Music venue is a concert organizer that is evaluating the installation if portab
ID: 470069 • Letter: M
Question
Music venue is a concert organizer that is evaluating the installation if portable toilets in the arena where a concert is going to take place. In particular they are considering giving service to a part of the complex that is far away from the existing restrooms. Assume there will be one single queue to access all the portable toilets in this area and that only one person uses a toilet at a time. There are 2 arrivals per minute on average, the standard deviation of the time between arrivals is 15 seconds, and the average time a person spends in the toilet is 4 minutes (with standard deviation of 1 minute). MV1. What is the average utilization of 10 toilets? MV2. What is the expected time a customer will have to wait? MV3. How long will the average toilet be empty during a 5-hour concert? MV4. What is going to be the average number of people in the queue? (your answer does not have to be an integer).
Explanation / Answer
Arrival rate, = 2 per minute
Service Rate, µ = 1/4 per minute
MV1. Given, m = 10 Average utilization of the server, = /mµ = 2 / 10 / 0.25 = 0.8 = 80%
MV2. Coefficient of variation of interarrival time, Ca = std dev / mean of interarrival times = 15 sec / 30 sec = 0.5
Coefficient of variation of service time, Cs = std dev / mean of service times = 1/4 = 0.25
(Ca2 + Cs2)/2 = (0.52 + 0.252)/2 = 0.15625
Average length of queue (number of people waiting in queue), Lq = 2/(1 - )*(Ca2 + Cs2)/2 = 0.82/(1-0.8)*0.15625 = 3.2*0.15625 = 0.5
Average waiting time = Lq / = 0.5/2 = 0.25 minute = 15 seconds
MV3. Average utilisation = 0.8, as computed in part 1. So idle time = 1 - 0.8 = 0.2 . Out of 5 hour , the average time the toilet will be empty = 0.2*5 = 1 hour
MV4. Average number of people waiting in line, Lq = 2/(1 - )*(Ca2 + Cs2)/2 = 0.82/(1-0.8)*0.15625 = 3.2*0.15625 = 0.5
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