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14.4 Markov Analysis of Machine Operations

Paul Tolsky, owner of Tolsky Works, has recorded the operation of his milling machine for several years. Over the past 2 years, 80% of the time the milling machine functioned correctly during the current month if it had functioned correctly in the preceding month. This also means that only 20% of the time did the machine not function correctly for a given month when it was functioning correctly during the preceding month. In addition, it has been observed that 90% of the time the machine remained incorrectly adjusted for any given month if it was incorrectly adjusted the preceding month. Only 10% of the time did the machine operate correctly in a given month when it did not operate correctly during the preceding month. In other words, this machine can correct itself when it has not been functioning correctly in the past, and this happens 10% of the time. These values can now be used to construct the matrix of transition probabilities. Again, state 1 is a situation in which the machine is functioning correctly, and state 2 is a situation in which the machine is not functioning correctly. The matrix of transition probabilities for this machine is

P = [\begin{array}{l} 0.8 & 0.2 \\ 0.1 & 0.9 \end{array}]

$P = [\begin{array}{l} 0.8 & 0.2 \\ 0.1 & 0.9 \end{array}]$

where

\begin{array}{l} P_{11} & = & 0.8 & = & \begin{array}{l} probability that the machine will be functioning c o r r e c t l y this month \\ given it was functioning c o r r e c t l y last month \end{array} \\ P_{12} & = & 0.2 & = & \begin{array}{l} probability that the machine will not be functioning c o r r e c t l y this month \\ given it was functioning c o r r e c t l y last month \end{array} \\ P_{21} & = & 0.1 & = & \begin{array}{l} probability that the machine will be functioning c o r r e c t l y this month \\ given it was n o t functioning correctly last month \end{array} \\ P_{22} & = & 0.9 & = & \begin{array}{l} probability that the machine will n o t be functioning correctly this month \\ given it was n o t functioning correctly last month \end{array} \end{array}

$\begin{array}{l} P_{11} & = & 0.8 & = & \begin{array}{l} probability that the machine will be functioning c o r r e c t l y this month \\ given it was functioning c o r r e c t l y last month \end{array} \\ P_{12} & = & 0.2 & = & \begin{array}{l} probability that the machine will not be functioning c o r r e c t l y this month \\ given it was functioning c o r r e c t l y last month \end{array} \\ P_{21} & = & 0.1 & = & \begin{array}{l} probability that the machine will be functioning c o r r e c t l y this month \\ given it was n o t functioning correctly last month \end{array} \\ P_{22} & = & 0.9 & = & \begin{array}{l} probability that the machine will n o t be functioning correctly this month \\ given it was n o t functioning correctly last month \end{array} \end{array}$

Look at this matrix for the machine. The two probabilities in the top row are the probabilities of functioning correctly and not functioning correctly given that the machine was functioning correctly in the last period. Because these are mutually exclusive and collectively exhaustive, the row probabilities again sum to 1.

What is the probability that Tolsky’s machine will be functioning correctly 1 month from now? What is the probability that the machine will be functioning correctly in 2 months? To answer these questions, we again apply Equation 14-3:

\begin{array}{l} π (1) & = & π (0) P \\ = & (1.0) [\begin{array}{l} 0.8 & 0.2 \\ 0.1 & 0.9 \end{array}] \\ = & [(1) (0.8) + (0) (0.1), (1) (0.2) + (0) (0.9)] \\ = & (0.8, .2) \end{array}

$\begin{array}{l} π (1) & = & π (0) P \\ = & (1.0) [\begin{array}{l} 0.8 & 0.2 \\ 0.1 & 0.9 \end{array}] \\ = & [(1) (0.8) + (0) (0.1), (1) (0.2) + (0) (0.9)] \\ = & (0.8, .2) \end{array}$

Therefore, the probability that the machine will be functioning correctly 1 month from now, given that it is now functioning correctly, is 0.80. The probability that it will not be functioning correctly in 1 month is 0.20. Now we can use these results to determine the probability that the machine will be functioning correctly 2 months from now. The analysis is exactly the same:

\begin{array}{l} π (2) & = & π (1) P \\ = & (0.8, 0.2) [\begin{array}{l} 0.8 & 0.2 \\ 0.1 & 0.9 \end{array}] \\ = & [(0.8) (0.8) + (0.2) (0.1), (0.8) (0.2) + (0.2) (0.9)] \\ = & (0.66, 0.34) \end{array}

$\begin{array}{l} π (2) & = & π (1) P \\ = & (0.8, 0.2) [\begin{array}{l} 0.8 & 0.2 \\ 0.1 & 0.9 \end{array}] \\ = & [(0.8) (0.8) + (0.2) (0.1), (0.8) (0.2) + (0.2) (0.9)] \\ = & (0.66, 0.34) \end{array}$

This means that 2 months from now there is a probability of 0.66 that the machine will still be functioning correctly. The probability that the machine will not be functioning correctly is 0.34. Of course, we could continue this analysis as many times as we want in computing state probabilities for future months.

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Table of Contents for 14.4 Markov Analysis of Machine Operations

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Table of Contents for
14.4 Markov Analysis of Machine Operations