Answer to question a : It is given that customer arrival rate = 30 per hour Cust
ID: 373611 • Letter: A
Question
Answer to question a :
It is given that customer arrival rate = 30 per hour
Customer service rate ( @1.5 minutes per customer ) = 40 per hour
Since the system has more capacity to serve customers than arrival rate, the system is considered to be stable
Answer to question b:
Let arrival rate = a = 30/ hour
Service rate = S = 45/ hour
Therefore, Number of customers waiting in the line
= a^2/ s x ( s – a )
= 30 x 30 / 45 x ( 45 – 30)
= 900/45 x 15
= 1.33
ON AVERAGE 1.33 CUSTOMERS ARE WAITING IN LINE
Answer to question c :
Average waiting time in the line by any customer
= 30/45x(45 – 30)
= 30/(45 x 15) hour
= 2/45 hours
= 2.66 minutes
AVERAGE WAITING TIME IN LINE BY ANY CUSTOMER =2.66 MINUTES
Answer to question d:
Average time customer spends at restaurant
= Average waiting time in the line + 1 / s
= 2/45 + 1/45
= 3/45
= 1/15 hours
= 4 minutes
ON AVERAGE CUSTOMER SPENDS 4 MINUTES AT THE RESTAURANT
Answer to question e :
Probability that there are ZERO cars in the system = Po = 1 – a/s 1 – 30/45 = 1 – 0.666 = 0.334
Probability that there is 1 car in the system = P1 = (a/s)x P0 = 0.666x 0.334 = 0.222
Probability that there are 2 cars in the system = P2 = (a/s)^2 x Po = 0.666x0.666x0.334 = 0.1481
Probability that there are 3 cars in the system = P3 = ( a/s)^3 x P0 = 0.666x0.666x0.666x0.334 =0.0986
Therefore, Probability that there are maximum 3 cars in the system
= P0 + P1 + P2 + P3
= 0.334 +0.222 + 0.1481 + 0.0986
= 0.8027
Therefore,
Probability of more than 3 cars in the drive through line
= 1 – Probability of maximum 3 cars in the system
= 1 – 0.8027
= 0.1973
Therefore , for 0.1973 fraction of time there are more than 3 cars in the drive through system
FOR 0.1973 FRACTION OF TIME , THERE ARE MORE THAN 3 CARS IN THE DRIVE THROUGH SYSTEM
ON AVERAGE 1.33 CUSTOMERS ARE WAITING IN LINE
Explanation / Answer
A busy fast-food restaurant has one drive-through window. An average of 30 customers arrive at the window per hour. It takes an average of 1.5 minutes to serve a customer. Assume that inter- arrival and service times are exponentially distributed. The following questions are concerned with the steady state behavior of this queueing system. (a) Is the system stable? Explain how you found your answer. b) On average, how many customers are waiting in line? (c) On average, how long does each customer wait in line? (d) On average, how long does a customer spend at the restaurant (from the time of arrival to the time the purchase is completed)? (e) What fraction of the time are more than 3 cars in the drive-through lane? This includes the car, if any, being served.
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.