Question

1. Visitors arrive at the ticket booth at the Metropolitan Museum at the rate of 5 every 6 minutes. The average service time is 1 minute. The arrival rate is assumed to follow Poisson distribution and service times are exponentially distributed. e) What is the probability that there are no visitors at the window? f) What percentage of the time is the ticketing clerk busy? g) What is the probability that there are exactly 2 visitors in the ticketing area? h) By how much would your answer to part (a)-(d) be reduced if a second ticketing clerk were added?

Answer 1

Arrival time (l)= 5 visitors every 6 minutes = 50 visitors per hour

Service time (m) = 1 visitor per minute = 60 visitors per hour

a) Average time spent in system Ws= 1/m-l = 1/60-50= 1/10 = 0.1 hr = 6 mins

b) Average no of visitors in system Ls=l/m-l = 50/(60-50)= 50/10 = 5 visitors

c) Average visitors waiting in queue Lq = l*Ls/m = 50*5/60 = 25/6 = 4.167 visitors

d) Average time spent in line Wq= l*Ws/m = (50*0.1)/60 = 0.083 hr = 5 mins

Utilization rate (p) = l/m = 50/60 = 0.833 = 83.3%

(1-p) = 1-0.833 =0.167

e)Probability that there is no visitor in system = (1-p)(p^0) = 0.167* 1 = 0.167

f) Percentage of time the ticketing clerk is busy = utilization rate = 83.3%

g) Probability that there are exactly 2 visitors in system = (p^2)*(1-p) = (0.833^2)*0.167 = 0.1158

h) If no of ticketing clerks (s) =2

Given,

M|M|2 model

Utilization p= l/sm = 50/(2*60)= 0.4167

1-p = 1-0.4167 = 0.5833

Also l/m= 50/60 = 0.833

Probability that no customer is in system Po= [(0.833)^0/0! + (0.833)^1/1! + (0.833)^2/2! (1/(1-0.4167))]^-1 = 0.4119

Utilization p= l/sm = 10/(2*15)= 0.33

Also l/m= 10/15 = 2/3 = 0.66

Probability that no customer is in system Po= [(0.66)^0/0! + (0.66)^1/1! + (0.66)^2/2! (1/(1-0.33))]^-1 = 0.507

Average visitors waiting in queue Lq = Po (l/m)^s*0.4167/ (2)!(1-0.4167)^2 = (0.4119*(0.833^2)*0.4167) / (2*(0.833^2)) = 0.085 visitors

Average time spent in line Wq= Lq/l = 0.085/50 = 0.0017 hr = 0.102 mins

Average time spent in system Ws = Wq+ (1/m) = 0.0017 + (1/60) = 0.01836 hr = 1.102 mins

Average no of visitors in system Ls = l*Ws = 50*1.102 = 55.1 visitors