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Published byFrancis Spencer Modified over 9 years ago
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Delays Deterministic Assumes “error free” type case Delay only when demand (known) exceeds capacity (known) Stochastic Delay may occur any time Random arrivals and departures per stat function
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Deterministic Delay
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Delay Estimation (1/2) Greatest Delay Greatest Queue Total Delayed Aircraft Delay Period Runway Capacity Delay Period
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Delay Estimation (2/2) Area A/C-Hours of delay Demand-Capacity D-C Time
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Example (1/2) End hour OperationsCapacity D-CCumul. 72530 -50 830 00 94030 10 5030 2030 114530 1545 121530 -1530 131030 -2010 141530 -150 152030 -100
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Example (2/2) Area under the curve = ½*1*10+½*(10+30)*1+½*(30+40)*1+ +½*(40+30)*1 +½*(30+10)*1+½*10*1=120 AC-hr Avg Delay to All AC = 120/250 = 28.8 min/AC Avg Delay to Delayed = 120/(40+50+45+15+10+15) = 41.1 min/AC
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Stochastic Delay Queuing theory concepts Probability function Arrival rate Service time Required data Arrival pattern Service pattern Service method Queue discipline Number of servers
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Delay Equations (random/poisson arrivals, uniform service dist means variance = 0) See p. 304 To use for HW prob 16, must compute average hourly demand (blows up if demand>supply) part c should more properly be worded “increased” to 8 minutes, not “limited” to 8 minutes.
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Mathematical Formulation of Delay
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M/D/1 Queuing Model
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q -- Arrival (or departure) rate = λ Q -- Service rate (utilization) = Average waiting time in queue M/D/1 Queuing Model
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M/D/1 Equations Average Time in System (hours) Average Queue Length (number) Average Time in service (hours)
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M/M/1 Queuing Models M -- Exponentially distributed arrival and departure times and one departure channel (server, e.g., runway) 1 – One runway q – Arrival (or departure) rate Q -- Service rate From statistics recall: Exponential distribution:
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M/M/1 Queuing Models Average waiting time in queue Average time in system Average queue length Probability of k units in system: P(k)= (q/Q) k [1-(q/Q)]
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Example (note different terminology) (1/2) Arrival rate q = 250/9 = 27.8 A-C/hr Service rate Q = 30 A-C/hr Use M/M/1 model End hour OperationsCapacity D-CCumul. 72530 -50 830 00 94030 10 5030 2030 114530 1545 121530 -1530 131030 -2010 141530 -150 152030 -100
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Example (2/2) Average wait time E(w)=q/[Q(Q-q)]=27.8/[30(30-27.8)] = 0.42 hr/A-C Average queue length E(m)=q 2 /[Q(Q-q)]=27.8 2 /[30(30-27.8)] = 11.7 A-C Probability of no plane in the system P(0) = (q/Q) 0 [1-(q/Q)] = 0.073 Probability of one in the system (no line) P(1) = (q/Q) 1 [1-(q/Q)] = 0.068 Probability of two in the system (one in line) P(1) = (q/Q) 2 [1-(q/Q)] = 0.063
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