# Math Help - A Problem in Queuing Theory

1. ## A Problem in Queuing Theory

A company gets an average of five calls an hour from prospective clients. It takes a company representative an average of twenty minutes to handle one call (the distribution of time to handle one call is exponential). A prospective client who cannot immediately talk to a representative never calls again. For each prospective client that talks to a representative the company makes one thousand dollars. How many representatives should the company maintain if each is paid ten dollars an hour?

2. Originally Posted by akbar
A company gets an average of five calls an hour from prospective clients. It takes a company representative an average of twenty minutes to handle one call (the distribution of time to handle one call is exponential). A prospective client who cannot immediately talk to a representative never calls again. For each prospective client that talks to a representative the company makes one thousand dollars. How many representatives should the company maintain if each is paid ten dollars an hour?

You have to write the problem in terms of a Markov process, and look for an invariant distribution.

Using the memoryless property of exponential random variables, the fact that if $X\sim\mathcal{E}(\lambda)$, $Y\sim\mathcal{E}(\mu)$, then $Z=\min(X,Y)\sim\mathcal{E}(\lambda+\mu)$ and $P(Z=X)=\frac{\mu}{\lambda+\mu}$, you can get the following (up to possible errors).

Suppose there are $N$ representatives, that the arrivals of clients are given by a Poisson process of parameter $\lambda$ and that the length of a call is an exponential random variable of parameter $\mu$. The queuing process is described by a Markov process $(X_t)_{t\geq 0}$ on state space $\{0,\ldots,N\}$, corresponding to the number of simultaneously busy representatives, with $X_0=0$, holding times at $k=0,\ldots,N-1$ being exponential with parameter $q(k)=\lambda+k\mu$ and with parameter $q(N)=N\mu$ at $N$, and transition probabilities between "neighbours" only, given by $p_{k,k+1}=\frac{\lambda}{\lambda+k\mu}=1-p_{k,k-1}$.

If $\pi$ is the invariant distribution, then the average number of people on the phone per amount of time is the expectation of $\pi$. This leads to the answer.

3. My first question is: Why do you assume that the arrival of customers is described by a Poisson process? Apart from a remote allusion ("five calls an hour from prospective clients"), nothing in the problem outline gives an idea about the distribution of the client calls.

Is this a common assumption in queuing theory problems? True that once such an assumption is made (clearly the most manageable since the inter-arrival times are exponentially distributed, and we can then use its memoryless property), the problem becomes much more manageable. I guess this relates more to the general way of approaching queuing problems.

4. Originally Posted by akbar
My first question is: Why do you assume that the arrival of customers is described by a Poisson process?
I don't know one queuing problem where arrival times don't make a Poisson process. Anyway, how did you expect to find an answer without knowing the distribution of arrivals?? If all the phone calls arrive at the same time, you'll obviously need more representatives than if the phone calls are evenly spaced. Thus, without an implicit assumption, the problem wouldn't be manageable at all... this should ring bells

5. I certainly didn't expect to solve the problem without an assumption, but maybe another (just as realistic) distribution would have been possible. Nothing in the problem outline prevented it.

Originally Posted by Laurent
I don't know one queuing problem where arrival times don't make a Poisson process.
Knowing from an expert that one can confidently make such assumption in general problems in Queuing Theory is particularly helpful. I haven't done as many problems involving queues as you certainly did, but this was the first one I came across without any reference to the distribution of arrivals. This is the reason of my doubts.

In any case, the idea of finding an "equilibrium" distribution on the states of representatives makes perfectly sense.

Thanks a lot for your wisdom .