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3. At a border inspection station, vehicles arrive at the rate of 10 per bour in

ID: 388849 • Letter: 3

Question

3. At a border inspection station, vehicles arrive at the rate of 10 per bour in a Poisson distribution. For simplicity in this problem, assume that there is only one lane and one inspector, who can inspect vehicles at the rate of 12 per hour in an exponentially distributed fashion (USE MODEL 1) (10 points) a b. c. d. What is the average length of the waiting line? What is the average time that a vehicle must wait to get through the system? What is the utilization of the inspector? What is the probability that when you arrive there will be three or more vehicles ahead of you?

Explanation / Answer

Following details are provided :

Arrival rate of vehicles = a = 10 per hour

Service ( i.e. inspection ) rate of vehicles = 12 per hour

= Average length of people in the queue + average length of people being served

= a^2/ S x ( s -a ) + a/s

= 10 x 10 / 12 x ( 12 – 10 ) + 10/12

= 100/ 24+ 10/12

= 120/24

= 5

Average length of the waiting line = 5

= average time in the waiting queue + average time of being served

= 10/12 x ( 12 – 10) + 1/12 hour

= 5/12 + 1/12 hour

= 6/12 hour

= 0.5 hour

= 30 minutes

Average time a vehicle must wait to go through the system = 30 minutes

= a/s x 100

= 10/12 x 100

= 83.33%

Utilization of the inspector = 83.33%

Probability that there is 1 vehicle in the queue = P1 = ( a/s) x Po = ( 10/12) x 0.1666 = 0.8333 x 0.1666 = 0.1388

Probability that there will be 2 vehicles in the queue = P2 = ( 10/12)^2 x 0.1666 = 0.8333 x 0.8333 x 0.1666 = 0.1157

Therefore,

Probability that there will be less than 3 vehicles ahead of you

= Po + P1 + P2

= 0.1666 + 0.1388 + 0.1157

= 0.4211

Probability that there will be 3 or more vehicles ahead of you

= 1 – Probability that there will be less than 3 vehicles ahead of you

= 1 – 0.4211

= 0.5789

Probability that there will be 3 or more vehicles ahead of you = 0.5789

Average length of the waiting line = 5

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