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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

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 , \(j=0, \ldots, n\)

Customers arrive at a single-server station in accordance with a Poisson process with rate \(\lambda\). All arrivals that find the server free immediately enter service. All service times are exponentially distributed with rate \(\mu\). An arrival that finds the server busy will leave the system and roam around "in orbit" for an exponential time with rate \(\theta\) at which time it will then return. If the server is busy when an orbiting customer returns, then that customer returns to orbit for another exponential time with rate \(\theta\) before returning again. An arrival that finds the server busy and \(N\) other customers in orbit will depart and not return. That is, \(N\) is the maximum number of customers in orbit. (a) Define states. (b) Give the balance equations. In terms of the solution of the balance equations, find (c) the proportion of all customers that are eventually served. (d) the average time that a served customer spends waiting in orbit.

Consider a single-server exponential system in which ordinary customers arrive at a rate \(\lambda\) and have service rate \(\mu\). In addition, there is a special customer who has a service rate \(\mu_{1}\). Whenever this special customer arrives, she goes directly into service (if anyone else is in service, then this person is bumped back into quete). When the special customer is not being serviced, she spends an exponential amount of time (with mean \(1 / \theta\) ) out of the system. (a) What is the average arrival rate of the special customer? (b) Define an appropriate state space and set up balance equations. (c) Find the probability that an ordinary customer is bumped \(n\) times.

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?

There are two types of customers. Type 1 and 2 customers arrive in accordance with independent Poisson processes with respective rate \(\lambda_{1}\) and \(\lambda_{2} .\) There are two servers. A type 1 arrival will enter service with server 1 if that server is free; if server 1 is busy and server 2 is free, then the type 1 arrival will enter service with server \(2 .\) If both servers are busy, then the type 1 arrival will go away, Atype 2 customer can only be served by server \(2 ;\) if server 2 is free when a type 2 customer arrives, then the customer enters service with that server. If server 2 is busy when a type 2 arrives, then that customer goes away. Once a customer is served by either server, he departs the system. Service times at server \(i\) are exponential with rate \(\mu_{i}, i=1,2\) Suppose we want to find the average number of customers in the system. (a) Define states. (b) Give the balance equations. Do not attempt to solve them. In terms of the long-run probabilities, what is (c) the average number of customers in the system? (d) the average time a customer spends in the system?

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