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The manager of a market can hire either Mary or Alice. Mary, who gives service at an exponential rate of 20 customers per hour, can be hired at a rate of \(\$ 3\) per hour. Alice, who gives service at an exponential rate of 30 customers per hour, can be hired at a rate of \(\$ C\) per hour. The manager estimates that, on the average, each customer's time is worth \(\$ 1\) per hour and should be accounted for in the model. If customers arrive at a Poisson rate of 10 per hour, then (a) what is the average cost per hour if Mary is hired? if Alice is hired? (b) find \(C\) if the average cost per hour is the same for Mary and Alice.

Short Answer

Expert verified
The average cost per hour is \$3.5 when Mary is hired and \(\$1/6 + C\) when Alice is hired. To make the average cost per hour the same for both Mary and Alice, the hiring rate for Alice should be around \$3.\overline{3} per hour.

Step by step solution

01

Calculate the average number of customers in the system for Mary

Let's first compute the average number of customers in the system (\(L_s\)) when Mary is serving customers. We are given that customers arrive at a Poisson rate of 10 per hour, and Mary provides service at an exponential rate of 20 customers per hour. To compute the average number of customers in the system (\(L_s\)) when the service and arrival rates are \(\lambda\) and \(\mu\), respectively, we use the formula: \[L_s = \frac{\lambda^2}{\mu(\mu - \lambda)}\] For Mary: - \(\lambda = 10\) (customers arrival rate) - \(\mu = 20\) (service rate) Plug the values into the formula: \[L_s = \frac{10^2}{20(20 - 10)} = \frac{100}{200} = 0.5\text{ customers}\]
02

Calculate the average number of customers in the system for Alice

Now we'll compute the average number of customers in the system when Alice is serving customers. We're given that Alice provides service at an exponential rate of 30 customers per hour. For Alice: - \(\lambda = 10\) (customers arrival rate) - \(\mu = 30\) (service rate) Plug the values into the formula: \[L_s = \frac{10^2}{30(30 - 10)} = \frac{100}{600} = \frac{1}{6}\text{ customers}\]
03

Calculate the average cost per hour for Mary and Alice

We can now compute the average cost per hour for both Mary and Alice using the formula: \[Average\, Cost\, per\, Hour = L_s(\$1) + \$ Hiring\, Rate\] For Mary, \(L_s = 0.5\) and Hiring Rate \(= \$3\): \[Average\, Cost\, per\, Hour = 0.5(\$1) + \$3 = \$3.5\] For Alice, \(L_s = \frac{1}{6}\) and Hiring Rate \(= \$C\): \[Average\, Cost\, per\, Hour = \frac{1}{6}(\$1) + \$C = \frac{1}{6} + C\]
04

Find the value of \(C\) that makes the average cost per hour the same for both Mary and Alice

To find the value of \(C\), we equate the average cost per hour for Mary and Alice and solve for \(C\): \[\$3.5 = \frac{1}{6} + C\] Subtract \(\frac{1}{6}\) from both sides of the equation: \[3.5 - \frac{1}{6} = C\] \[\implies C = \$3.\overline{3}\] Thus, if the manager wants the average cost per hour to be the same for Mary and Alice, the hiring rate for Alice should be around \( \$3.\overline{3}\) per hour.

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Key Concepts

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

Poisson Process
The Poisson Process is a statistical model used to describe a series of events that occur independently of one another, at a constant average rate. In queuing theory, this can refer to the random arrival of customers to a service station, such as in the manager’s market scenario. The arrivals are described as a Poisson process when two conditions are met: first, the number of events occurring in non-overlapping intervals of time is independent, and second, the probability of an event occurring in a small time interval is proportional to the length of that interval.

Using a Poisson process allows us to model and calculate the likelihood of events such as customer arrivals per unit of time. For instance, in our exercise, the arrival rate is given as 10 customers per hour. This means that, on average, 10 customers are expected to arrive every hour, and the time between arrivals is governed by an exponential distribution, which leads us to our next core concept.
Exponential Service Rate
The exponential service rate is a key component of queueing theory where service times follow an exponential distribution. This property allows us to describe situations where the length of time required to service one customer doesn’t depend on how long previous customers have been serviced. In the educational platform's context, where either Mary or Alice can be hired, they have different service rates – 20 and 30 customers per hour, respectively.

An exponential service rate also implies that the probability of service completion in a short time interval is the same throughout the service process. This memoryless property is typical of exponential random variables and is appropriate for service systems like the one described in the original exercise. When we say Mary has an exponential service rate of 20 customers per hour, it means that, on average, she can serve one customer every 3 minutes (\frac{1}{20} of an hour). Similarly, Alice's rate of 30 customers per hour means she can serve one customer every 2 minutes. These service rates are essential for determining the average number of customers in the system, represented by the formula \(L_s = \frac{\text{arrival rate}^2}{\text{service rate} (\text{service rate} - \text{arrival rate})}\).
Cost Analysis
Cost analysis in the context of queuing theory involves calculating the total cost associated with operating a service system, including both direct and indirect costs. Direct costs may include wages paid to employees, while indirect costs can be the opportunity cost of customers’ time spent waiting.

In our exercise involving Mary and Alice, cost analysis is used to determine the economic feasibility of hiring either employee based on their service rates and corresponding hourly wages. The manager must account for the cost of customers' time (valued at \$1 per hour) in addition to the hiring rate of the service provider. The average cost per hour formula, \(Average\text{ Cost per Hour} = L_s(\$1) + \$ Hiring\text{ Rate}\), combines the indirect cost of customers waiting (the product of the average number of customers in the system, \(L_s\), and the cost of a customer's time) with the direct cost of hiring the service provider. By performing cost analysis, the manager can make data-driven decisions that balance customer satisfaction (minimizing waiting time) with operational efficiency and affordability. In conclusion, Mary or Alice's hire should align with the market's overall strategy for maintaining budget while ensuring customer service quality.

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Most popular questions from this chapter

Suppose in Exercise 17 we want to find out the proportion of time there is a type 1 customer with server \(2 .\) In terms of the long-run probabilities given in Exercise 17 , what is (a) the rate at which a type 1 customer enters service with server \(2 ?\) (b) the rate at which a type 2 customer enters service with server \(2 ?\) (c) the fraction of server 2 's customers that are type \(1 ?\) (d) the proportion of time that a type 1 customer is with server \(2 ?\)

Consider a network of three stations. Customers arrive at stations 1 , 2, 3 in accordance with Poisson processes having respective rates \(5,10,15\). The service times at the three stations are exponential with respective rates \(10,50,100\). A customer completing service at station 1 is equally likely to (a) 8o to station 2, (b) go to station 3 , or \((\mathrm{c})\) leave the system. A customer departing service at station 2 always goes to station 3. A departure from service at station 3 is equally likely to either go to station 2 or leave the system. (i) What is the average number of customers in the system (consisting of all three stations)? (ii) What is the average time a customer spends in the system?

Consider a \(M / G / 1\) system with \(\lambda E[S]<1\). (a) Suppose that service is about to begin at a moment when there are \(n\) customers in the system. (i) Argue that the additional time until there are only \(n-1\) customers in the system has the same distribution as a busy period. (ii) What is the expected additional time until the system is empty? (b) Suppose that the work in the system at some moment is \(A\). We are interested in the expected additional time until the system is empty- call it \(E[T] .\) Let \(N\) denote the number of arrivals during the first \(A\) units of time. (i) Compute \(E[T \mid N]\). (ii) Compute \(E[T]\).

Consider an \(M / G / 1\) system in which the first customer in a busy period has service distribution \(G_{1}\) and all others have distribution \(G_{2}\). Let C denote the number of customers in a busy period, and let \(S\) denote the service time of a customer chosen at random. Argue that (a) \(a_{0}=P_{0}=1-\lambda E[S]\). (b) \(E[S]=a_{0} E\left[S_{1}\right]+\left(1-a_{0}\right) E\left[S_{2}\right]\) where \(S_{i}\) has distribution \(G_{i}\). (c) Use (a) and (b) to show that \(E[B]\), the expected length of a busy period, is given by $$ E[B]=\frac{E\left[S_{1}\right]}{1-\lambda E\left[S_{2}\right]} $$ (d) Find \(E[C]\).

Carloads of customers arrive at a single-server station in accordance to a Poisson process with rate 4 per hour. The service times are exponentially distributed with rate 20 per hour. If each carload contains cither 1,2 , or 3 customers with respective probabilities \(\frac{1}{8}, \frac{1}{2}, \frac{1}{4}\), compute the average customer delay in queue.

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