Question

A group of \(n\) processors is arranged in an ordered list. When a job arrives, the first processor in line attempts it; if it is unsuccessful, then the next in line tries it; if it too is unsuccessful, then the next in line tries it, and so on. When the job is successfully processed or after all processors have been unsuccessful, the job leaves the system. At this point we are allowed to reorder the processors, and a new job appears. Suppose that we use the one- closer reordering rule, which moves the processor that was successful one closer to the front of the line by interchanging its position with the one in front of it. If all processors were unsuccessful (or if the processor in the first position was successful), then the ordering remains the same. Suppose that each time processor \(i\) attempts a job then, independently of anything else, it is successful with probability \(p_{i}\). (a) Define an appropriate Markov chain to analyze this model. (b) Show that this Markov chain is time reversible. (c) Find the long-run probabilities.

Short answer

In this analysis, we defined a Markov chain \((X_t)\), where \(X_t\) represents the ordering of the \(n\) processors at time \(t\). The state-space for \(X_t\) has n! states due to the n! possible orderings of the processors. The transition probabilities were specified for successful processing and unsuccessful processing. The Markov chain was proven to be time reversible as the detailed balance equation is satisfied. Finally, the long-run probabilities were found to be the stationary distribution, which is given by \(\pi(i) = \frac{1}{n!}\) for all states \(i\).

Step-by-step solution

Defining a Markov Chain for the Processor Model

Let us define a Markov chain \((X_t)\), where \(X_t\) represents the ordering of the \(n\) processors at time \(t\). The state-space for \(X_t\) will be the set of all possible permutations of the \(n\) processors. Since the system has n! possible orderings, our Markov chain will have n! states. Now, let's specify the transition probabilities. Suppose the processors are arranged as \(p_1, p_2, ..., p_n\). If processor \(i\) processes the job successfully, its position changes with the processor in front of it \((i-1)\). The transition probabilities can be defined as: - If processor i is successful (for \(1<i \leq n\)): \(P(X_{t+1}=p_1,..,p_{i-2},p_i,p_{i-1},..,p_n|X_t = p_1,..,p_n) = p_i(1-p_1)...(1-p_{i-1})\) - If all processors are unsuccessful or processor 1 is successful: \(P(X_{t+1}=p_1,..,p_n|X_t = p_1,..,p_n) = (1-p_1)p_2(1-p_3)...(1-p_n) + p_1\)

Prove that the Markov Chain is Time Reversible

A Markov Chain is time reversible if the detailed balance equation is satisfied. The detailed balance equation for a Markov chain is given by: \( \pi(i)P_{ij} = \pi(j)P_{ji},\) where \(\pi(i)\) and \(\pi(j)\) are the long-run probabilities of states \(i\) and \(j\), and \(P_{ij}\) is the transition probability from state \(i\) to state \(j\). Given that the processors are arranged in an ordered list, we know that there exists a stationary distribution where every possible ordering is equally likely. Let \(\pi\) denote the stationary distribution of this Markov chain. Therefore, we can write: \(\pi(i)=\frac{1}{n!}\) Now, we can prove this Markov chain is time reversible by examining the detailed balance equation: \(\frac{1}{n!}P_{ij} = \frac{1}{n!}P_{ji} \implies P_{ij} = P_{ji}\) Since the detailed balance equation is satisfied, our Markov chain is time reversible.

Finding the Long-run Probabilities

As we have proven that the Markov chain is time reversible, we can now compute its long-run probabilities. The stationary distribution (\(\pi\)) for this Markov chain is given by: \(\pi(i) = \frac{1}{n!}\) Since every possible ordering is equally likely in the long-run, the long-run probabilities of the Markov chain are simply the stationary distribution: Long-run probabilities: \(\pi(i) = \frac{1}{n!}\) for all states \(i\).