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Machines in a factory break down at an exponential rate of six per hour. There is a single repairman who fixes machines at an exponential rate of eight per hour. The cost incurred in lost production when machines are out of service is \(\$ 10\) per hour per machine. What is the average cost rate incurred due to failed machines?

Short Answer

Expert verified
The average cost rate incurred due to failed machines is \(\$30\) per hour.

Step by step solution

01

Identify the Failure Rate and Repair Rate

In the problem, we are given the failure rate and repair rate for machines. The failure rate \(\lambda\) is six per hour, and the repair rate \(\mu\) is eight per hour.
02

Calculate Average Number of Failed Machines

Using the fact that the average number of failed machines can be found using the formula \(\frac{\lambda}{\mu-\lambda}\), we can plug in our failure and repair rates: \[\text{Average Number of Failed Machines} = \frac{6}{8-6} = \frac{6}{2} = 3\]
03

Calculate the Average Cost Rate

Now that we have the average number of failed machines, we can calculate the average cost rate. Given that the cost incurred per hour per failed machine is \(\$10\), the average cost rate is: \[ \text{Average Cost Rate} = 3 \cdot 10 = \$ 30 \text{ per hour} \] Thus, the average cost rate incurred due to failed machines is \(\$30\) per hour.

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Key Concepts

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

Understanding Exponential Rate and Failure Rate
When discussing the reliability of machines or systems, the term failure rate, denoted as \( \lambda \), is crucial. It represents the frequency at which an engineered system or component fails, expressed in failures per time unit, often an hour.

In the context of our example, the failure rate of the factory's machines is six failures per hour. This number represents an exponential distribution of failures meaning that the time between each failure is random, but the average number of failures per hour remains constant. This kind of rate is typical in electronics and mechanical systems where failures occur independently and at a constant average rate over time.

Understanding the failure rate is foundational when calculating the impact of breakdowns on production costs and scheduling maintenance or repairs. For students, it's important to grasp that the failure rate doesn't necessarily mean six machines will fail every single hour on the dot but that this is the average over a larger time span.
Analyzing the Repair Rate
The repair rate, often denoted as \( \mu \), is the average number of repairs that can be completed in a given time frame by a service agent—in this case, the repairman. For our factory, the repairman fixes machines at a rate of eight per hour. This rate is also modeled exponentially, which means the repair times are random but on average, eight machines are serviced every hour.

It is essential to note the repair rate should exceed the failure rate to avoid an unmanageable accumulation of broken machines. In our case, the repair rate is higher, which allows for a steady flow of machines being fixed and returned to service. Students should take away that a reliable repair process is vital for minimizing production downtime and associated costs.
Calculating the Average Cost Rate
The average cost rate is a crucial figure that quantifies the financial impact of intermittent machine failures on the production process. To compute it, multiply the average number of failed machines by the cost incurred per machine per hour. In our given example, the calculation yielded an average cost rate of \(30 per hour, arising from the three machines that are, on average, down at any given time.

To further break it down, every hour that passes with a machine out of operation represents a loss of \)10 in production value. By understanding the average number of failures and the cost associated with each, businesses can plan better and allocate resources effectively to mitigate these losses. For students, comprehending how failure and repair rates translate into direct costs is beneficial for managing operational efficiency in any automated or machine-reliant environment.

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Most popular questions from this chapter

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.

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?

In an \(M / G / 1\) queue, (a) what proportion of departures leave behind 0 work? (b) what is the average work in the system as seen by a departure?

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 thea "idle" until \(K\) new arrivals have occumed. (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?

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 busy-period 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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