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Consider a system where the interarrival times have an arbitrary distribution \(F\), and there is a single server whose service distribution is \(G\). Let \(D_{n}\) denote the amount of time the \(n\) th customer spends waiting in queue. Interpret \(S_{n}, T_{n}\) so that $$ D_{n+1}=\left\\{\begin{array}{ll} D_{n}+S_{n}-T_{n}, & \text { if } D_{n}+S_{n}-T_{n} \geqslant 0 \\ 0, & \text { if } D_{n}+S_{n}-T_{n}<0 \end{array}\right. $$

Short Answer

Expert verified
In the given equation for the waiting time of customers in a single server system with interarrival time distribution \(F\) and service time distribution \(G\), the variables \(S_n\) and \(T_n\) can be interpreted as follows: - \(S_n\) represents the service time for the \(n\)th customer. - \(T_n\) represents the interarrival time between the arrival of the \(n\)th customer and the arrival of the \((n+1)\)th customer.

Step by step solution

01

The variable \(D_n\) represents the amount of time the \(n\)th customer spends waiting in the queue. Our goal is to interpret the variables \(S_n\) and \(T_n\) in the given equation. #Step 2: Identify the role of Sn and Tn in the equation#

The equation is: $$ D_{n+1}=\left\{\begin{array}{ll} D_{n}+S_{n}-T_{n}, & \text { if } D_{n}+S_{n}-T_{n} \geqslant 0 \\\ 0, & \text { if } D_{n}+S_{n}-T_{n}<0 \end{array}\right. $$ This is a piecewise function, which means that the equation for \(D_{n+1}\) depends on whether or not the expression \(D_n + S_n - T_n\) is non-negative. #Step 3: Analyzing the two cases#
02

We have two cases: 1. If \(D_n + S_n - T_n \geq 0:\) In this case, the waiting time for the \((n+1)\)th customer is equal to the waiting time of the \(n\)th customer, plus some additional time \(S_n\), minus some other time \(T_n\). Here, the \((n+1)\)th customer's waiting time depends on the \(n\)th customer's waiting time, which means the server is still working on the \(n\)th customer when the \((n+1)\)th customer arrives. Hence, \(S_n\) would represent the service time of the \(n\)th customer, and \(T_n\) would represent the interarrival time between the \(n\)th and \((n+1)\)th customers. 2. If \(D_n + S_n - T_n < 0:\) In this scenario, the \((n+1)\)th customer's waiting time is zero, meaning they have no delay. This tells us that the server finished working on the \(n\)th customer before the \((n+1)\)th customer arrives. In this case, \(S_n\) represents the service time of the \(n\)th customer, and \(T_n\) is the interarrival time between the \(n\)th and \((n+1)\)th customers, similar to the previous case. #Step 4: Interpreting Sn and Tn#

Based on the analysis in step 3, we can interpret \(S_n\) and \(T_n\) as follows: - \(S_n\) is the service time for the \(n\)th customer. - \(T_n\) is the interarrival time between the arrival of the \(n\)th customer and the arrival of the \((n+1)\)th customer.

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Most popular questions from this chapter

Carloads of customers arrive at a single-server station in accordance with a Poisson process with rate 4 per hour. The service times are exponentially distributed with rate 20 per hour. If each carload contains either 1,2, or 3 customers with respective probabilities \(\frac{1}{4}, \frac{1}{2}\), and \(\frac{1}{4}\), compute the average customer delay in queue.

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.

In a queue with unlimited waiting space, arrivals are Poisson (parameter \(\lambda\) ) and service times are exponentially distributed (parameter \(\mu\) ). However, the server waits until \(K\) people are present before beginning service on the first customer; thereafter, he services one at a time until all \(K\) units, and all subsequent arrivals, are serviced. The server is then "idle" until \(K\) new arrivals have occurred. (a) Define an appropriate state space, draw the transition diagram, and set up the balance equations. (b) In terms of the limiting probabilities, what is the average time a customer spends in queue? (c) What conditions on \(\lambda\) and \(\mu\) are necessary?

A group of \(m\) customers frequents a single-server station in the following manner. When a customer arrives, he or she either enters service if the server is free or joins the queue otherwise. Upon completing service the customer departs the system, but then returns after an exponential time with rate \(\theta\). All service times are exponentially distributed with rate \(\mu\). (a) Find the average rate at which customers enter the station. (b) Find the average time that a customer spends in the station per visit.

Customers arrive at a single-server station in accordance with a Poisson process having rate \(\lambda .\) Each customer has a value. The successive values of customers are independent and come from a uniform distribution on \((0,1)\). The service time of a customer having value \(x\) is a random variable with mean \(3+4 x\) and variance \(5 .\) (a) What is the average time a customer spends in the system? (b) What is the average time a customer having value \(x\) spends in the system?

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