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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}

j =0

When the CAMEL phase 3 service is triggered in this manner, it will not obtain information elements such as location information Typically, the CAMEL service, when triggered in this manner, should not change any message data, ie it should not use CAP CON-SMS

(4211) (4212)

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A few words of explanation are in order Intuitively, (4211) may be obvious by noting that pj gives the long-run fraction of time the stock on hand is j The longt run average stock on hand is de ned as limt (1/t) 0 X(u) du This long-run average can be seen as a long-run average cost per time unit by imagining that a cost at rate j is incurred when the stock on hand is j Using this interpretation, the result (4211) can be seen as a consequence of Theorem 422, which will be discussed below The result (4212) uses the PASTA property: in the long run the fraction of customers who nd the system out of stock upon arrival equals the fraction of time the system is out of stock Further, we have the long-run average number of stock replenishments per time unit = p0 This result follows from (426) by noting that the average replenishment frequency equals the average number of transitions from state 0 to state Q per time unit Example 412 (continued) Unloading ships with an unreliable unloader In this example we need a regularity condition to ensure that Assumption 421 is satis ed (Assumption 412 trivially holds) Let denote the expected amount of time needed to complete the unloading of a ship It is not dif cult to verify that = 1 (1 + / ); see (A5) in Appendix A In order to satisfy Assumption 421 it should be required that the arrival rate of ships is less than the reciprocal of the expected completion time That is, the assumption < +

.

should be made The proof is omitted that under this condition the expected cycle length in Assumption 421 is nite (take state (0, 1) for the regeneration state r) Denote the equilibrium probabilities by p(j, 0) and p(j, 1) The probability p(j, 1) gives the long-run fraction of time that j ships are present and the unloader is available and the probability p(j, 0) gives the long-run fraction of time that j ships are present and the unloader is in repair Using the transition rate diagram in Figure 412 and applying the rate in = rate out principle, we obtain the equilibrium equations: p(0, 1) = p(1, 1), ( + + )p(i, 1) = p(i 1, 1) + p(i + 1, 1) + p(i, 0), ( + )p(1, 0) = p(1, 1), ( + )p(i, 0) = p(i 1, 0) + p(i, 1),

i = 1, 2, ,

CAP ETC[ assisting SSP IP Routing Address = CDS Address; O-CdPN = B-MSISDN; CgPN = A-party] ISUP IAM[MSRN] VMSC To called party Content delivery system

i = 2, 3,

p(i, 1) = 1

has a unique solution A brute-force method for solving the equilibrium equations is to truncate this in nite system through a suf ciently large integer N (to be found by trial and error) such that +1 [p(i, 0) + p(i, 1)] for some prespeci ed i=N accuracy number In Section 44 we discuss a more sophisticated method to solve the in nite system of linear equations Once the state probabilities have been computed, we nd

i=1

i[p(i, 0) + p(i, 1)], p(i, 0),

ISUP IAM[ CdPN = MSISDN-B; CgPN = A-party]

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