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The economy alternates between good and bad periods. During good times customers arrive at a certain single-server queueing system in accordance with a Poisson process with rate \(\lambda_{1}\), and during bad times they arrive in accordance with a Poisson process with rate \(\lambda_{2} .\) A good time period lasts for an exponentially distributed time with rate \(\alpha_{1}\), and a bad time period lasts for an exponential time with rate \(\alpha_{2}\). An arriving customer will only enter the queueing system if the server is free; an arrival finding the server busy goes away. All service times are exponential with rate \(\mu .\) (a) Define states so as to be able to analyze this system. (b) Give a set of linear equations whose solution will yield the long-run proportion of time the system is in each state. In terms of the solutions of the equations in part (b), (c) what proportion of time is the system empty? (d) what is the average rate at which customers enter the system?

Short Answer

Expert verified
The queueing system can be analyzed by defining 4 states: \(S(0,0)\), \(S(1,0)\), \(S(0,1)\), and \(S(1,1)\), representing the number of customers in the system (0 or 1) and the economy condition (good or bad). Linear equations can be set up to find the long-run proportion of time spent in each state: 1. \(P_{00}(\lambda_2 + \alpha_1) = P_{10}\mu + P_{01}\alpha_2\) 2. \(P_{10}(\mu + \alpha_1) = P_{00}\lambda_2\) 3. \(P_{01}(\lambda_1 + \alpha_2) = P_{11}\mu + P_{00}\alpha_1\) 4. \(P_{11}(\mu + \alpha_2) = P_{01}\lambda_1\) 5. \(P_{00} + P_{10} + P_{01} + P_{11} = 1\) The proportion of time the system is empty is given by \(P_E = P_{00} + P_{01}\). The average customer arrival rate can be found as \(\overline{\lambda} = P_{00}\lambda_2 + P_{01}\lambda_1\). Solve the system of linear equations for the long-run probabilities and substitute the values into the corresponding equations to find the proportion of time the system is empty and the average customer arrival rate.

Step by step solution

01

(Define States)

To analyze this queueing system, we need to define the states. Let's denote the state by \(S(n,g)\), where \(n\) represents the number of customers in the system (either 0 or 1, since new customers only join if the server is free), and \(g\) is a binary indicator for the economy condition (1 for good times, and 0 for bad times). This gives us 4 possible states: - State \(S(0,0)\): No customer in the system, and the economy is in a bad state. - State \(S(1,0)\): One customer in the system, and the economy is in a bad state. - State \(S(0,1)\): No customer in the system, and the economy is in a good state. - State \(S(1,1)\): One customer in the system, and the economy is in a good state.
02

(Set up Linear Equations)

We will now set up linear equations representing the flow of probabilities between these states. Let \(P_{ij}\) denote the long-run proportion of time spent in state \(S(i,j)\). For state \(S(0,0)\): \(P_{00}(\lambda_2 + \alpha_1) = P_{10}\mu + P_{01}\alpha_2\) For state \(S(1,0)\): \(P_{10}(\mu + \alpha_1) = P_{00}\lambda_2\) For state \(S(0,1)\): \(P_{01}(\lambda_1 + \alpha_2) = P_{11}\mu + P_{00}\alpha_1\) For state \(S(1,1)\): \(P_{11}(\mu + \alpha_2) = P_{01}\lambda_1\) Additionally, we have the normalization constraint: \(P_{00} + P_{10} + P_{01} + P_{11} = 1\)
03

(Proportion of Time System is Empty)

To find the proportion of time the system is empty, we need to consider the probabilities of states \(S(0,0)\) and \(S(0,1)\), which represent the cases when no customers are in the system. Therefore, we'll add the probabilities as follows: \[P_E = P_{00} + P_{01}\]
04

(Average Customer Arrival Rate)

To find the average customer arrival rate, we need to calculate the weighted average of the arrival rates in good and bad times, based on the proportion of time in each state. That is: \(\overline{\lambda} = \frac{P_{00}\lambda_2 + P_{01}\lambda_1}{P_{00} + P_{01} + P_{10} + P_{11}}\) Since we are given that \(P_{00} + P_{01} + P_{10} + P_{11} = 1\), this simplifies to: \(\overline{\lambda} = P_{00}\lambda_2 + P_{01}\lambda_1\) To calculate the proportion of time system is empty and the average customer arrival rate, we must first solve the system of linear equations for the long-run probabilities (\(P_{00}, P_{01}, P_{10}, P_{11}\)) and then substitute the values into the corresponding equations.

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Most popular questions from this chapter

Consider a sequential-service system consisting of two servers, \(A\) and \(B\). Arriving customers will enter this system only if server \(A\) is free. If a customer does enter, then he is immediately served by server \(A\). When his service by \(A\) is completed, he then goes to \(B\) if \(B\) is free, or if \(B\) is busy, he leaves the system. Upon completion of service at server \(B\), the customer departs. Assume that the (Poisson) arrival rate is two customers an hour, and that \(A\) and \(B\) serve at respective (exponential) rates of four and two customers an hour. (a) What proportion of customers enter the system? (b) What proportion of entering customers receive service from B? (c) What is the average number of customers in the system? (d) What is the average amount of time that an entering customer spends in the system?

Potential customers arrive to a single-server hair salon according to a Poisson process with rate \(\lambda .\) A potential customer who finds the server free enters the system; a potential customer who finds the server busy goes away. Each potential customer is type \(i\) with probability \(p_{i}\), where \(p_{1}+p_{2}+p_{3}=1\). Type 1 customers have their hair washed by the server; type 2 customers have their hair cut by the server; and type 3 customers have their hair first washed and then cut by the server. The time that it takes the server to wash hair is exponentially distributed with rate \(\mu_{1}\), and the time that it takes the server to cut hair is exponentially distributed with rate \(\mu_{2}\). (a) Explain how this system can be analyzed with four states. (b) Give the equations whose solution yields the proportion of time the system is in each state. In terms of the solution of the equations of (b), find (c) the proportion of time the server is cutting hair; (d) the average arrival rate of entering customers.

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 $$ W_{Q}^{2}=W_{Q}^{2} \text { (nonpreemptive) }+\frac{\lambda_{1} E\left[S_{1}\right] E\left[S_{2}\right]}{1-\lambda_{1} E\left[S_{1}\right]} $$

Consider the \(\mathrm{M} / \mathrm{M} / 1\) system in which customers arrive at rate \(\lambda\) and the server serves at rate \(\mu .\) However, suppose that in any interval of length \(h\) in which the server is busy there is a probability \(\alpha h+o(h)\) that the server will experience a breakdown, which causes the system to shut down. All customers that are in the system depart, and no additional arrivals are allowed to enter until the breakdown is fixed. The time to fix a breakdown is exponentially distributed with rate \(\beta\). (a) Define appropriate states. (b) Give the balance equations. In terms of the long-run probabilities, (c) what is the average amount of time that an entering customer spends in the system?

Compare the \(M / G / 1\) system for first-come, first-served queue discipline with one of last-come, first-served (for instance, in which units for service are taken from the top of a stack). Would you think that the queue size, waiting time, and busyperiod distribution differ? What about their means? What if the queue discipline was always to choose at random among those waiting? Intuitively, which discipline would result in the smallest variance in the waiting time distribution?

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