The Bijou Theater shows vintage movies. Customers arrive at the theater line at
ID: 442915 • Letter: T
Question
The Bijou Theater shows vintage movies. Customers arrive at the theater line at the rate of 100 per hour. The ticket seller averages 24 seconds per customer, which includes placing validation stamps on customers’ parking lot receipts and punching their frequent watcher cards. (Because of these added services, many customers don’t get in until after the feature has started.) Use Exhibit 10.9.
What would be the effect on customer time in the system of having a second ticket taker doing nothing but validations and card punching, thereby cutting the average service time to 20 seconds? (Round your answer to 3 decimal places.)
What would be the customer time in the system if a second window was opened with each server doing all three tasks? (Use closest /µ value from Exhibit 10.9. Do not round intermediate calculations. Round your answer to 3 decimal places.)
References
The Bijou Theater shows vintage movies. Customers arrive at the theater line at the rate of 100 per hour. The ticket seller averages 24 seconds per customer, which includes placing validation stamps on customers’ parking lot receipts and punching their frequent watcher cards. (Because of these added services, many customers don’t get in until after the feature has started.) Use Exhibit 10.9.
Explanation / Answer
Customers arrival rate is 100 per hour whereas service time is 24 seconds per customer, therefore service rate is 150 (60*60 /24) per hour.
a. Average customer time in the queue system = 1 / (service rate - arrival rate)
That is= 1 /(150-100) = 1/50 hour =(60/50) 1.2 minutes
b. In case average service time is 20 seconds then service rate is (60*60/20) 180 per hour
Therefore Average time in the queue system = 1 / (180-100) = 1/80 hours = (60/80) 0.75 minutes
c. In case second window (similiar to first) is opened, there are two similiar servers then closest /µ value is 100/300(2*150} that is system utilization
Average time in the system = average time in the queue + 1/service rate
Average time in queue = Average number in queue / arrival rate
Average number in queue = [(arrival rate*service rate *P0 )/((c-1)!(c*service rate-arrival rate)2 ](arrival rate/service rate)c where P0 is probability no one in the queue and c represents number of servers
P0 = [ 1 + (2/3) + (1/2)*(2/3)2 (3/2)]-1 =1/2
therefore average number in queue = [(100*150*.5)/(1*200*200)](2/3)2 =25/300
hence average waiting time in queue = 1/(12*100) hours = 3seconds
Therefore average time spend in the queue system = (3+24) 27 seconds
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