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Terminal Design Passenger Processing. errata  Consider TSA impacts.

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Presentation on theme: "Terminal Design Passenger Processing. errata  Consider TSA impacts."— Presentation transcript:

1 Terminal Design Passenger Processing

2 errata  Consider TSA impacts

3 Passenger Space Guidelines (1/2)  Personal space Transit 18x24 inches Airports 5-10 ft 2 (30 inch circle)  Lateral movement 30 inches between “traffic” lanes  Longitudinal movement 8-10 feet per person  Net pedestrian area 20-30 ft 2

4 Passenger Space Guidelines (2/2)  Queuing space 5-10 ft 2  Stairs 10-20 ft 2 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)

5 Passenger System (1/5)  Entryways Passenger and visitors Enplaning and deplaning Auto doors20-30 pax/min Manual doors10-15 pax/min  Lobby areas All persons using airport Seating capacity 15-25% of enplaning Space 20 sf/pax

6 Passenger System (2/5)  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

7 Passenger System (3/5)  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’

8 Passenger System (4/5)  Departure lounge Estimate 80% of pax need seating Space 10-15 ft 2 /pax Walking corridors should be provided  Boarding corridors 10 ft. wide Service rate 2-4 pax/min

9 Passenger System (4/5)  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

10 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)

11 Queuing Equations (2/4)  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) = q 2 /[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

12 Queuing Equations (2/4)  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(t s ) = 1/Q Average queue length: E(m) = q 2 /[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

13 Queuing Equations (3/4)  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(t 2 ) +NT/(N+1) –E(t 1 ) Where  t 2 = time when 1 st bag shows up  t 1 = time when passengers arrive  N = number of bags per person  T = between first and last bags

14 Queuing Equations (4/4)  Total passenger processing time E(T) = E(w) + E(s) + E(t) where E(w) average wait in queue time E(s) average service time E(t) average walk time

15 Passenger Flow - Enplaning DA1A1 A2A2 ETTXXSS L L JJ

16 Enplaning Flow Example DETXSS L J 500 225 100 300 17 5 200 Device Service time Doors 10 sec Express 90 sec Ticket 180 sec Security (X) 30 sec Seat Select 25 sec Jetway 20 sec n Servers 2 6 5 4 Pax/hr/n 360 40 20 120 144 180 500 pax/hr Pax/hr 720 240 120 600 576 720

17 Enplaning Flow Example Device q Q Wait (min) Service (sec) Gate 500 7200.19 10 Express225 2401.88 90 Ticket 100 1201.25 180 Security 500 6000.50 30 Seat Select 300 5760.06 25 Jetway 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.

18 Enplaning Flow Example AT O Concession Stands 60 40 50 30 60 30 75 100 40 60

19 Enplaning Flow Example AT O Concession Stands 60 40 50 30 60 30 75 100 40 60 395 305 185 300 235 295 Walk dist (ft)

20 Enplaning Flow Example 380 629 150 60 X 120 760

21 Enplaning Flow Example Wait time E(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 time E(s)=1(10)+0.45(90)+0.20(180)+1(30)+0.60(25)+1(10)= 151.2 s. = 2.52 min Walk time E(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 time E(T)= 2.01 + 2.52 + 6.83 = 11.36 min

22 Passenger Flow Deplaning DBECRJS

23 Deplaning Flow Example (1/8) DBECRJS 500 pax/hr 1.5 bags/pax 1 visit/pax Device Service time Doors 10 sec Escalator 5 sec Security exit 3 sec Car rental 240 sec Jetway 10 sec 35 60 25 75 100

24 Deplaning Flow Example (2/8) Device Service time Doors 10 sec Escalator 5 sec Security exit 3 sec Car rental 240 sec Jetway 10 sec Servers 4 1 14 2 Pax/hr 360 720 1200 15 360 DBECRJS 175 105 125 94 500 200 31 70 Bags: 1.5 bags/pax = 309 bags, 2 servers

25 Deplaning Flow Example (3/8) Device q Q Wait (min) Service (min) Doors 10001440 0.05 10 Escalator 500 720 0.19 5 Security exit 500 1200 0.04 3 Car rental 195 2101.86 240 Jetway 500 7200.19 10

26 Enplaning Flow Example (4/8) Car Rentals 75 100 40 30 50 40 35 40 Incoming Bags 40 80 70

27 Deplaning Flow Example (5/8) 315 405 295 345 415 650 Car Rentals 75 100 40 30 50 35 40 Incoming Bags 50 75 70 220 210

28 Deplaning Flow Example (6/8) 380 629 150 60 X 120 900

29 Deplaning Flow Example (7/8) For bags E(w)= 0.19+0.19+ 0.04+0.06(1.86) = 0.53 min E(s)= 10+5+3+0.06(240)=0.54 min E(t)= [900+0.35(415)+0.06(405)]/2.5=7.13 min Avg arrival time=0.53+0.54+7.13 =8.20 Bags/device 309/2 = 155 bags Load time 155/10 = 15.5 min E( b ) = E(t 2 )+nT/(n+1)–E(t 1 )=10+[1.5(15.5)/2.5]-8.20= 11.10 min

30 Deplaning Flow Example (8/8) Wait time E(w)=1(0.19+0.19+0.04)+0.41(11.1)+0.39(1.86)+0.05 = 5.74 min Service time E(s)=1(10+5+3)+0.39(240)+1(10)= 2.03 min Walk time E(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 min Total time E(T)= 5.74 + 2.03 + 9.41= 17.18 min

31 Terminal Footprint

32 Airport Roadway Circulation Deplaning Enplaning Terminal Frontage Road Terminal Access Road Terminal Exit Road Short Term Parking Long Term Parking

33 Gate Configuration  Large airlines have their own  Smaller typically combine/share  May need to have “airline” terminals  Wide bodies occupy outside gates


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