Let's break down and solve the queueing problem step-by-step. ## 1. **Identify the Queueing Model** - **Single-server, Poisson arrivals (rate = 10/hr)** - **Service time:** - Before training: Exponential, mean = 5 min - After training: General, mean = 5.5 min, std dev = 4 min This is an **M/G/1 queue** (Poisson arrivals, General service time, 1 server). --- ## 2. **Convert Units for Consistency** - **Arrival rate (\(\lambda\)):** 10 per hour = \( \frac{10}{60} = 1/6 \) per minute - **Service time (\(E[S]\)):** - Before: 5 min - After: 5.5 min --- ## 3. **Utilization (\(\rho\))** \[ \rho = \lambda E[S] \] - **Before:** \( \rho_{before} = \frac{1}{6} \times 5 = 0.833 \) - **After:** \( \rho_{after} = \frac{1}{6} \times 5.5 = 0.917 \) --- ## 4. **M/G/1 Mean Number in System (\(L\)) and Mean Waiting Time in Queue (\(W_q\))** The **Pollaczek–Khinchine formula** for M/G/1 queue: \[ W_q = \frac{\lambda E[S^2]}{2(1 - \rho)} \] where \(E[S^2]\) is the **second moment** of service time. \[ E[S^2] = (\text{Std Dev})^2 + (E[S])^2 \] --- ### **Before Training (Exponential Service, Std Dev = 5 min):** - \(E[S] = 5\) - Std Dev = 5 - \(E[S^2] = 5^2 + 5^2 = 25 + 25 = 50\) \[ W_{q,before} = \frac{(1/6) \times 50}{2(1 - 0.833)} = \frac{8.333}{2 \times 0.167} = \frac{8.333}{0.334} = 24.96\ \text{min} \] --- ### **After Training (General Service, Std Dev = 4 min):** - \(E[S] = 5.5\) - Std Dev = 4 - \(E[S^2] = 4^2 + 5.5^2 = 16 + 30.25 = 46.25\) \[ W_{q,after} = \frac{(1/6) \times 46.25}{2(1 - 0.917)} = \frac{7.708}{2 \times 0.083} = \frac{7.708}{0.166} = 46.43\ \text{min} \] --- ## 5. **Conclusion:** - **Before training:** Mean wait in queue = **24.96 min** - **After training:** Mean wait in queue = **46.43 min** ### **Recommendation:** **Do NOT** have the server undergo the training, as it will **increase the mean waiting time in the queue** due to the higher mean service time, even though the variance is lower. --- ### **Summary Table:** | Scenario | Mean Service (min) | Std Dev (min) | Utilization (\(\rho\)) | \(E[S^2]\) | Mean Wait in Queue (min) | |-----------------|-------------------|---------------|------------------------|------------|--------------------------| | Before Training | 5 | 5 | 0.833 | 50 | 24.96 | | After Training | 5.5 | 4 | 0.917 | 46.25 | 46.43 | --- ### **Final Answer:** > **Management should NOT have the server undergo further training, since it will significantly increase the mean waiting time in the queue.**