1. Delays Deterministic Stochastic 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
There is not too much content in the textbook about delay, ( basically the textbook mentions only one fundamental equation). However, problems at the end of the textbook require use of specific models, such as MM1, MD1, etc. Complimentary material has been added to this ppt to aid students in solving the homework problems. 
Formula derivation: all the queuing equations for stochastic models are derived from one basic equation, which is in the text book. Thus as long as you know about what's the standard deviation and expected value, you can derive the queuing equations yourself, which is much less confusing I believe, and also helps people easily understand. There are no queuing equations in the chapter on delay in the textbook, so formula deviation could help with consistency. We may not have time to derive all the equation in class, but students can take a look at the equations to see what's going on. 

2. Deterministic Delay

3. Queuing Diagram (1/2) Greatest Delay Total delay time Delayed Aircraft
Greatest Queue
Runway Capacity
Delay Period

4. Delay Estimation (2/2) Area Demand-Capacity A/C-Hours of delay D-C
Time

5. Example (1/2) End hour Operations Capacity D-C Cumul. (Queue) 7 25 30-5 8 9 40 10 50 20 11 45 15 12
-15 13
-20 14
-10
sum
250

6. 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/( ) = 41.1 min/AC

7. Stochastic Delay Queuing theory concepts Required data
Probability function
Arrival rate
Service time
Required data
Arrival pattern
Service pattern
Service method
Queue discipline
Number of servers

8. 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.

9. Mathematical Formulation of Delay*
Where W= mean delay to arriving (or departing) aircraft
λ = mean arrival (or departure) rate of aircraft (number per hour)
𝜇 = mean service rate for arrivals (or departures),
= reciprocal of mean service time (number per hour)
𝜎 = standard deviation of mean service time of arriving (or
departing) aircraft (number per hour)
Can be used in all stochastic delay models.
*Note: the book uses λ and μ; some of these and other lecture’s notes use q and Q

10. M/D/1 Queuing Model
M -- exponentially distributed times between arrivals of successive vehicles (Poisson arrivals)
D -- Service times are fixed and constant
(no variance: 𝜎=0 )
1 – One runway
“Traffic intensity” term is used to define the ratio of average arrival to departure rates:

11. M/D/1 Queuing Model q -- Arrival (or departure) rate = λ
Q -- Service rate (utilization) = 𝜇
Average waiting time in queue

12. M/D/1 Equations Average Time in System (hours)
Average Queue Length (number)
Average Time in service (hours)

13. 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:

14. 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)]

15. Example (note different terminology) (1/2)
End hour
Operations
Capacity
D-C
Cumul. 7 25 30 -5 8 9 40 10 50 20 11 45 15 12
-15 13
-20 14
-10
Arrival rate
q = 250/9 = 27.8 A-C/hr
Service rate
Q = 30 A-C/hr
Use M/M/1 model

16. Example (2/2) Average wait time Average queue length
E(w)=q/[Q(Q-q)]=27.8/[30( )] = 0.42 hr/A-C
Average queue length
E(m)=q2/[Q(Q-q)]=27.82/[30( )] = 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
Continue, in same fashion