1. **State the problem:** We have 3 servers (workers) each with their own queue. Customers arrive at a total rate of 140 per minute, and each customer takes on average 1.1 seconds to be served. The coefficients of variation for arrival and service times are 1.4 and 1 respectively. We want to find the average number of customers waiting in the queue. 2. **Convert units and define variables:** - Arrival rate $\lambda = 140$ customers/minute. - Service time $S = 1.1$ seconds = $\frac{1.1}{60}$ minutes. - Number of servers $c = 3$. - Coefficient of variation of interarrival times $C_a = 1.4$. - Coefficient of variation of service times $C_s = 1$. 3. **Calculate service rate per server:** $$\mu = \frac{1}{S} = \frac{1}{\frac{1.1}{60}} = \frac{60}{1.1} \approx 54.5455 \text{ customers per minute}$$ 4. **Calculate traffic intensity $\rho$:** $$\rho = \frac{\lambda}{c \mu} = \frac{140}{3 \times 54.5455} = \frac{140}{163.6365} \approx 0.855$$ 5. **Calculate utilization per server:** $$\rho < 1 \text{ so system is stable}$$ 6. **Calculate average waiting time in queue $W_q$ using the Pollaczek-Khinchine formula for $M/G/c$ with variability:** The formula for average waiting time in queue for $G/G/c$ is approximated by: $$W_q \approx \frac{C_a^2 + C_s^2}{2} \times W_q^{M/M/c}$$ Where $W_q^{M/M/c}$ is the average waiting time in queue for an $M/M/c$ system: $$W_q^{M/M/c} = \frac{L_q^{M/M/c}}{\lambda}$$ 7. **Calculate $L_q^{M/M/c}$ (average number in queue for $M/M/c$):** First, calculate $P_0$ (probability system is empty): $$a = \frac{\lambda}{\mu} = \frac{140}{54.5455} \approx 2.566$$ Calculate sum term: $$\sum_{n=0}^{c-1} \frac{a^n}{n!} = \frac{a^0}{0!} + \frac{a^1}{1!} + \frac{a^2}{2!} = 1 + 2.566 + \frac{2.566^2}{2} = 1 + 2.566 + 3.292 = 6.858$$ Calculate last term: $$\frac{a^c}{c!} = \frac{2.566^3}{3!} = \frac{16.9}{6} = 2.817$$ Calculate denominator for $P_0$: $$D = 6.858 + \frac{2.817}{1 - \rho} = 6.858 + \frac{2.817}{1 - 0.855} = 6.858 + \frac{2.817}{0.145} = 6.858 + 19.424 = 26.282$$ So, $$P_0 = \frac{1}{D} = \frac{1}{26.282} = 0.03805$$ Calculate $L_q^{M/M/c}$: $$L_q = P_0 \times \frac{a^c \rho}{c! (1-\rho)^2} = 0.03805 \times \frac{2.817 \times 0.855}{6 \times 0.145^2}$$ Calculate denominator: $$6 \times 0.145^2 = 6 \times 0.021025 = 0.12615$$ Calculate numerator: $$2.817 \times 0.855 = 2.408$$ So, $$L_q = 0.03805 \times \frac{2.408}{0.12615} = 0.03805 \times 19.08 = 0.726$$ 8. **Calculate $W_q^{M/M/c}$:** $$W_q^{M/M/c} = \frac{L_q}{\lambda} = \frac{0.726}{140} = 0.00519 \text{ minutes} = 0.3114 \text{ seconds}$$ 9. **Calculate $W_q$ for $G/G/c$ system:** $$W_q = \frac{C_a^2 + C_s^2}{2} \times W_q^{M/M/c} = \frac{1.4^2 + 1^2}{2} \times 0.00519 = \frac{1.96 + 1}{2} \times 0.00519 = 1.48 \times 0.00519 = 0.00768 \text{ minutes}$$ 10. **Calculate average number waiting in queue $L_q$ for $G/G/c$:** $$L_q = \lambda \times W_q = 140 \times 0.00768 = 1.075$$ **Final answer:** On average, there are approximately **1.075** customers waiting in the queue.