Question

Problem #4  Mike Dreskin manages a large Los Angeles movie theater complex called Cinema I, II, III, and IV. Each of the four auditoriums plays a different film; the schedule is set so that starting times are staggered to avoid the large crowds that would occur if all four movies started at the same time. The theater has a single ticket booth and a cashier who can maintain an average service rate of 280 movie patrons per hour. Service times are assumed to follow an exponential distribution. Arrivals on a typically active day are Poisson distributed and average 210 per hour.

To determine the efficiency of the current ticket operation, Mike wishes to examine several queue operating characteristics.

(a) Find the average number of moviegoers waiting in line to purchase a ticket.

(b) What percentage of the time is the cashier busy?

(c) What is the average time that a customer spends in the system?

(d) What is the average time spent waiting in line to get to the ticket window?

(e) What is the probability that there are more than two people in the system?

Answer 1

Answer:

Given that:

The theater has a single ticket booth and a cashier who can maintain an average service rate of 280 movie patrons per hour. Service times are assumed to follow an exponential distribution. Arrivals on a typically active day are Poisson distributed and average 210 per hour.

(a) Find the average number of moviegoers waiting in line to purchase a ticket.

Average number of moviegoers waiting in line to pruchase a ticket

(b) What percentage of the time is the cashier busy?

Percentage of the time the cashier is busy

= 100 - Percentage of the time the cashier is idle

= 100 – {100 x P0) = 100 - 100(1 - ρ)

= 100 x ρ = 75%

(c)  What is the average time that a customer spends in the system?

Average time that a customer spends in the system

= E(v) = 1/70 hour

= 6/7 minute

(d) What is the average time spent waiting in line to get to the ticket window?

Average time spent waiting in line to get to the ticket window

= E(w) = 210/(280 x 70) hour

= 9/14 minute

(e)  What is the probability that there are more than two people in the system?

Probability that there are more than two people in the system

= 1 – P(n ≤ 2) = 1 – (P0 + P1 + P2)

= 1- P0(1 + 0.75 + 0.752)

= 1- (37/64) = 27/64