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Unformatted text preview: SOLUTION HW 12. P4. (a) Yes. The kernel is G 01 ( x ) = 1 e λx G 10 ( x ) = G ( t ) , where G is the servicetime cdf. (b) No. It does not have the Markov property at downward jumps. (c) No. Same reason as for (b). P5. First solve π = πP , ∑ k ∈ I π k = 1, for the stationary probabilities, π k , k ∈ I , for the embedded DTMC . Then use the SMP formula, p j = π j μ j ∑ k ∈ I π k μ k , to derive the limiting probabilities p j for the SMP : p 1 = 0 . 0274 , p 2 = 0 . 0454 , p 3 = 0 . 2016 , p 4 = 0 . 7256 . P6. Let X ( t ) denote the number of down components at time t ≥ 0. (We could just as easily work with the number of up components.) Since all the random variables have exponential distributions, there are many possible choices for the embedded Markov renewal sequence. For example, each of the following is possible: 1. S n = time at which repair person returns to the system for the n th time 2. S n = time at which repair person goes away from the system for the n th time 3. S n = time of n th failure of a component 4. S n = time of n th completion of a component repair In the first and second cases, we can take Y n = X ( S n ), since the definition of S n includes the information as to whether the repair person is on duty or not just after S n , which is needed to define the kernel. In cases (3) and (4), we would need a twodimensional state variable, Y n = ( X ( S n ) ,m ), where m = 1 if the repair person is on duty just after S n and m = 0 otherwise. The lackofmemory property of the exponential distribution then ensures that { ( Y n ,S n ) } is a Markov renewal sequence in each case. In each case, we define the associated semiMarkov process in the usual way: Z ( t ) := Y N ( t ) . (Various combinations of the four cases are also possible.) However, if we want to compute the longrun probability that the repair person is on duty in terms of the limiting probabilities for the semiMarkov process { Z ( t ) ,t ≥ } , then we are further restricted in our definition of the embedded Markov renewal sequence: the process Z ( t ) must tell us unambiguously whether or not the repair person is on duty at time t , not just at the most recent S n . This rules out cases (1) and (2) above (with the onedimensional state variable, Y n = X ( S n )). In case (1), for example, Z ( t ) = i tells us that there were i down components at the most recent instant when the repair person returned to the system (and became “on duty”), but we don’t know if the repair person is still on duty at t . Similarly, in case (2), Z ( t ) = i tells us that there were i down components at the most recent instant 1 when the repair person left the system (and became “off duty”), but we don’t know if the...
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This note was uploaded on 04/24/2011 for the course IE 2084 taught by Professor Bailey,m during the Spring '08 term at Pittsburgh.
 Spring '08
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