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Chapter 6: Problem 34

Four workers share an office that contains four telephones. At any time, each worker is either "working" or "on the phone." Each "working" period of worker \(i\) lasts for an exponentially distributed time with rate \(\lambda_{i}\), and each "on the phone" period lasts for an exponentially distributed time with rate \(\mu_{i}, i=1,2,3,4\). (a) What proportion of time are all workers "working"? Let \(X_{i}(t)\) equal 1 if worker \(i\) is working at time \(t\), and let it be 0 otherwise. Let \(\mathrm{X}(t)=\left(X_{1}(t), X_{2}(t), X_{3}(t), X_{4}(t)\right)\) (b) Argue that \(\\{\mathrm{X}(t), t \geqslant 0\\}\) is a continuous-time Markov chain and give its infinitesimal rates. (c) Is \(\\{\mathrm{X}(t)\\}\) time reversible? Why or why not? Suppose now that one of the phones has broken down. Suppose that a worker who is about to use a phone but finds them all being used begins a new "working" period. (d) What proportion of time are all workers "working"?

Short Answer

Expert verified
In summary, to find the proportion of time all workers are working, we define the state space of the continuous-time Markov chain, compute its stationary distribution, and use it to find the probability of all workers working. We compute the infinitesimal rates using the given parameters and analyze whether the Markov Chain maintains time reversibility by checking the detailed balance equation. Finally, we analyze a modified scenario with a broken phone, adjust the transition rates accordingly, and recompute the stationary distribution and proportion of time in which all workers are working.

Step by step solution

01

(1) Define the State Space

The state space of the continuous-time Markov Chain consists of all possible combinations of workers' statuses, i.e., whether they are working or on the phone. Since there are four workers, the state space can be represented as a set of 4-tuples: \[S = \{(X_1(t), X_2(t), X_3(t), X_4(t)) \mid X_i(t) \in \{0, 1\}, \forall i \in \{1, 2, 3, 4\}\}\] There are \(2^4 = 16\) possible states in the state space \(S\).
02

(2) Find the Proportion of Time All Workers are Working (Part a)

We are asked to find the proportion of time that all workers are working, that is, \(P(X_i(t) = 1, \forall i \in \{1, 2, 3, 4\})\). This probability can be computed by finding the stationary distribution of the continuous-time Markov Chain. The transition rates between states can be written in matrix form, where the diagonal elements represent the rates of leaving the state, and the off-diagonal elements represent the rates of transitioning to other states. Let's call this matrix \(Q\). Compute the stationary distribution, represented as a probability vector \(\pi\), by solving the following system of linear equations: \[\pi Q = 0\] and \[\sum_{s \in S} \pi(s) = 1\] Once we find the stationary distribution, the proportion of time all workers are working is \(\pi(1, 1, 1, 1)\).
03

(3) Give the Infinitesimal Rates (Part b)

The infinitesimal rates of the continuous-time Markov chain are determined by the transition rates between states. Recall that: 1. Each "working" period of worker \(i\) lasts for an exponentially distributed time with rate \(\lambda_i\), 2. Each "on the phone" period lasts for an exponentially distributed time with rate \(\mu_i\), for \(i = 1, 2, 3, 4\). The infinitesimal rate of transitioning from state \(s\) to state \(s'\) can be written as: \[q_{ss'} = \begin{cases} \lambda_i & \text{if s and s' differ in the } i\text{-th component and } X_i(s) = 0 \\ \mu_i & \text{if s and s' differ in the } i\text{-th component and } X_i(s) = 1 \\ -\sum_{s' \neq s} q_{ss'} & \text{if } s = s' \\ 0 & \text{otherwise} \end{cases}\]
04

(4) Time Reversibility (Part c)

A continuous-time Markov chain is time reversible if the following detailed balance equation holds for all pairs of states \(s\) and \(s'\): \[\pi(s)q_{ss'} = \pi(s')q_{s's}\] Since we have computed the stationary distribution \(\pi\) and the infinitesimal rates \(Q\), we can check whether the detailed balance equation is satisfied for all pairs of states. If it holds, then the Markov chain is time reversible; otherwise, it is not.
05

(5) Proportion of Time All Workers are Working with a Broken Phone (Part d)

Let's consider the new scenario where one phone is broken, and a worker who tries to use a phone but finds them all in use starts a new working period. In this case, we need to recompute the transition rates and update the matrix \(Q\). The updated rates can be written as: \[q_{ss'} = \begin{cases} \lambda_i & \text{if s and s' differ in the } i\text{-th component and } X_i(s) = 0 \\ \mu_i & \text{if s and s' differ in the } i\text{-th component and } X_i(s) = 1 \text{ and a phone is available} \\ \lambda_i & \text{if s and s' differ in the } i\text{-th component and } X_i(s) = 1 \text{ and all phones are in use} \\ -\sum_{s' \neq s} q_{ss'} & \text{if } s = s' \\ 0 & \text{otherwise} \end{cases}\] Solve the system of linear equations with the new transition rates to find the updated stationary distribution \(\pi'\): \[\pi' Q' = 0\] and \[\sum_{s \in S} \pi'(s) = 1\] The proportion of time all workers are working when one phone is broken is given by \(\pi'(1, 1, 1, 1)\).

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Key Concepts

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

Exponential Distribution
Understanding the exponential distribution is crucial when dealing with continuous-time Markov chains. This distribution plays a key role in modelling the time between events in a process where events occur independently at a constant average rate. For the office problem, the exponential distribution is used to model the time lengths of both 'working' and 'on the phone' periods for each worker.

The main characteristic of the exponential distribution is its memoryless property, which means that the probability of an event occurring in the future is independent of how much time has already passed. Mathematically, if the time until an event occurs follows an exponential distribution with rate \(\lambda_i\) for worker \(i\), then the probability that the event occurs within time \(t\) is given by \(1 - e^{-\lambda_i t}\).

When we say that a worker's 'working' period follows an exponential distribution with a rate \(\lambda_i\), we're stating that the expected time for the worker to stay working before switching to 'on the phone' is \(1/\lambda_i\) and vice versa for the 'on the phone' period with rate \(\mu_i\). This concept is applied directly in building the transition rate matrix for the continuous-time Markov chain.
Stationary Distribution
Stationary distribution provides valuable insight into the long-term behavior of a Markov chain. In the context of our office scenario, the stationary distribution tells us the long-run proportion of time the system spends in each state, essentially revealing how often all workers are working or on the phone.

The stationary distribution \(\pi\) is a probability distribution over the states that remains unchanged as time progresses. It's found by solving the equation \(\pi Q = 0\) together with the normalization condition \(\sum_{s \in S} \pi(s) = 1\), where \(Q\) is the transition rate matrix. These equations ensure that the distribution does not change from one moment to the next. In other words, it is 'stationary'.

For the problem at hand, finding the stationary distribution involves determining the probability vector \(\pi\) such that the long-run proportion of time all workers are working is \(\pi(1, 1, 1, 1)\). This quantification is crucial for operational planning such as allocating tasks or estimating productivity.
Time Reversibility
Time reversibility is a special property that can make analyzing continuous-time Markov chains easier. A Markov chain is said to be time reversible if the process looks the same whether time moves forwards or backwards. In mathematical terms, a Markov chain is time reversible if for any two states \(s\) and \(s'\), the rate of transitioning from \(s\) to \(s'\) under the stationary distribution is the same as transitioning from \(s'\) to \(s\).

To check if the workers' office scenario Markov chain is time reversible, one utilizes the detailed balance equations: \(\pi(s)q_{ss'} = \pi(s')q_{s's}\). If these hold true for all state pairs \(s\) and \(s'\), the Markov chain is confirmed to be time reversible. Time reversibility implies that the process's behavior is naturally balanced over time between any two states, a property that can simplify the analysis and potentially offer insights into the symmetry of the system's behavior.

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Most popular questions from this chapter

There are two machines, one of which is used as a spare. A working machine will function for an exponential time with rate \(\lambda\) and will then fail. Upon failure, it is immediately replaced by the other machine if that one is in working order, and it goes to the repair facility. The repair facility consists of a single person who takes an exponential time with rate \(\mu\) to repair a failed machine. At the repair facility, the newly failed machine enters service if the repairperson is free. If the repairperson is busy, it waits until the other machine is fixed; at that time, the newly repaired machine is put in service and repair begins on the other one. Starting with both machines in working condition, find (a) the expected value and (b) the variance of the time until both are in the repair facility. (c) In the long run, what proportion of time is there a working machine?

Consider a graph with nodes \(1,2, \ldots, n\) and the \(\left(\begin{array}{l}n \\\ 2\end{array}\right) \operatorname{arcs}(t, j), i \neq j, i, j,=1, \ldots, n\) (See Section 3.6.2 for appropriate definitions.) Suppose that a particle moves along this graph as follows: Events occur along the arcs \((i, j)\) according to independent Poisson processes with rates \(\lambda_{i j} .\) An event along arc \((i, j)\) causes that arc to become excited. If the particle is at node \(i\) at the moment that \((i, j)\) becomes excited, it instantaneously moves to node \(j, i, j=1, \ldots, n .\) Let \(P_{j}\) denote the proportion of time that the particle is at node \(j .\) Show that $$ P_{j}=\frac{1}{n} $$ Hint: Use time reversibility.

Potential customers arrive at a full-service, one-pump gas station at a Poisson rate of 20 cars per hour. However, customers will only enter the station for gas if there are no more than two cars (including the one currently being attended to) at the pump. Suppose the amount of time required to service a car is exponentially distributed with a mean of five minutes. (a) What fraction of the attendant's time will be spent servicing cars? (b) What fraction of potential customers are lost?

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 $$ P_{i j}(t)=\left(\begin{array}{l} j-1 \\ i-1 \end{array}\right) e^{-\lambda t i}\left(1-e^{-\lambda t}\right)^{j-i} $$

Each individual in a biological population is assumed to give birth at an exponential rate \(\lambda\), and to die at an exponential rate \(\mu .\) In addition, there is an exponential rate of increase \(\theta\) due to immigration. However, immigration is not allowed when the population size is \(N\) or larger. (a) Set this up as a birth and death model. (b) If \(N=3,1=\theta=\lambda, \mu=2\), determine the proportion of time that immigration is restricted.
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