This problem involves Markov Processes. The backgroundinfo is that machines can

ID: 3227154 • Letter: T

Question

This problem involves Markov Processes. The backgroundinfo is that machines can either be in the running state or thedown (broken) state. All periods are 1-hour inlength. Here are the relevant tables from the problem. Table of probabilities prior to new part. From/To Running Down Running .9 .1 Down .3 .7 Table of Probabilities after the new part From/To Running Down Running .95 .05 Down .6 .4 The Question is: If the total cost of the system being downfor any period is $500 dollars, what is the breakeven cost for thenew hardware component on a time-period basis? This problem involves Markov Processes. The backgroundinfo is that machines can either be in the running state or thedown (broken) state. All periods are 1-hour inlength. Here are the relevant tables from the problem. Table of probabilities prior to new part. From/To Running Down Running .9 .1 Down .3 .7 Table of Probabilities after the new part From/To Running Down Running .95 .05 Down .6 .4 The Question is: If the total cost of the system being downfor any period is $500 dollars, what is the breakeven cost for thenew hardware component on a time-period basis? This problem involves Markov Processes. The backgroundinfo is that machines can either be in the running state or thedown (broken) state. All periods are 1-hour inlength. Here are the relevant tables from the problem. Table of probabilities prior to new part. From/To Running Down Running .9 .1 Down .3 .7 Table of Probabilities after the new part From/To Running Down Running .95 .05 Down .6 .4 The Question is: If the total cost of the system being downfor any period is $500 dollars, what is the breakeven cost for thenew hardware component on a time-period basis? From/To Running Down Running .95 .05 Down .6 .4 The Question is: If the total cost of the system being downfor any period is $500 dollars, what is the breakeven cost for thenew hardware component on a time-period basis?

Explanation / Answer

Let the long term running period for the new hardware component on a time-period basis be P and therefore the long term breakdown period would be (1-P)

Now from the first column of the new matrix we get:

P = 0.95P + 0.6(1-P)

P = 0.95P + 0.6 - 0.6P

P = 0.35P + 0.6

0.65P = 0.6

P = 0.6/0.65 = 0.9231

Therefore long term breakdown period = 1 - P = 1 - 0.9231 = 0.0769

Therefore the long term cost of the breakdown would be: 500*(1-P) = 500*0.0769 = $38.45

Therefore the long term cost of the breakdown in stationary condition would be $38.45 per hour.

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This problem involves Markov Processes. The backgroundinfo is that machines can

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