12-13 Mike Dreskin manages a large Los Angeles movie theater complex called Cine

Question

12-13 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 poison distributed and average 210 per hour. To determine the efficiency of the current ticktet operating, Mike wishes to examine several queue operating characteristics.

(a) Find the average number of moviegoers waiting in line to pruchase 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 people?

Explanation / Answer

Solution

This is the case of an M/M/1 (Poisson arrival-exponential service-single service channel) Queue system.

Back-up Theory

Let n = number of customers in the system and m = number of customers in the queue.

[Trivially, n = m + number of customers under service.]

Let (/µ) =

The steady-state probability of n customers in the system is given by Pn = n(1 - ) …………(1)

The steady-state probability of no customers in the system is given by P0 = (1 - ) …………(2)

Average queue length = E(m) = (2)/{µ(µ - )} ……………………………………………..(3)

Average number of customers in the system = E(n) = ()/(µ - )……………………………..(4)

Average waiting time = E(w) = ()/{µ(µ - )} ……………………………………………..(5)

Average time spent in the system = E(v) = {1/(µ - )}……………………………………..(6)

Proportion of idle time of service channel = P0 = (1 - ) …………………………………….(7)

Now, to work out solution,

Given, = 210/hr, /µ = 280/hr, = 210/280 = 0.75…………………………………………(8)

Part (a)

Average number of moviegoers waiting in line to pruchase a ticket

= E(m) = 2102/(280 x 70) [vide (3) under Back-up Theory]

= 2.25. ANSWER

Part (b)

Percentage of the time the cashier is busy

= 100 - Percentage of the time the cashier is idle

= 100 – {100 x P0) = 100 - 100(1 - ) [vide (7) under Back-up Theory]

= 100 x = 75% {from (12)}ANSWER

Part (c)

Average time that a customer spends in the system

= E(v) = 1/70 hour [vide (6) under Back-up Theory]

= 6/7 minute ANSWER

Part (d)

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

= E(w) = 210/(280 x 70) hour [vide (5) under Back-up Theory]

= 9/14 minute ANSWER

Part (e)

Sub-part (i)

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) [vide (1) and (2) under Back-up Theory]

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

Part (e)

Sub-part (ii)

Probability that there are more than three people in the system

= 1 – P(n 3) = 1 – (P0 + P1 + P2 + P3)

= 1- P0(1 + 0.75 + 0.752 + 0.753) [vide (1) and (2) under Back-up Theory]

= 1- (175/256) = 81/256 ANSWER

Part (e)

Sub-part (iii)

Probability that there are more than four people in the system

= 1 – P(n 4) = 1 – (P0 + P1 + P2 + P3   + P4)

= 1- P0(1 + 0.75 + 0.752 + 0.753 + 0.754) [vide (1) and (2) under Back-up Theory]

= 1- (781/1024) = 243/1024 ANSWER

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