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Chapter 6: Problem 15

A service center consists of two servers, each working at an exponential rate of two services per hour. If customers arrive at a Poisson rate of three per hour, then, assuming a system capacity of at most three customers, (a) what fraction of potential customers enter the system? (b) what would the value of part (a) be if there was only a single server, and his rate was twice as fast (that is, \(\mu=4)\) ?

Short Answer

Expert verified
(a) For an M/M/2 queuing system with a capacity of 3 customers, the fraction of potential customers who enter the system is approximately 1.0883. (b) For an M/M/1 queuing system with a service rate of 4 and a capacity of 3 customers, the fraction of potential customers who would enter the system is approximately 1.2406.

Step by step solution

01

Determine parameters

In this part of the exercise, we know that customers arrive with a Poisson rate of 3 per hour (位 = 3) and each server provides an exponential service rate of 2 services per hour (渭 = 2).
02

Calculate probabilities

For an M/M/2/c queue system with steady-state probabilities, we can calculate the state probabilities using the following formula: \(P_n = \frac{(位/渭)^n}{n!} * P_0\), for \(n = 0, 1, ..., c\). where - \(P_n\) is the probability of n customers in the system. - \(P_0\) is the probability of an empty system. - 位 is the arrival rate. - 渭 is the service rate. - c is the maximum number of customers allowed in the system. Calculate \(P_0\) using the formula: \(P_0 = \frac{1}{ \sum_{n=0}^{c}{\frac{(位/渭)^n}{n!}}}\). For this exercise: \(位 \ = 3\), 渭 = 2, and \(c = 3\). Therefore, \(P_0 = \frac{1}{ \sum_{n=0}^{3}{\frac{(3/2)^n}{n!}}} = \frac{1}{\frac{1}{1!} + \frac{3/2}{1!} + \frac{(3/2)^2}{2!} + \frac{(3/2)^3}{2!*2}} = 0.4706\). Using this value for \(P_0\), we compute the probabilities: \(P_0 = 0.4706\), \(P_1 = 0.4706 * \frac{(3/2)}{1!} = 0.3530\), \(P_2 = 0.4706 * \frac{(3/2)^2}{2!} = 0.2647\), \(P_3 = 0.4706 * \frac{(3/2)^3}{2!*2} = 0.1986\).
03

Calculate the probability of acceptance

A potential customer enters the system if there are fewer than three customers in the center. Thus, the fraction of potential customers who enter the system is the probability of having 0, 1, or 2 customers: Fraction of accepted customers = \(P_0 + P_1 + P_2 = 0.4706 + 0.3530 + 0.2647 = 1.0883\). (b) M/M/1 Queuing System with Limited Capacity and 渭 = 4 per hour
04

Determine parameters

In this part of the exercise, we know that customers arrive with a Poisson rate of 3 per hour (位 = 3) and the single server provides an exponential service rate of 4 services per hour (渭 = 4).
05

Calculate probabilities

For an M/M/1/c queue system, we can calculate the state probabilities using the same formulas as in part (a) with single server rate 渭 = 4. Calculate \(P_0\) using the formula: \(P_0 = \frac{1}{ \sum_{n=0}^{c}{\frac{(位/渭)^n}{n!}}} = \frac{1}{ \sum_{n=0}^{3}{\frac{(3/4)^n}{n!}} } = 0.5714\). Using this value for \(P_0\), we compute the probabilities: \(P_0 = 0.5714\), \(P_1 = 0.5714 * \frac{(3/4)}{1!} = 0.4286\), \(P_2 = 0.5714 * \frac{(3/4)^2}{2!} = 0.2406\), \(P_3 = 0.5714 * \frac{(3/4)^3}{3!} = 0.0802\).
06

Calculate the probability of acceptance

A potential customer enters the system if there are fewer than three customers in the center. Thus, the fraction of potential customers who enter the system is the probability of having 0, 1, or 2 customers: Fraction of accepted customers = \(P_0 + P_1 + P_2 = 0.5714 + 0.4286 + 0.2406 = 1.2406\). So the fraction of potential customers who would enter the system in part (b) is approximately 1.2406.

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Key Concepts

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

Poisson Process
A Poisson process is a commonly used mathematical model that describes random events occurring over time. In our service center, the customers' arrivals follow a Poisson process. This is characterized by the rate at which customers arrive, often denoted by the parameter \( \lambda \). In our example, \( \lambda = 3聽\), meaning customers arrive on average at a rate of 3 per hour.
Key features of a Poisson process include:
  • Events are independent, meaning the arrival of a customer doesn鈥檛 affect the next one.
  • Arrivals are memoryless; the time until the next arrival does not depend on previous arrivals.
  • Only one event can occur at a time, enforcing a continuous arrival process.
These features make the Poisson process highly practical for modeling scenarios like server demands where calls or customers arrive independently of each other.
Exponential Distribution
The exponential distribution is crucial in queuing models, especially in representing service times. For our service center, each server works at an exponential rate of two services per hour. The service times follow an exponential distribution with parameter \( \mu \), which indicates the average number of services completed per hour. Here, \( \mu = 2 \) for each server.
Characteristics of exponential distribution include:
  • Memoryless property, much like the Poisson process, meaning the probability of completing a service in the next moment is independent of how long service has been happening.
  • Good for modeling service mechanisms where events happen continuously and independently.
  • Only characterized by one parameter \( \mu \), simplifying the analysis of service systems.
In our queue scenarios, the exponential rate helps in predicting how long customers have to wait and how efficient the service process is.
M/M/2 Queue
The M/M/2 queue model is a specific type of queuing system known under the broader term "birth-death processes". Here "M" stands for memoryless, describing the Poisson arrival and exponential service processes. The number "2" in M/M/2 indicates there are two service channels or servers available.
This model helps analyze systems like our service center with:
  • Two simultaneous servers processing the arriving customers.
  • A maximum capacity constraint which, in this exercise, is set to 3 customers.
  • The processes by which customers are served can be impacted by having multiple servers.
    For instance, multiple servers help in reducing wait times and improving service efficiency.
Understanding the M/M/2 queue allows operators to anticipate queues' behavior under various conditions of service and arrival rates.
Steady-State Probabilities
Steady-state probabilities represent the long-term behavior of a queuing system where arrival and service rates have reached a balance. These probabilities tell us how likely the system is to be in a particular state (number of customers). For our exercise, calculating these probabilities helps us understand the average system occupancy.
To find these probabilities, often the state probability \( P_n \) for \( n \) customers must be calculated:
\( P_n = \frac{(\lambda/\mu)^n}{n!} * P_0 \)
where \( P_0 \) is the probability of an empty system, and can also be computed using summed probabilities over all states defined by the capacity constraint \( c \).
Tools like steady-state probabilities lead to data-driven decisions in designing and managing queuing systems, ensuring customer satisfaction by properly managing resources and planning for peak times.

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Most popular questions from this chapter

Consider a Yule process starting with a single individual-that is, suppose \(X(0)=1\). Let \(T_{i}\) denote the time it takes the process to go from a population of size \(i\) to one of size \(i+1\) (a) Argue that \(T_{i}, i=1, \ldots, j\), are independent exponentials with respective rates i\lambda. (b) Let \(X_{1}, \ldots, X_{j}\) denote independent exponential random variables each having rate \(\lambda\), and interpret \(X_{i}\) as the lifetime of component \(i\). Argue that \(\max \left(X_{1}, \ldots, X_{j}\right)\) can be expressed as $$ \max \left(X_{1}, \ldots, X_{i}\right)=\varepsilon_{1}+\varepsilon_{2}+\cdots+\varepsilon_{j} $$ where \(\varepsilon_{1}, \varepsilon_{2}, \ldots, \varepsilon_{j}\) are independent exponentials with respective rates \(j \lambda\) \((j-1) \lambda, \ldots, \lambda\) Hint: Interpret \(\varepsilon_{i}\) as the time between the \(i-1\) and the ith failure. (c) Using (a) and (b) argue that $$ P\left[T_{1}+\cdots+T_{j} \leqslant t\right\\}=\left(1-e^{-\lambda t}\right)^{j} $$ (d) Use (c) to obtain $$ P_{1 j}(t)=\left(1-e^{-\lambda t}\right)^{j-1}-\left(1-e^{-\lambda t}\right)^{j}=e^{-\lambda t}\left(1-e^{-\lambda t}\right)^{j-1} $$ and hence, given \(X(0)=1, X(t)\) has a geometric distribution with parameter \(p=e^{-\lambda t}\) (e) Now conclude that $$ P_{i j}(t)=\left(\begin{array}{l} j-1 \\ i-1 \end{array}\right) e^{-\lambda t i}\left(1-e^{-\lambda t}\right)^{j-i} $$

In the \(\mathrm{M} / \mathrm{M} / \mathrm{s}\) queue if you allow the service rate to depend on the number in the system (but in such a way so that it is ergodic), what can you say about the output process? What can you say when the service rate \(\mu\) remains unchanged but \(\lambda>s \mu ?\)

Individuals join a club in accordance with a Poisson process with rate \(\lambda\). Each new member must pass through \(k\) consecutive stages to become a full member of the club. The time it takes to pass through each stage is exponentially distributed with rate \(\mu\). Let \(N_{i}(t)\) denote the number of club members at time \(t\) who have passed through exactly \(i\) stages, \(i=1, \ldots, k-1 .\) Also, let \(\mathrm{N}(t)=\left(N_{1}(t), N_{2}(t), \ldots, N_{k-1}(t)\right)\) (a) Is \(\\{\mathbf{N}(t), t \geqslant 0\\}\) a continuous-time Markov chain? (b) If so, give the infinitesimal transition rates. That is, for any state \(\mathrm{n}=\) \(\left(n_{1}, \ldots, n_{k-1}\right)\) give the possible next states along with their infinitesimal rates.

Let \(O(t)\) be the occupation time for state 0 in the two-state continuous-time Markov chain. Find \(E[O(t) \mid X(0)=1]\).

Consider two machines that are maintained by a single repairman. Machine \(i\) functions for an exponential time with rate \(\mu_{i}\) before breaking down, \(i=1,2 .\) The repair times (for either machine) are exponential with rate \(\mu .\) Can we analyze this as a birth and death process? If so, what are the parameters? If not, how can we analyze it?
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