Using the results in 423 again the number of

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Unformatted text preview: next departure will be exponential(10). However, if the output process is Poisson, then the number of departures in disjoint intervals are independent. Proof for s = 1. Our first step in making the result in Theorem 4.8 seem reasonable is to check by hand that if there is one server and the queue is in equilibrium, then the time of the first departure, D, has an exponential distribution with rate . There are two cases to consider. Case 1. If there are n 1 customers in the queue, then the time to the next departure has an exponential distribution with rate µ, i.e., fD (t) = µe µt Case 2. If there are n = 0 customers in the queue, then we have to wait an exponential( ) amount of time until the first arrival, and then an independent exponential(µ) for that customer to depart. If we let T1 and T2 be the waiting 142 CHAPTER 4. CONTINUOUS TIME MARKOV CHAINS times for the arrival and for the departure, then breaking things down according to the value of T1 = s, the density of D = T1 + T2 in this case is Zt Zt fD (t) = e s · µe µ(t s) ds = µe µt e ( µ)s ds 0 = µt µe µ ⇣ 1 ( e µ)t...
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