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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

Expert verified
(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

01

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): \[p_i = \frac{\lambda^i}{\mu^i} p_0, \quad i=1,\ldots,k\]
02

Normalize probabilities

Now, normalize probabilities so that the sum of all probabilities is equal to 1: \[p_0 + p_1 + \ldots + p_k = 1 \quad \Rightarrow \quad p_0 \left(1 + \frac{\lambda}{\mu} + \left(\frac{\lambda}{\mu}\right)^2 + \ldots + \left(\frac{\lambda}{\mu}\right)^k\right) = 1\]
03

Calculate \(p_0\)

Notice that this is a geometric series. Evaluate \(p_0\) by using the formula for the geometric series sum: \[p_0 = \frac{1}{1 + \frac{\lambda}{\mu} + \left(\frac{\lambda}{\mu}\right)^2 + \ldots + \left(\frac{\lambda}{\mu}\right)^k} = \frac{1 - (\frac{\lambda}{\mu})^{k+1}}{1 - \frac{\lambda}{\mu}}\]
04

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
05

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 \(1/\mu\).
06

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
07

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
08

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: \[\text{Avg. items on shelf + customers in queue} = \sum_{i=1}^{k+n} i \frac{\lambda^i}{\mu^i} p_0\]
09

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: \[\text{Avg. customers in system} = \text{Avg. items on shelf + customers in queue} = \sum_{i=1}^{k+n} i \frac{\lambda^i}{\mu^i} p_0\]

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Most popular questions from this chapter

Arrivals to a three-server system are according to a Poisson process with rate \(\lambda\). Arrivals finding server 1 free enter service with \(1 .\) Arrivals finding 1 busy but 2 free enter service with \(2 .\) Arrivals finding both 1 and 2 busy do not join the system. After completion of service at either 1 or 2 the customer will then either go to server 3 if 3 is free or depart the system if 3 is busy. After service at 3 customers depart the system. The service times at \(i\) are exponential with rate \(\mu_{i}, i=1,2,3\). (a) Define states to analyze the above system. (b) Give the balance equations. (c) In terms of the solution of the balance equations, what is the average time that an entering customer spends in the system? (d) Find the probability that a customer who arrives when the system is empty is served by server 3 .

Customers arrive at a two-server station in accordance with a Poisson process with a rate of two per hour. Arrivals finding server 1 free begin service with that server. Arrivals finding server 1 busy and server 2 free begin service with server 2. Arrivals finding both servers busy are lost. When a customer is served by server 1 , she then either enters service with server 2 if 2 is free or departs the system if 2 is busy. \(\mathrm{A}\) customer completing service at server 2 departs the system. The service times at server 1 and server 2 are exponential random variables with respective rates of four and six per hour. (a) What fraction of customers do not enter the system? (b) What is the average amount of time that an entering customer spends in the system? (c) What fraction of entering customers receives service from server \(1 ?\)

Consider a single-server queue with Poisson arrivals and exponential service times having the following variation: Whenever a service is completed a departure occurs only with probability \(\alpha .\) With probability \(1-\alpha\) the customer, instead of leaving, joins the end of the queue. Note that a customer may be serviced more than once. (a) Set up the balance equations and solve for the steady-state probabilities, stating conditions for it to exist. (b) Find the expected waiting time of a customer from the time he arrives until he enters service for the first time. (c) What is the probability that a customer enters service exactly \(n\) times, \(n=\) \(1,2, \ldots ?\) (d) What is the expected amount of time that a customer spends in service (which does not include the time he spends waiting in line)? Hint: Use part (c). (e) What is the distribution of the total length of time a customer spends being served? Hint: Is it memoryless?

Potential customers arrive to a single-server hair salon according to a Poisson process with rate \(\lambda .\) A potential customer who finds the server free enters the system; a potential customer who finds the server busy goes away. Each potential customer is type \(i\) with probability \(p_{i}\), where \(p_{1}+p_{2}+p_{3}=1\). Type 1 customers have their hair washed by the server; type 2 customers have their hair cut by the server; and type 3 customers have their hair first washed and then cut by the server. The time that it takes the server to wash hair is exponentially distributed with rate \(\mu_{1}\), and the time that it takes the server to cut hair is exponentially distributed with rate \(\mu_{2}\). (a) Explain how this system can be analyzed with four states. (b) Give the equations whose solution yields the proportion of time the system is in each state. In terms of the solution of the equations of (b), find (c) the proportion of time the server is cutting hair; (d) the average arrival rate of entering customers.

A group of \(m\) customers frequents a single-server station in the following manner. When a customer arrives, he or she either enters service if the server is free or joins the queue otherwise. Upon completing service the customer departs the system, but then returns after an exponential time with rate \(\theta\). All service times are exponentially distributed with rate \(\mu\). (a) Find the average rate at which customers enter the station. (b) Find the average time that a customer spends in the station per visit.

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