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other customers present The process {X(t)} makes a downcrossing from state j to state j 1 if the service of a customer is completed and j 1 other customers are left behind Observation 1 Since customers arrive singly and are served singly, the long-run average number of upcrossings from j 1 to j per time unit equals the long-run average number of downcrossings from j to j 1 per time unit This follows by noting that in any nite time interval the number of upcrossings from j 1 to j and the number of downcrossings from j to j 1 can differ at most by 1 Observation 2 The long-run fraction of customers seeing j 1 other customers upon arrival is equal to the long-run average number of upcrossings from j 1 to j per time unit the long-run average number of arrivals per time unit for j = 1, 2, In other words, the long-run average number of upcrossings from j 1 to j per time unit equals j 1 The latter relation for xed j is in fact a special case of the Little relation (241) by assuming that each customer nding j 1 other customers present upon arrival pays $1 (using this reward structure observation 2 can also be obtained directly from the renewal-reward theorem) Observations 1 and 2 do not use the assumption of exponential services and apply in fact to any regenerative queueing process in which customers arrive singly and are served singly Observation 3 For exponential services, the long-run average number of downcrossings from j to j 1 per time unit equals pj with probability 1 for each j 1 The proof of this result relies heavily on the PASTA property To make this clear, x j and note that service completions occur according to a Poisson process with rate as long as the server is busy Equivalently, we can assume that an exogenous Poisson process generates events at a rate of , where a Poisson event results in a service completion only when there are j customers present Thus, by part (a) of Theorem 241, E[Ij (t)] = E[Dj (t)] for t > 0 (272).

for any j 1, where Ij (t) is de ned as the amount of time that j customers are present during (0, t] and Dj (t) is de ned as the number of downcrossings from j to j 1 in (0, t] Letting the constant dj denote the long-run average number of downcrossings from j to j 1 per time unit, we have by the renewal-reward theorem that limt Dj (t)/t = dj with probability 1 Similarly, limt Ij (t)/t = pj with probability 1 The renewal-reward theorem also holds in the expected-value version Thus, for any j 1,

gsmSCF 2 (D-CSI)

Hence relation (272) gives that dj = pj for all j 1 By observations 1 and 2 we have dj = j 1 This gives j 1 = pj for all j 1, as was to be proved

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In 3 the method of embedded Markov chains will be used to derive an explicit expression for the customer-average probabilities j

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21 A street lamp is replaced by a new one upon failure and upon scheduled times T , 2T , There is always a replacement at the scheduled times regardless of the age of the street lamp in use The lifetimes of the street lamps are independent random variables and have a common Erlang (2, ) distribution What is the expected number of street lamps used in a scheduling interval 22 The municipality of Gotham City has opened a depot for temporarily storing chemical waste The amount of waste brought in each week has a gamma distribution with given shape parameter and scale parameter The amounts brought in during the successive weeks are independent of each other (a) What is the expected number of weeks until the total amount of waste in the depot exceeds the critical level L (b) Give an asymptotic estimate for the expected value from question (a) 23 Limousines depart from the railway station to the airport from the early morning till late at night The limousines leave from the railway station with independent interdeparture times that are uniformly distributed between 10 and 20 minutes Suppose you plan to arrive at the railway station at 3 o clock in the afternoon What are the estimates for the mean and the standard deviation of your waiting time at the railway station until a limousine leaves for the airport 24 Consider the expression (214) for the renewal function M(t) (a) Prove that for any k = 0, 1,

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