3. “ The theoretical study of waiting lines, expressed in mathematical terms--including components such as number of waiting lines, number of servers, average wait time, number of queues or lines, and probabilities of queue times' either increasing or decreasing ” .

4. Why queues form? Queues or waiting lines arise when the demand for a service facility exceeds the capacity of that facility, that is, the customers do not get service immediately upon request but must wait, or the service facilities stand idle and wait for customers. Some customers wait when the total number of customers requiring service exceeds the number of service facilities, some service facilities stand idle when the total number of service facilities exceeds the number of customers requiring service.

7. Service Distribution

8. Number of servers

9. Total Capacity (infinite if not specified)

10. Population Size (infinite)

11. Service Discipline (FCFS/FIFO)

13. D: Deterministic (e.g. fixed constant)

14. E k : Erlang with parameter k

15. H k : Hyperexponential with param. k

16. G: General (anything)

20. L q :average number in the queue

21. W :average waiting time in whole system

22. W q :average waiting time in the queue

26. L q =  Wq (steady-state argument)

28. Finding L is hard or easy depending on the type of system. In general:

31. Step 2 :

33. Step 4 :

37. Traffic control

38. Determining the sequence of computer operations

39. Predicting computer performance

40. Health services (eg. control of hospital bed assignments)

41. Airport traffic, airline ticket sales

42. Layout of manufacturing systems.

48. What is the probability of n packets in the gateway?

49. mean number of packets in the gateway?

50. The number of buffers so P(overflow) is <10 -6 ?

52. Service rate μ = 1/0.002 = 500 pps

53. Gateway utilization ρ = λ/μ = 0.25

54. Prob. of n packets in gateway =

55. Mean number of packets in gateway =

56. Solution... Probability of buffer overflow: = P(more than 13 packets in gateway) = ρ 13 = 0.25 13 = 1.49x10 -8 = 15 packets per billion packets To limit the probability of loss to less than 10 -6 :

57. Solution... To limit the probability of loss to less than 10 -6 : or = 9.96