PizzaTime Restaurants is building a new pizza place and needs to determine how big to make the various parts of its facility. It wants to be able to accommodate a maximum of 300 customers per hour at its peak times. PizzaTime has collected the following information: the average time to place and receive an order is 1.05 minutes, 30 percent of the customers have cars and require parking spots, and the average length of time at the restaurant is 50 minutes per customer. Assuming a capacity cushion of 20 percent. a. Find the number of cash registers required. (assume an average of 4 customers per group) (Round up your answer to the next whole number.) Number of cash registers 2 b. Find the number of parking spaces needed. (Do not round intermediate calculations. Round up your answer to the next whole number.) Number of parking spaces c. Find the number of seats/tables needed. (assume 4 seats per table) (Round up your answer to the next whole number.) Number of seats/tables

1.) If 300 customers arrive at peak hour, then on average 75 orders will be placed ( considering group of 4 each).
average number of orders per hour =60/1.05 =57.14 orders, hence number of cash registers required=75/57.14=approximately 2 registers.
2.) This problem will be solved using principle of little ‘s law which states that
rate of flow in aprocess xthroughput time of the process= in process inventory
since the restaurant can accomodate 300 customers per hour, rate of flow=300/60 = 5 per minute.
Throughput time of the process =average length of time at restaaurant + average time to place and recieve an order
= 50 +1.05 = 51.05 minute .
There fore in process inventory , number of customers at any time =rate flow in process xThroughput time of the process
=5 x51.05=255.25 customers.
Since 30 % of customers will have cars , hence numbers of cars expected to be parked at any point of time = 255.25×30%= 76.575
capacity cushion = 20 %
hence ,number f parking space needed =numbers of cars expected to be parked at any point of timex(1+ capacity cushion)
=76.575×1.2
=91.89
=92
3.) Numbers of tables required=255.25 /4 = 64 tables ( 4 customer per table)
capacity cushion of 20 %
Hence, number of tables =64×1.2
=78 tables
 
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