SOLUTION: At Wazalendo SACCOS there is only one window for front office operations. The window is always opened from 7:00 am to 03:00 pm. It has been discovered that the average number of cl

Question 1203388: At Wazalendo SACCOS there is only one window for front office operations. The window is always opened from 7:00 am to 03:00 pm. It has been discovered that the average number of clients is 72 during the day and that the average service time is 5 minutes per person.
a) What is the average number of clients in the system?
b) Calculate the length of the queue.
c) What is the utilization factor of these SACCOS?
d) For how long will a customer be in this system?
e) Comment on the efficiency of this SACCOS in servicing its clients on the basis of calculations above.
Answer by asinus(45) (Show Source): You can put this solution on YOUR website!
This is a classic **queueing theory problem** modeled as an **M/M/1 queue**, where:
- **Arrival rate** (\( \lambda \)) is the average number of clients per unit time.
- **Service rate** (\( \mu \)) is the average number of clients that can be served per unit time.
### **Given Data**:
1. Total operation time: \( 8 \, \text{hours} = 480 \, \text{minutes} \)
2. Average number of clients per day: \( \lambda = 72 \)
- Arrival rate: \( \lambda = 72 / 480 = 0.15 \, \text{clients per minute} \)
3. Average service time per client: \( 5 \, \text{minutes} \)
- Service rate: \( \mu = 1 / 5 = 0.2 \, \text{clients per minute} \)
---
### **Formulas**:
1. **Utilization factor** (\( \rho \)):
\[
\rho = \frac{\lambda}{\mu}
\]
2. **Average number of clients in the system** (\( L \)):
\[
L = \frac{\lambda}{\mu - \lambda}
\]
3. **Average number of clients in the queue** (\( L_q \)):
\[
L_q = L - \rho = \frac{\lambda^2}{\mu(\mu - \lambda)}
\]
4. **Average time spent in the system per client** (\( W \)):
\[
W = \frac{1}{\mu - \lambda}
\]
5. **Average time spent in the queue per client** (\( W_q \)):
\[
W_q = W - \frac{1}{\mu} = \frac{\lambda}{\mu(\mu - \lambda)}
\]
---
### **Calculation**:
Let's calculate the required values.
### **Results**:
1. **Utilization factor** (\( \rho \)): \( 0.75 \) (75% utilization)
2. **Average number of clients in the system** (\( L \)): \( 3.00 \)
3. **Average number of clients in the queue** (\( L_q \)): \( 2.25 \)
4. **Average time spent in the system** (\( W \)): \( 20.00 \, \text{minutes} \)
5. **Average time spent in the queue** (\( W_q \)): \( 15.00 \, \text{minutes} \)
---
### **Efficiency Analysis**:
- **Utilization**: The SACCOS operates at 75% utilization, which is efficient and avoids overloading the system. However, it suggests limited capacity for handling surges in demand.
- **Time in system**: On average, a customer spends 20 minutes in the system, of which 15 minutes are waiting in the queue.
- **Queue Length**: The average queue length is 2.25 clients, indicating that customers experience some waiting.
### **Comment**:
The system is moderately efficient but could improve service time or add more service windows to reduce queueing time further.
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