Question

A facility produces items according to a Poisson process with rate \(\lambda\). However, it has shelf space for only \(k\) items and so it shuts down production whenever \(k\) items are present. Customers arrive at the facility according to a Poisson process with rate \(\mu\). Each customer wants one item and will immediately depart either with the item or empty handed if there is no item available. (a) Find the proportion of customers that go away empty handed. (b) Find the ayerage time that an item is on the shelf. (c) Find the average number of items on the shelf. Suppose now that when a customer does not find any available items it joins the "customers' queue" as long as there are no more than \(n-1\) other customers waiting at that time. If there are \(n\) waiting customers then the new arrival departs without an item. (d) Set up the balance equations. (c) In terms of the solution of the balance equations, what is the average number of customers in the system.

Short answer

(a) The proportion of customers that go away empty handed is: \[P_{empty-handed} = p_0 = \frac{1 - (\frac{\lambda}{\mu})^{k+1}}{1 - \frac{\lambda}{\mu}}\] (b) The average time that an item is on the shelf is: \[E[T] = \frac{1/\mu}{\sum_{i=1}^{k} i \frac{\lambda^i}{\mu^i} p_0}\] (c) The average number of items on the shelf is: \[\text{Avg. items on shelf} = \sum_{i=1}^{k} i \frac{\lambda^i}{\mu^i} p_0\] (d) The balance equations when there is a queue can be expressed as: \[\lambda p_i = \mu p_{i+1}\] (e) The average number of customers in the system is: \[\text{Avg. customers in system} = \sum_{i=1}^{k+n} i \frac{\lambda^i}{\mu^i} p_0\]

Step-by-step solution

Determine the equilibrium probabilities

To find the proportion of customers that go away empty handed, we first find the equilibrium probabilities of the number of items on the shelf. Denote these probabilities as \(p_0\) (no items), \(p_1\) (one item), ..., \(p_k\) (\(k\) items): \[p_i = \frac{\lambda^i}{\mu^i} p_0, \quad i=1,\ldots,k\]

Normalize probabilities

Now, normalize probabilities so that the sum of all probabilities is equal to 1: \[p_0 + p_1 + \ldots + p_k = 1 \quad \Rightarrow \quad p_0 \left(1 + \frac{\lambda}{\mu} + \left(\frac{\lambda}{\mu}\right)^2 + \ldots + \left(\frac{\lambda}{\mu}\right)^k\right) = 1\]

Calculate \(p_0\)

Notice that this is a geometric series. Evaluate \(p_0\) by using the formula for the geometric series sum: \[p_0 = \frac{1}{1 + \frac{\lambda}{\mu} + \left(\frac{\lambda}{\mu}\right)^2 + \ldots + \left(\frac{\lambda}{\mu}\right)^k} = \frac{1 - (\frac{\lambda}{\mu})^{k+1}}{1 - \frac{\lambda}{\mu}}\]

Calculate proportion of empty-handed customers

Now that we know \(p_0\), the proportion of customers that go away empty handed is the probability of the system being empty when a customer arrives, which is equal to \(p_0\): \[P_{empty-handed} = p_0 = \frac{1 - (\frac{\lambda}{\mu})^{k+1}}{1 - \frac{\lambda}{\mu}}\] (b) Average time that an item is on the shelf

Calculate time between customer arrivals

Since customers arrive at the facility according to a Poisson process with rate \(\mu\), the expected time between customer arrivals is \(1/\mu\).

Compute average item shelf time

The average time an item is on the shelf is the time between customer arrivals divided by the average number of items on the shelf. To find the average number of items on the shelf, compute the weighted sum of \(p_i\): \[\text{Avg. items on shelf} = \sum_{i=0}^{k} i p_i = \sum_{i=1}^{k} i \frac{\lambda^i}{\mu^i} p_0\] The average time an item is on the shelf is \(E[T] = \frac{1/\mu}{\sum_{i=1}^{k} i \frac{\lambda^i}{\mu^i} p_0}\). (c) Average number of items on the shelf The result has already been computed in part (b): \[\text{Avg. items on shelf} = \sum_{i=1}^{k} i \frac{\lambda^i}{\mu^i} p_0\] (d) Set up the balance equations when there is a queue

Determine state probabilities using balance equation

The balance equation for the system can be written as follows for \(i = 0, 1, \ldots\): \[\lambda p_i = \mu p_{i+1}\] Solving this equation recursively, we can express all probabilities \(p_i\) in terms of \(p_0\): \[p_i = \left(\frac{\lambda}{\mu}\right)^i p_0, \quad i = 0, 1, \ldots\] (e) Average number of customers in the system

Calculate average number of items on shelf and customers in queue

Determine how long the queue is allowed to be (\(n\)). Then calculate the weighted sum of \(p_i\), considering both the items on the shelf and the customers in the queue: \[\text{Avg. items on shelf + customers in queue} = \sum_{i=1}^{k+n} i \frac{\lambda^i}{\mu^i} p_0\]

Compute average number of customers in system

Since we count both items on the shelf and customers in the queue, the average number of customers in the system is simply the average number of items on the shelf plus the average number of customers in the queue: \[\text{Avg. customers in system} = \text{Avg. items on shelf + customers in queue} = \sum_{i=1}^{k+n} i \frac{\lambda^i}{\mu^i} p_0\]