Students arrive at the Administrative Services Office at an average of one every 12 minutes, and their requests take, on average, 10 minutes to be processed. The service counter is staffed by only one clerk, Judy Gumshoes, who works eight hours per day. Assume Poisson arrivals and exponential service times.

What percentage of time is Judy idle?

**Note: Round your answer to 1 decimal place.**


How much time, on average, does a student spend waiting in line?

**Note: Do not round intermediate calculations. Round your answer to 1 decimal place.**


How long is the (waiting) line, on average?

**Note: Round your answer to 2 decimal places.**


What is the probability that an arriving student (just before entering the Administrative Services Office) will find at least one other student waiting in line?

**Note: Do not round intermediate calculations. Round your answer to 4 decimal places.**



This problem deals with a single-server queue model, which can be analyzed using the **M/M/1 queue** system in queuing theory. The key parameters of an M/M/1 queue are:

- **Arrival rate (\(\lambda\))**: the average number of students arriving per unit time.
- **Service rate (\(\mu\))**: the average number of students that can be served per unit time.

### Step 1: Find the percentage of time Judy is idle.

#### Given:
- Average time between arrivals = 12 minutes → Arrival rate \(\lambda = \frac{1}{12}\) students per minute.
- Average service time per student = 10 minutes → Service rate \(\mu = \frac{1}{10}\) students per minute.

Now, we calculate the **utilization rate** (\(\rho\)):

\[
\rho = \frac{\lambda}{\mu} = \frac{1/12}{1/10} = \frac{10}{12} = 0.8333
\]

The percentage of time Judy is busy is \( \rho \times 100 = 83.33\%\).

The percentage of time Judy is idle is \(100\% - 83.33\%\):

\[
\text{Idle percentage} = 100 - 83.33 = 16.7\%
\]

### Step 2: Find the average time a student spends waiting in line.

The **average time spent in the queue (waiting time in line)**, denoted \(W_q\), is given by:

\[
W_q = \frac{\rho}{\mu (1 - \rho)}
\]

Substituting the known values:

\[
W_q = \frac{0.8333}{1/10 \times (1 - 0.8333)} = \frac{0.8333}{\frac{1}{10} \times 0.1667} = \frac{0.8333}{0.01667} = 49.9985 \text{ minutes}
\]

Rounding to 1 decimal place, the average time a student spends waiting in line is:

\[
W_q \approx 50.0 \text{ minutes}
\]

### Step 3: Find the average length of the waiting line.

The **average number of students in the queue (waiting line length)**, denoted \(L_q\), is given by:

\[
L_q = \lambda \times W_q
\]

Substituting the known values:

\[
L_q = \frac{1}{12} \times 50.0 = 4.1667 \text{ students}
\]

Rounding to 2 decimal places, the average length of the waiting line is:

\[
L_q \approx 4.17 \text{ students}
\]

### Step 4: Find the probability that an arriving student will find at least one other student waiting.

The **probability that an arriving student will find at least one student in the queue** is the same as the **probability that the system is not empty**. This is given by:

\[
P(\text{at least one waiting}) = \rho = 0.8333
\]

Therefore, the probability that an arriving student will find at least one other student waiting in line is:

\[
P(\text{at least one waiting}) \approx 0.8333
\]

### Summary of Answers:
1. **Percentage of time Judy is idle**: \(16.7\%\)
2. **Average time a student spends waiting in line**: \(50.0\) minutes
3. **Average length of the waiting line**: \(4.17\) students
4. **Probability of finding at least one other student waiting**: \(0.8333\)

Would you like further details or have any questions?

---

Here are 5 related questions:
1. How would the results change if the service time increased to 12 minutes?
2. What would the idle percentage be if two clerks worked instead of one?
3. How does the arrival rate affect the average waiting time in line?
4. Can the queue length be reduced by improving the service rate to 8 minutes?
5. What happens to the system's performance if the arrival rate doubles?

**Tip**: Understanding queuing models helps in optimizing service processes, especially in reducing wait times and improving efficiency.

Students arrive at the Administrative Services Office at an average of one every 12 minutes, and their requests take, on average, 10 minutes to be processed. The service counter is staffed by only one clerk, Judy Gumshoes, who works eight hours per day. Assume Poisson arrivals and exponential service times.

What percentage of time is Judy idle?

How much time, on average, does a student spend waiting in line?

How long is the (waiting) line, on average?

What is the probability that an arriving student will find at least one other student waiting in line?

Solve an M/M/1 queue problem involving Poisson arrivals and exponential service times at the Administrative Services Office. Find the percentage of time the clerk is idle, the average waiting time, the queue length, and the probability of at least one student waiting. Perfect for undergraduate students studying queuing theory and operations research.

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