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After being repaired, a machine functions for an exponential time with rate \(\lambda\) and then fails. Upon failure, a repair process begins. The repair process proceeds sequentially through \(k\) distinct phases. First a phase 1 repair must be performed, then a phase 2, and so on. The times to complete these phases are independent, with phase \(i\) taking an exponential time with rate \(\mu_{i}, i=1, \ldots, k\) (a) What proportion of time is the machine undergoing a phase \(i\) repair? (b) What proportion of time is the machine working?

Short Answer

Expert verified
The proportion of time the machine undergoes a phase i repair is: \[P_i = \frac{\mu_i}{\lambda + \sum_{i=1}^{k} \mu_i}\] The proportion of time the machine is working is: \[P_w = \frac{\lambda}{\lambda + \sum_{i=1}^{k} \mu_i}\]

Step by step solution

01

Determine overall rates

Let's first find out the overall rate for the whole system, which includes the machine's functioning time (rate λ) and the sum of repair processes' times (sum of rates μi). The overall rate is given by: \[\Omega = \lambda + \sum_{i=1}^{k} \mu_i\]
02

(a) Proportion of time spent on phase i repair

To find the proportion of time the machine undergoes a phase i repair, we can calculate it using the rate for that phase and the overall rate. The proportion of time spent on phase i repair is: \[P_i = \frac{\mu_i}{\Omega}\]
03

(b) Proportion of time the machine is working

Similarly, to find the proportion of time the machine is working, we need to divide the rate λ by the overall rate Ω. The proportion of time spent working is: \[P_w = \frac{\lambda}{\Omega}\] Now you can use these formulas to calculate the proportion of time spent on each repair phase and the proportion of time the machine is working, given the rate values λ and μi for the machine functioning and repair processes.

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Key Concepts

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

Machine Repair
Imagine a machine that is vital in a factory setup. Over time, like any machine, it might encounter failures. When a failure occurs, the machine requires repairs before it can go back to performing its tasks. This cycle of work and repair defines the machine repair process.

Understanding this repair dynamic is crucial. It helps to predict how long the machine will be out of service when it fails. Typically, the duration a machine functions before breaking down can be modeled using an exponential distribution. This is characterized by a rate \(\lambda\), which describes how quickly the breakdowns occur.

Therefore, the machine repair process involves predicting and managing the cycle of functioning and breaking down, making it easier to schedule maintenance and reduce downtime.
Phase Repair Process
Repairing a machine isn’t a single-step process. Instead, it involves moving through multiple phases, each requiring a specific type of repair. In our context, these phases are represented by different rates \(\mu_i\). Each rate corresponds to how quickly that part of the repair can be completed.

For instance, if a machine requires three phases of repair, represented by \(\mu_1, \mu_2, \mu_3\), each phase has its own exponential distribution time frames. Understanding each phase's individual rate helps plan the repair process more efficiently. Knowing the rates allows technicians to predict which phases might become bottlenecks in getting the machinery back to operational readiness.
  • Phase 1: Basic repairs with rate \(\mu_1\)
  • Phase 2: Intermediate repairs with rate \(\mu_2\)
  • Phase 3: Final repairs with rate \(\mu_3\)
When repairs are sequential, knowing these phases can enhance control over maintenance operations.
Proportion of Time
To effectively plan maintenance and operations, it's essential to know how much time a machine spends in each activity. It is crucial to know what proportion of time the machine spends undergoing each phase of the repair process or simply functioning.

The formula \(P_i = \frac{\mu_i}{\Omega}\) helps determine the proportion of time the machine is specifically in phase \(i\) of its repair. This proportion gives insight into how much of the total machine cycle is occupied by each repair phase. By identifying these percentages, businesses can allocate resources more efficiently to phases potentially impacting production the most.

Similarly, \(P_w = \frac{\lambda}{\Omega}\) gives the proportion of time the machine is operational. Knowing these proportions helps in understanding the machine's availability and reliability, guiding adjustments to improve overall production efficiency.
Continuous Random Variables
In mathematical terms, the repair process and machine functioning times are modeled as continuous random variables. This means that they can take any value within a range, allowing more flexible and precise modeling of real-world behaviors.

An exponential distribution is often used because it describes the time until the next event (like a breakdown or completion of a repair phase) nicely. Each phase of repair and the machine's operational time are modeled as separate yet interlinked exponential distributions:
  • The time until the machine breaks down is described by \(\lambda\).
  • The time until a particular repair phase is completed is described by \(\mu_i\).
These continuous random variables allow for a robust analysis of the machine's performance over time. They help with precise predictions about the machine’s uptime and the efficiency of the repair process.

Understanding continuous random variables and their behavior is key to improving the management and reliability of machines within any industrial setting.

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Most popular questions from this chapter

Consider a taxi station where taxis and customers arrive in accordance with Poisson processes with respective rates of one and two per minute. A taxi will wait no matter how many other taxis are present. However, an arriving customer that does not find a taxi waiting leaves. Find (a) the average number of taxis waiting, and (b) the proportion of arriving customers that get taxis.

Four workers share an office that contains four telephones. At any time, each worker is either "working" or "on the phone." Each "working" period of worker \(i\) lasts for an exponentially distributed time with rate \(\lambda_{i}\), and each "on the phone" period lasts for an exponentially distributed time with rate \(\mu_{i}, i=1,2,3,4\). (a) What proportion of time are all workers "working"? Let \(X_{i}(t)\) equal 1 if worker \(i\) is working at time \(t\), and let it be 0 otherwise. Let \(\mathrm{X}(t)=\left(X_{1}(t), X_{2}(t), X_{3}(t), X_{4}(t)\right)\) (b) Argue that \(\\{\mathrm{X}(t), t \geqslant 0\\}\) is a continuous-time Markov chain and give its infinitesimal rates. (c) Is \(\\{\mathrm{X}(t)\\}\) time reversible? Why or why not? Suppose now that one of the phones has broken down. Suppose that a worker who is about to use a phone but finds them all being used begins a new "working" period. (d) What proportion of time are all workers "working"?

There are two machines, one of which is used as a spare. A working machine will function for an exponential time with rate \(\lambda\) and will then fail. Upon failure, it is immediately replaced by the other machine if that one is in working order, and it goes to the repair facility. The repair facility consists of a single person who takes an exponential time with rate \(\mu\) to repair a failed machine. At the repair facility, the newly failed machine enters service if the repairperson is free. If the repairperson is busy, it waits until the other machine is fixed; at that time, the newly repaired machine is put in service and repair begins on the other one. Starting with both machines in working condition, find (a) the expected value and (b) the variance of the time until both are in the repair facility. (c) In the long run, what proportion of time is there a working machine?

In Example \(6.20\), we computed \(m(t)=E[O(t)]\), the expected occupation time in state 0 by time \(t\) for the two-state continuous-time Markov chain starting in state 0\. Another way of obtaining this quantity is by deriving a differential equation for it. (a) Show that $$ m(t+h)=m(t)+P_{00}(t) h+o(h) $$ (b) Show that $$ m^{\prime}(t)=\frac{\mu}{\lambda+\mu}+\frac{\lambda}{\lambda+\mu} e^{-(\lambda+\mu) t} $$ (c) Solve for \(m(t)\).

Customers arrive at a service station, manned by a single server who serves at an exponential rate \(\mu_{1}\), at a Poisson rate \(\lambda .\) After completion of service the customer then joins a second system where the server serves at an exponential rate \(\mu_{2} .\) Such a system is called a tandem or sequential queueing system. Assuming that \(\lambda<\mu_{i}\), \(i=1,2\), determine the limiting probabilities. Hint: Try a solution of the form \(P_{n, m}=C \alpha^{n} \beta^{m}\), and determine \(C, \alpha, \beta\).

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