A fast food restaurant has one drive-in window. An average of 40 customers per h

Question

A fast food restaurant has one drive-in window. An average of 40 customers per hour arrive at the window. It takes an average of three minutes to serve a customer. Assume 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 the time the customer arrives to time service is completed)? What fraction of time are more than three cars waiting for service (this includes the car, if any, at the window)?

Explanation / Answer

lamda = 40

u = 1/ 1 min ( 1 hr / 60 mins ) = 60

p = 40 / 60 = 0.66667

Now,

a) average = (lamda2) / u (u-lamda) = (40)2 / (60 (60 - 40)) = 1.3333

b) lamda / (u - lamda) = (40) / (60 - 40) = 2

c) since more than 3 cars is hard to copmute we always use [1 - (p <= 3 )]

so we have:

p3 = (1 - (40/60)) (40/60)3 + (1 - (40/60)) (40/60)2 + (1 - (40/60)) (40/60) = 0.1481 + 0.2222 + 0.0988 = 0.4691

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