A fast-food restaurant has one drive-through window. An average of 40 customers

Question

A fast-food restaurant has one drive-through window. An average of 40 customers per hour arrive at the window. It takes an average of 1 minute to serve a customer. Assume that interarrival and service times are exponential. On the average, how many customers are waiting in line? On the average, how long does a customer spend at the restaurant (from time of arrival to time service is completed)? What fraction of the time are more than 3 cars waiting for service (this includes the car (if any) at the window)?

Explanation / Answer

arrival rate 40 customers per hour service time 1 minute per customer service rate 60 customers per hour a) avg no in the queue = (arrival rate ^ 2) / (service rate * (service rate - arrval rate)) Avg number of customers waiting in line = 1 number (value rounded off) b) avg time in the system = 1 / (service rate - arrival rate) Avg time spent by customer In the restaurant 0.05 hours c) Probability that there are n customers in the system = [1-(arrival rate/service rate)][(arrival rate/service rate)^n] Fraction of time 3 cars are waiting for service 0.10 Probability that there are more than 3 cars in the system = 1 - probability that there are 3 or less than 3 cars in the system Fraction of time that more than 3 cars are waiting in service = 0.90

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