t_{j} }} \end{align} $$

I can't tell if your way is wrong, but it's different from what I learned.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.