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A proof of this theorem can be found in Wolff (1989) In applying this theorem, the dif culty is usually to nd the constant In speci c applications one might use simulation to nd As a special case, Theorem 225 includes a central limit theorem for the renewal process {N (t)} studied in Section 21 Taking the rewards Rn equal to 1 it follows that the renewal process {N (t)} is asymptotically N (t/ 1 , 2 t/ 3 ) distributed with 2 = 2 2 1 1 Next we give two illustrative examples of the renewal-reward model Example 222 A stochastic clearing system In a communication system messages requiring transmission arrive according to a Poisson process with rate The messages are temporarily stored in a buffer having ample capacity Every T time units, the buffer is cleared from all messages present The buffer is empty at time t = 0 A xed cost of K > 0 is incurred for each clearing of the buffer Also, for each message there is a holding cost of h > 0 for each time unit the message has to wait in the buffer What is the value of T for which the long-run average cost per time unit is minimal We rst derive an expression for the average cost per time unit for a given value of the control parameter T To do so, observe that the stochastic process describing the number of messages in the system regenerates itself each time the buffer is cleared from all messages present This fact uses the lack of memory of the Poisson arrival process so that at any clearing epoch it is not relevant how long ago the last message arrived Taking a cycle as the time interval between two successive clearings of the buffer, we have the expected length of one cycle = T To specify the expected cost incurred during one cycle, we need an expression for the total waiting time of all messages arriving during one cycle It was shown in Example 114 that E[total waiting time in (0, T )] = This gives 1 E[cost incurred during one cycle] = K + h T 2 2 1 2 T 2.

with probability 1 When K = 0 and h = 1, the system incurs a cost at rate j whenever there are j messages in the buffer, in which case the average cost per time unit gives the average number of messages in the buffer Hence the long-run average number of messages in the buffer = 1 T 2

OPTIONAL, OPTIONAL, OPTIONAL, OPTIONAL, OPTIONAL, OPTIONAL, OPTIONAL, OPTIONAL, OPTIONAL, OPTIONAL, OPTIONAL,

Putting the derivative of the cost function equal to 0, it follows that the long-run average cost is minimal for T = 2K h

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Example 223 A reliability system with redundancies An electronic system consists of a number of independent and identical components hooked up in parallel The lifetime of each component has an exponential distribution with mean 1/ The system is operative only if m or more components are operating The non-failed units remain in operation when the system as a whole is in a non-operative state The system availability is increased by periodic maintenance and by putting r redundant components into operation in addition to the minimum number m of components required Under the periodic maintenance the system is inspected every T time units, where at inspection the failed components are repaired The repair time is negligible and each repaired component is again as good as new The periodic inspections provide the only repair opportunities The following costs are involved For each component there is a depreciation cost of I > 0 per time unit A xed cost of K > 0 is made for each inspection and there is a repair cost of R > 0 for each failed component How can we choose the number r of redundant components and the time T between two consecutive inspections such that the long-run average cost per time unit is minimal subject to the requirement that the probability of system failure between two inspections is no more than a prespeci ed value We rst derive the performance measures for given values of the parameters r and T The stochastic process describing the number of operating components is regenerative Using the lack of memory of the exponential lifetimes of the components, it follows that the process regenerates itself after each inspection Taking a cycle as the time interval between two inspections, we have E(length of one cycle) = T

j =0

.

j =0

OPTIONAL, OPTIONAL}

uj zj ,

Equating coef cients yields j 1 pj = r0 z0 + uj ,

j 0

Since uj bR1 for some constant b and R1 > z0 , the coef cient uj tends to j zero faster than z0 Hence we can conclude the asymptotic expansion (C7) It is noted that Theorem C1 does not require that {pj } is a probability distribution The theorem applies to any sequence {pj , j = 0, 1, } with pj 0 for all j and j =0 pj < The asymptotic expansion (C7) is very useful for both theoretical and computational purposes It appears that in many applications the asymptotic expansion for pj can be used for relatively small values of j To illustrate this, consider the generating function (C5) for the problem of success runs This generating function P (z) is the ratio of the two analytic functions N (z) = ps zs and D(z) = 1 s pk 1 (1 p)zk whose domains of de nition can be extended k=1 to the whole complex plane (R = ) It is readily veri ed that the equation

Figure 81 Extension capability for CAP Initial DP Argument Reproduced from 3GPP TS 29078, v640, Section 611, by permission of ETSI

has a unique root z0 on the interval (1, ) Hence condition C1 is satis ed The veri cation of the technical conditions C2 and C3 is omitted and is left to the interested reader The unique root z0 of the above equation must be numerically

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