Question

1. At Baywatch City Airport 3 passengers per minute arrive on average. Inter-arrival times are exponentially distributed. Arriving (actually departing) passengers go through a checkpoint consisting of a metal detector and baggage X-ray machine. Whenever a checkpoint is in operation, two employees are required. These two employees work simultaneously to check a single passenger. A checkpoint can check an average of 3 passengers per minute, where the time to check a passenger is also exponentially distributed.

1. Why is this an M/M/1, not an M/M/2, queueing system?
2. What is the probability that a passenger will have to wait before going through the checkpoint?
3. On average, how many passengers are waiting in line to enter the checkpoint?
4. On average, how long will a passenger spend at the checkpoint (including waiting time in line)?
5. What percentage of the time is the line longer than 4 people?
6. Baywatch City is experiencing a tourism boom and arrival rates at the security checkpoint are increasing. Lines are getting longer. Once the probability that the line will be longer than 5 people exceeds 0.90, Gotham city will consider opening a second checkpoint. At what arrival rate will this happen?

Answer 1

Let , = Arrival rate or interarrival rate

=Average number of passenger arrive at Baywatch city airport .

  per minute = 180 per hour

Let , = Service rate

= Average number of passengers a checkpoint can check

per minute .= 180 per hours

n = Number of customers in the system .( includes one who is taking service )

m= the number of customers in the queue ( excludes one who is taking service )

n follows geometric distribution with pmf as follows ,

P(n=r) = P(there are r passenger inthe system ) = ; r = 0,1,..... :  

a. Altough they are two employees to check passenger , but they check single passenger simulteneously . i.e single service station is there so this is M/M/1 model not M/M2 model

b. Probability that a passenger will have to wait before going through the checkpoint =

Probability that system is busy = p[ n 3 ] = P[ n > 0] = 1- P[n=0 ] 1-[1- ] = =

Probability that a passenger will have to wait before going through the checkpoint = 1

c.  On average, how many passengers are waiting in line to enter the checkpoint is given by

  

d. On average, how long will a passenger spend at the checkpoint (including waiting time in line)=

Average time spent by a customer in the system = E(V) =

e .

What percentage of the time is the line longer than 4 people

P[ m>4] = P[n > 5 ] = 1- P[n<5] = 1 -{ P[n=0] +P[n=1] +P[n=2] +P[n=3] +P[n=4]}

Here ,

P[ m>4] = P[n > 5 ] = 1- P[n<5] = 1 -0 = 1

100%  percentage of the time is the line longer than 4 people .