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Full question below:

You are the manager of the customer support division in your company. Your division uses 3 telephone lines operated by 3 separate customer service representatives. A customer is put on hold if their call arrives while all 3 customer service representatives are busy serving other customers. You observe that customer calls arrive at a Poisson rate of 5 per hour, and that the length of the customer calls is exponentially distributed. You also observe that 75% of the time, a customer is not put on hold, while the remaining 25% of the time, a customer can expected to be put on hold for an average of 12 minutes. You wish to improve service in the division by making sure that 90% of the time, a customer is not put on hold, while 10% of the time, a customer can expect to be put on hold for an average of only 4 minutes. How many telephone lines will you add to your division to achieve your goal?


So I think the biggest problem here is that I don't know $\mu$. I do know $\rho=\frac{\lambda}{c\mu}=\frac{5}{3\mu}$ for this problem. I understand that "time on hold" refers to to time waiting in the queue. With $W$ is time waiting in the queue, I know: $$E[W]=\frac{\rho}{\lambda(1-\rho)}P(W>0)$$

With 3 operators, I used the fact that "75% of the time, a customer is not put on hold, while the remaining 25% of the time, a customer can expected to be put on hold for an average of 12 minutes" to calculate: $$E[W]=.75(0) + .25(12min)=3min$$ Then using $E[W]$ along with $P(W>0)=.25$, I solved the first equation to find $\mu=\frac{10}{3}$.

Knowing $u$, I used "90% of the time, a customer is not put on hold, while 10% of the time, a customer can expect to be put on hold for an average of only 4 minutes" to find the new $E[W]=1min$ and $P(W>0)=.1$.

To solve for $c$(number of servers) I again plugged these numbers into the original equation for $E[W]$ and found $c=3.3$. You can obviously only have an integer number of servers, so this would be $c=4$, and minus the original 3 would give the addition of just 1 server as the answer.

Sorry for the long question, but am I doing this right? I feel like I messed up along the way or made some wrong assumptions (mostly that $E[W] can be calculated from the information in the problem).

Thanks for looking.

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  • $\begingroup$ Are we assuming that there is no limit to the number of customers that can be placed on hold? Or, for example, can each phone line only handle two customers (one on the line and one on hold)? $\endgroup$ – Math1000 Nov 27 '15 at 13:07
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I wished to add this as a comment, but I am not allowed to.

We are dealing with an $M/M/c$ queueing system. You calculated the service rate $\mu$ correctly, but then continue incorrectly.

I'd suggest you look at these lecture notes, page 44, equations (5.1) and (5.3) and increase the number of servers $c$ until we have that the probability to wait $\Pi_W := \mathbb{P}(W > 0) \le 0.1$ and the expected waiting time $\mathbb{E}[W] \le 4$ minutes.

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Let $X(t)$ be the number of customers in the system at time $t$ (including those on hold). Then $\{X(t):t\geqslant 0\}$ is a continuous-time Markov chain on $\{0,1,2,\ldots\}$. Assuming $\rho :=\frac\lambda{3\mu}$ we can find the stationary distribution $\pi$ with the global balance equations \begin{align} \lambda\pi_0 &= \mu\pi_1\\ \lambda\pi_1 &= 2\mu\pi_2\\ \lambda\pi_n &= 3\mu\pi_n, n\geqslant 2. \end{align} This yields $$\begin{align}\pi_1 &= \frac\lambda\mu\pi_0\\ \pi_2 &= \left(\frac\lambda\mu\right)\left(\frac\lambda{2\mu}\right)\pi_0\\ \pi_{3+i} &= \rho^i\left(\frac\lambda\mu\right)\left(\frac\lambda{2\mu}\right)\pi_0,\ i=0,1,2,\ldots \end{align} $$ From $$\sum_{n=0}^\infty \pi_n = 1$$ we obtain $$\pi_0 = \left[1 + \left(\frac\lambda\mu\right) + \left(\frac\lambda\mu\right)\left(\frac\lambda{2\mu}\right)+ \frac1{3!}\left(\frac\lambda\mu\right)^3\left(\frac1{1-\rho}\right) \right]^{-1}, $$ and thus the probability of being put on hold is $$\sum_{i=3}^\infty \pi_i = \frac{\pi_3}{1-\rho} = \frac{\frac1{3!}\left(\frac\lambda\mu\right)^3}{1-\rho}\left[1 + \left(\frac\lambda\mu\right) + \left(\frac\lambda\mu\right)\left(\frac\lambda{2\mu}\right)+ \frac1{3!}\left(\frac\lambda\mu\right)^3\left(\frac1{1-\rho}\right) \right]^{-1}. $$ A standard result in queueing theory (e.g. Halfin, Whitt (1981) is that the probability of delay can be approximated for a large number of servers $n$ by $$HW(\beta) = \frac1{1+\beta\Phi(\beta)/\varphi(\beta)}\tag 1 $$ where $\Phi$ is the CDF of the standard normal distribution, $\varphi$ the PDF of the standard normal distribution, and $\beta>0$ satisfies $$\frac\lambda\mu = n - \beta\sqrt n.\tag 2 $$ To obtain $\mathbb P(\mathrm{Delay})<\frac1{10}$, we find $\beta\approx 1.4202$ from $(1)$. Solving $(2)$ for $n$ and substituting $\beta$, $\lambda$ and $\mu$ yields $$n\sim\frac5{10/3} + 1.4202\sqrt{10/3} = 4.0929. $$ Therefore we should have at least $\lceil 4.0929\rceil=5$ servers. Given that the service rate is $\mu=\frac5{144}$ per minute, the expected delay time with $5$ servers is $$\mathbb E[\mathrm{Delay}]= \mathbb P(\mathrm{Delay})\frac{\rho}{\lambda(1-\rho)} = \frac{72}{65}. $$

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