Let's solve the problem step-by-step: --- **Given Data:** - Number of machines (M): **8** - Each machine operates for **18 hours** between breakdowns. - One person repairs machines at a **service rate of 0.35 per hour** (μ = 0.35/hr). - Arrival rate of breakdowns: Each machine breaks down every 18 hours, so for 8 machines: \[ \lambda = \frac{8}{18} \text{ breakdowns per hour} \approx 0.4444 \text{ per hour} \] - Each employee is paid **$21/hour**. - Machine downtime is valued at **$94.5/hour**. --- ## (a) **Single Employee (Single-Server Queuing System, M/M/1)** **Step 1: Arrival Rate** \[ \lambda = \frac{8}{18} = 0.4444\ \text{machines/hr} \] **Step 2: Service Rate** \[ \mu = 0.35\ \text{machines/hr} \] **Step 3: System Utilization** \[ \rho = \frac{\lambda}{\mu} = \frac{0.4444}{0.35} = 1.27 \] But, for an M/M/1 queue, the system is only stable if \(\rho < 1\). Since \(\rho > 1\), the system is **unstable** with just one server. This means the queue will grow indefinitely, which is not practical. Let's still calculate the other values for completeness, but this indicates that a single employee **cannot handle** the load. --- ## (b) **Two Employees (M/M/2 System)** - **Service Rate per server**: \( \mu = 0.35 \) - **Number of servers**: \( c = 2 \) - **Total Service Rate:** \( c\mu = 2 \times 0.35 = 0.70 \) **System Utilization (\(\rho\)):** \[ \rho = \frac{\lambda}{c\mu} = \frac{0.4444}{0.70} = 0.6349 \] ### Step 1: Probability that a machine waits (Erlang C formula) Let’s use the **M/M/c queue formulas**: #### Probability that an arriving machine must wait (\(P_w\)): \[ P_0 = \left[ \sum_{n=0}^{c-1} \frac{(\lambda/\mu)^n}{n!} + \frac{(\lambda/\mu)^c}{c!} \cdot \frac{1}{1 - \rho} \right]^{-1} \] \[ P_{wait} = \frac{(\lambda/\mu)^c}{c!} \cdot \frac{1}{1-\rho} \cdot P_0 \] With \(c = 2\), \(\lambda/\mu = 0.4444/0.35 = 1.27\): \[ P_0 = \left[ \frac{(1.27)^0}{0!} + \frac{(1.27)^1}{1!} + \frac{(1.27)^2}{2!} \cdot \frac{1}{1-0.6349}\right]^{-1} \] Calculate each term: - \(\frac{(1.27)^0}{0!} = 1\) - \(\frac{(1.27)^1}{1!} = 1.27\) - \(\frac{(1.27)^2}{2!} = \frac{1.6129}{2} = 0.80645\) So, \[ P_0 = \left[1 + 1.27 + 0.80645 \cdot \frac{1}{1-0.6349}\right]^{-1} \] \[ \frac{1}{1-0.6349} = \frac{1}{0.3651} = 2.738 \] \[ 0.80645 \cdot 2.738 = 2.208 \] \[ P_0 = \left[1 + 1.27 + 2.208\right]^{-1} = (4.478)^{-1} = 0.2233 \] **Now for \(P_{wait}\):** \[ P_{wait} = \frac{(1.27)^2}{2!} \cdot \frac{1}{1-0.6349} \cdot P_0 \] \[ = 0.80645 \cdot 2.738 \cdot 0.2233 = 0.4925 \] --- **Average number in queue (\(L_q\)):** \[ L_q = P_{wait} \cdot \frac{\rho}{1-\rho} \] \[ = 0.4925 \cdot \frac{0.6349}{1-0.6349} = 0.4925 \cdot \frac{0.6349}{0.3651} = 0.4925 \cdot 1.739 = 0.8567 \] **Average number in system (\(L\)):** \[ L = L_q + \frac{\lambda}{\mu} \] \[ = 0.8567 + \frac{0.4444}{0.35} = 0.8567 + 1.27 = 2.1267 \] **Average waiting time in queue (\(W_q\)):** \[ W_q = \frac{L_q}{\lambda} = \frac{0.8567}{0.4444} = 1.928\ \text{hours} \] **Average time in system (\(W\)):** \[ W = W_q + \frac{1}{\mu} \] \[ = 1.928 + \frac{1}{0.35} = 1.928 + 2.857 = 4.785\ \text{hours} \] --- ## (c) **Cost Calculations** ### **For 1 Employee (k=1):** - **System is unstable (\(\rho > 1\)), so cost is extremely high (theoretically infinite).** ### **For 2 Employees (k=2):** - **Employee Cost:** \(2 \times \$21 = \$42\) per hour - **Downtime Cost:** \(L \times \$94.5 = 2.1267 \times 94.5 = \$201.98\) per hour - **Total Cost per hour:** \[ TC(k=2) = 42 + 201.98 = \$243.98 \] --- ### **Summary Table:** | Option | Number of Employees | \(L_q\) | \(L\) | \(W_q\) | \(W\) | Total Cost per hour | |----------------|--------------------|--------|--------|--------|--------|---------------------| | Single Server | 1 | -- | -- | -- | -- | Infinite (unstable) | | Two Servers | 2 | 0.8567 | 2.1267 | 1.928 | 4.785 | \$243.98 | --- ## **Final Answers:** ### (a) **Single Employee:** - **System is unstable (\(\rho = 1.27 > 1\)); queue will grow without bound.** ### (b) **Two Employees:** - \(L_q = 0.8567 \text{ machines}\) - \(L = 2.1267 \text{ machines}\) - \(W_q = 1.928 \text{ hours}\) - \(W = 4.785 \text{ hours}\) ### (c) **Total Cost:** - \(TC(k=1) = \text{Infinite (system is unstable)}\) - \(TC(k=2) = \$243.98 \text{ per hour}\) **The total cost to operate all eight machines favors the use of:** - **Two employees** to maintain and repair them. --- If you need the answers filled in the boxes, let me know!

Let's approach this problem systematically, providing detailed calculations for each part. --- ## **Given Data:** - Number of machines, \( M = 8 \) - Each machine operates for an **average of 18 hours** between breakdowns. - Repair personnel rate: **each employee repairs at a rate of \(\mu = 0.35\) machines/hour**. - Employee cost: **\$21/hour**. - Downtime cost per machine per hour: **\$94.5/hour**. --- ## **Step 1: Calculate the arrival rate of breakdowns \(\lambda\).** Each machine breaks down every 18 hours, so: \[ \text{Breakdowns per machine per hour} = \frac{1}{18} \approx 0.0556 \] Total breakdowns for all 8 machines per hour: \[ \lambda = 8 \times 0.0556 = \frac{8}{18} \approx 0.4444 \text{ breakdowns/hour} \] --- ## **Part (a): Compute the cost if there is **only 1 employee** (single server model).** ### **Step 2: Check system stability** - Service rate per employee: \(\mu = 0.35\) per hour. - Arrival rate: \(\lambda = 0.4444\) per hour. Calculate utilization: \[ \rho = \frac{\lambda}{\mu} = \frac{0.4444}{0.35} \approx 1.27 \] Since \(\rho > 1\), **the system is unstable**. The queue will grow indefinitely, leading to infinite waiting times and costs. **Thus, with only 1 employee, the system cannot operate efficiently.** --- ## **Part (b): Compute the cost with **2 employees** (multi-server queue).** ### **Step 3: Determine the parameters** - Number of servers: \( c = 2 \) - Total service rate: \[ c \times \mu = 2 \times 0.35 = 0.70 \text{ machines/hour} \] - System utilization: \[ \rho = \frac{\lambda}{c \mu} = \frac{0.4444}{0.70} \approx 0.635 \] Since \(\rho < 1\), the system is **stable**. --- ### **Step 4: Calculate probability that an arriving breakdown has to wait (Erlang C formula)** **Step 4.1: Compute \(\frac{\lambda}{\mu}\) for the traffic intensity:** \[ a = \frac{\lambda}{\mu} = \frac{0.4444}{0.35} \approx 1.27 \] **Step 4.2: Compute \(P_0\), the probability that the system is empty:** \[ P_0 = \left[ \sum_{n=0}^{c-1} \frac{a^n}{n!} + \frac{a^{c}}{c!} \times \frac{1}{1 - \rho} \right]^{-1} \] Calculations: - \(a^0/0! = 1\) - \(a^1/1! = 1.27\) - \(a^2/2! = \frac{(1.27)^2}{2} = \frac{1.6129}{2} = 0.80645\) Sum: \[ S = 1 + 1.27 + 0.80645 = 3.07645 \] Calculate: \[ \frac{a^c}{c!} = 0.80645 \] \[ P_0 = \left[ 3.07645 + \frac{0.80645}{1 - 0.635} \right]^{-1} \] Since: \[ 1 - \rho = 1 - 0.635 = 0.365 \] \[ \frac{0.80645}{0.365} \approx 2.208 \] Total sum: \[ P_0 = \left[ 3.07645 + 2.208 \right]^{-1} = (5.28445)^{-1} \approx 0.1892 \] --- ### **Step 5: Probability that an arriving breakdown has to wait (\(P_{wait}\))** \[ P_{wait} = \frac{\frac{a^{c}}{c!}}{1 - \rho} \times P_0 = \frac{0.80645}{0.365} \times 0.1892 \approx 2.208 \times 0.1892 \approx 0.4179 \] --- ### **Step 6: Calculate average number in queue (\(L_q\))** \[ L_q = P_{wait} \times \frac{\rho}{1 - \rho} = 0.4179 \times \frac{0.635}{0.365} \approx 0.4179 \times 1.741 \approx 0.727 \] ### **Step 7: Calculate total number of breakdowns in the system (\(L\))** \[ L = L_q + \text{average number being served} = L_q + c \times \text{probability that servers are busy} \] But more straightforwardly, the average number in the system: \[ L = \frac{a}{1} \quad \text{(for M/M/c)} \text{, but better to use Little's Law:} \] Use Little's Law: \[ L = \lambda \times W \] Where \(W\) is the average total time in the system: \[ W = W_q + \frac{1}{\mu} \] --- ### **Step 8: Calculate average waiting time in queue (\(W_q\))** \[ W_q = \frac{L_q}{\lambda} = \frac{0.727}{0.4444} \approx 1.635 \text{ hours} \] ### **Step 9: Total average time in system (\(W\))** \[ W = W_q + \frac{1}{\mu} = 1.635 + \frac{1}{0.35} \approx 1.635 + 2.857 = 4.492 \text{ hours} \] ### **Step 10: Calculate the average number of breakdowns in the system (\(L\))** \[ L = \lambda \times W = 0.4444 \times 4.492 \approx 2.0 \text{ machines} \] --- ## **Part (c): Cost calculation** ### **Employee cost:** \[ \text{Number of employees} = 2 \] \[ \text{Employee cost per hour} = 2 \times \$21 = \$42 \] ### **Downtime cost:** \[ \text{Total downtime} = L \times \$94.5 = 2.0 \times 94.5 = \$189 \] ### **Total cost per hour:** \[ \boxed{ \text{Total Cost} = \text{Employee cost} + \text{Downtime cost} = \$42 + \$189 = \boxed{\$231} } \] --- ## **Final Summary:** | Part | Description | Result | |--------|--------------|---------| | (a) | With 1 employee | **System unstable** (cost infinite) | | (b) | With 2 employees | \(\boxed{\$231}\) per hour | | (c) | Conclusion | To minimize costs, **2 employees** are needed, costing about **\$231/hour** to repair and maintain 8 machines efficiently. --- **Note:** If you need further calculations for more employees or optimization, please specify!