The local Quick Burger fast food restaurant has a drive-through window. Customer

Question

The local Quick Burger fast food restaurant has a drive-through window. Customers in cars arrive at the window at the rate of 10 per hour (Poisson distributed). It requires an average of four minutes (exponentially distributed) to take and fill an order. The restaurant chain has a service goal of an average waiting time of three minutes. 1. Will the current system meet the restaurant’s service goal? 2. If the restaurant is not meeting its service goal, it can add a second drive-in window that will reduce the service time per customer to 2.5 minutes. Will the additional window enable the restaurant to meet its service goal? 3. During the two-hour lunch period the arrival rate of drive-in customers increases to 20 per hour. Will the two-window system be able to achieve the restaurant’s service goal during the rush period?

Explanation / Answer

(1)

Since the problem starts with a single window operation, this problem can be modeled as an M/M/1 queuing system

= average arrival rate = 10 per hour
= average service rate per server = 1 in 4 min = 15 in 60 min = 15 per hour

The average waiting time per car (Wq) is given by -

Wq = / (*( - )) = 10 / (15*(15 - 10)) = 0.133 hours = 8 min. So, the goal of 3 minutes waiting is not achieved.

(2)

For the two-window system, the work per customer gets distributed. So, it is NOT an M/M/2 situation but rather an M/M/1 situation with reduced service time only.

= average arrival rate = 10 per hour
= average service rate per server = 1 in 2.5 min = 24 in 60 min = 24 per hour

Wq = / (*( - )) = 10 / (24*(24 - 10)) = 0.0298 hours = 1.79 min. So, the service goal is achieved as 1.79 < 3.0

(3)

During the lunch period, = 20; = 24

Wq = / (*( - )) = 20 / (24*(24 - 20)) = 0.208 hours = 12.5 min. So, the goal will not be achieved during the lunch hours even with the second window working.

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