Solution. The state is \(0\) (idle) or \(1\) (busy). The system is idle at \(t + h\) if it was idle and no customer arrived, or busy and the service finished: \[P_0(t + h) = P_0(t)(1 - \lambda h) + P_1(t)\mu h + o(h),\qquad P_0'(t) = -\lambda P_0(t) + \mu P_1(t) .\] In equilibrium \(0 = -\lambda p_0 + \mu p_1\), so \(p_1 = \frac \lambda \mu p_0\), and with \(p_0 + p_1 = 1\), \[p_0 = \frac {\mu }{\lambda + \mu },\qquad p_1 = \frac {\lambda }{\lambda + \mu } .\] By the Poisson arrivals, a fraction \(p_1 = \frac {\lambda }{\lambda + \mu }\) of all customers is lost.

Solution. Here \(\lambda = 10\) and \(\mu = 10\) per hour, and the states are \(0\), \(1\) and \(2\). The balance equations, obtained by setting the derivatives to zero, are \[-10p_0 + 10p_1 = 0,\qquad 10p_0 - 20p_1 + 10p_2 = 0,\] so \(p_1 = p_0\) and \(p_2 = 2p_1 - p_0 = p_0\). With \(p_0 + p_1 + p_2 = 1\), the equilibrium distribution is \(\left (\frac 13, \frac 13, \frac 13\right )\). Arrivals who find two customers present are lost, so one third of the customers go away.

Solution. With \(n\geq 1\) customers present, the next change is an arrival at rate \(\lambda \) or a departure at rate \(\mu \), so \[P_0'(t) = -\lambda P_0(t) + \mu P_1(t),\qquad P_n'(t) = -(\lambda + \mu )P_n(t) + \lambda P_{n-1}(t) + \mu P_{n+1}(t),\quad n\geq 1 .\] In equilibrium, \(\mu p_1 = \lambda p_0\) and \(\mu p_{n+1} - \lambda p_n = \mu p_n - \lambda p_{n-1}\) for \(n\geq 1\). The quantity \(\mu p_{n+1} - \lambda p_n\) is therefore the same for all \(n\), and it is \(0\) at \(n = 0\); so \(p_{n+1} = \rho p_n\) with \(\rho = \frac \lambda \mu \), and \(p_n = \rho ^np_0\). These sum to \(1\) only if \(\rho <1\), and then \[p_n = (1 - \rho )\rho ^n,\qquad n = 0, 1, 2, \dots ,\] a geometric distribution. The mean number in the system is \(L = \sum _nnp_n = \frac {\rho }{1 - \rho } = \frac {\lambda }{\mu - \lambda }\). The server is busy a fraction \(1 - p_0 = \rho \) of the time, \(\rho \) is called the traffic intensity, and the queue is stable only when customers arrive more slowly than they can be served.

For example, with \(\lambda = 8\) and \(\mu = 10\) per hour, \(\rho = 0.8\), the server is idle \(20\%\) of the time, and the mean number present is \(\frac {0.8}{0.2} = 4\). Raising the arrival rate to \(9.5\) per hour, a rise of under \(20\%\), gives \(\rho = 0.95\) and a mean of \(19\): congestion grows sharply as \(\rho \to 1\).

Solution. The balance equations are \(\mu p_1 = \lambda p_0\) and, as for the M/M/1 queue but with departure rate \((n + 1)\mu \) from state \(n + 1\), \((n + 1)\mu p_{n+1} = \lambda p_n\). So \(p_n = \frac {(\lambda /\mu )^n}{n!}p_0\), and summing, \(p_0 = e^{-\lambda /\mu }\): \[p_j = \frac {e^{-\lambda /\mu }(\lambda /\mu )^j}{j!},\qquad j = 0, 1, 2, \dots .\] The number of customers present is Poisson with mean \(\frac \lambda \mu \), the arrival rate times the mean service time.