Question

M 6 _ A 2 . The pe rson staffin g the ticket booth at an aquari um is able to distribute tickets and brochures to 325 patrons every hour. On a typical day, and average of 278 people arrive every hour top gain entrance to the aquarium. The aquarium’s manager wants to make the arrival process as convenient as possible for the patrons and so wishes to examine several queue operating characteristics. a) What type of system is this? b) What is the average number of patrons waiting in line to purchase tickets? (Enter two decimal places) c) What percentage of time is the ticket window busy ? (Enter three decimal places) d) What is the average time (in minutes) that a visitor to the aquarium spends in the ticketing system? (Enter two decimal places) e) What is the average time spent waiting in line (in minutes) to get to the ticket window? (Enter tw o decimal places) f) What is the probability there are more than two or more people waiting in line to get a ticket ? (Enter three decimal places)

Answer 1

Solution: The person staffing the ticket booth at an aquarium is able to distribute the tickets and brouchers 325 patrons every year On a typical day and average of 278 people arrive every hour top gain

A] Type of system M/M/L

b]

The average number of patrons µ = 325, ? = 278

Lq = (r*r) / µ(µ - ?) = 5.06

c]

percentage of time is the ticket window busy

Traffic Intensity = ? / µ = 0.85538

d]The average time =

Ws = 1 / (µ - ?) = 1 / 47 = 0.0213 hrs

e]

average time spent waiting in line

Wq = ? / µ(µ - ? = 278 / 15275 = 0.1092 min

f] We assume that arrival rate follows pattern of Poisson Distribution and Service rate ( i.e. distribution of tickets and brochures ) follow a negative exponential distribution

Accordingly.

Arrival rate = lambda = 278/ hour

Service rate = Alpha = 325/ hour

Probability that there are ZERO people waiting in line to get a ticket

= P0 = 1 – lamda/ alpha = 1 – 278/325 = 1 – 0.855 = 0.145

Probability that there are ONE person waiting in line to get a ticket

P1 = ( lambda/ alpha)^1 x Po

= (278/325)^1 x 0.145

= 0.855 x 0.145

= 0.124 ( rounded to 3 decimal places)

Probability that there are Zero or 1 person waiting in line to get a ticket

= Po + P1 = 0.145 + 0.124 = 0.269

Therefore,

Probability that there are two or more people waiting in the line

= 1 – ( probability that there are less than 2 people waiting in the line)

= 1 – ( P0 + P1)

= 1 – 0.269

= 0.731