terminal design passenger processing. errata consider tsa impacts

Terminal DesignPassenger Processing

errata

Consider TSA impacts

Passenger Space Guidelines (1/2)

Personal space Transit 18x24 inches Airports 5-10 ft2 (30 inch circle)

Lateral movement 30 inches between “traffic” lanes

Longitudinal movement 8-10 feet per person

Net pedestrian area 20-30 ft2

Passenger Space Guidelines (2/2)

Queuing space 5-10 ft2

Stairs 10-20 ft2 Escalators can be smaller

Pedestrian flow f = s/a Where f pedestrian flow, s speed, a area per

pedestrian (note analog to vehicular traffic flow density relationship: flow = density * velocity)

Passenger System (1/5)

Entryways Passenger and visitors Enplaning and deplaning Auto doors 20-30 pax/min Manual doors 10-15 pax/min

Lobby areas All persons using airport Seating capacity 15-25% of enplaning Space 20 sf/pax


Ticket counter Check in and baggage drop Estimate 10% of peak hour originating pax with 5

pax in line max Spacing: 10-12 ft. between counters without bags 12-16 ft. between counters for

regular Queue space 3 ft./pax = 15 ft. Provide 20-35 ft. circulation area behind queues 10 ft. of depth for the counter itself


Security Service rate 300-450 pax/hr (lower

than book says) 15-20 ft wide; 30-60 ft long Deplaning exit corridor 15-20 ft wide,

revolving door or guards

20-40’ 10’-20’

15-20’


Departure lounge Estimate 80% of pax need seating Space 10-15 ft2/pax Walking corridors should be provided

Boarding corridors 10 ft. wide Service rate 2-4 pax/min


Corridors 20 ft wide minimum 40-50 ft desirable for maneuvering

Stairs 30 inches minimum per lane Speed 50-300 ft/min; average 100

ft/min

Baggage claim Special procedures

Queuing Equations (1/4)

Arrivals Poisson rate q

Service Exponential rate Q

Ratio ρ= q/Q < 1.0

More than one server (n) Ratio ρ = q/(Qn)


M/M/1 Wait time in queue: E(w) = q/[Q(Q-

q)] Average time in system: E(t) = 1/(Q-q) Average queue length: E(m) =

q2/[Q(Q-q)] Probability of k “units” in system: P(k)= (q/Q)k

[1-(q/Q)]

Used for flow through processes Entrance gates Security Jetways


M/D/1 Wait time in queue: E(w) = q/[2Q(Q-q)] Average time in system: E(t) =

[2Q-q]/[2Q(Q-q)] Average time in service: E(ts) = 1/Q Average queue length: E(m) = q2/[2Q(Q-

q)]

For multiple servers (n), approximate Q as nQ Be careful – service time is not affected by n Check equations above!

Used for processes with fixed service Ticket services Car rental


If ρ= q/Q > 1.0 Wait time in queue: E(w’) = E(w)0.9 +E(e)

where E(w)0.9 is the E(w) when ρ =0.9 and E(e)=T(q-nQ)/(2nQ)

where T is the time that demand exceeds service

n=number of servers Average queue length: E(m’) = [E(w’)+1/Q)]q

Baggage claims Average delay E(b) = E(t2) +NT/(N+1) –E(t1) Where

t2 = time when 1st bag shows up t1 = time when passengers arrive N = number of bags per person T = between first and last bags


Total passenger processing timeE(T) = E(w) + E(s) + E(t)where

E(w) average wait in queue time

E(s) average service timeE(t) average walk time

Passenger Flow - Enplaning

D

A1

A2

E

T

T

X

X

SS

SS

L

L

J

J

Enplaning Flow Example

D

E

T

X

SS

L J500

225

100

300

175 200

Device Service timeDoors 10 sec Express 90 sec Ticket 180 sec Security (X) 30 secSeat Select 25 sec Jetway 20 sec

n Servers

266544

Pax/hr/n360 40 20120144180

500 pax/hr

Pax/hr720 240 120600576720


Device q Q Wait (min) Service (sec)Gate 500 720 0.19 10Express 225 240 1.88 90Ticket 100 120 1.25 180Security 500 600 0.50 30Seat Select 300 576 0.06 25Jetway 500 720 0.19 20

Use the average wait in queue time equations to get wait. Remember to use the right queuing

equation for the right device.


ATO

Con

cessio

n S

tan

ds

60

40

50

30 60 30 3075 100

40 60

60


ATO

Con

cessio

n S

tan

ds

60

40

50

30 60 30 3075 100

40 60

60

395

305

185

300235

295

Walk dist (ft)


380

629 150

60

X

120

760


Wait timeE(w)=1(0.19)+0.45(1.88)+0.20(1.25)+1(0.50)+0.60(0.06)+1(0.19) = 2.01 min

Service timeE(s)=1(10)+0.45(90)+0.20(180)+1(30)+0.60(25)+1(10)= 151.2 s. = 2.52 min

Walk timeE(t)= [0.45[(235+295)/2]+0.20[(395+305)/2]+ 0.35[0.75(185)+0.25(300)]+1(760)]/2.5 = 408 s. = 6.8 min

Total timeE(T)= 2.01 + 2.52 + 6.83 = 11.36 min

Passenger Flow Deplaning

D

B

E

CR

J S

Deplaning Flow Example (1/8)

D

B

E

CR

J S

500 pax/hr1.5 bags/pax1 visit/pax

Device Service timeDoors 10 secEscalator 5 secSecurity exit 3 secCar rental 240 secJetway 10 sec

35 60

25 75

100


Device Service timeDoors 10 sec Escalator 5 secSecurity exit 3 sec Car rental 240 secJetway 10 sec

Servers411142

Pax/hr360720120015360

D

B

E

CR

J S

175 105

125 94

500200 31

70

Bags: 1.5 bags/pax = 309 bags, 2 servers


Device q Q Wait (min) Service (min)Doors 1000 1440 0.05 10Escalator 500 720 0.19 5Security exit 500 1200 0.04 3Car rental 195 210 1.86 240Jetway 500 720 0.19 10

Enplaning Flow Example (4/8)C

ar

Ren

tals

75

10

04

0

30 50 5040 35

40Incoming Bags

4080

70


315

405

295

345415

650

Car

Ren

tals

75

10

04

0

30 50 5050 35

40Incoming Bags

5075

70220

210


380

629 150

60

X

120

900


For bagsE(w)= 0.19+0.19+ 0.04+0.06(1.86) = 0.53 minE(s)= 10+5+3+0.06(240)=0.54 minE(t)= [900+0.35(415)+0.06(405)]/2.5=7.13 minAvg arrival time=0.53+0.54+7.13 =8.20

Bags/device 309/2 = 155 bagsLoad time 155/10 = 15.5 minE(b) = E(t2)+nT/(n+1)–E(t1)=10+[1.5(15.5)/2.5]-8.20= 11.10 min


Wait timeE(w)=1(0.19+0.19+0.04)+0.41(11.1)+0.39(1.86)+0.05 = 5.74 min

Service timeE(s)=1(10+5+3)+0.39(240)+1(10)= 2.03 min

Walk timeE(t)= [1(900)+0.40[(295+345)/2]+0.21(415+210)+ 0.14(415+650)+0.19(315)+0.06(405+210)]/2.5 = 9.41 minTotal timeE(T)= 5.74 + 2.03 + 9.41= 17.18 min

Terminal Footprint

Airport Roadway Circulation

Deplaning Enplaning

Terminal Frontage Road

Term

inal A

ccess R

oad

Term

inal Exit

Road

Short Term Parking

Long Term Parking

Gate Configuration

Large airlines have their own Smaller typically combine/share May need to have “airline”

terminals Wide bodies occupy outside

gates

terminal design passenger processing. errata consider tsa impacts

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