The capacity of a fast-food outlet is 150 customers an hour during peak hours, and 75 customers an h

The capacity of a fast-food outlet is 150 customers an hour during peak hours, and 75 customers an hour during normal hours. The outlet operates 24 hours a day, seven days a week. The rate of demand is 180 customers per hour during the peak demand period 12:00 to 1:00 PM (noon) daily, and 60 customers per hour in other times.

a. How many customers are at the fast-food outlet at 1:00PM?

b. How long does the customer that arrives at 1 PM should wait before being served? answer in minutes.

c. At what time (in PM) will the peak hour congestion be over, i.e., inventory reduces to zero?

d. What is the average flow time of a customer in minutes from 12:00 to 3:00 PM? Answer in minutes

e. The outlet manager decided to maintain the capacity at 150 customers per hour for one hour beyond the peak hour (i.e., until 2:00PM). What is the (new) maximum wait time in minutes?

a.)

Given: 

Total capacity of customers = 150

Demand = 180 customers

Formula:

Customers remain unserved during peak hours = Demand - Total capacity

Solution:

Customers remain unserved during peak hours = 180 - 150

Answer:

Customers remain unserved during peak hours = 30

Therefore, number of customers at the fast-food outlet at 1 PM is 30

b.)

Given:

Capacity of serving during normal hours = 75

Capacity of serving per minute = 60/75 = 0.8 minute per customer

Solution:

As per previous answer, there are already 30 customers remain unserved. So, the new customers arriving at 1PM have to wait for:

= 30*0.8 = 24  

Answer:

Customer should wait 24 minutes before they can be served.

c.)

Given:

Capacity of serving during normal hours = 75

Solution:

From 1 PM to 2 PM, there is extra 30 customers. 

After 2 PM, 60 new customers will come as per given data. 

So, there will be around 90 (30 + 60) customers to be served. 

We can serve only 75 customers between 1PM - 2PM, so there will be remaining 15 customers in the line. 15 = (90 - 75)

Next time slot is 2PM- 3PM, 60 new customers will come as per given data. Now, we have 60+15= 75 customers to be served. As we can serve 75 customers during normal hours. So, we will be serving all customers of 75 between timeslot of 2PM-3PM and there will no waiting customers after 3 PM. 

Answer:

Therefore, peak of congestion shall be over by 3 PM

d.)

Given:

From 12-1, number of customers = 180

From 1-3, number of customers = 60 per hour

Solution:

So total customers are = 60+60+180

Total customers = 300 customers

Time available from 12 to 3 is 180 minutes.

So flow time = Time available / total customers

flow time = 180 / 300

Answer:

Flow time = 0.6 minutes

e.)

Given:

Capacity to serve one customer = 60/150 = 0.4 minutes

Solution:

There will be excess of 30 customers after 1 PM, these 30 customers can be served in 12 minutes by solving:

= (0.4*30)

= 12

Therefore, new maximum wait time is 12 minutes

Answer:

New maximum wait time is 12 minutes