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Consider a train station to which customers arrive in accordance with a Poisson process having rate \(\lambda\). A train is summoned whenever there are \(N\) customers waiting in the station, but it takes \(K\) units of time for the train to arrive at the station. When it arrives, it picks up all waiting customers. Assuming that the train station incurs a cost at a rate of \(n c\) per unit time whenever there are \(n\) customers present, find the long-run average cost.

Short Answer

Expert verified
The long-run average cost of the train station can be found by interpreting it as an M/G/1 queue system. Considering the arrival rate λ, service rate μ = N/K, and using Little's Law, we have L = λK, where L is the average number of customers in the system. The long-run average cost per unit time is then C = L*c = λKc.

Step by step solution

01

Analyze the problem as an M/G/1 queue system

The problem can be interpreted as an M/G/1 queue, where Poisson arrivals are served in batches whenever the train arrives (every K units of time). A train is summoned with N customers in the system, meaning there needs to be N-1 customers when the train is arriving.
02

Determine the arrival rate and service rate

The arrival rate is given, and it is λ. The service rate can be calculated as μ = N/K, as a train takes K units of time to arrive and picks up all N waiting customers.
03

Calculate the utilization factor

The utilization factor, ρ, in an M/G/1 queue can be calculated as the ratio between arrival rate and service rate, which is ρ = λ/μ. In this case, ρ = λK/N.
04

Apply Little's Law to find average number of customers in the system

Using Little's Law, we have L = λW, where L is the average number of customers in the system and W is the average time a customer spends in the system. Since W = K, we have L = λK.
05

Calculate the long-run average cost per unit time

We know that there is a cost at a rate of n*c per unit time for n customers present in the system. The long-run average cost per unit time, denoted by C, can be calculated as C = L*c, where L is the average number of customers in the system and c is the cost per customer per unit time. In this case, C = λKc. Thus, the long-run average cost of the train station is λKc per unit time.

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Most popular questions from this chapter

Consider a renewal process with mean interarrival time \(\mu .\) Suppose that each event of this process is independently "counted" with probability \(p\). Let \(N_{C}(t)\) denote the number of counted events by time \(t, t>0\). (a) Is \(N_{C}(t), t \geqslant 0\) a renewal process? (b) What is \(\lim _{t \rightarrow \infty} N_{C}(t) / t ?\)

Consider a 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 server 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 $$ E[W]=\lim _{n \rightarrow \infty}\left(W_{1}+\cdots+W_{n}\right) / n $$ 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.

To prove Equation ( \(7.24)\), define the following notation: \(X_{i}^{j} \equiv\) time spent in state \(i\) on the \(j\) th visit to this state; \(N_{i}(m) \equiv\) number of visits to state \(i\) in the first \(m\) transitions In terms of this notation, write expressions for (a) the amount of time during the first \(m\) transitions that the process is in state \(i ;\) (b) the proportion of time during the first \(m\) transitions that the process is in state \(i\) Argue that, with probability 1 , (c) \(\sum_{j=1}^{N_{i}(m)} \frac{X_{i}^{j}}{N_{i}(m)} \rightarrow \mu_{i}\) as \(m\) (d) \(\mathrm{N}_{i}(m) / m \rightarrow \pi_{i} \quad\) as \(m \rightarrow \infty\). (e) Combine parts (a), (b), (c), and (d) to prove Equation (7.24).

Consider a single-server bank for which customers arrive in accordance with a Poisson process with rate \(\lambda .\) If a customer will enter the bank only if the server is free when he arrives, and if the service time of a customer has the distribution \(G\), then what proportion of time is the server busy?

A truck driver regularly drives round trips from \(\mathrm{A}\) to \(\mathrm{B}\) and then back to \(\mathrm{A}\). Each time he drives from \(A\) to \(B\), he drives at a fixed speed that (in miles per hour) is uniformly distributed between 40 and \(60 ;\) each time he drives from \(\mathrm{B}\) to \(\mathrm{A}\), he drives at a fixed speed that is equally likely to be either 40 or 60 . (a) In the long run, what proportion of his driving time is spent going to \(\mathrm{B}\) ? (b) In the long run, for what proportion of his driving time is he driving at a speed of 40 miles per hour?

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