Furthermore the limiting fraction of time the server

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Unformatted text preview: g fraction of time in state 1 is µF µF + µG 3.1. LAWS OF LARGE NUMBERS 105 To see that this is reasonable and to help remember the formula, consider the nonrandom case. If the machine always works for exactly µF days and then needs repair for exactly µG days, then the limiting fraction of time spent working is µF /(µF + µG ). Proof. In order to compute the limiting fraction of time the machine is working we let ti = si + ui be the duration of the ith cycle, and let the reward ri = si , the amount of time the machine was working during the ith cycle. In this case, Theorem 3.3 implies that R(t) Eri µF ! = t Eti µF + µG which gives the desired result. For a concrete example of alternating renewal processes, consider Example 3.5. Poisson janitor. A light bulb burns for an amount of time having distribution F with mean µF then burns out. A janitor comes at times of a rate Poisson process to check the bulb and will replace the bulb if it is burnt out. (a) At what rate are bulbs replaced? (b) What...
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This document was uploaded on 03/06/2014 for the course MATH 4740 at Cornell.

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