s service time) with equilibrium Laplace transform T (s). is 1 to allow for the departure of customer n-1 except if customer By making an analogous argument to the one that we made for the n-1 departs before customer n arrives, in which case I is (), because distribution of the number of customers which arrive during a Qn_1=0. service, Q (z) may be found. This is just the number of arrivals during the customers system time, SO that Equation 3.35 may be obtained. (3.30) (3.35) The probability generating function for In+1 is given by Equation 3.31. However the service time and waiting time are independent, giving Equation 3.36. (3.36) Putting these two together, and replacing U-UZ with S gives Equation 3.37 and Equation 3.38. (3.31) w"(s)H`(s)=7'(s) Note that the number of arrivals during each service are inde- (3.37) pendent of the current queue length, allowing this distribution to be factored out at the start of (Equation 3.31). By taking limits as n-x an equation for the equilibrium generating function, Q, for the number of customers left behind by one departing may be found (1-)-H*(s) from Equation 3.32. w"(s)= (1-p)s (3.38) vH (s)+s-v 4*(v-uz)(2-1)9, (3.32) This is the equilibrium transform for the waiting time. z-H"(U-UZ) exists whenever the queue is stable. However -H* (0) - His 3.3.4 The G/M/m queue in equilibrium the mean service duration and SO this condition is The analysis is very similar to the above but instead work with the queue immediately after each arrival. The waiting distribution fol- Setting z=(0 shows that 4 is the equilibrium probability that a lows immediately from this because the remaining time to complete customer departs leaving the queue empty. Dividing by z-1, taking the service of the customer at the front of the queue is exponential. limits as Z -> I and applying L'Hopital's rule determines 4 in (See the "memoryless" property of the exponential above.) Thus if Equation 3.33 and hence Equation 3.34. r(z) is the generating function of customers in queue ahead of an arrival, the trans