Stochastic

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Unformatted text preview: is the limiting fraction of time that the light bulb works? (c) What is the limiting fraction of visits on which the bulb is working? Solution. Suppose that a new bulb is put in at time 0. It will last for an amount of time s1 . Using the lack of memory property of the exponential distribution, it follows that the amount of time until the next inspection, u1 , will have an exponential distribution with rate . The bulb is then replaced and the cycle starts again, so we have an alternating renewal process. To answer (a), we note that the expected length of a cycle Eti = µF +1/ , so if N (t) is the number of bulbs replaced by time t, then it follows from Theorem 3.1 that 1 N (t) ! t µF + 1/ In words, bulbs are replaced on the average every µF + 1/ units of time. To answer (b), we let ri = si , so Theorem 3.4 implies that in the long run, the fraction of time the bulb has been working up to time t is Eri µF = Eti µF + 1/ To answer (c), we note that if V (t) is the number of visits the janitor has made by time t, then by the law of large numbers for the Poisson process we have...
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This document was uploaded on 03/06/2014 for the course MATH 4740 at Cornell University (Engineering School).

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