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and solve for the numerical data = 1, 1/ 1 = 2, 1/ 2 = 5, s1 = 5 and s2 = 10 (Answer: 00464) Verify experimentally that the loss probability is nearly insensitive to the distributional form of the service times (eg compute the loss probability 00470 for the above data when the service time in group 1 has an E2 distribution and the service time in group 2 has an H2 distribution with balanced means and a squared coef cient of variation of 4) 519 Customers of the types 1, , m arrive at a service centre according to independent Poisson processes with respective rates 1 , , m The service centre has c identical servers An arriving customer of type j requires bj servers and is lost when there are no bj servers available A customer of type j has an exponentially distributed service time with mean 1/ j for j = 1, , m The customer keeps all of the assigned bj servers busy during his service time and upon completion of the service time the bj servers are simultaneously released Let p(n1 , , nm ) be the long-run fraction of time that nj groups of bj servers are handling type j customers for j = 1, , m (a) Verify from the equilibrium equations for the probabilities p(n1 , , nm ) that, for some constant C > 0, m ( j / j )nj p(n1 , , nm ) = C nj !

.

j =1

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for all (n1 , , nm ) with n1 b1 + + nm bm c (b) What is the long-run fraction of type j customers who are lost The above product-form solution can also be proved by considering the process {(X1 (t), , Xm (t))} in the in nite-server model (c = ) with Xj (t) denoting the number of type j customers present at time t The processes {X1 (t)}, , {Xm (t)} are independent of each other and each separate process {Xj (t)} constitutes an M/M/ queueing process having a Poisson distribution with mean j / j as equilibrium distribution Noting that the process {(X1 (t), , Xm (t))} is time reversible, it can be concluded from the result in Exercise 58 that the above product-form solution holds The normalization constant C can be computed as follows Let {pj , 0 j c} denote the equilibrium distribution of the numbers of busy servers in the loss model with c servers and let {pj } denote the equilibrium distribution of the number of busy servers in the in nite-server model Then pj = pj

( ) ( )

The normalization constant C is given by p0 It is left to the reader to verify that {pj } can be computed as the convolution of m compound Poisson distributions The j th compound Poisson distribution represents the limiting distribution of the numbers of busy servers in a batch arrival M X /G/ queue with group service, where the arrival rate of batches is j , each batch consists of bj customers and the mean service time of the customers from the same batch is 1/ j ; see part (b) of Exercise 115 Finally, it is noted that the loss model has the insensitivity property 520 Batches of containers arrive at a stockyard according to a Poisson process with a rate of = 15 batches per day Each batch consists of two or three containers with respective probabilities of 2 and 1 The stockyard has space for only 50 containers An arriving batch 3 3 nding not enough space is lost and is brought elsewhere Containers from the same batch are removed simultaneously after a random time The holding times of the batches are independent random variables and have a lognormal distribution with a mean of 1 day and a standard deviation of 2 days for batches of size 3 and a mean of 1 day and a standard deviation of 1 day for batches of size 3 Calculate the long-run fraction of batches of size 2 2 that are lost and the long-run fraction of batches of size 3 that are lost

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