Homework: HW#18-Module D-Waiting Lines-Part 02 Score: 0 of 5 pts Problem D.10 Sa
ID: 357933 • Letter: H
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Homework: HW#18-Module D-Waiting Lines-Part 02 Score: 0 of 5 pts Problem D.10 Save C Elof 7(4 complete)"/>] Hw score: 15%, 3 of 20 pts Question Help Beate Klingenberg manages a Poughkeepsie, New York, movie theater complex called Cinema 8 Each of the eight auditoriums plays a different film; the schedule staggers starting times to avoid the large crowds that would occur if all eight movies started at the same time. The theater has a single ticket booth and a cashier who can maintain an average service rate of 300 patrons per hour. Service times are assumed to follow a negative exponential distribution. Arrivals on a normally active day are Poisson distributed and average 200 per hour To determine the efficiency of the current ticket operation, Beate wishes to examine several queue-operating characteristics a) The average number of moviegoers waiting in line to purchase a ticket 1.33 customers (round your response to two decimal places) b) The percentage of time that the cashier is busy 67 percent (round your response to the nearest whole number). c) The average time that a customer spends in the system 60 minutes (round your response to two decimal places) d) The average time spent waiting in the line to get to the ticket window 40 minutes (round your response to two decimal places). e) The probability that there are more than two people in the system 0296 (round your response to three decimal places) The probability that there are more than three people in the system (round your response to three decimal places) Enter your answer in the answer box and then click Check Answer Check Ansy remaining 7:15 PM a403/31/2018Explanation / Answer
This is single server M/M/1 queue model
Arrival rate, ? = 200 per hour
Service rate, ? = 300 per hour
a) Average number of customers waiting in line (Lq) = ?2/(?*(?-?)) = 2002/(300*(300-200)) = 1.33
b) Percentage of time that cashier is busy = ?/? = 200/300 = 67 percent
c) Average time a customer spends in the system (W) = 1/(?-?) = 1/(300-200) = 0.01 hour = 0.60 minutes
d) Average time spent waiting in line (Wq) = Lq/? = 1.33/200 = 0.0067 hour = 0.40 minutes
e) Probabilty of n people in system, Pn =(1-?)*?n , where ? = ?/? = 200/300 = 0.67
P0 =(1-0.67)*0.670 = 0.3333
P1 =(1-0.67)*0.671 = 0.2222
P2 =(1-0.67)*0.672 = 0.1481
Probability of more than two people in system = 1 - (P0+P1+P2) = 1 - (0.3333+0.2222+0.1481) = 0.2963
P3 =(1-0.67)*0.673 = 0.0988
Probability of more than three people in system = 1 - (P0+P1+P2+P3) = 1 - (0.3333+0.2222+0.1481+0.0988) = 0.1976
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