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.
 Spring '10
 DURRETT
 The Land

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