/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 3 The manager of a market can hire... [FREE SOLUTION] | 91影视

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The manager of a market can hire either Mary or Alice. Mary, who gives service at an cxponential 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
If the average cost per hour is the same for both Mary and Alice, Alice's hourly wage \(C = \$3.333\).

Step by step solution

01

Find the average cost per hour for Mary

To find the average cost per hour for Mary, we can first calculate the rate at which her service will be utilized. Mary serves customers at an exponential rate of 20 per hour and customers arrive at a Poisson rate of 10 per hour. So the utilization rate for Mary is: Mary's Utilization rate (蟻) = arrival rate (位) / service rate (渭) = \(\frac{10}{20}\) = 0.5 Now we account for customer's hourly time worth: Number of people in the queue (Lq) = \(\frac{蟻^2}{(1 - 蟻)}\) = \(\frac{0.5^2}{(1 - 0.5)}\) = 0.5 Cost of waiting customers per hour = Lq 脳 value of customer's time per hour = 0.5 脳 \(1 = \)0.50 Total average cost per hour for Mary = hourly wage + cost of waiting customers = \(3 + \)0.50 = $3.50
02

Find the average cost per hour for Alice

Now, let's compute Alice's average cost per hour, in a similar manner: Alice's Utilization rate (蟻) = arrival rate (位) / service rate (渭) = \(\frac{10}{30}\) = \(\frac{1}{3}\) Number of people in the queue (Lq) = \(\frac{蟻^2}{(1 - 蟻)}\) = \(\frac{(\frac{1}{3})^2}{(1 - \frac{1}{3})}\) = \(\frac{1}{6}\) Cost of waiting customers per hour = Lq 脳 value of customer's time per hour = \(\frac{1}{6}\) 脳 \(1 = \)0.167 Total average cost per hour for Alice = hourly wage + cost of waiting customers = \(C + \)0.167
03

Find C when average costs per hour are equal

Now we will set the average cost per hour for both Mary and Alice equal and solve for C: \(3.50 = \)C + $0.167 C = \(3.50 - \)0.167 = $3.333 So, if the average cost per hour is to be the same for both Mary and Alice, Alice's hourly wage C should be $3.333.

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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 fundamental concept used to model the rate of random events that occur independently over time. It is frequently applied in queuing theory to describe phenomena such as the arrival of customers in a system. The arrivals are usually spread out over time in such a way that the number of occurrences in any given time interval can be predicted by a Poisson distribution.

For example, in the provided exercise, customers arrive at a market at a Poisson rate of 10 per hour. This means the arrivals are random but with a predictable average rate. Understanding the Poisson process is crucial as it lays the foundation for calculating other queuing metrics, like queue utilization rate and service times.
Queue Utilization Rate
The queue utilization rate, often denoted by the symbol \(\rho\), measures the proportion of time the service provider is busy. This rate is determined by the ratio of the arrival rate of customers (\(\lambda\)) to the service rate (\(\mu\)). It's important as it gives us insight into how often the service provider, such as Mary or Alice in our exercise, is actively engaged in service versus being idle.

Another utility of calculating \(\rho\) is in identifying the efficiency of the system. If the value is close to 1, it means that the system is nearly always busy and may be at risk of overloading. Values much smaller than 1 signal underutilization, meaning that resources may not be used effectively. In the calculation step for Mary and Alice, the utilization rates help us determine the cost of customer waiting time, a critical component in overall cost assessment.
Average Cost Per Hour Calculation
Calculating the average cost per hour for a service provider in a queuing system combines operational costs with customer waiting costs. For service providers like Mary and Alice in the given exercise, the operational cost is their hourly wage. The customer waiting cost is estimated by the average number of people in the queue (\(L_q\)) multiplied by the value of each customer's time per hour.

By determining the average cost per hour, an organization can assess the immediate financial implications of hiring a service provider. It's also a key figure in making decisions on the cost-effectiveness of service operations, taking into account not only the direct expenses but also the potential loss due to customer wait times.
Cost-Benefit Analysis
Cost-benefit analysis is a decision-making tool that compares the costs of an action or investment to the benefits that result from it. By assigning monetary values to both costs and benefits, it helps in choosing the option that maximizes profit or minimizes losses.

In queuing systems, this analysis might involve weighing the cost of hiring a service provider against the benefits of reduced waiting times for customers, which could translate to higher customer satisfaction and potential increased sales. In our exercise, finding the wage for Alice where the average cost per hour is the same for both Mary and Alice engages the principles of cost-benefit analysis to make a strategic staffing decision.

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Most popular questions from this chapter

Customers arrive at a two-server system according to a Poisson process having rate \(\lambda=5\). An arrival finding server 1 free will begin service with that server. An arrival finding server 1 busy and server 2 free will enter service with server 2. An arrival finding both servers busy goes away. Once a customer is served by either server, he departs the system. The service times at server \(i\) are exponential with rates \(\mu_{i}\), where \(\mu_{1}=4, \mu_{2}=2\) (a) What is the average time an entering customer spends in the system? (b) What proportion of time is server 2 busy?

Consider the \(M / M / 1\) 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? (d) what proportion of entering customers complete their service? (c) what proportion of customers arrive during a breakdown?

A group of \(n\) customers moves around among two servers. Upon completion of service, the served customer then joins the queue (or enters service if the server is free) at the other server. All service times are exponential with rate \(\mu\). Find the proportion of time that there are \(j\) customers at server 1 , \(j=0, \ldots, n\)

Consider a single-server queue with Poisson arrivals and exponential service times having the following variation: Whenever a service is completed a departure occurs only with probability \(\alpha\). With probability \(1-\alpha\) the customer, instead of leaving, joins the end of the queue. Note that a customer may be serviced more than once. (a) Set up the balance equations and solve for the steady-state probabilities, stating conditions for it to exist. (b) Find the expected waiting time of a customer from the time he arrives until he enters service for the first time. (c) What is the probability that a customer enters service exactly \(n\) times, \(n=\) \(1,2, \ldots ?\) (d) What is the expected amount of time that a customer spends in service (which does not include the time he spends waiting in line)? Use part (c). (c) What is the distribution of the total length of time a customer spends being served? 7 Is it memoryless?

The economy alternates between good and bad periods. During good times customers arriye 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?

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