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Ein Unternehmen möchte s Maschinen desselben Typs einsetzen. Diese Maschinen können ausfallen, wobei eine gegebene Wahrscheinlichkeitsverteilung zugrundeliegt. Um die ausgefallenen Maschinen zu ersetzen, bestellt das Unternehmen zu Beginn einer jeden Woche so viele, daß man damit wieder insgesamt s Stück hätte; es dauert aber bei jeder neuen Bestellung eine Woche, bis sie ausgeliefert wird. Sei \(\mathrm{X}_{\mathrm{n}}\) die Anzahl der betriebsfähigen Maschinen zu Beginn der n-ten Woche und \(\mathrm{Y}_{\mathrm{n}}\) die Anzahl der Maschinen, die während dieser Woche ausfallen. Zeigen Sie, daß die Rekursionsformel $$ \mathrm{X}_{\mathrm{n}+1}=\mathrm{s}-\mathrm{Y}_{\mathrm{n}} $$ gilt und \(\mathrm{da\beta}\left\\{\mathrm{x}_{\mathrm{n}}, \mathrm{n} \geqslant 1\right\\}\) eine Markow-Kette ist. Nehmen Sie an, da \(\beta\) die Anzahl der ausfallenden Maschinen gleichverteilt ist, das heißt $$ \mathrm{P}\left\\{\mathrm{Y}_{\mathrm{n}}=\mathrm{j} \mid \mathrm{x}_{\mathrm{n}}=\mathrm{i}\right\\}=\frac{1}{\mathrm{i}+1} \text { für } \mathrm{j}=0,1, \ldots, \mathrm{i} $$ Bestimmen Sie die Übergangsmatrix der Kette, die stationäre Verteilung und den Erwartungswert für die Anzahl der laufenden Maschinen bei stationärem Zustand!

Step by step solution

01

- Understand the Recursion Formula

The given recursion formula is \[ \mathrm{X}_{\mathrm{n}+1}=\mathrm{s}-\mathrm{Y}_{\mathrm{n}} \] where \(\mathrm{X}_n\) is the number of operational machines at the beginning of the \(n\)-th week and \(\mathrm{Y}_n\) is the number of machines that fail during the \(n\)-th week. To show that this formula is valid, note that each week we start with \(s\) machines, and \(\mathrm{Y}_n\) machines fail during the week.

02

- Prove the Recursion Formula

Since the company orders new machines to ensure there are always \(s\) machines at the start of each week, the number of machines at the start of week \(n+1\) will be the total machines ordered at week \(n\) minus the failed machines of the same week, which gives us \[ \mathrm{X}_{\mathrm{n}+1}=\mathrm{s}-\mathrm{Y}_{\mathrm{n}}. \]

03

- Markov Chain

A process is a Markov Chain if the probability of transitioning to the next state depends only on the current state and not on the history of states. The number of failing machines, \(\mathrm{Y}_n\), depends only on the current number of operational machines, \(\mathrm{X}_n\), hence the process described by \(\left\{\mathrm{X}_{\mathrm{n}}, \mathrm{n} \geqslant 1\right\}\) is a Markov Chain.

04

- Transition Matrix

Given \[ \mathrm{P}\left(\mathrm{Y}_{\mathrm{n}} = j \mid \mathrm{X}_{\mathrm{n}} = i\right) = \frac{1}{i+1} \quad \text{for} \quad j=0,1,\ldots,i, \] we derive the transition probability as: \[ P_{ij} = \mathrm{P}\left(\mathrm{X}_{n+1} = j \mid \mathrm{X}_n = i\right) = \mathrm{P}\left(\mathrm{s} - \mathrm{Y}_n = j \mid \mathrm{X}_n = i\right) = \mathrm{P}\left(\mathrm{Y}_n = \mathrm{s}-j \mid \mathrm{X}_n = i\right) = \frac{1}{i+1} \quad \text{if} \quad \mathrm{j} = \mathrm{s}- \mathrm{Y}_n.\]

05

- Stationary Distribution

Let \( \pi_i \) be the stationary distribution such that it satisfies \( \sum_i \pi_i = 1 \), and \[ \pi_j = \sum_{i} \pi_i P_{ij} .\] Solving this equation gives the stationary distribution for the Markov Chain. This step involves solving the linear system formed by the balance equations \( \pi_j = \sum_{i} \pi_i P_{ij} \) and \( \sum_i \pi_i = 1 \).

06

- Expected Value

The expected number of operational machines at equilibrium is \[ \sum_{i=0}^s i \pi_i. \] Solving for \(\pi_i\) from the balance equations and computing this sum gives the expected value.

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Key Concepts

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

Recursion Formula

The recursion formula in reliability engineering helps define the relationship between the state of machines from one time period to the next. In this exercise, it is given by: \[ \mathrm{X}_{\mathrm{n}+1}=\mathrm{s}-\mathrm{Y}_{\mathrm{n}} \] where:

• \( \mathrm{X}_n \) represents the number of operational machines at the beginning of the \(n\)-th week
• \( \mathrm{Y}_n \) is the number of machines that fail during the \(n\)-th week.

To understand this formula: each week starts with \(s\) machines. Some machines fail during the week. The company orders new machines so that it starts the next week with \(s\) machines again. This means the number of working machines at the beginning of week \(n+1\) equals the total number ordered minus the failed machines from the previous week. Hence, the formula

Thus, \( \mathrm{X}_{\mathrm{n}+1}=\mathrm{s}-\mathrm{Y}_{\mathrm{n}} \) holds.

Stationary Distribution

The stationary distribution is central to understanding long-term behavior in Markov Chains. This distribution provides the probabilities that the system will be in each possible state after a large number of time steps.

Let \( \pi_i \) be the stationary probability distribution for state \(i\). It must satisfy two key conditions:

• All \( \pi_i \) should sum to 1: \[ \sum_i \pi_i = 1 \]
• Each \(\pi_i\) should be stable under transitions: \[ \pi_j = \sum_{i} \pi_i P_{ij} \] where \(P_{ij}\) represents the transition matrix.

Solving these equations is typically done using linear algebra techniques and yields the stationary probabilities for each state in the Markov Chain.

Transition Matrix

The transition matrix \( P \) describes the probabilities of moving from one state to another in a Markov Chain. For a given state \(i\), the element \(P_{ij}\) represents the probability of transitioning to state \(j\).

Given in the exercise:

• \[ \mathrm{P}\left( \mathrm{Y}_{\mathrm{n}} = j \mid \mathrm{X}_{\mathrm{n}} = i \right) = \frac{1}{i+1} \]
• This implies: \[ P_{ij} = \mathrm{P}\left( \mathrm{X}_{n+1} = j \mid \mathrm{X}_{n} = i \right) = \frac{1}{i+1} \quad \text{for} \quad j = 0,1,\ldots,i \]

In summary, constructing the transition matrix involves determining the probability of moving from each possible current state \(i\) to each future state \(j\). These probabilities together form the transition matrix \( P \).

Expected Value

The expected value is a key concept in probability and statistics and refers to the average outcome of a random variable over many trials. In the context of Markov Chains, it is the expected number of operational machines in the long run.

To find the expected value of the operational machines at equilibrium:

• First, determine the stationary distribution \(\pi_i\).
• Then calculate: \[ \sum_{i=0}^s i \pi_i \] where \( i \pi_i \) represents the probability-weighted number of machines.

The expected value essentially tells us the average number of operational machines we can expect after many weeks when the system has reached its steady-state.

Probability Distribution

Understanding probability distribution is crucial for defining the behavior of Markov Chains. In this exercise, the distribution of failing machines is given as:

• \[ \mathrm{P} \left( \mathrm{Y}_{\mathrm{n}} = j \mid \mathrm{X}_{\mathrm{n}} = i \right) = \frac{1}{i+1} \quad \text{for} \quad j=0,1,\ldots, i \]

This defines a uniform distribution over the number of failing machines when the number of operational machines is \(i\).

In general, a probability distribution gives the likelihood of various possible outcomes. In a Markov process, it helps to understand how the state transitions occur and predicts future states based on current conditions.