Question

People arrive at an automatic transaction machine (ATM) according to a Poisson process with intensity \(\lambda\). The service time required at the ATM is constant, \(a\) seconds. Unfortunately, this machine does not allow for any waiting customers (i.e., no queue is allowed), which means that persons who arrive while the ATM is busy have to leave. When the \(a\) seconds of a customer have elapsed, the ATM is free to serve again, and so on. Suppose that the ATM is free at time 0 , and let \(T_{n}\) be the time of the arrival of the \(n\)th customer. Find the distribution of \(T_{n}\), and compute \(E T_{n}\) and \(\operatorname{Var} T_{n}\). Remark. Customers arriving (and leaving) while the ATM is busy thus do not affect the service time.

Short answer

\(T_n\) is \((n-1)a + \sum_{i=1}^{n} \text{Exponential}(\lambda)\). \(E[T_n] = (n-1)a + \frac{n}{\lambda}\), \(\operatorname{Var}[T_n] = \frac{n}{\lambda^2}\).

Step-by-step solution

Understanding the Problem

This problem involves a Poisson process with rate \(\lambda\) and deterministic service time of \(a\) seconds. The ATM has no waiting line, so only arrivals when the ATM is free are served.

Arrival Distribution

In a Poisson process, the time between consecutive arrivals follows an exponential distribution with rate \(\lambda\). Thus, the time until the next arrival after the ATM is free is \(\frac{1}{\lambda}\).

Arrival Time Calculation

Since customers can only be served if the ATM is free, and the service time is \(a\) seconds, each successful service is followed by a free period during which a customer must arrive. Therefore, the time of the \(n\)th arrival that gets served is a sum of deterministic and exponential random variables.

Distribution of \(T_n\)

Each \(T_n\) results from \(n-1\) successful services, taking \(a\) seconds each, plus the time until the \(n\)-th arrival after the \(n-1\)-th customer has finished. The total time \(T_n = (n-1)a + \sum_{i=1}^{n}\text{Exponential}(\lambda)\). The sum of \(n\) exponentials has gamma distribution, \(\Gamma(n, \lambda)\), which is equivalent to \(\text{Erlang}(n, \lambda)\).

Finding \(E[T_n]\) and \(\operatorname{Var}[T_n]\)

For a Gamma random variable \(\Gamma(n, \lambda)\), \(E[T] = \frac{n}{\lambda}\) and \(\operatorname{Var}[T] = \frac{n}{\lambda^2}\). Including \(n-1\) deterministic durations of \(a\), we have:\(E[T_n] = (n-1)a + \frac{n}{\lambda}\)\(\operatorname{Var}[T_n] = \frac{n}{\lambda^2}\), as the deterministic times do not contribute to variance.

Key concepts

Exponential Distribution

The exponential distribution is crucial in understanding the time between events in a Poisson process. It models the time until the next event occurs. For the ATM problem, each customer arrival is considered an event. The time between these events is random and follows an exponential distribution.
The rate parameter of an exponential distribution is denoted as \( \lambda \). This rate indicates how frequently the events occur. The larger the \( \lambda \), the more frequent the events. In mathematical terms, the waiting time \( X \) for an event with an exponential distribution is expressed as \( f(x) = \lambda e^{-\lambda x} \) for \( x \geq 0 \).

• The mean of an exponential distribution is \( \frac{1}{\lambda} \).
• The variance is also \( \frac{1}{\lambda^2} \).
  

Understanding this distribution helps clarify when the next customer will arrive at the ATM when it is free.

Gamma Distribution

The Gamma distribution extends the exponential distribution to model the waiting time for several events. When adding up multiple independent exponential variables (like waiting for several customer arrivals), the result is a gamma distribution.
For the n-th arrival at the ATM, each arrival after a free period follows an exponential distribution. However, when considering all arrivals together, the waiting time cumulatively forms a Gamma distribution.
The Gamma distribution is expressed as \( \Gamma(k, \theta) \), where \( k \) is the shape parameter, representing the number of events, and \( \theta \) is the scale parameter equivalent to \( \frac{1}{\lambda} \). In this context, \( k \) equals \( n \), the number of customers served.

• The mean of a gamma distribution is \( k \times \theta \).
• The variance is \( k \times \theta^2 \).
  

This explains why the time to the n-th arrival, involving n exponential variables, is modeled by a \( \Gamma(n, \lambda) \) distribution.

Erlang Distribution

The Erlang distribution is a specialized form of the gamma distribution, where the shape parameter \( k \) is a whole number. It particularly describes the sum of \( k \) exponential variables with the same rate \( \lambda \).
In the ATM scenario, since we often deal with integer numbers of customer arrivals or services, the Erlang distribution precisely describes the distribution of the waiting time for these arrivals. When calculating the time \( T_n \) for the n-th arrival after considering \( n-1 \) previously served customers, we're looking at an Erlang distribution.

• The mean of the Erlang distribution remains as \( \frac{n}{\lambda} \).
• The variance also is \( \frac{n}{\lambda^2} \).
  

The Erlang distribution aids in precisely identifying the timing when dealing with these integer-based processes.

Arrival Distribution

Arrival distribution refers to the way customer arrivals are spread over time. In a Poisson process, arrivals are random and independent, but their inter-arrival times follow an exponential distribution.
Considering an ATM setting, the arrival distribution helps in understanding how frequently a new customer arrives when the ATM becomes free. This stochastic process dictates many real-world scenarios, from call centers to service queues.
The arrival distribution in the context of a Poisson process depends heavily on its rate \( \lambda \). The higher this rate, the more clustered the arrivals will be. Each arrival is only successful if the ATM is free, and as such, this distribution is pivotal for calculating how likely a customer is to be served without having to leave.

• Each arrival during a free ATM period follows the exponential distribution law.
• The time till the nth successful arrival follows a gamma distribution.
  

Understanding this arrival distribution clarifies how customer interactions with the ATM naturally self-organize over time.