Question

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 300 300 patrons per hour. Service times are assumed to follow a negative exponential distribution. Arrivals on a normally active day are Poisson distributed and average 200 200 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​ = 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​ = minutes ​(round your response to two decimal​ places).

​d) The average time spent waiting in the line to get to the ticket window​ = minutes ​(round your response to two decimal​ places).

​e) The probability that there are more than two people in the system​ = ​(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).

The probability that there are more than four people in the system​ = ​(round your response to three decimal​ places).

Answer 1

Arrival rate =  λ = 200 customers per hour

Service rate = µ = 300 customers per hour

Utilization = p = λ / µ = 200 / 300 = 2/3

a)

The average number of moviegoers waiting in line to purchase a ticket =  ( λ x λ)   / (µ(  µ -  λ)) = (200 x 200) / ( 300 x ( 300 - 200)

= (200 x 200) / ( 300 x 100) = 1.33 customers

b)

The percentage of time that the cashier is busy = λ / µ = 200 / 300 = 0.6666 = 67% (Roundin off to nearest whole number)

c)

The average time that a customer spends in the system​ = 1 / ( µ - λ) = 1 / (300 - 200) = 1 / 100 = 0.01 hours = 0.01 x 60 = 0.6 minutes

d)

The average time spent waiting in the line to get to the ticket window = λ  / (µ(  µ -  λ)) = 200 / ( 300 x (300 -200)

= 200 / ( 300 x 100) = 0.0067 hours = 0.4 minutes

e)

Probability that there are no customers in the system = (1 - p) (p^0) = ( 1 - (2/3)) ((2/3)^0) = 0.333

Probability that there is 1 customer in the system= ( 1 - p) (p^1) = ( 1 - (2/3)) ((2/3)^1) = 0.222

Probability that there are 2 customers in the system = (1 -p) (p^2) = ( 1 -(2/3)) ((2/3)^2) = 0.148

So, Probability that there are more than 2 customers in the system = 1 - ( Probability that there are no customers in the system + Probability that there is 1 customer in the system + Probability that there are 2 customer in the system)

= 1 - (0.333 + 0.222 + 0.148)

= 1 - 0.703

= 0.297

Probability that there are 3 customers in the system = (1 -p) (p^3) = ( 1 -(2/3)) ((2/3)^3) = 0.0987

So, Probability that there are more than 3 customers in the system = 1 - ( Probability that there are no customers in the system + Probability that there is 1 customer in the system + Probability that there are 2 customers in the system + Probability that there are 3 customers in the system)

= 1 - (0.333 + 0.222 + 0.148 + 0.0987)

= 1 - 0.8017

= 0.1983 = 0.198 (Rounding off to three decimal places)

Probability that there are 4 customers in the system = (1 -p) (p^4) = ( 1 -(2/3)) ((2/3)^4) = 0.0658

So, Probability that there are more than 4 customers in the system = 1 - ( Probability that there are no customers in the system + Probability that there is 1 customer in the system + Probability that there are 2 customers in the system + Probability that there are 3 customers in the system + Probability that there are 4 customers in the system)

= 1 - (0.333 + 0.222 + 0.1481+ 0.0987 + 0.0658)

= 1 - 0.8676

= 0.1324 = 0.132 (Rounding off to three decimal places)