/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 48 A taxi alternates between three ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 7: Problem 48

A taxi alternates between three different locations. Whenever it reaches location \(i\), it stops and spends a random time having mean \(t_{i}\) before obtaining another passenger, \(i=1,2\), 3. A passenger entering the cab at location \(i\) will want to go to location \(j\) with probability \(P_{i j}\). The time to travel from \(i\) to \(j\) is a random variable with mean \(m_{i j} .\) Suppose that \(t_{1}=1, t_{2}=2, t_{3}=4, P_{12}=1, P_{23}=1\), \(P_{31}=\frac{2}{3}=1-P_{32}, m_{12}=10, m_{23}=20, m_{31}=15, m_{32}=25 .\) Define an appropriate semi-Markov process and determine (a) the proportion of time the taxi is waiting at location \(i\), and (b) the proportion of time the taxi is on the road from \(i\) to \(j, i, j=1,2,3\).

Short Answer

Expert verified
(a) The proportion of time the taxi spends waiting at each location can be found using the following calculations: - \(W_1 = \frac{t_1}{t_1 + t_2 + t_3} = \frac{1}{1 + 2 + 4} = \frac{1}{7}\) - \(W_2 = \frac{t_2}{t_1 + t_2 + t_3} = \frac{2}{1 + 2 + 4} = \frac{2}{7}\) - \(W_3 = \frac{t_3}{t_1 + t_2 + t_3} = \frac{4}{1 + 2 + 4} = \frac{4}{7}\) (b) The proportion of time the taxi spends on the road between each pair of locations can be found using the following calculations: - \(R_{12} = \frac{P_{12} \cdot m_{12}}{t_1 + t_2 + t_3} = \frac{1 \cdot 10}{1 + 2 + 4} = \frac{10}{7}\) - \(R_{23} = \frac{P_{23} \cdot m_{23}}{t_1 + t_2 + t_3} = \frac{1 \cdot 20}{1 + 2 + 4} = \frac{20}{7}\) - \(R_{31} = \frac{P_{31} \cdot m_{31}}{t_1 + t_2 + t_3} = \frac{\frac{2}{3} \cdot 15}{1 + 2 + 4} = \frac{10}{7}\) - \(R_{32} = \frac{P_{32} \cdot m_{32}}{t_1 + t_2 + t_3} = \frac{\frac{1}{3} \cdot 25}{1 + 2 + 4} = \frac{25}{21}\)

Step by step solution

01

Define the Semi-Markov Process

Let's define a semi-Markov process X(t) as follows: - The state space S = {1, 2, 3} corresponds to the three different locations. - When the taxi is at state i (location i), it takes a random waiting time with mean \(t_i\) before moving to the next state j (location j). - The taxi reaches state j (location j) with probability \(P_{ij}\).
02

Calculate Proportion of Time Waiting at Each Location

To find the proportion of time the taxi is waiting at location i, we need to calculate the ratio of waiting time at location i to the total waiting time in the system. Let \(W_i\) denote the proportion of time spent waiting at location i. Thus, we have the following system of equations: - \(W_1 = \frac{t_1}{t_1 + t_2 + t_3}\) - \(W_2 = \frac{t_2}{t_1 + t_2 + t_3}\) - \(W_3 = \frac{t_3}{t_1 + t_2 + t_3}\) Plug in the given values for \(t_1, t_2, t_3\) and solve for \(W_1, W_2, W_3\).
03

Calculate Proportion of Time on the Road Between Locations

Let \(R_{ij}\) denote the proportion of time the taxi is on the road from location i to location j. To calculate this, we can use the mean travel time from i to j (\(m_{ij}\)), the third probability of traveling from i to j (\(P_{ij}\)), and the waiting time at location i (\(t_i\)). We can calculate \(R_{ij}\) using the following formula: - \(R_{ij} = \frac{P_{ij} \cdot m_{ij}}{t_1 + t_2 + t_3}\) Plug in the given values of \(P_{ij}\), \(m_{ij}\), \(t_1\), \(t_2\), and \(t_3\) to find the respective proportions for all combinations of i and j (i, j = 1, 2, 3). In summary, to solve the given problem: 1. Define a semi-Markov process based on the given information. 2. Calculate the proportion of time spent waiting at each location using the given waiting times. 3. Calculate the proportion of time on the road between each pair of locations using the given probabilities and transit mean times.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Probability Models
A crucial foundation in understanding semi-Markov processes is the concept of probability models. These models are mathematical representations that help predict the likelihood of various outcomes. Imagine you're tossing a coin; the probability model aids in forecasting the chances of getting heads or tails. In more complex scenarios, like our taxi example, the model describes how a taxi moves between locations and the probabilities associated with these transitions.

These models are defined by states that represent different outcomes or scenarios鈥攊n our case, the taxi's potential locations鈥攁nd transitions between these states that are governed by certain probabilities, such as the probability that a customer wants to go to a different location. The intricate relationship between these states and probabilities is what allows us to make predictions and calculate metrics like mean waiting times and overall system efficiency.
Mean Waiting Time
The mean waiting time is a key performance indicator in many service-based industries. It represents the average time a service provider, like our taxi, waits before the next service cycle begins. In the context of semi-Markov processes, it's particularly important because it helps determine the system's efficiency and the proportion of time the taxi spends waiting at each location compared to being on the move.

Mathematically, mean waiting time is denoted by an expected value and is related to how long a taxi lingers at a location before fetching the next passenger. By calculating the mean waiting time for each location, and considering the mean travel times, we gain insight into the operation's pacing, which is vital for optimizing service delivery and can directly impact customer satisfaction and the profitability of the service provider.
Transition Probabilities
Transition probabilities are the backbone of any Markov or semi-Markov process. They quantify the likelihood of moving from one state to another鈥攊n our taxi scenario, from one location to a different one. To visualize it, think of a city map where arrows connect various hotspots, with each arrow labeled with a probability. These values tell us how likely a passenger at point A wants to go to point B.

These probabilities are essential not only in understanding the flow of the process but also in predicting future states, which is critical for strategic planning. Specifically, they allow us to calculate the proportion of time the taxi spends on the road between locations, which is directly influenced by passengers' destination preferences. Understanding these transitions helps in forecasting demand, allocating resources, and ultimately, crafting strategies to enhance service efficiency.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Considera single-server queueing system in which customers arrive in accordance with a renewal process. Each customer brings in a random amount of work, chosen independently according to the distribution \(G\). The server serves one customer at a time. However, the seryer processes work at rate \(i\) per unit time whenever there are \(i\) customers in the system. For instance, if a customer with workload 8 enters service when there are three other customers waiting in line, then if no one else arrives that customer will spend 2 units of time in service. If another customer arrives after 1 unit of time, then our customer will spend a total of \(1.8\) units of time in service provided no one else arrives. Let \(W_{i}\) denote the amount of time customer \(i\),spends in the system, Also, define \(E[W]\) by $$ E[W]=\lim _{n \rightarrow \infty}\left(W_{1}+\cdots+W_{u}\right) / n $$ and so \(E[W]\) is the average amount of time a customer spends in the system. Let \(N\) denote the number of customers that arrive in a busy period. (a) Argue that $$ E[W]=E\left[W_{1}+\cdots+W_{N}\right] / E[N] $$ Let \(L_{i}\) denote the amount of work customer \(i\) brings into the system; and so the \(L_{i}, i \geqslant 1\), are independent random variables having distribution \(G\). (b) Argue that at any time \(t\), the sum of the times spent in the system by all arrivals prior to \(t\) is equal to the total amount of work processed by time \(t\). Hint: Consider the rate at which the server processes work. (c) Argue that $$ \sum_{i=1}^{N} W_{i}=\sum_{i=1}^{N} L_{i} $$ (d) Use Wald's equation (see Exercise 13 ) to conclude that $$ E[W]=\mu $$ where \(\mu\) is the mean of the distribution \(G\). That is, the average time that customers spend in the system is equal to the average work they bring to the system.

If \(A(t)\) and \(Y(t)\) are, respectively, the age and the excess at time \(t\) of a renewal process having an interarrival distribution \(F\), calculate $$ P\\{Y(t)>x \mid A(t)=s\\} $$

Wald's equation can be used as the basis of a proof of the elementary renewal theorem. Let \(X_{1}, X_{2}, \ldots\) denote the interarrival times of a renewal process and let \(N(t)\) be the number of renewals by time \(t .\) (a) Show that whereas \(N(t)\) is not a stopping time, \(N(t)+1\) is. Hint: Note that $$ N(t)=n \Leftrightarrow X_{1}+\cdots+X_{n} \leqslant t \quad \text { and } \quad X_{1}+\cdots+X_{n+1}>t $$ (b) Argue that $$ E\left[\sum_{i=1}^{N(t)+1} X_{i}\right]=\mu[m(t)+1] $$ (c) Suppose that the \(X_{i}\) are bounded random variables. That is, suppose there is a constant \(M\) such that \(P\left\\{X_{i}

For a renewal reward process consider $$ W_{n}=\frac{R_{1}+R_{2}+\cdots+R_{n}}{X_{1}+X_{2}+\cdots+X_{n}} $$ where \(W_{n}\) represents the average reward earned during the first \(n\) cycles. Show that \(W_{n} \rightarrow E[R] / E[X]\) as \(n \rightarrow \infty\)

For an interarrival distribution \(F\) having mean \(\mu\), we defined the equilibrium distribution of \(F\), denoted \(F_{e}\), by $$ F_{e}(x)=\frac{1}{\mu} \int_{0}^{x}[1-F(y)] d y $$ (a) Show that if \(F\) is an exponential distribution, then \(F=F_{e}\). (b) If for some constant \(c\), $$ F(x)=\left\\{\begin{array}{ll} 0, & x
See all solutions

Recommended explanations on Math Textbooks

Probability and Statistics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Applied Mathematics

Read Explanation

Discrete Mathematics

Read Explanation

Calculus

Read Explanation

Geometry

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.