Let's solve each part step by step: --- ## **Given:** - **Arrival rate, λ = 4 boats/hour** - **Service rate per server, μ = 8 boats/hour** - **Number of servers, s = 2** This is an **M/M/s** queue (specifically, M/M/2). --- ## **Formulas for M/M/s (s = 2):** - **Traffic intensity:** \( \rho = \frac{\lambda}{s\mu} = \frac{4}{2 \times 8} = .25 \) - **Probability that the system is idle (P₀):** \[ P_ = \left[ \sum_{n=}^{s-1} \frac{(\lambda/\mu)^n}{n!} + \frac{(\lambda/\mu)^s}{s!} \cdot \frac{1}{1 - \rho} \right]^{-1} \] Plugging in values: \[ \lambda/\mu = 4/8 = .5; \quad s = 2 \] \[ \sum_{n=}^{1} \frac{.5^n}{n!} = \frac{.5^}{!} + \frac{.5^1}{1!} = 1 + .5 = 1.5 \] \[ \frac{.5^2}{2!} \cdot \frac{1}{1-.25} = \frac{.25}{2} \cdot \frac{1}{.75} = .125 \cdot 1.3333 = .1667 \] \[ P_ = \left[1.5 + .1667\right]^{-1} = (1.6667)^{-1} = .6 \] **[This matches the answer you already have.]** --- ### **(a) Probability that the dock will be idle:** **.600** --- ### **(b) Average number of boats waiting for service (\(L_q\)):** \[ L_q = P_ \cdot \frac{(\lambda/\mu)^s \cdot \rho}{s! \cdot (1 - \rho)^2} \] \[ L_q = .6 \cdot \frac{(.5)^2 \cdot .25}{2 \cdot (1 - .25)^2} \] \[ = .6 \cdot \frac{.25 \cdot .25}{2 \cdot .75^2} \] \[ = .6 \cdot \frac{.0625}{2 \cdot .5625} \] \[ = .6 \cdot \frac{.0625}{1.125} \] \[ = .6 \cdot .0556 = .0333 \] **Answer:** \[ \boxed{.0333} \] --- ### **(c) Average time a boat will spend waiting for service (\(W_q\)):** \[ W_q = \frac{L_q}{\lambda} = \frac{.0333}{4} = .0083 \text{ hours} \] **Answer:** \[ \boxed{.0083} \text{ hours} \] --- ### **(d) Average time a boat will spend at the dock (\(W\)):** \[ W = W_q + \frac{1}{\mu} \] \[ = .0083 + \frac{1}{8} \] \[ = .0083 + .125 = .1333 \text{ hours} \] **Answer:** \[ \boxed{.1333} \text{ hours} \] --- ### **(e) Would you be satisfied with the service level?** - Both channels will be empty approximately **60%** of the time. - When customers show up, they **are not** likely to have to wait for more than 5 minutes (.0083 hrs ≈ .5 min). - On the whole, the expansion **may not** be worth the needed investment, since waiting time is already extremely low. --- ## **Final Summary** 1. **Idle probability:** .600 2. **Average number waiting:** .0333 boats 3. **Average waiting time:** .0083 hours (about .5 min) 4. **Average time at dock:** .1333 hours (about 8 minutes) 5. **Conclusion:** The new system will have very low waiting and high idle time, so the investment **may not be justified**.

# Analysis of Fore and Aft Marina's Docking Facility ## **Given Information** - **Arrival rate, \( \lambda \)**: 4 boats/hour - **Service rate per server, \( \mu \)**: 8 boats/hour - **Number of servers, \( s \)**: 2 This is an **M/M/s** queue (specifically, M/M/2). --- ## **Calculations** ### **(a) Probability that the dock will be idle (\( P_0 \))** 1. **Traffic intensity:** \[ \rho = \frac{\lambda}{s \mu} = \frac{4}{2 \times 8} = 0.25 \] 2. **Probability that the system is idle:** \[ P_0 = \left[ \sum_{n=0}^{s-1} \frac{(\lambda/\mu)^n}{n!} + \frac{(\lambda/\mu)^s}{s!} \cdot \frac{1}{1 - \rho} \right]^{-1} \] - Calculate components: \[ \lambda/\mu = \frac{4}{8} = 0.5 \] \[ \sum_{n=0}^{1} \frac{(0.5)^n}{n!} = 1 + 0.5 = 1.5 \] \[ \frac{(0.5)^2}{2!} \cdot \frac{1}{1 - 0.25} = \frac{0.25}{2} \cdot \frac{1}{0.75} = 0.125 \cdot 1.3333 = 0.1667 \] - Combine: \[ P_0 = \left[1.5 + 0.1667\right]^{-1} = (1.6667)^{-1} \approx 0.6 \] **Answer:** \[ \boxed{0.600} \] --- ### **(b) Average number of boats waiting for service (\( L_q \))** \[ L_q = P_0 \cdot \frac{(\lambda/\mu)^s \cdot \rho}{s! \cdot (1 - \rho)^2} \] 1. Plug in values: \[ L_q = 0.6 \cdot \frac{(0.5)^2 \cdot 0.25}{2 \cdot (1 - 0.25)^2} \] \[ = 0.6 \cdot \frac{0.25 \cdot 0.25}{2 \cdot 0.75^2} = 0.6 \cdot \frac{0.0625}{2 \cdot 0.5625} \] \[ = 0.6 \cdot \frac{0.0625}{1.125} \approx 0.6 \cdot 0.0556 = 0.0333 \] **Answer:** \[ \boxed{0.0333} \] --- ### **(c) Average time a boat will spend waiting for service (\( W_q \))** \[ W_q = \frac{L_q}{\lambda} = \frac{0.0333}{4} = 0.0083 \text{ hours} \] **Answer:** \[ \boxed{0.0083} \text{ hours} \] --- ### **(d) Average time a boat will spend at the dock (\( W \))** \[ W = W_q + \frac{1}{\mu} \] \[ = 0.0083 + \frac{1}{8} = 0.0083 + 0.125 = 0.1333 \text{ hours} \] **Answer:** \[ \boxed{0.1333} \text{ hours} \] --- ### **(e) Manager's Satisfaction with Service Level** - **Idle Probability:** Approximately **60%** of the time, the dock will be empty. - **Waiting Time:** Customers are unlikely to wait more than **0.5 minutes**. - **Conclusion:** Given the low waiting time, the current service level seems satisfactory, but the high idle time suggests that expansion may not be worth the investment. --- ## **Final Summary** 1. **Idle Probability:** \( \boxed{0.600} \) 2. **Average Number Waiting:** \( \boxed{0.0333} \) boats 3. **Average Waiting Time:** \( \boxed{0.0083} \) hours (about 0.5 minutes) 4. **Average Time at Dock:** \( \boxed{0.1333} \) hours (about 8 minutes) 5. **Management Conclusion:** The service level is satisfactory, but expansion may not be justified.

# Summary of Fore and Aft Marina's Docking Facility Analysis ## **(a) Probability that the dock will be idle** The probability that the dock will be idle is approximately **0.600**. This indicates that the dock is empty 60% of the time. ## **(b) Average number of boats waiting for service** The average number of boats waiting for service is **0.0333** boats. This means that, on average, there are very few boats in the queue at any given time. ## **(c) Average time a boat will spend waiting for service** The average waiting time for a boat is **0.0083 hours** (about 0.5 minutes). This suggests that customers experience minimal delays before receiving service. ## **(d) Average time a boat will spend at the dock** The average time a boat spends at the dock is **0.1333 hours** (about 8 minutes). This includes both waiting and service time, reflecting efficient processing. ## **(e) Manager's Satisfaction with Service Level** The service level appears satisfactory due to low waiting times, but the high idle time indicates that expansion might not be necessary. Therefore, the manager may reconsider investment in additional capacity.