Page 55 - A First Course In Stochastic Models

46 RENEWAL-REWARD PROCESSES

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Theorem 2.2.5 Assume that R(t) ≥ 0 with E(C ) < ∞ and E(R ) < ∞. Then
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x
R(t) − gt 1 − y
1 2
lim P √ ≤ x = √ e 2 dy, x ≥ 0,
t→∞ ν t/µ 1 2π −∞
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2
where µ 1 = E(C 1 ), µ 2 = E(C ), g = E(R 1 )/E(C 1 ) and ν = E(R 1 − gC 1 ) .
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A proof of this theorem can be found in Wolff (1989). In applying this theo-
rem, the difficulty is usually to find the constant ν. In specific applications one
might use simulation to find ν. As a special case, Theorem 2.2.5 includes a central
limit theorem for the renewal process {N(t)} studied in Section 2.1. Taking the
rewards R n equal to 1 it follows that the renewal process {N(t)} is asymptotically
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2
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N(t/µ 1 , σ t/µ ) distributed with σ = µ 2 − µ .
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Next we give two illustrative examples of the renewal-reward model.
Example 2.2.2 A stochastic clearing system
In a communication system messages requiring transmission arrive according to
a Poisson process with rate λ. 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 fixed 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 first 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 1.1.4 that
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E[total waiting time in (0, T )] = λT .
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This gives
1 2
E[cost incurred during one cycle] = K + hλT .
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