Don't you put me on hold!

Let us represent the state of the system by a discrete probability distribution, $p_n$ , the probability of the call server having $n$ of the $N$ representatives occupied by a call.

At each moment in time, an un-occupied representative can start a new call, or an occupied representative can end the call they're currently on.

Shifting focus to the state of the whole call center, the number of occupied representatives can increase of decrease by one, unless none or all of the representatives are occupied, in which case it can only increase or decrease by one, respectively.

Let's consider a few cases in detail:

• For the probability of no representatives being occupied, we have

$\frac{\partial p_0(t)}{\partial t} = \beta p_1(t) - \lambda p_0(t)$

The $n=0$ state can gain probability from systems in the $n=1$ state freeing up their lone occupied representative, and it can lose probability by any one of its unoccupied representatives taking a call.

• For the probability of just one representative being occupied, we have

$\displaystyle\frac{\partial p_1(t)}{\partial t} = \lambda \left[p_0(t) - p_1(t)\right] + \beta\left[2 p_2(t) - p_1(t)\right]$

Note the fundamental difference between $\lambda$ , and $\beta$ . $\lambda$ is the probability of a new call attempt being made per unit time. It is independent of the number of people who have already called. No matter what the state of the call server, new people call on average once every $\lambda^{-1}$ minutes.

On the other hand, $\beta$ is the probability per unit time that a given call will end. Thus, the probability per unit time that any of $n$ calls will end is given by $\beta n\times dt$ .

Generally, we have (with $\beta = \langle t \rangle^{-1}$ for ease of writing)

$\frac{\partial p_n(t)}{\partial t} = \begin{cases} \beta p_1(t) - \lambda p_0(t)& n = 0\\ \lambda \left[p_{n-1}(t) - p_n(t)\right] + \beta \left[(n+1)p_{n+1}(t) - n p_n(t)\right] & 1 \leq n \leq N-1 \\ -\beta N p_N(t) + \lambda p_{N-1}(t) & n = N \end{cases}$

Now, when the call center opens, it is obviously in the $n = 0$ state, so that $p_0(0) = 1$ , and $p_n(0)=0$ for $n \geq1$ . As time goes on, the probability distribution of the call center will evolve until it is equal to the steady state distribution, $p_n(\infty)$ .

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To find this distribution, we set the time derivative equal to zero.

First, we have the boundary case $\dot{p}_0 = 0 = -\lambda p_0 + \beta p_1$ which implies $p_1 = \frac{\lambda}{\beta} p_0$ .

Let's analyze the intermediate values of $n$ . We have

$0=\lambda \left[p_{n-1}(t) - p_n(t)\right] + \beta \left[(n+1)p_{n+1}(t) - n p_n(t)\right]$

We can simplify things somewhat if we use the step operator $\mathbb{E}^{\pm1}$ defined by its effect $\mathbb{E}f(n) = f(n\pm 1)$ .

Then the last equation becomes

$0 = \left(\mathbb{E}^{-1}-1\right)\lambda p_n + \left(\mathbb{E}-1\right)\beta p_n$

Notice that $1=\mathbb{E}^{-1}\mathbb{E}$ . We have

$0 = \left(\mathbb{E}-1\right)\left(n\beta p_n - \mathbb{E}^{-1}\lambda p_n\right)$

Now, $\left(\mathbb{E} - 1\right)\neq 0$ , so we must have $\left(n\beta p_n - \mathbb{E}^{-1}\lambda p_n\right) = 0$ , or $p_n = \frac{1}{n}\frac{\lambda}{\beta}p_{n-1}$ .

Following this through to the base case, we have $\displaystyle p_n = \frac{1}{n!}\left(\frac{\lambda}{\beta}\right)^n p_0$ .

Now, we have everything in terms of the known parameters $\lambda$ and $\beta$ , and the unknown value $p_0$ . We can find this using the normalization condition $\sum_i p_i = 1$ :

$p_0^{-1} = \sum_i \frac{1}{i!}\left(\frac{\lambda}{\beta}\right)^i$

Therefore, the steady state distribution is given by

$\displaystyle p_n = \frac{\frac{1}{n!}\left(\frac{\lambda}{\beta}\right)^n}{ \sum_i \frac{1}{i!}\left(\frac{\lambda}{\beta}\right)^i}$

which is plotted below

the probability of not reaching a representative is then $p_{10}$ , the probability of all representatives being occupied.

Since each call lasts on average for $\langle t \rangle = 10 \ \text{min}$ and calls arrive at random at a rate of $\lambda=0.7 \ \text{calls/min}$ . So, on average, we will find a free operator every $\langle t \rangle$ , or, in terms of 'number of calls', every $\lambda'=\lambda \cdot \langle t \rangle=7 \ \text{calls}$ . The number of incoming calls follows a Poisson distribution , i.e.:

$\displaystyle \mathbb{P}(k,\lambda')=\frac{(\lambda') ^{k} e^{-\lambda'}}{k!}$

where $k$ are the number of calls. Since there are just $N=10$ representatives and that customers who call when all representatives are busy are put on hold and end their calls immediately, we have that $0 \leq k \leq 10$ and our distribution becomes

$\displaystyle \mathbb{P}(k,\lambda' \ | \ 0 \leq k \leq 10)=\frac{(\lambda') ^{k} e^{-\lambda'}}{k!} \bigg[\sum_{i=0}^{10} \frac{(\lambda') ^{i} e^{-\lambda'}}{i!}\bigg]^{-1}$

In other terms, the distribution is normalized for $0 \leq k \leq 10$ . This is a particular case of Zero-truncated Poisson distribution . We will be put on hold if all the $N$ representatives are busy, i.e. when there will be $k=10$ calls. Hence

$\displaystyle \mathbb{P}(k=10,\lambda'=7\ | \ 0 \leq k \leq 10)=\frac{(7) ^{10} e^{-7}}{10!} \bigg[\sum_{i=0}^{10} \frac{(7) ^{i} e^{-7}}{i!}\bigg]^{-1}=\boxed{0.0787409}$