1. Queuing Models

2. © 2002 Prentice-Hall, IncCh 9-2 Stay in Queue: Short Video http://www.youtube.com/watch?v=IPxBKxU8GIQ &feature=related http://www.youtube.com/watch?v=IPxBKxU8GIQ &feature=related 2

3. Introduction Queues are everyday occurrence: affect people - – Shopping for groceries. – Buying gasoline. – Making bank deposit. – Waiting on telephone. Discuss how analytical models of queues help managers evaluate cost effectiveness of service systems. Look at queuing system costs. Describe characteristics of queues and underlying mathematical assumptions used to develop queuing decision models. 3

4. Queuing System Costs Most queuing problems are involved with finding ideal level of services firm should provide. – Supermarkets decide how many cash register checkout positions should be opened. – Gasoline stations decide how many pumps should be available. – Manufacturing plants determine optimal number of mechanics on duty each shift to repair machines that break down. Most managers recognize trade-off: cost of providing good service vs. cost of customer waiting time. 4

5. Queuing Costs and Service Levels 5

6. Three Rivers Shipping Company Example Three Rivers runs huge docking facility. Approximately five ships arrive to unload cargoes every 12-hour work shift. Each hour ship sits idle in line costs firm $1,000. If one team on duty to handle unloading work, each ship waits average 7 hours to be unloaded. – If two teams are working, average waiting time drops to 4 hours. – For three teams, it is 3 hours. – For four teams, it is only 2 hours. 6

7. Waiting Line Cost Analysis Team salaries are computed as: number of people in team (assumed to be 50), times number of hours each person works per day (12 hours), times hourly salary of $10 per hour. If two teams are employed, rate is doubled. 7 Three Rivers Shipping

8. Waiting Line Cost Analysis 8

9. Characteristics Of Queuing System Three primary parts of a queuing system. (1) Arrivals or inputs to system referred to as calling population. (2) Queue or waiting line itself. (3) Service facility. These three components have certain characteristics that need to be examined prior to development of mathematical queuing models. 9

10. Arrival Characteristics Major three characteristics of input source that generates arrivals or customers for service system. (1) Size of arrival population. (2) Pattern of arrivals or arrival distribution at queuing system. (3) Behavior of arrivals. 10

11. Size of the Arrival Population Population sizes are unlimited (essentially infinite) or (limited) finite. When number of arrivals at any given moment is small portion of potential arrivals, arrival population is considered unlimited. Examples of unlimited populations include - – Cars arriving at highway tollbooth. – Shoppers arriving at supermarket. – Students arriving to register for classes. Most queuing models assume infinite arrival population. 11

12. Arrival Distribution Customers either arrive at service facility according to some known schedule (i.e., one patient every 15 minutes or one student for advising every half hour) or they arrive randomly. For any given arrival rate, such as two customers per hour, or four trucks per minute, a discrete Poisson distribution can be established: P(X) = (e -  X ) / X !for X = 0, 1, 2, … where: X = number of arrivals per unit of time. P(X) = probability of exactly X arrivals.  = average arrival rate (average number of arrivals per unit of time). e = 2.7183 (exponential constant). 12

13. Examples Poisson Distribution for Arrival Times 13

14. Behavior of Arrivals Most queuing models assume arriving customer is patient customer. Patient customers are people or machines wait in queue until served and do not switch between lines. Life and decision models are complicated by fact people balk or renege. Balking refers to customers who refuse to join queue because it is too long to suit their needs. Reneging customers enter queue but become impatient and leave without completing transaction. 14

15. Queue Characteristics Length of queue either limited (finite) or unlimited (infinite). Queue limited when it cannot increase to infinite length due to physical or other restrictions. Queue defined as unlimited when size is unrestricted. Assume queue lengths are unlimited. Queue discipline. Rule by which customers in line receive service. Most systems use queue discipline known as First-In, First-Out rule (FIFO). 15

16. Service Facility Characteristics Important to examine two basic properties: (1) Configuration of service facility. (2) Pattern of service times. Number of service channels in queuing system is number of servers. Single-phase means customer receives service at only one station before leaving system. Multiphase implies two or more stops before leaving system. 16

17. Queuing System Configurations 17

18. Queuing System Configurations 18

19. Service Time Distribution Assume random service times are described by exponential probability distribution. For any given service rate, exponential distribution can be established using formula: P(t) = e -  t for t  0 where: t = service time. P(t) = probability service time will be greater than t.  = average service rate (average number of customers served per unit of time). e = 2.7183 (exponential constant). 19

20. Examples Exponential Distribution for Service Times 20

21. Measuring Queue Performance Queuing models help manager obtain performance measures (operating characteristics) of waiting line system. Some measures commonly used in practice.  = Utilization factor of system L q = Average length of queue. L= Average number of customers in system. W q = Average time each customer spends in queue. W= Average time each customer spends in system. P 0 = Probability no customers in system. P n = Probability exactly n customers in system. 21

22. Kendall’s Notation for Queuing Systems Queuing theory commonly uses three-symbol notation known as Kendall’s notation: A/B/S where: Adenotes arrival probability distribution choices are: M (Markovian) for a Poisson distribution, D for constant or deterministic distribution, or G for general distribution with known mean and variance. B denotes service time probability distribution choices are: M for exponential service times, D for constant or deterministic service times, or G for general service times with known mean and variance. S denotes the number of servers (or channels). 22

23. Queuing Models Reviewed 23

24. Single-server Queuing System (M/M/1 Model) Assumptions M/M/1 Queuing Model Poisson arrivals served on FIFO (first-in, first-out) basis. Every arrival waits to be served regardless of line length. Arrivals independent of preceding arrivals, but average number of arrivals (the arrival rate) does not change over time. Arrivals are described by Poisson probability distribution and come from infinite or very large population. Exponential service times vary from one customer to next and are independent of one another, but average rate is known. Service times occur according to the exponential probability distribution. Average service rate is greater than average arrival rate. 24

25. Operating Characteristics Equations M/M/1 Queuing System Operating characteristic equations for the M/M/1 queuing system are as follows:  = average number of arrivals per time period.  = average number of people or items served per time period. 1. Average server utilization.  =  /  2.Average number of customers waiting in line. L q =   (  - ) ] 25

26. Operating Characteristics Equations M/M/1 Queuing System 3. Average number of customers or units in system. L = L q +  /  4. Average time customer or unit spends waiting in line for service. W q = L q / = L q =    (  - ) ] 5. Average time a customer or unit spends in system. W = W q + 1 /  26

27. Operating Characteristics Equations M/M/1 Queuing System 6. Probability zero customers or units in system. P 0 = 1 -  /  7. Probability n customers or units in system. P n = (  /  ) n P o 27

28. © 2002 Prentice-Hall, IncCh 9-28 Example-1 Our local food court uses an average of 10,000 pounds of potatoes per week. The average number of pounds of potatoes on hand is 5000 pounds. On the average, how long do potatoes stay in the restaurant before being used?

29. © 2002 Prentice-Hall, IncCh 9-29 Solution We are given that L=5000 pounds and λ = 10,000 pounds/week. Therefore W = 5000 pounds/(10,000 pounds/week)=0.5 weeks.

30. © 2002 Prentice-Hall, IncCh 9-30 Example-2 Suppose that all car owners fill up when their tanks are exactly half full. At the present time, an average of 7.5 customers per hour arrive at a single-pump gas station. It takes an average of 4 minutes to service a car. Assume that inter arrival and service times are both exponential. For the present situation, compute L and W.

31. © 2002 Prentice-Hall, IncCh 9-31 Suppose that a gas shortage occurs and panic buying takes place. – To model the phenomenon, suppose that all car owners now purchase gas when their tank are exactly three-fourths full. – Since each car owner is now putting less gas into the tank during each visit to the station, we assume that the average service time has been reduced to 3.333 minutes. – How has panic buying affected L and W?

32. © 2002 Prentice-Hall, IncCh 9-32 We have an M/M/1/GD/∞/∞ system with λ = 7.5 cars per hour µ = 15 cars per hour.  = 7.5/15 =0.50. L = 0.50/1-0.50 = 1 W = L/λ = 1/7.5 = 0.13 hour. Hence, in this situation, everything is under control, and long lines appear to be unlikely.

33. © 2002 Prentice-Hall, IncCh 9-33 We now have an M/M/1/GD/∞/∞ system with λ = 2(7.5) = 15 cars per hour. Now µ = 60/3.333 = 18 cars per hour, and  = 15/18 = 5/6. Then Thus, panic buying has cause long lines.

34. © 2002 Prentice-Hall, IncCh 9-34 Example-3 An average of 10 machinists per hour arrive seeking tools. At present, the tool center is staffed by a clerk who is paid $6 per hour and who takes an average of 5 minutes to handle each request for tools. Since each machinist produces $10 worth of goods per hour, each hour that a machinists spends at the tool center costs the company $10. The company is deciding whether or not it is worthwhile to hire (at $4 per hour) a helper for the clerk. If the helper is hired the clerk will take an average of only 4 minutes to process requirements for tools. Assume that service and arrival times are exponential. Should the helper be hired?

35. © 2002 Prentice-Hall, IncCh 9-35 Solution Goal: Minimize the sum of the hourly service cost and expected hourly cost due to the idle times of machinists Delay cost is the component of cost due to customers waiting in line Goal: Minimize Expected cost/hour = service cost/hour + expected delay cost/hour Expected delay cost/hour = (expected delay cost/customer) (expected customers/hour) Expected delay cost/customer = ($10/machinist-hour)(average hours machinist spends in the system) = 10W Expected delay cost/hour = 10W Now compute expected cost/hour if the helper is not hired and also compute the same if the helper is hired

36. © 2002 Prentice-Hall, IncCh 9-36 If the helper is not hired = 10 machinists per hour and  = 12 machinists per hour W = 1/(  - ) for M/M/1/GD/  / . Therefore, W = 1/(12-10) = ½ = 0.5 hour Service cost /hour = $6/hour and expected delay cost/hour = 10(0.5)(10) = $50 Without the helper, the expected hourly cost is $6 + $50 = $56 With the helper,  = 15 customers/hour. Then W = 1/(  - )= 1/(15-10) = 0.2 hour and the expected delay cost/hour = 10(0.2)(10) = $20 Service cost/hour = $6 + $4 = $10/hour With the helper, the expected hourly cost is $10 + $20 = $30

37. Introducing Costs Into the Model Cost of customer waiting time is $10 per hour. Cost of providing service is mechanic’s salary of $12 per hour. Total cost is sum of waiting cost and service cost. Total cost = C w x L + C s x s where: C w = customer waiting cost per unit time. L = average number of customers in system. C s = cost of providing service per server per unit time period. s = number of servers in queuing system. 37

38. Introducing Costs Into Model Total cost = C w x L + C s x s where: C w = customer waiting cost per unit time. L = average number of customers in system. C s = cost of providing service per server per unit time. s = number of servers in queuing system. Total cost = ($10)(2) + ($12)(1) = $32 per hour. 38 With C w = $12 per hour, L = 2 C s = $12 per hour, s = 1

39. Multiple-server Queuing System (M/M/S Model) Multiple-server system assumes Poisson arrivals follow Poisson probability distribution and exponential service times follow exponential probability distributions. Service is first come, first served and all servers are assumed to perform at same rate. Other assumptions listed earlier for single-channel model apply. Operating Characteristics Equations for M/M/s Queuing System = average number of arrivals per time.  = average number of customers served per time per server. s = number of servers. 39

40. Operating Characteristic Equations M/M/s Queuing System Operating characteristic equations for M/M/s queuing system are as follows: 1. Average server utilization in the system.  =  / ( s  ) 2. Probability zero customers or units in system. 40

41. Operating Characteristic Equations M/M/s Queuing System 3. Average number of customers or units waiting in line for service. 41 4. Average number of customers or units in system. L = L q +  /  5. Average time a customer or unit spends waiting in line for service. W q = L q /

42. Operating Characteristic Equations M/M/s Queuing System 6. Average time customer or unit spends in system. W = W q + 1 /  7. Probability n customers or units in system. 42

43. © 2002 Prentice-Hall, IncCh 9-43 Example-4 Bank Staffing Example: The manager of a bank must determine how many tellers should work on Fridays. For every minute a customer stands in line, the manager believes that a delay cost of 5 cents is incurred. An average of 2 customers per minute arrive at the bank. On the average it takes, a teller 2 minutes to complete a customer’s transaction. It costs the bank $9 per hour to hire a teller. Inter-arrival times and service times are exponential. To minimize the sum of service costs and delay costs, how many tellers should the bank have working on Fridays? = 2 customer per minute and  = 0.5 customer per minute, /s  requires that 4/s < 1. Thus, there must be at least 5 tellers, or the number of customers present will “blow up.” Now compute for s = 5, 6…. Expected service cost/minute + expected delay cost/minute

44. © 2002 Prentice-Hall, IncCh 9-44 Solution Each teller is paid $9/60 = 15 cents per minute. Expected service cost/minute = 0.15s Expected delay cost/minute = (expected customers/minute) (expected delay cost/customer) Expected delay cost/customer = 0.05Wq Expected delay cost/minute = 2(0.05) Wq = 0.10 Wq For s = 5,  = /s  = 2/.5(5) = 0.8

45. © 2002 Prentice-Hall, IncCh 9-45 P(j  5)=0.55 W q =.55/(5(.5)-2) = 1.1 minutes For s = 5, expected delay cost/minute = 0.10(1.1) = 11 cents For s = 5, total expected cost/minute = 0.15(5) + 0.11 = 86 cents Since s = 6 has a service cost per minute of 6(0.15) = 90 cents, 6 tellers cannot have a lower total cost than 5 tellers. Hence, having 5 tellers serve is optimal

46. Multiple-server Queuing System (M/M/S/ Finite /N Model) 1. There are s servers with identical service time distributions. 2. Finite population size of units seeking service is size N. 3. Poisson arrival distribution of each customer in population follows Poisson distribution with mean rate of. 4. Exponential service times are exponentially distributed with mean rate of . 5. Both and  are specified for same time period. 6. Customers are served on first-come, first-served basis. 46

47. Operating Characteristics Equations Finite Population Queuing System = average number of arrivals per time period.  = average number of people or items served per time. s = number of servers. N = size of the population. Operating characteristic equations follow. 1. Probability zero customers or units in system. 47

48. Operating Characteristics Equations Finite Population Queuing System 2. Probability n customers in system. 48

49. Operating Characteristics Equations Finite Population Queuing System 3. Average number of customers or units in line waiting for service. 4. Average number of customers or units in system. 49

50. Operating Characteristics Equations Finite Population Queuing System 5.Average time a customer or unit spends in the queue waiting for service. 6. Average time a customer or unit spends in the system. 50