s the first queue on route is g, the last j and the queue transition is from queue h position k to queue i position e. (3.66) Consider the probability flux of arrivals of customers on route I to their first node j, r(1)=j, it is easily verified that Equation 3.72 holds and, by symmetry, a similar relationship holds for departures, as in Equation 3.73. [I,(k) U,8(k,n,+1)x(C) (3.72) (3.67) The following result makes it easy to verify that Equation 3.64 is (3.73) indeed the equilibrium queue distribution. Theorem 1 Similarly for transitions between queues from node i position e to node h position k following route r, Equation 3.74 is also easily verified. Suppose that numbers q'(C,D) > () for each pair of states C,D can be found SO that Equation 3.68 is obtained. (3.68) (C)) (3.74) By adding Equations 3.72 to 3.74 and summing over all possible Furthermore, suppose numbers J (C) > () for each state C can transitions out of state C, Equation 3.68 may be obtained. be found satisfying Equation 3.69 for each pair of states C,D. Then JT is the equilibrium probability distribution for both Markov pro- The total rate out of state C in both networks is given by Equation 3.75. cess whose transition rates are determined by q(C,D), q'(C,D) respectively. (3.75) r(C)q(C,D) = (3.69) It is sufficient to obtain the equilibrium equations from Equation 3.68, as in Equation 3.70, but the left hand side is given by Equation Thus the conditions of theorem 1 are verified and Equation 3.64 3.71. is indeed the equilibrium distribution of the network. As described SO far the customers have to call at each node in turn Er(C)q(c,D)-Er(D)q(D,c) (3.70) along their route. However it is possible to model customer classes in which each new customer is assigned one from a possible set of D & C D & C routes at random. This is achieved as follows. Let S be a route r(C)q(C)r(C)a(C) which can be followed. Suppose the probability that this route is followed is f Each such route S is then offered traffic u = us Table 3.5 Transition rates for the reversed 