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Chapter 8: Problem 17

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 \(1 ?\)

Short Answer

Expert verified
(a) The fraction of customers who do not enter the system is \( \frac{1}{8}\) (b) The average amount of time that an entering customer spends in the system is \( \frac{49}{384} \) hours. (c) The fraction of entering customers receiving service from server 1 is \( \frac{3}{8}\).

Step by step solution

01

Model the system

First, we need to model the system using a M/M/c/C model where c = 2, since we have two servers. The arrival rate (位) is two per hour and the service rates at server 1 (渭1) and server 2 (渭2) are four and six per hour, respectively.
02

Calculate the state probabilities

Next, we need to calculate the state probabilities using the steady-state balance equations: \(P_0 = \frac{1}{1 + \frac{\lambda}{\mu_1} + \frac{\lambda^2}{\mu_1 \mu_2}}\) \(P_1 = \frac{\lambda}{\mu_1} P_0\) \(P_2 = \frac{\lambda^2}{\mu_1 \mu_2} P_0\) We have: \(P_0 = \frac{1}{1 + \frac{2}{4} + \frac{2^2}{4 \cdot 6}} = \frac{1}{1 + 0.5 + \frac{1}{6}} = \frac{1}{\frac{16}{12}} = \frac{3}{4}\) \(P_1 = \frac{2}{4} P_0 = \frac{1}{2} \cdot \frac{3}{4} = \frac{3}{8}\) \(P_2 = \frac{2^2}{4 \cdot 6} P_0 = \frac{1}{6} \cdot \frac{3}{4} = \frac{1}{8}\)
03

Calculate system performance measures

(a) The fraction of customers who do not enter the system is represented by the probability of the system being in state 2 (both servers are busy). Thus, we have: Fraction of customers who do not enter the system = \(P_2 = \frac{1}{8}\) (b) To calculate the average amount of time that an entering customer spends in the system, we first need to find the average time spent in each server. For an M/M/c/C model, the expected service time for server i is given by: \(E[T_i] = \frac{1}{\mu_i}\) \(E[T_1] = \frac{1}{4} = 0.25h\) \(E[T_2] = \frac{1}{6} = \frac{1}{6}h\) Now, we can find the average amount of time that an entering customer spends in the system by considering the probabilities of going through each server: Average time in the system = \(E[T_1] P_1 + E[T_2] P_2 = 0.25 \cdot \frac{3}{8} + \frac{1}{6} \cdot \frac{1}{8} = \frac{3}{32} + \frac{1}{48} = \frac{49}{384} h\) (c) The fraction of entering customers who receive service from server 1 is represented by the probability of the system being in state 1 (only server 1 is busy): Fraction of entering customers receiving service from server 1 = \(P_1 = \frac{3}{8}\) Final answers: (a) The fraction of customers who do not enter the system is \( \frac{1}{8}\) (b) The average amount of time that an entering customer spends in the system is \( \frac{49}{384} \) hours. (c) The fraction of entering customers receiving service from server 1 is \( \frac{3}{8}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Poisson Process
In queueing theory, a Poisson process is a fundamental concept used to model random events occurring over time. An intuitive way to think of it is as a stream of events happening independently at a constant average rate. In our exercise, customers arrive at a two-server station according to a Poisson process at a rate of two customers per hour. - **Constant Rate**: This means that, on average, there are two arrivals each hour, but the exact timing between arrivals can vary. - **Independence**: Each customer's arrival is independent of the others. The arrival of one customer does not affect the likelihood of another arriving soon after. - **Probability**: The Poisson process can be used to determine the probability of a certain number of arrivals in a fixed period. Poisson processes are particularly useful in scenarios like this, where events happen randomly but at a predictable average rate. This makes the Poisson process a key tool in analyzing and predicting customer arrivals in queueing systems.
Exponential Distribution
The exponential distribution often pairs with the Poisson process due to its simplicity and useful properties in modeling the time between events in a Poisson process. In our scenario, the service times for servers are modeled using exponential random variables, which follow the exponential distribution. - **Memoryless Property**: One of the most important features of the exponential distribution is its memoryless property. This means that the time until the next event (e.g., service completion) is independent of the time elapsed since the last event.- **Rate Parameter (\(\mu\))**: The rate parameter in the exponential distribution represents how often events occur. For server 1, \(\mu_1 = 4\) per hour meaning it serves one customer approximately every 15 minutes, and for server 2, \(\mu_2 = 6\) per hour meaning approximately every 10 minutes.- **Applications**: The exponential distribution is typically used to model the duration of events, such as service times in queues or system lifetimes in reliability studies.This distribution simplifies calculations and helps in analyzing the dynamics of queues, providing insights into expected service times and waiting periods.
M/M/c Queue
The M/M/c queue is a classic model in queueing theory used to analyze multi-server systems. It captures the behavior of queues where arrivals follow a Poisson process, service times are exponentially distributed, and there are multiple servers available.- **M/M/2 Configuration**: In the exercise, the M/M/c model is specifically an M/M/2 queue, meaning two servers are available to handle service requests.- **Arrival Rate (\(\lambda\))**: The arrival rate of customers is denoted by \(\lambda = 2\) per hour, reflecting the incoming flow of customers modeled by a Poisson process.- **Service Rates (\(\mu_1, \mu_2\))**: Service rates for the servers are given as \(\mu_1 = 4\) and \(\mu_2 = 6\), representing exponentially distributed service times.- **Blocking**: Because there's no waiting line, when both servers are busy, incoming customers are 'lost', highlighting the importance of balancing service rate and arrival rate in queue systems.The M/M/c model is crucial for system planners and managers to understand queue dynamics, optimize server utilization, and improve customer satisfaction.
Steady-State Probabilities
Within queueing theory, steady-state probabilities are essential for understanding long-term behavior of queues when the system reaches equilibrium. These probabilities help determine the proportion of time the system is in various states.- **Equilibrium Criteria**: At steady state, the properties of the system remain constant over time, meaning the rate of entries equals the rate of exits in the queueing model.- **Probability Calculations**: With the equations shown, the probabilities \(P_0\), \(P_1\), and \(P_2\) were calculated to determine how often each state occurs: - \(P_0\) (neither server busy): \(\frac{3}{4}\) - \(P_1\) (only server 1 busy): \(\frac{3}{8}\) - \(P_2\) (both servers busy): \(\frac{1}{8}\)- **Insights**: These probabilities directly influence performance metrics like: - The fraction of customers who do not enter the system. - The likelihood of a server being busy or free. - Customer wait times and service levels.Understanding steady-state probabilities supports better decision-making for resources allocation and enhancing service processes in multi-server queueing systems.

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Most popular questions from this chapter

For the \(M / G / 1\) queue, let \(X_{n}\) denote the number in the system left behind by the nth departure. (a) If $$ X_{n+1}=\left\\{\begin{array}{ll} X_{n}-1+Y_{n}, & \text { if } X_{n} \geqslant 1 \\ Y_{n}, & \text { if } X_{n}=0 \end{array}\right. $$ what does \(Y_{n}\) represent? (b) Rewrite the preceding as $$ X_{n+1}=X_{n}-1+Y_{n}+\delta_{n} $$ where $$ \delta_{n}=\left\\{\begin{array}{ll} 1, & \text { if } X_{n}=0 \\ 0, & \text { if } X_{n} \geqslant 1 \end{array}\right. $$ Take expectations and let \(n \rightarrow \infty\) in Equation ( \(8.64\) ) to obtain $$ E\left[\delta_{\infty}\right]=1-\lambda E[S] $$ (c) Square both sides of Equation \((8.64)\), take expectations, and then let \(n \rightarrow \infty\) to obtain $$ E\left[X_{\infty}\right]=\frac{\lambda^{2} E\left[S^{2}\right]}{2(1-\lambda E[S])}+\lambda E[S] $$ (d) Argue that \(E\left[X_{\infty}\right]\), the average number as seen by a departure, is equal to \(L\).

In a queue with unlimited waiting space, arrivals are Poisson (parameter \(\lambda\) ) and service times are exponentially distributed (parameter \(\mu\) ). However, the server waits until \(K\) people are present before beginning service on the first customer; thereafter, he services one at a time until all \(K\) units, and all subsequent arrivals, are serviced. The server is then "idle" until \(K\) new arrivals have occurred. (a) Define an appropriate state space, draw the transition diagram, and set up the balance equations. (b) In terms of the limiting probabilities, what is the average time a customer spends in queue? (c) What conditions on \(\lambda\) and \(\mu\) are necessary?

Customers arrive at a single-server station in accordance with a Poisson process with rate \(\lambda\). All arrivals that find the server free immediately enter service. All service times are exponentially distributed with rate \(\mu\). An arrival that finds the server busy will leave the system and roam around "in orbit" for an exponential time with rate \(\theta\) at which time it will then return. If the server is busy when an orbiting customer returns, then that customer returns to orbit for another exponential time with rate \(\theta\) before returning again. An arrival that finds the server busy and \(N\) other customers in orbit will depart and not return. That is, \(N\) is the maximum number of customers in orbit. (a) Define states. (b) Give the balance equations. In terms of the solution of the balance equations, find (c) the proportion of all customers that are eventually served; (d) the average time that a served customer spends waiting in orbit.

Suppose that a customer of the \(M / M / 1\) system spends the amount of time \(x>0\) waiting in queue before entering service. (a) Show that, conditional on the preceding, the number of other customers that were in the system when the customer arrived is distributed as \(1+P\), where \(P\) is a Poisson random variable with mean \(\lambda\). (b) Let \(W_{Q}^{*}\) denote the amount of time that an \(M / M / 1\) customer spends in queue. As a by-product of your analysis in part (a), show that $$ P\left[W_{\mathrm{Q}}^{*} \leqslant x\right]=\left\\{\begin{array}{ll} 1-\frac{\lambda}{\mu} & \text { if } x=0 \\ 1-\frac{\lambda}{\mu}+\frac{\lambda}{\mu}\left(1-e^{-(\mu-\lambda) x}\right) & \text { if } x>0 \end{array}\right. $$

Consider the priority queueing model of Section \(8.6 .2\) but now suppose that if a type 2 customer is being served when a type 1 arrives then the type 2 customer is bumped out of service. This is called the preemptive case. Suppose that when a bumped type 2 customer goes back in service his service begins at the point where it left off when he was bumped. (a) Argue that the work in the system at any time is the same as in the nonpreemptive case. (b) Derive \(W_{Q}^{1}\) Hint: How do type 2 customers affect type 1 s? (c) Why is it not true that $$ V_{Q}^{2}=\lambda_{2} E\left[S_{2}\right] W_{Q}^{2} $$ (d) Argue that the work seen by a type 2 arrival is the same as in the nonpreemptive case, and so \(W_{Q}^{2}=W_{Q}^{2}\) (nonpreemptive) \(+E[\) extra time \(]\) where the extra time is due to the fact that he may be bumped. (e) Let \(N\) denote the number of times a type 2 customer is bumped. Why is $$ E[\text { extra time } \mid \mathrm{N}]=\frac{N E\left[S_{1}\right]}{1-\lambda_{1} E\left[S_{1}\right]} $$ Hint: When a type 2 is bumped, relate the time until he gets back in service to a "busy period." (f) Let \(S_{2}\) denote the service time of a type \(2 .\) What is \(E\left[N \mid S_{2}\right] ?\) (g) Combine the preceding to obtain $$ W_{Q}^{2}=W_{Q}^{2} \text { (nonpreemptive) }+\frac{\lambda_{1} E\left[S_{1}\right] E\left[S_{2}\right]}{1-\lambda_{1} E\left[S_{1}\right]} $$
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