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Chapter 6: Problem 24

Consider a taxi station where taxis and customers arrive in accordance with Poisson processes with respective rates of one and two per minute. A taxi will wait no matter how many other taxis are present. However, an arriving customer that does not find a taxi waiting leaves. Find (a) the average number of taxis waiting, and (b) the proportion of arriving customers that get taxis.

Short Answer

Expert verified
In conclusion, (a) The average number of waiting taxis is \( \frac{1}{3}\). (b) The proportion of arriving customers that get taxis is \( \frac{2}{3}\).

Step by step solution

01

Identify the birth-death process components

In this case, let's identify the birth and death rates of the queue at any state i: - Birth rate (位): The arrival rate of customers, which is given as 2 customers per minute. - Death rate (渭): The arrival rate of taxi, which is given as 1 taxi per minute. Let's denote the probability that there are i taxis waiting as P(i).
02

Formulate balance equations for the steady state

Now we will write balance equations to ensure traffic conservation in and out of the states: 1. For state 0, the balance equation would be 位P(0) = 渭P(1), since there are no taxis waiting. 2. For states i 鈮� 1, the balance equation would be 位P(i - 1) = 渭P(i), since taxis can arrive and depart.
03

Solve balance equations

We shall solve the balance equations to find P(i) for all states. 1. For state 0, the equation \(位P(0) = 渭P(1)\) becomes \(2P(0) = P(1)\). 2. For states i 鈮� 1, the equation \( 位P(i - 1) = 渭P(i)\) becomes \(2P(i - 1) = P(i)\). Now, notice that we have a recursive relation in P(i), so the probabilities for states i 鈮� 1 can be found as follows: \(P(1) = 2P(0)\) \(P(2) = 2P(1) = 4P(0)\) \(P(3) = 2P(2) = 6P(0)\) \(P(i) = 2i P(0)\), for all i.
04

Normalize probabilities

We need to normalize the probabilities to ensure:\( \sum_{i = 0}^{\infty} P(i) = 1\). Summation formula for i = 0 to infinity, 1 = P(0) + \(2P(0)\) + \(4P(0)\) + \(6P(0)\) + ... = P(0)(1 + 2 + 4 + 6 + ...) The sum of the series can be represented as: \( \sum_{i = 1}^{\infty} 2iP(0) = 2P(0) \sum_{i = 1}^{\infty} i \) The sum of positive integers up to infinity is given by: \(\sum_{i = 1}^{\infty} i = \frac{1}{4}\) Now, we have: 1 = P(0)(1 + 2 * \(\frac{1}{4}\)) = P(0)\(\frac{3}{2}\) Thus, we find the probability for state 0 as, \( P(0) = \frac{2}{3}\).
05

Find the average number of waiting taxis (a)

The average number of waiting taxis (L) can be found as: \( L = \sum_{i = 1}^{\infty} iP(i)\) Using our summation relation from step 4, we get: \( L = \sum_{i = 1}^{\infty} i(2iP(0)) = 2P(0) \sum_{i = 1}^{\infty} i^2\) Knowing that the sum of the squares of positive integers up to infinity is given by: \(\sum_{i = 1}^{\infty} i^2 = \frac{1}{12}\) We can now calculate the average number of waiting taxis: \(L = 2P(0) * \frac{1}{12} = 2 * \frac{2}{3} * \frac{1}{12} = \frac{1}{3}\) So, the average number of waiting taxis is 1/3.
06

Find the proportion of arriving customers that get taxis (b)

The proportion of arriving customers who get taxis is given by the probability of not finding a waiting taxi (i.e., in state 0, which is P(0)): Proportion = P(0) = \(\frac{2}{3}\) So, 2/3 of the arriving customers get taxis. In conclusion, (a) The average number of waiting taxis is 1/3. (b) The proportion of arriving customers that get taxis is 2/3.

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Most popular questions from this chapter

Let \(Y\) denote an exponential random variable with rate \(\lambda\) that is independent of the continuous-time Markov chain \(\\{X(t)\\}\) and let $$ \bar{P}_{i j}=P\\{X(Y)=j \mid X(0)=i\\} $$ (a) Show that $$ \bar{P}_{i j}=\frac{1}{v_{i}+\lambda} \sum_{k} q_{i k} \bar{P}_{k j}+\frac{\lambda}{v_{i}+\lambda} \delta_{i j} $$ where \(\delta_{i j}\) is 1 when \(i=j\) and 0 when \(i \neq j\). (b) Show that the solution of the preceding set of equations is given by $$ \overline{\mathbf{P}}=(\mathbf{I}-\mathbf{R} / \lambda)^{-1} $$ where \(\overline{\mathbf{P}}\) is the matrix of elements \(\bar{P}_{i j}, \mathbf{I}\) is the identity matrix, and \(\mathbf{R}\) the matrix specified in Section \(6.8\). (c) Suppose now that \(Y_{1}, \ldots, Y_{n}\) are independent exponentials with rate \(\lambda\) that are independent of \(\\{X(t)\\}\). Show that $$ P\left\\{X\left(Y_{1}+\cdots+Y_{n}\right)=j \mid X(0)=i\right\\} $$ is equal to the element in row \(i\), column \(j\) of the matrix \(\overline{\mathbf{p}}^{n}\). (d) Explain the relationship of the preceding to Approximation 2 of Section \(6.8\).

In Example \(6.20\), we computed \(m(t)=E[O(t)]\), the expected occupation time in state 0 by time \(t\) for the two-state continuous-time Markov chain starting in state \(0 .\) Another way of obtaining this quantity is by deriving a differential equation for it. (a) Show that $$ m(t+h)=m(t)+P_{00}(t) h+o(h) $$ (b) Show that $$ m^{\prime}(t)=\frac{\mu}{\lambda+\mu}+\frac{\lambda}{\lambda+\mu} e^{-(\lambda+\mu) t} $$ (c) Solve for \(m(t)\).

Suppose that a one-celled organism can be in one of two states-either \(A\) or \(B\). An individual in state \(A\) will change to state \(B\) at an exponential rate \(\alpha ;\) an individual in state \(B\) divides into two new individuals of type \(A\) at an exponential rate \(\beta\). Define an appropriate continuous-time Markov chain for a population of such organisms and determine the appropriate parameters for this model.

A total of \(N\) customers move about among \(r\) servers in the following manner. When a customer is served by server \(i\), he then goes over to server \(j, j \neq i\), with probability \(1 /(r-1)\). If the server he goes to is free, then the customer enters service; otherwise he joins the queue. The service times are all independent, with the service times at server \(i\) being exponential with rate \(\mu, i=1, \ldots, r .\) Let the state at any time be the vector \(\left(n_{1}, \ldots, n_{r}\right)\), where \(n_{i}\) is the number of customers presently at server \(i, i=1, \ldots, r, \sum_{i} n_{i}=N\) (a) Argue that if \(X(t)\) is the state at time \(t\), then \(\\{X(t), t \geqslant 0\\}\), is a continuoustime Markov chain. (b) Give the infinitesimal rates of this chain. (c) Show that this chain is time reversible, and find the limiting probabilities.

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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