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(D x + y)m(y) dy M(y) dy, 0 x D
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ALTERNATING RENEWAL PROCESSES
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Table 821
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Approximate and exact values for D = 08 Dapp Dopt error (%) 000 000 024 156
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2911 2911 2847 2847 2259 2298 1142 1588
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Using this result and the formulas (829), (817) and (812), the above expression for the long-run average cost can be worked out as g(D) = K (1 1 ) h D + 1 + M(D)
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The function g(D) is minimal for the unique solution of the equation D+
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M(y) dy =
K (1 1 ) h
(8210)
In general it is computationally demanding to nd an exact solution of this equation Except for special cases, one needs numerical Laplace inversion to compute x 0 M(y) dy; see Appendix F However, an approximate solution to (8210) is easily calculated when it is assumed that the optimal value of D is suf ciently large compared to 1 Then, by Theorem 823,
M(y) dy
D2 +D 2 1
2 2 3 1 + 23 2 2 1 6 2 4 1 1
Table 821 gives for several examples the optimal value Dopt and the approximate value Dapp together with the relative error 100 g(Dapp ) g(Dopt )/g(Dopt ) In all examples we take 1 = 1, h = 1 and K = 25 The arrival rate is 05 and 08 2 The squared coef cient of variation cB of the batch size is 1 , 1 , 1 1 and 3, where 3 2 2 the rst two values correspond to an Erlang distribution and the latter two values to an H2 distribution with balanced means Can you give a heuristic explanation why the optimal value of D decreases when the coef cient of variation of the batch size increases
ALTERNATING RENEWAL PROCESSES
An alternating renewal process is a two-state process alternating between an onstate and an off-state The on-times and the off-times are independent and identically distributed random variables The two sequences of on-times and off-times are
ADVANCED RENEWAL THEORY
mutually independent For any s > 0, let Pon (s) = P {the process is in the on-state at time s} and U (s) = P {the amount of time the process is in the on-state during [0, s]} Theorem 831 Suppose that the on-times and off-times have exponential distributions with respective means 1/ and 1/ Then, assuming that an on-time starts at epoch 0, Pon (s) = and
+ e ( + )s , + +
(831)
P {U (s) x} =
e (s x)
(s x) n!
e x
( x)k , k!
0 x < s (832)
The distribution function P {U (s) x} has a mass of e s at x = s
Proof Let Poff (s) = P {the process is in the off-state at time s} By considering what may happen in the time interval (s, s + s] with s small, it is straightforward to derive the linear differential equation Pon (s) = Poff (s) Pon (s), s > 0
Since Poff (s) = 1 Pon (s), we nd Pon (s) = ( + )Pon(s), s > 0 The solution of this equation is given by (831) The proof of (832) is more complicated The random variable U (s) is equal to s only if the rst on-time exceeds s Hence P {U (s) x} has mass e s at x = s Now x 0 x < s By conditioning on the lengths of the rst on-time and the rst off-time, we obtain P {U (s) x} =
e y dy
P {U (s y u) x y} e u du
Noting that P {U (s y u) x y} = 1 if s y u x y, we next obtain P {U (s) x} = e (s x) (1 e x ) +
e y dy
P {U (s y u) x y} e u du
Substituting this equation repeatedly into itself leads to the desired result (832)
ALTERNATING RENEWAL PROCESSES
Corollary 832 Suppose that the on-times and off-times have exponential distributions with respective means 1/ and 1/ Then, for any t0 > 0 and 0 x < t0 ,