A customer care company receives an average of 4.5 calls every 5 minutes. If each customer executive can handl

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Q. A customer care company receives an average of 4.5 calls every 5 minutes. If each customer executive can handle one of these calls over the 5-minute period. But if an executive is not unavailable to take the call, then the call is put on hold. Assuming that the calls received by the customer care company follows a Poisson distribution, what is the minimum number of customer executives are needed on duty so that calls received are placed on hold for at the most 10% of the time?

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Concept understanding — Poisson Queueing Theory

Poisson Queueing Theory: A First Look

Imagine a coffee shop. Customers arrive at random times — sometimes three come in a minute, sometimes none for five minutes. The barista serves them one at a time, and each service takes a random duration. You want to answer questions like: How long will the next customer wait? How many customers will be in line on average? That is queueing theory.

Poisson queueing theory is the simplest, most powerful model for such systems. It rests on two key ideas: arrivals happen according to a Poisson process, and service times follow an exponential distribution. Together, they make the mathematics clean and the predictions sharp.

Intuition: What makes arrivals "Poisson"?

Think of a Poisson process as completely random arrivals — no memory, no pattern. If you watch the coffee shop door, the chance that a customer arrives in the next second is always the same, regardless of how long it has been since the last customer. This is called the memoryless property.

Concretely, if the average arrival rate is $\lambda$ customers per hour, then:

The number of customers arriving in any fixed time interval (say, 10 minutes) follows a Poisson distribution.
The time between consecutive arrivals follows an exponential distribution with mean $1/\lambda$ .

Note

The exponential distribution is the only continuous distribution that is memoryless. That is why it pairs naturally with the Poisson process.

The precise statement

A queueing system is described by Kendall's notation: $A/B/c$ , where $A$ is the arrival process, $B$ is the service-time distribution, and $c$ is the number of servers. The classic Poisson queue is $M/M/1$ :

M = Markovian (Poisson) arrivals
M = Markovian (exponential) service times
1 = one server

\text{Arrival rate} = \lambda, \quad \text{Service rate} = \mu

The system is stable only if $\lambda < \mu$ — otherwise the queue grows without bound.

Key results for $M/M/1$

Let $\rho = \lambda / \mu$ be the traffic intensity (the fraction of time the server is busy). Then:

Quantity	Formula	Meaning
Probability system is empty	$P_0 = 1 - \rho$	Fraction of time no customers
Average number in system	$L = \frac{\rho}{1 - \rho}$	Customers in queue + being served
Average number in queue	$L_q = \frac{\rho^2}{1 - \rho}$	Customers waiting only

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Calls are held whenever the number of calls exceeds the number of executives $n$ . Requiring the hold probability $P(X>n)\le0.10$ , i.e. $P(X\le n)\ge0.90$ , with $\lambda=4.5$ gives $n=7.$

P(X=k)=\dfrac{e^{-\lambda}\lambda^{k}}{k!}

Steps

$\lambda=4.5$ ; $e^{-4.5}=0.011109.$
Compute term-by-term Poisson probabilities and their cumulative sum:

$k$	$P(X=k)$	$P(X\le k)$
0	0.01111	0.01111
1	0.04999	0.06110
2	0.11248	0.17358
3	0.16872	0.34230
4	0.18981	0.53210
5	0.17083	0.70293
6	0.12812	0.83105
7	0.08236	0.91341

The company can serve all calls if $X\le n$ ; a call is held only when $X>n$ . …

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