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of the busy period density is determined by the functional equation (s) = b (s + (s)),
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(9225)
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where b (s) = 0 e sx b(x) dx is the Laplace transform of the probability density b(x) of the service time of a customer By relation (E8) in Appendix E, the Laplace transform of P {B > x} is given by
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e sx P {B > x} dx =
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1 (s) s
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(9226)
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The key to the proof of (9225) is the assertion that the amount of time needed to empty the system when the system starts with n customers present is distributed as the sum of the lengths of n independent busy periods B1 , , Bn To see this, note rst that the order of service has no effect on the amount of time needed to empty the system Following Tak cs (1962), imagine now the following service discipline a The initial n customers C1 , , Cn are separated Customer C1 is served rst, after which all customers (if any) are served who have arrived during the service time of customer C1 , and this way of service is continued until the system is free of all customers but C2 , , Cn Next this procedure is repeated with customer C2 , etc This veri es the above assertion The remainder of the proof is now simple Let the random variables S1 and 1 denote the length of the service initiating the busy period and the number of customers arriving during that rst service time Then, by conditioning on S1 and 1 , we nd E(e sB ) =
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0 n=0 0 n=0
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( t)n E(e sB | S1 = t, 1 = n) b(t) dt n! ( t)n E(e s(t+B0 + +Bn ) ) b(t) dt, n!
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where B0 = 0 and B1 , , Bn are independent random variables each having the same distribution as the busy period B Thus we nd (s) =
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( t)n st e [ (s)]n b(t) dt n!
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b(t) dt = b (s + (s)),
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as was to be proved In the same way as (9225) was derived, we can derive the generating function of the random variable N which is de ned as the number of customers served in one busy period Letting F (z) = P {N = k}zk , it is left k=0 to the reader to verify that F (z) = zb ( F (z)), |z| 1 (9227)
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Using relation (E2) in Appendix E, it easily follows from (9225) that the rst two moments of the length of a busy period are given by E(B) = E(S) 1 and E(B 2 ) = E(S 2 ) , (1 )3 (9228)
where the random variable S denotes the service time of a customer The result (9228) shows that the squared coef cient of variation of the length of a busy period 2 2 2 equals cB = (1+cS )/(1 ), where cS is the squared coef cient of variation of the 2 explodes when approaches 1 Consequently, the service time S The value of cB density of the busy period has a very long tail for close to 1 As an illustration, 2 consider the case of gamma services with E(S) = 1 and cS = 2 Then the tail probability P {B > 1000} has the respective values 470 10 4 , 363 10 3 and 115 10 2 for = 090, 095 and 099 These values have been computed by using the general formula
P {B x} =
n=1 0
e y
( y)n 1 bn (y) dy, n!
x 0,
(9229)
where bn (x) denotes the probability density of the sum S1 + + Sn of n service a times S1 , , Sn The reader is referred to Tak cs (1962) for a proof of this formula The numerical evaluation of this in nite series offers no dif culties when the service time has a gamma distribution Then bn (x) is a gamma density as well, so that each term of the series can be written as an incomplete gamma integral; see Appendix B Fast codes for the numerical evaluation of an incomplete gamma integral are widely available If the service times are not gamma distributed, one has to resort to numerical inversion of the Laplace transform (9226) for the computation of P {B > x} In inverting this Laplace transform, the problem is that (s) is not explicitly given but is given in the form of a functional equation However, the value of (s) for a given point s can be simply computed by an iterative procedure Iterative procedure for (s) For a given point s, the function value (s) can be seen as a xed point of the equation z = b (s + z) It was shown in Abate and Whitt (1992) that this equation can be solved by repeated substitution Starting with z0 = 1, compute the (complex) number zn from zn = b (s + zn 1 ), n = 1, 2,
The sequence {zn } converges to the desired value (s)