Question

Q3: A Security Scanner at a Small Airport (Pollaczek-Khinchin)

Outer Moosewood Regional Airport, a relatively small rural facility, has a single body scanner for security checks. During the airport’s busy hours, outgoing passengers arrive at the scanner facility in a memoryless manner at an average rate of 3 per minute (or 0.05 passengers per second). They line up to pass through the body scanner. The current scanner must be replaced with one of two possible models, called “A” and “B”. Model A takes 12 seconds to scan a passenger and has a 32% chance of referring passengers for manual screening. Model B takes 15 seconds to scan a passenger but has only an 8% chance of referring passengers for manual screening. We model this system as an M/G/1 queue in which the service facility is the scanner plus the manual screening (if needed). Assume that every manual screening takes exactly 15 seconds and a customer cannot enter the scanner until the previous customer’s manual screening (when needed) is complete.

(a) For each kind of scanner, what is the expected service time E[S]?

b) For each kind of scanner, what is the server loading factor, rho?

(c) For each kind of scanner, what is the variance^2 of the service time?

(d) For each kind of scanner, use the Pollaczek-Khinchin formula to predict the average number of customers Lq waiting to enter the scanner during the airport’s busy hours.

(e) For each kind of scanner, what is the average number of seconds each busy-hour customer will spend passing through security (including the line, the scan, and manual screening if that is necessary)?

Answer 1

Hello,

Given information in the Problem is

λ=0.05 passengers per second.

Average time taken by the scanner is 15 seconds. that is μ = 15 second.

Solution

a) expected service time E[S] for A

using the below formula E[S] = 1/μ

E[S] = 1/12 = 0.08333

expected service time E[S] for B

using the below formula E[S] = 1/μ

E[S] = 1/15 = 0.06666

b) what is the server loading factor, rho?

for A    λ=0.05 and μ =12

ρ=λμ = 0.6

for B

λ=0.05 and μ =15

ρ=λμ = 0.75

c) what is the variance^2 of the service time

for A

Formula of variance is = E[X2] - E[X]2

E[X] = 1/μ; E[X2] = 2/μ2

μ =12

E[X] = 1/μ = 0.08333

E[X2] = 2/μ2 = 0.0138

Variance = 0.0138 - ( 0.08333)^2 = 0.006857

for B

Formula of variance is = E[X2] - E[X]2

E[X] = 1/μ; E[X2] = 2/μ2

μ =15

E[X] = 1/μ = 0.06666

E[X2] = 2/μ2 = 0.008888

Variance = 0.008888 - ( 0.06666)^2 = =0.00444

d) average number of customers= λ2E[S2]/2(1−ρ)

For A

=(0.05^2)*(0.0138)/2*(1-0.6)

=0.0013 / 0.8 = 0.001625

for B

=(0.05^2)*(0.008888) /2*(1- 0.75)

= 0.00001144/0.5= 7.62666666666667E-06