# Stochastic

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

## This document was uploaded on 03/06/2014 for the course MATH 4740 at Cornell University (Engineering School).

Ask a homework question - tutors are online