Question

In: Statistics and Probability

"Mike Dreskin manages a large Los Angeles movie theater complex called Cinema I, II, III, and...

"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? More than three people? More than four?"

Please use an Excel Model. (MD1?)

Solutions

Expert Solution

answer) Given data

λ = 210,

µ = 225,

(a). Average number of moviegoers waiting in line to purchase a ticket, Lq = / (1- )

we know that

ρ = λ/μ =210 / 225= 0.933

Lq =0.93333332 / (1 - 0.9333333)

Lq=13.07

(b)

Percentage of the time is the cashier busy, = /

=210 / 225 = 0.9333333

    = 93.33%

(c)

Average time that a customer spends in the queue, Wq = Lq /

Wq= 13.07 / 210

   Wq= 0.062 hour

Wq=0.062 * 60 min

   Wq= 3.72 minutes

Average time that a customer spends in the system, W = Wq + 1/

W= 0.062 + 1/225 = 0.062 + 0.00444 = 0.067 hour

W= 0.067 * 60 min = 4.02 minutes

(d).

Average time spent waiting in line to get to the ticket window (Wq)=3.72 minutes

(e)

More than two people in the system:

require probability=1 - [(1 - ) + (1 - ) ]

=1 - [ (1 - 0.933) + (1 - 0.933) * 0.933]

= 1 - [0.0666 + 0.0622]

= 1- 0.1289

= 0.8711

More than fiure people in the system:

Require probability=1 - [(1 - ) + (1 - ) + (1 - ) + (1 - ) + (1 - ) ]

=1 - [ (1 - 0.933) + (1 - 0.933) * 0.933 + (1 -0.933) * 0.9332 (1 - 0.933) * 0.9333 + (1 - 0.933) *0.9334]

= 1- 0.2917545

require probability= 0.7082


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