THE POLLACZEK KHINTCHINE FORMULA in .NET framework

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THE POLLACZEK KHINTCHINE FORMULA
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Consequently, j = limn Ln /n = limn Qn /n = qj for all j We are now ready to prove that
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(257)
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with b(t) denoting the probability density of the service time of a customer Before proving this result, we note that the unknown q0 is determined by the fact that the left-hand side of (257) equals 1 for z = 1 By applying L Hospital s rule, we nd q0 = 1 , in agreement with Little s formula 1 p0 = By the bounded convergence theorem in Appendix A, lim E(zQn ) = lim
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Hence, by (256) and (257),
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pj zj =
(1 )(1 z)A(z) A(z) z
(258)
Since the long-run average queue size Lq is given by
Lq =
j =1
(j 1)pj =
j =0
jpj (1 p0 )
(see Exercise 228), the Pollaczek Khintchine formula for Lq follows by differentiating the right-hand side of (258) and taking z = 1 in the derivative It remains to prove (257) To do so, note that Qn = Qn 1 (Qn 1 ) + An , n = 1, 2, ,
where (x) = 1 for x > 0, (x) = 0 for x = 0 and An is the number of customers arriving during the nth service time By the law of total probability, P {An = k} =
e t
( t)k b(t) dt, k!
k = 0, 1,
62 and so
k=0
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P {An = k}zk =
e t (1 z) b(t) dt
Since the random variables Qn 1 (Qn 1 ) and An are independent of each other, E(zQn ) = E(zQn 1 (Qn 1 ) )E(zAn ) We have E(zQn 1 (Qn 1 ) ) = P {Qn 1 = 0} +
j =1
(259)
zj 1 P {Qn 1 = j }
1 = P {Qn 1 = 0} + [E(zQn 1 ) P {Qn 1 = 0}] z Substituting this in (259), we nd zE(zQn ) = E(zQn 1 ) (1 z)P {Qn 1 = 0} A(z) Letting n , we next obtain the desired result (257) This completes the proof Before concluding this section, we give an amusing application of the Pollaczek Khintchine formula Example 251 Ladies in waiting Everybody knows women spend on average more time in the loo than men As worldwide studies show, women typically take 89 seconds to use the loo about twice as long as the 39 seconds required by the average man However, this does not mean that the queue for the women s loo is twice as long as the queue for the men s The sequence for the women s loo is usually far longer To explain this using the Pollaczek Khintchine formula, let us make the following reasonable assumptions: 1 Men and women arrive at the loo according to independent Poisson processes with the same rates 2 The expected amount of time people spend in the loo is twice as large for women as for men 3 The coef cient of variation of the time people spend in the loo is larger for women than for men 4 There is one loo for women only and one loo for men only
This application is based on the article Ladies Waiting by Robert Matthews in New Scientist, Vol 167, Issue 2249, 29 July 2000
THE POLLACZEK KHINTCHINE FORMULA
Let w and m denote the average arrival rates of women and men Let w and cw denote the mean and the coef cient of variation of the amount of time a woman spends in the loo Similarly, m and cm are de ned for men It is assumed that w w < 1 Using the assumptions w = m , w = 2 m and cw cm , it follows from (252) and the Pollaczek Khintchine formula (253) that the average queue size for the women s loo =
2 2 1 1 2 ( w w ) 2 (2 m m ) (1 + cw ) (1 + cm ) 2 1 w w 2 1 2 m m 2 1 2 ( m m ) 4 (1 + cm ) 2 1 m m
Hence the average queue size for the women s loo 4 (the average queue size for the men s loo) The above derivation uses the estimate 1 2 m m 1 m m and thus shows that the relative difference actually increases much faster than a factor 4 when the utilization factor w w becomes closer to 1 Laplace transform of the waiting-time probabilities The generating-function method enabled us to prove the Pollaczek Khintchine formula for the average queue size Using Little s formula we next found the Pollaczek Khintchine formula for the average delay in queue of a customer The latter formula can also be directly obtained from the Laplace transform of the waiting-time distribution This Laplace transform is also of great importance in itself The waiting-time probabilities can be calculated by numerical inversion of the Laplace transform; see Appendix F A simple derivation can be given for the Laplace transform of the waiting-time distribution in the M/G/1 queue when service is in order of arrival The derivation parallels the derivation of the generating function of the number of customers in the system Denote by Dn the delay in queue of the nth arriving customer and let the random variables Sn and n denote the service time of the nth customer and the time elapsed between the arrivals of the nth customer and the (n+1)th customer Since Dn+1 = 0 if Dn + Sn < n and Dn+1 = Dn + Sn n otherwise, we have Dn+1 = (Dn + Sn n )+ , n = 1, 2, , (2510) where x + is the usual notation for x = max(x, 0) From the recurrence formula (2510), we can derive that for all s with Re(s) 0 and n = 1, 2, ( s)E e sDn+1 = E e sDn b (s) sP {Dn+1 = 0},