1. QUEUEING THEORY OPERES3

2. QUEUEING THEORY I think I shall never see a queue as long as this. -Any Customer, Anytime, Anywhere

3. A queue is a waiting line of "customers" requiring service from one or more servers. A queue forms whenever existing demand exceeds the existing capacity of the service facility; that is whenever arriving customers cannot receive immediate service due to busy servers. Queueing Theory is the study of the waiting line systems. QUEUEING THEORY

4. STRUCTURE OF A QUEUEINGMODEL Queueing System

5. 1.0. The Arrival Process A. The calling source may be single or multiple populations. B. The calling source may be finite or infinite. C. Single or bulk arrivals may occur. D. Total, partial or no control of arrivals can be exercised by the queueing system. E. Units can emanate from a deterministic or probabilistic generating process. F. A probabilistic arrival process can be described either by an empirical or a theoretical probability distribution. G. A stationary arrival process may or may not exist. COMPONENTS OF THE QUEUEING PROCESS

6. 2.0 The Queue Configuration The queue configuration refers to the number of queues in the system, their relationship to the servers and spatial consideration. A. A queue may be a single queue or a multiple queue. B. Queues may exist B.1 physically in one place B.2 physically in disparate locations B.3 conceptually B.4 not at all C. A queueing system may impose restriction on the maximum number of units allowed. COMPONENTS OF THE QUEUEING PROCESS

7. 3.0 Queue Discipline The following disciplines are possible: A. lf the system is filled to capacity, the arriving unit is rejected B. Balking - a customer does not join the queue C. Reneging - a customer joins the queue and subsequently decides to leave D. Collusion - customers collaborate to reduce waiting time E. Jockeying - a customer switching between multiple queues F. Cycling - a customer returning to the queue after being given service Customers who do not balk, renege, collude, jockey, cycle, non-randomly select from among multiple queues are said to be patient. COMPONENTS OF THE QUEUEING PROCESS

8. 4.0 Service Discipline A. First-Come, First-Served (FCFS) B. Last-Come, First-Served (LCFS) C. Service in Random Order (SIRO) D. Round Robin Service E. Priority Service preemptive non-preemptive COMPONENTS OF THE QUEUEING PROCESS

9. 5.0. Service Facility A. The service facility can have none, one, or multiple servers. B. Multiple servers can be parallel, in series (tandem) or both. COMPONENTS OF THE QUEUEING PROCESS Single Queue, Single Server Multiple Queue, Multiple Servers

10. 5.0. Service Facility COMPONENTS OF THE QUEUEING PROCESS Multiple Servers in Series Single Queue, Multiple Servers Channels in parallel may be cooperative or uncooperative. By policy, channels can also be variable. Multiple Servers, both in series and parallel

11. 5.0. Service Facility C. Service times can be deterministic or probabilistic. Random variables may be specified by an empirical or theoretical distribution. D. State: Dependent state parameters refer to cases where the parameters refer to cases where the parameters are affected by a change of the number of units in the system. E. Breakdowns among servers can also be considered. COMPONENTS OF THE QUEUEING PROCESS

12. 1.0. Taxonomy of Queueing Models A model may be represented using the Kendall- Lee Notation: CLASSIFICATIONS OF MODELS AND SOLUTIONS (a/b/c):( d/e/f) where: a = arrival rate distribution b = service rate distribution c = no. of parallel service channels (identical service) d = service discipline e = maximum no. allowed in the system f = calling source Common Notations: M – Poissonl/Exponential rates G - General Distribution of Service Times Ek- Erlangian Distribution

13. 2.0 Methods of Solution A. Analytical: The use of standard queueing models yields analytical results. B. Simulation: Some complex queueing systems cannot be solved analytically. (non-Poisson models) 3.0 Transient vs. Steady State A. A solution in the transient state is one that is time dependent. B. A solution is in the steady state when it is in statistical equilibrium (time independent) CLASSIFICATIONS OF MODELS AND SOLUTIONS

14. 4.0 Analytical Queueing Models - Information Flow In steady state systems, the operating characteristics do not vary with time. CLASSIFICATIONS OF MODELS AND SOLUTIONS

15. Notations: λc = effective mean arrival rate λ = λc if queue is infinite λe = λ - [expected number who balk if the queue is finite] W = expected waiting time of a customer in the system Wq = expected waiting time of a customer in the queue L = expected no. of customers in the system Lq = expected number of customers in the queue Po = probability of no customers in the system Pn = probability of n customers in the system ρ = traffic intensity= λ/μ ρc= effective traffic intensity= λe/μ CLASSIFICATIONS OF MODELS AND SOLUTIONS

16. The following expressions are valid for all queueingmodels.These relationships were developed by J. Little Note: lf the queue is finite, λ is replaced by λe GENERAL RELATIONSHIPS: (LITTLE'S FORMULA) L=λW Lq = λWq W =Wq+1/μ L= Lq + λ/ μ

17. In the models that will be presented the following assumptions hold true for any model: 1. The customers of the queueing system are patient customers. 2. The service discipline is general discipline (GD), which means that the derivations do not consider any specific type of service discipline. EXPONENTIAL QUEUEING MODELS

18. The derivation of the queueing models involve the use of a set of difference-differential equations which allow the determination of the state probabilities. These state probabilities can also be calculated by the use of the following principle: Rate-Equality Principle: The rate at which the process enters state n equals the rate at which it leaves state n. EXPONENTIAL QUEUEING MODELS

19. SINGLE CHANNEL-POISSON/EXPONENTIAL MODEL [(M/M/1):(GD/ α /α)] Characteristics: 1. Input population is infinite. 2. Arrival rate has a Poisson Distribution 3. There is only one server. 4. Service time is exponentially distributed with mean1/μ. [λ<μ] 5. System capacity is infinite. . 6. Balking and reneging are not allowed. CASE 1: SINGLE CHANNEL-POISSON/EXPONENTIAL MODEL

20. Using the rate-equality principle, we obtain our first equation for this type of system: To understand the above relationship, consider state 0. When in state 0, the process can leave this state only by an arrival. Since the arrival rate is λ and the proportion of the time that the process is in state 0 is given by Po, it follows that the rate at which the process leaves state 0 is λPo. On the other hand, state 0 can only be reached from state 1 via a departure. That is, if there is a single customer in the system and he completes service, then the system becomes empty. Since the service rate is μ and the proportion of the time that the system has exactly once customer is P1, it follows that the rate at which the process enters 0 is μPl. CASE 1: SINGLE CHANNEL-POISSON/EXPONENTIAL MODEL λPo=μP1

21. The balance equations using this principle for any n can now be written as: CASE 1: SINGLE CHANNEL-POISSON/EXPONENTIAL MODEL State Rate at which the process leaves = rate at which it enters 0λP0=μP1 n ≥1 (λ + μ)Pn = λPn-1 + μPn+l

22. In order to solve the above equations, we rewrite them to obtain Solving in terms of P0 yields: CASE 1: SINGLE CHANNEL-POISSON/EXPONENTIAL MODEL

23. In order to determine P0, we use the fact that the Pn must sum to 1, and thus CASE 1: SINGLE CHANNEL-POISSON/EXPONENTIAL MODEL or

24. Note that for the above equations to be valid, it is necessary for to be less than 1 so that the sun of the geometric progression customers in the system at any time, we use The last equation follows upon application of the algebraic identity CASE 1: SINGLE CHANNEL-POISSON/EXPONENTIAL MODEL

25. The rest of the steady state queueing statistics can be calculated using the expression for L and Little's Formula. A summary of the queueing formulas for Case I is given below. CASE 1: SINGLE CHANNEL-POISSON/EXPONENTIAL MODEL

26. SUMMARY OF CASE 1 FORMULAS

27. Example 1: Joe's Service Station operates a single gas pump. Cars arrive according to a Poisson distribution at an average rate of 15 cars per hour. Joe can service cars at the rate of 20 cars per hour with service time following an exponential distribution. a) What fraction of the time is Joe busy servicing cars? b) How many cars can Joe expect to find at his station? c) What is the probability that there are at least 2 cars in the station? d) How long can a driver expect to wait before his car enters the service facility? e) If you need to put in gas at Joe's service station, how long will it take you on the average? CASE 1: SINGLE CHANNEL-POISSON/EXPONENTIAL MODEL

28. Example 2: In the previous example, Joe is planning to set an area of land near the service facility to park cars waiting to be serviced. If each car requires 200 square feet on the average, how much space would be required? CASE 1: SINGLE CHANNEL-POISSON/EXPONENTIAL MODEL

29. MULTIPLE SERVER, POISSON/EXPONENTIAL MODEL [(M/M/C):(GD/∞/∞ )] The assumptions of Case II are the same as Case 1 except that the number of service channels is more than one. For this case, the service rate of the system is given by: CASE II : MULTIPLE SERVER, POISSON/EXPONENTIAL MODEL cµ η ≥ c ηµ η < c

30. Thus, a multiple server model is equivalent to a single-server system with service rate varying with η. Using the equality rate principle we have the following balance equations: CASE II : MULTIPLE SERVER, POISSON/EXPONENTIAL MODEL λη = λ & µη = ηµ η < c µη = cµ η ≥ c = 0

31. CASE II : MULTIPLE SERVER, POISSON/EXPONENTIAL MODEL

32. CASE II : MULTIPLE SERVER, POISSON/EXPONENTIAL MODEL But: Therefore: η≥c If ρ=λ/µ

33. CASE II : MULTIPLE SERVER, POISSON/EXPONENTIAL MODEL . Hence: To solve for , we note that To solve for Lq: (k=n-c)

34. CASE II : MULTIPLE SERVER, POISSON/EXPONENTIAL MODEL To simplify:

35. SUMMARY OF CASE 2 FORMULAS:

36. Example: Students arriving at a student registration center of a university must have their registration materials processed by an operator seated at a computer terminal. The systems design calls for 4 operators to be on duty, each operator performing an identical service. Students arrive according to a Poisson process at an average rate of 100 per hour. Each operator can process 40 students per hour with service time being exponentially distributed. a) What fraction of the time that there are no students in the registration center? b) If the waiting area inside the center building will comfortably accommodate five students, what percentage of time will there be students lined up outside the building? c) How long on the average does student spend in the center building waiting? d) On the average, can we expect a student arriving 3 minutes before closing time to make it just before closing? (He is able to register) e) How long is the line of students waiting to register? CASE II : MULTIPLE SERVER, POISSON/EXPONENTIAL MODEL

37. SINGLE CHANNEL,POISSON ARRIVALS, ARBITRARY SERVICE TIME : Pollaczek - Khintchine Formula [(M/G/l): (GD/∞/∞)] This case is similar to Case 1 except that the service rate distribution is arbitrary. Let: N = no. of units in the queueing system immediately after a unit departs T = the time needed to service the unit that follows the one departing (unit 1) at the beginning of the time count. K= no. of new arrivals units the system during the time needed to service the unit that follows the one departing (unit 1) Nl = no. of units left in the system when the unit (1) departs CASE III: SINGLE CHANNEL,POISSON ARRIVALS, ARBITRARY SERVICE TIME

38. CASE III: SINGLE CHANNEL,POISSON ARRIVALS, ARBITRARY SERVICE TIME Then: Nl = N + K – 1; if N = 0 = K Let: a = 1 if N = 0 a = 0 if N > 0 a*N =0 Then: Nl = N + K + a – 1

39. CASE III: SINGLE CHANNEL,POISSON ARRIVALS, ARBITRARY SERVICE TIME In a steady state system: ~. E(N) = E( ) E ( ) =E[ ] ) = E ( E (a) = -E (K) + 1 = a*N= 0 = Since a = 0 or 1: But

40. CASE III: SINGLE CHANNEL,POISSON ARRIVALS, ARBITRARY SERVICE TIME Therefore : but E(a) = 1 - E(K) Then:

41. CASE III: SINGLE CHANNEL,POISSON ARRIVALS, ARBITRARY SERVICE TIME If the arrival rate is Poisson E(K/t) = λt But in a Poisson Distribution: Mean = Variance E (K2/t) = λt + (λt)2

42. CASE III: SINGLE CHANNEL,POISSON ARRIVALS, ARBITRARY SERVICE TIME

43. CASE III: SINGLE CHANNEL,POISSON ARRIVALS, ARBITRARY SERVICE TIME Substituting and solving for E(N): The other quantities can be solved using the general relationships derived by Little.

44. SUMMARY OF CASE III QUEUEING FORMULAS

45. Example: Arrivals at the AAA Transmission Repair are observed to be random at the Poisson rate of 1.5 per day. The time required to repair transmission varies and the probability distribution is given below: Determine the steady state statistics of this system. CASE III: SINGLE CHANNEL,POISSON ARRIVALS, ARBITRARY SERVICE TIME

46. POISSON ARRIVAL AND SERVICE RATE, INFINITE NUMBER OF SERVERS: Self-Service Model [(M/M/∞): (GD/∞/∞)] Consider a multiple server system. The equivalent single server system if the number of servers is infinite would be: From the multiple server system: CASE IV: POISSON ARRIVAL AND SERVICE RATE, INFINITE NUMBER OF SERVERS

47. CASE IV: POISSON ARRIVAL AND SERVICE RATE, INFINITE NUMBER OF SERVERS But Therefore: Since the number of servers is infinite: Lq = Wq = 0.

48. CASE IV: POISSON ARRIVAL AND SERVICE RATE, INFINITE NUMBER OF SERVERS Solving for L:

49. SUMMARY OF CASE IV QUEUING FORMULAS (No Restriction) (Poisson Distributed)

50. Example: Supershopper Supermarket has 15 customers entering per hour. The customer spends an exponentially distributed amount of time shopping. The average time spent shopping is 20 minutes. a) On the average, how many customers can be found shopping in the aisles? b) What is the average length of time a customer spends shopping in the aisles? c) Find the probability of there being 6 or more customers in the aisles. CASE IV: POISSON ARRIVAL AND SERVICE RATE, INFINITE NUMBER OF SERVERS

51. SINGLE CHANNEL, POISSON/EXPONENTIAL MODEL, FINITE QUEUE [(M/M/1) : (GD/m/∞)] This case is similar to Case 1 except that the queue is finite, i.e., when the total number of customers in the system reaches the allowable limit, all arrivals balk. CASE V: SINGLE CHANNEL, POISSON/ EXPONENTIAL MODEL, FINITE QUEUE

52. Let m = maximum number allowed in system The balance equations are obtained in the same manner as before. n=0: n=1: CASE V: SINGLE CHANNEL, POISSON/ EXPONENTIAL MODEL, FINITE QUEUE

53. n=m-1: n=m: But Where is a finite geometric series with sum Therefore: CASE V: SINGLE CHANNEL, POISSON/ EXPONENTIAL MODEL, FINITE QUEUE

54. Now CASE V: SINGLE CHANNEL, POISSON/ EXPONENTIAL MODEL, FINITE QUEUE . . .

55. SUMMARY OF CASE V QUEUEING FORMULA

56. Example: The owner of a small full-service gas station operates it by himself. On Friday afternoon, customers arrive at the rate of 12 per hour. On the average, service takes 3 minutes and tends to be exponential. The space allows the accommodation of only 3 cars at one time (including the one being serviced). a) What is the mean number of customers in the station? b) What fraction of the time will the owner be idle? c) What fraction of customers will be lost? d) What is the mean customer waiting time in the queue? e) What is the probability that a customer waits? CASE V: SINGLE CHANNEL, POISSON/ EXPONENTIAL MODEL, FINITE QUEUE

57. MULTIPLE CHANNEL, POISSONEXPONENTIAL MODEL, FINITE QUEUE [(M/M/c):(GD/m/∞)] This case is an extension of Case V. We assume that the number of service channels is more than one. For this system: CASE VI : MULTIPLE CHANNEL, POISSONXPONENTIAL MODEL, FINITE QUEUE

58. The balance equations are similar to Case II. The manipulation of equations is basically the same. The following are the results. CASE VI : MULTIPLE CHANNEL, POISSONXPONENTIAL MODEL, FINITE QUEUE

59. As in Case V, we solve for: This expression is used in solving for the other statistics. CASE VI : MULTIPLE CHANNEL, POISSONXPONENTIAL MODEL, FINITE QUEUE

60. Example: Consider a barbershop with 3 barbers and 2 additional chairs for waiting customers. Customers arrive randomly at an average rate of 8 per hour, and haircuts take an average of 15 minutes, with an exponential distribution. Arriving customers who find the barbershop busy and 2 customers waiting leave immediately. a) What is the mean number of customers waiting at any time? b) What is the mean time a customer spends in the barbershop? c) Determine the mean number of customers in the barbershop? d) What fraction of the time are the barbers idle? CASE VI : MULTIPLE CHANNEL, POISSONXPONENTIAL MODEL, FINITE QUEUE

61. MACHINE SERVICING MODEL [(M/M/R):(GD/K/K)] This model assumes that R repairmen are available for servicing a total of K machines. Since a broken machine cannot generate new calls while in service, this model is an example of finite calling source. This model can be treated as a special case of the single server, infinite queue model. CASE VII : MACHINE SERVICING MODEL

62. Moreover, the arrival rate λ is defined as the rate of breakdown per machine. Therefore: CASE VII : MACHINE SERVICING MODEL

63. The balance equations yield the following formulas for the steady state system: The other measures are given by: CASE VII : MACHINE SERVICING MODEL

64. To solve for the effective arrival rate, we determine: CASE VII : MACHINE SERVICING MODEL

65. Example: Two repairmen are attending five machines in a workshop. Each machine breaks down according to a poisson distribution with a mean of 3 per hour. The repair time per machine is exponential with a mean of 15 minutes. a) Find the probability that the two repairmen are idle. That one repairman is idle. b) What is expected number of idle machines not being repaired? CASE VII : MACHINE SERVICING MODEL

66. COSTS INVOLVED IN THE QUEUEING SYSTEM 1. FACILITY COST - cost of (acquiring) services facilities Construction (capital investment) expressed by interest and amortization Cost of operation: labor, energy & materials Cost of maintenance & repair Other Costs such as insurance, taxes, rental of space 2. WAITING COST - may include ill-will due to poor service, opportunity loss of customers who get impatient and leave or a possible loss of repeat business due to dissatisfaction. ECONOMIC CONDITIONS

67. The total cost of the queueing system is given by: ECONOMIC CONDITIONS TC = SC+WC where: SC = facility (service cost)cost WC = waiting cost or the cost of waiting (in queue & while being served) per unit time TC = Total Cost

68. ECONOMIC CONDITIONS Let: Cw = cost of having 1 customer wait per unit time Then WCw = average waiting cost per customer But since λ customers arrive per unit time: WC = λ WCw = LCw

69. The behavior of the different cost Component is depicted in the following graph: ECONOMIC CONDITIONS

70. Management Objective: Cost Minimization or Achieving a Desired Service Level An example of a desired service level is the reduction of waiting time of customers. The minimization of cost would involve the minimization of the sum of service cost and waiting cost. The decision is a matter of organizational policy and influenced by competition and consumer pressure. ECONOMIC CONDITIONS

71. Example 1: A firm operates a single-server maintenance storeroom where electrical repair persons check out needed spare parts and equipment. Repair persons arrive at a rate of 8 per hour. The service rate is 10 per hour. Arrivals are Poisson distributed and service completion times follow the exponential model. The cost of waiting is Php 9 per hour. The company is considering giving the stock clerk a helper which would increase the service rate to 12 per hour. The cost of the helper is Php 6 per hour. What do you recommend? ECONOMIC CONDITIONS

72. Example 2: An engineering design firm is replacing its scientific computer, which is sued to solve problems encountered in design projects. The list of alternatives has been narrowed to 3 computers. Computation times are exponential and arrivals of problems at the computer center follow a poisson distribution at mean rate if 15 per day. Progress on a project is delayed until a problem is solved. The daily cost of delaying a project is Php250. Which computer should be selected? ECONOMIC CONDITIONS