I am seeking assistance with a queueing theory problem involving the M/M/c queue model from my textbook. I have attempted to solve the problem and would greatly appreciate it if someone could review my solutions for accuracy and provide any necessary corrections or insights.
Question: Consider a queueing system with $c$ tellers where customer arrivals follow a Poisson process with rate $\lambda$ customers per hour, and the service times are exponentially distributed with rate $\mu$ customers per hour. Calculate the average waiting time in the queue ($W_q$), the average time in the system ($W$), the mean queue length ($L_q$), and the mean number in the system ($L$) for the following cases:
a) One teller ($c = 1$), $\lambda = 8$ customers/hour, $\mu = 12$ customers/hour.
b) One teller ($c = 1$), mean number in the system $L = 5$ customers, $\mu = 12$ customers/hour.
c) Two tellers ($c = 2$), $\lambda = 8$ customers/hour, $\mu = 12$ customers/hour.
Solution:
Part a:
Given: $c = 1, \lambda = 8 \text{ customers/hour}, \mu = 12 \text{ customers/hour}$
- $W_q = \frac{\lambda}{\mu(\mu - \lambda)} = \frac{8}{12(12 - 8)} = \frac{8}{48} = \frac{1}{6} \text{ hours} = 10 \text{ minutes}$
- $W = \frac{1}{\mu - \lambda} = \frac{1}{12 - 8} = \frac{1}{4} \text{ hours} = 15 \text{ minutes}$
- $L_q = \frac{\lambda^2}{\mu(\mu - \lambda)} = \frac{8^2}{12(12 - 8)} = \frac{64}{48} = \frac{4}{3} \approx 1.33 \text{ customers}$
- $L = \frac{\lambda}{\mu - \lambda} = \frac{8}{12 - 8} = \frac{8}{4} = 2 \text{ customers}$
Part b:
Given: $c = 1, L = 5 \text{ customers}, \mu = 12 \text{ customers/hour}$
- $\lambda = \frac{L \times \mu}{L + 1} = \frac{5 \times 12}{5 + 1} = 10 \text{ customers/hour}$
- $W_q = \frac{\lambda}{\mu(\mu - \lambda)} = \frac{10}{12(12 - 10)} = \frac{10}{24} = \frac{5}{12} \text{ hours} = 25 \text{ minutes}$
- $W = \frac{1}{\mu - \lambda} = \frac{1}{12 - 10} = \frac{1}{2} \text{ hours} = 30 \text{ minutes}$
- $L_q = \frac{\lambda^2}{\mu(\mu - \lambda)} = \frac{10^2}{12(12 - 10)} = \frac{100}{24} = \frac{25}{6} \approx 4.17 \text{ customers}$
Part c:
Given: $c = 2, \lambda = 8 \text{ customers/hour}, \mu = 12 \text{ customers/hour}$
- $\rho = \frac{\lambda}{c \times \mu} = \frac{8}{2 \times 12} = \frac{1}{3}$
- $P_0 = \left[1 + 2 \times \frac{1}{3} + \frac{(2 \times \frac{1}{3})^2}{2 \times (1-\frac{1}{3})}\right]^{-1} = \frac{1}{2}$
- $P = \frac{(c\rho)^c}{c!(1-\rho)} \times P_0 = \frac{(2 \times \frac{1}{3})^2}{2 \times (1-\frac{1}{3})} \times \frac{1}{2} = \frac{1}{6}$
- $L_q = P \times \frac{\rho}{1 - \rho} = \frac{1}{6} \times \frac{\frac{1}{3}}{1 - \frac{1}{3}} = \frac{1}{12}$
- $W_q = \frac{L_q}{\lambda} = \frac{\frac{1}{12}}{8} = \frac{1}{96} \text{ hours} = 0.625 \text{ minutes}$
- $L = L_q + c \times \rho = \frac{1}{12} + 2 \times \frac{1}{3} = \frac{3}{4}$
- $W = \frac{L}{\lambda} = \frac{\frac{3}{4}}{8} = \frac{3}{32} \text{ hours} = 5.625 \text{ minutes}$