Question 4 (25 points) A busy fast-food restaurant has one drive-through window.

Question

Question 4 (25 points) 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.

Explanation / Answer

Given :;

Customer arrival rate = a = 30 / hour

Customer service rate ( @1.5 minute per customer =s = 60/1.5 = 40 / hour

For a system to be stable, it should be able to serve customer faster than their arrival rate

Since in this case,

Customer Service rate > Customer arrival rate, the system will be considered stable

THE SYSTEM IS STABLE

Number of customers waiting in line = a^2/ s x ( s – a ) = 30 x 30 / 40 x ( 40 – 30 ) = 900/400 = 2.25

2.25 CUSTOMERS ARE WAITING IN LINE

Average time a customer wait in line = a / s x ( s – a ) = 30/ 40 x ( 40 – 30) = 30/400 hours = 1800/400 minutes = 4.5 minutes

ON AVERAGE A CUSTOMER WAITS 4.5 MINUTES IN A LINE

Average time a customer waits in restaurant

= Average time a customer waits in line + 1 / s   hour

= Average time a customer waits in line + 60/40 minutes

= 4.5 + 1.5 minutes

= 6 minutes

ON AVERAGE CUSTOMER SPENDS 6 MINUTES AT THE RESTAURANT

Probability that there are ZERO cars in the line = P0 = 1 – a/s = 1 – 30/40 = 1 – 0.75 = 0.25

Probability that there is 1 customer waiting in the line = P1 = ( a/s) x P0 = 30/40 x 0.25 = 0.75 x 0.25 = 0.1875

Probability that there are 2 customers waiting in the line = ( a/s)^2 x Po = ( 30/40)^2 x 0.25 = 0.75 x 0.75 x 0.25 = 0.1406

Probability that there are 3 customers waiting in the line = ( a/s)^3 x Po = ( 30/40)^3 x 0.25 = 0.1054

Hence probability that there are maximum 3 cars in the drive through lane

= P0 + P1 + P2 + P3

= 0.25 + 0.1875 + 0.1406 + 0.1054

= 0.6835

Hence, probability that there are more than 3 cars in the drive through lane

= 1 – Probability that there are maximum 3 cars in the drive through lane

= 1 – 0.6835

=0.3165

Hence, for 0.3135 fraction of time there are more than 3 cars in the drive through lane

FOR 0.3135 FRACTION OF TIME THERE ARE MORE THAN 3 CARS IN THE DRIVE THROUGH LANE

THE SYSTEM IS STABLE

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