k+1 m customers Figure 3.8 Customers 'seen' on arrival and departure Transform methods 3/9 Now consider the arrival of customer m+k+1 and suppose that he A little care is needed in working with the Laplace transforms of too finds j m customers in the system. This means that the kth queuing variables since mixed distributions can occur e.g. waiting departure has already taken place and left behind at most m cus- times which are a mixture of an atom at the origin (no wait at all) tomers. (Equation 3.22.) and a continuous density (waiting time conditional on having to wait). (3.22) Moments can be obtained from these by taking derivatives at 1 and 0 respectively, as in Equation 3.27. The derivatives give factor- Hence Equation 3.23 can be obtained, and if we suppose that the ial moments in the case of the probability generating functions and equilibrium arrival distribution A(m) exists then Equation 3.24 moments about the origin (to within a sign) in the case of the follows. Laplace transform. (3.23) (3.27) (3.24) 3.3.3 The M/G/1 queue in equilibrium Similarly the equilibrium departure distribution can be shown to In general the M/G/1 queue is not Markov unless one includes the exist provided the equilibrium arrival distribution does and the two time to complete the customer in service as well as the number of are once again equal. customers in the queue. However an analysis can be conducted by examining the queue at special instants. In fact the number of 3.3.2 Outline of the method customers in queue between each departure is a Markov chain (referred to as the embedded Markov chain). This is SO because the The approach is based on a transformation of the Markov equations arrival process is Poisson. which describe the queuing process and SO obtaining algebraic Suppose the arrival rate of customers is V and the service distribu- equations for the transforms, and then solving these equations. This tion is H(x) with density h(x). The probability Pk that k customers latter step usually involves finding zeros of function