k=S0

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( 0 0 )k k!

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Armed with an explicit expression for W0 , we are able to give a formula for the long-run average number of back orders outstanding at the bases For each base j the failed items arriving at base j can be thought of as customers entering service in a queueing system with in nitely many servers Here the service time should be de ned as the repair time in case of repair at the base and otherwise as the time until receipt of a replacement from the depot Thus the average service time of a customer at base j is given by j = rj j + (1 rj )( j + W0 ), j = 1, , N

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The situation at base j can only be modelled approximately as an M/G/ queue The reason is that the arrival process of failed items interferes with the replacement times at the depot so that there is some dependency between the service times at base j Assuming that this dependency is not substantial, we nevertheless use the M/G/ queue as an approximating model and approximate the limiting distribution of the number of items in service at base j by a Poisson distribution with

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THE POISSON PROCESS

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mean j j for j = 1, , N In particular, the long-run average number of back orders outstanding at base j

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This expression and the expression for W0 enables us to calculate the total average number of outstanding back orders at the bases for a given assignment (S0 , S1 , , SN ) Next, by some search procedure, the optimal values of S0 , S1 , , SN can be calculated 114 The Poisson Process and the Uniform Distribution In any small time interval of the same length the occurrence of a Poisson arrival is equally likely In other words, Poisson arrivals occur completely randomly in time To make this statement more precise, we relate the Poisson process to the uniform distribution Lemma 114 For any t > 0 and n = 1, 2, ,

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P {Sk x | N (t) = n} =

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(117)

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for 0 x t and 1 k n In particular, for any 1 k n, E(Sk | N (t) = n) = Proof kt n+1 and E(Sk Sk 1 | N (t) = n) = t n+1 (118)

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Since the Poisson process has independent and stationary increments, P {Sk x, N (t) = n} P {N (t) = n} P {N (x) k, N (t) = n} P {N (t) = n} 1 P {N (t) = n} 1 e t ( t)n /n!

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P {Sk x | N (t) = n} = = =

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e x x t

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THE POISSON PROCESS AND RELATED PROCESSES

proving the rst assertion Since E(U ) = 0 P {U > u} du for any non-negative random variable U , the second assertion follows from (117) and the identity (p + q + 1)! p!q!

y p (1 y)q dy = 1,

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The right-hand side of (117) can be given the following interpretation Let U1 , , Un be n independent random variables that are uniformly distributed on the interval (0, t) Then the right-hand side of (117) also represents the probability that the smallest kth among U1 , , Un is less than or equal to x This is expressed more generally in Theorem 115 Theorem 115 For any t > 0 and n = 1, 2, , P {S1 x1 , , Sn xn | N (t) = n} = P {U(1) x1 , , U(n) xn }, where U(k) denotes the smallest kth among n independent random variables U1 , , Un that are uniformly distributed over the interval (0, t) The proof of this theorem proceeds along the same lines as that of Lemma 114 In other words, given the occurrence of n arrivals in (0, t), the n arrival epochs are statistically indistinguishable from n independent observations taken from the uniform distribution on (0, t) Thus Poisson arrivals occur completely randomly in time Example 115 A waiting-time problem In the harbour of Amsterdam a ferry leaves every T minutes to cross the North Sea canal, where T is xed Passengers arrive according to a Poisson process with rate The ferry has ample capacity What is the expected total waiting time of all passengers joining a given crossing The answer is E(total waiting time) = 1 2 T 2 (119)

To prove this, consider the rst crossing of the ferry The random variable N (T ) denotes the number of passengers joining this crossing and the random variable Sk represents the arrival epoch of the kth passenger By conditioning, we nd E(total waiting time)

E(total waiting time | N (T ) = n)P {N (T ) = n}