Answer:
1. c
2. f
3. d
Step-by-step explanation:
This type of queuing model is an example of "single-channel, single-phase" process.
However, from the information given:
The inter-arrival rate = 10 customers per hour
Arrival time = 6 minutes
Service time = 5 minutes
∴
utilization = service time/arrival time
= 5/6
Time on queue [tex]T_q = p \times \dfrac{utilization }{1 - utilization} \times \dfrac{Cv_a^2 -Cv_b^2}{2}[/tex]
[tex]T_q = 5 \times \dfrac{5/6 }{1 -5/6} \times \dfrac{(2.58^2 -0.75^2)}{2}[/tex]
[tex]T_q = 25 \times 3.60945[/tex]
[tex]T_q = 90.24 \\ \\ T_q \simeq 90 \ mins[/tex]
Thus, customer waiting in the queue line is;
[tex]= \dfrac{T_q}{q} \\ \\ = \dfrac{90}{6} \\ \\ \mathbf{=15}[/tex]
Thus, new customers will have to wait in a total time of;
= waiting time + service time
= 90 + 5
= 95 minutes.