Solution. \(\mu = 12\), \(\rho = \frac 56\). \(L = \rho /(1 - \rho ) = 5\), \(W = 1/(12 - 10) = 0.5\) hours, \(W_q = \rho /(\mu - \lambda ) = 0.417\) hours \(= 25\) minutes. \(P(N \geq 4) = \rho ^4 = 0.482\).

Solution. One fast server: \(W_q = 0.8/4 = 0.2\) hours, \(W = 0.25\) hours. Two slow servers: \(a = 1.6\), \(\rho = 0.8\), \(C(2, 1.6) = \frac {1.28/0.2}{1 + 1.6 + 1.28/0.2} = \frac {6.4}9 = 0.711\), \(W_q = 0.711/(20 - 16) = 0.178\) hours, \(W = 0.178 + 0.1 = 0.278\) hours. The pair has the shorter wait in the queue, but the fast server the shorter total time, because the slow servers take twice as long to serve each customer.

Solution. \(a = 4 \times 0.5 = 2\) erlangs. \(B(3, 2) = \frac {8/6}{1 + 2 + 2 + 8/6} = \frac {1.333}{6.333} = 0.211\): about one call in five is lost. A fourth line gives \(B(4, 2) = \frac {16/24}{6.333 + 0.667} = 0.095\).

Solution. By Little’s law \(W = L/\lambda = 30/90\) hours \(= 20\) minutes, whatever the distributions.

Solution. \(L(c) = \lambda W_q(c) + a\) with \(a = 2.4\). For \(c = 3\): \(W_q = 0.1079\) hours, \(L = 2.589 + 2.4 = 4.99\), cost \(60 + 149.7 = 209.7\). For \(c = 4\): \(W_q = 0.01794\), \(L = 2.83\), cost \(80 + 84.9 = 164.9\). For \(c = 5\): \(W_q = 0.00437\), \(L = 2.50\), cost \(100 + 75.1 = 175.1\). The minimum is at \(4\) tellers.

Solution. An arrival finds \(n\) customers with probability \((1 - \rho )\rho ^n\) (Poisson arrivals see time averages) and then waits for \(n\) exponential services, a gamma time. Summing, \(P(W_q > t) = \sum _{n \geq 1}(1 - \rho )\rho ^nP(\text {Gamma}(n, \mu ) > t) = \rho e^{-(\mu - \lambda )t}\), as can be checked by differentiating. For the numbers, \(0.8e^{-1} = 0.294\): nearly a third of customers wait more than an hour, although the mean wait is \(48\) minutes.