Answered

Westonci.ca offers fast, accurate answers to your questions. Join our community and get the insights you need now. Get immediate answers to your questions from a wide network of experienced professionals on our Q&A platform. Discover detailed answers to your questions from a wide network of experts on our comprehensive Q&A platform.

A new pop-up restaurant serving high-end street tacos just opened and is expected to have high demand based on the chef's prior restaurant experience. The pop-up restaurant has only one window where customers order and receive their food. It takes an average of 5 minutes to serve a customer. There is no seating available so the customer exits the system after receiving their order.
Due to the creative marketing strategy of "hiding" the restaurant location, only 10 customers arrive per hour. There is no specific pattern to the distribution of time between arrivals or service times, but from historical data from prior pop-ups, the coefficient of variation for time between arrivals is 2.58 and service time is 0.75.
1. What type of queuing model is this process?
a. single-channel, multi-phase.
b. multi-channel, multi-phase.
c. single-channel, single-phase.
d. multi-channel, single-phase.
2. How many customers would you expect to find waiting in line?
a. 3.
b. 18.
c. 10.
d. 0.
e. 7.
f. 15.
g. 12.
3. For new customers, how long do they have to wait in total before they leave with their food?
a. 45 minutes.
b. 20 minutes.
c. 50 minutes.
d. 95 minutes.
e. 15 minutes.
f. 100 minutes.
g. 40 minutes.
h. 10 minutes.
i. 0 minutes.
j. 90 minutes.

Sagot :

ajeigbeibraheem

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.

We appreciate your visit. Our platform is always here to offer accurate and reliable answers. Return anytime. Thank you for visiting. Our goal is to provide the most accurate answers for all your informational needs. Come back soon. Thank you for visiting Westonci.ca. Stay informed by coming back for more detailed answers.