1. Lecture 11 Queueing Models

2. 2 Queueing System  Queueing System:  A system in which items (or customers) arrive at a station, wait in a line (or queue), obtain some kind of service, an then leave the system.  Managerial Problem Related to Queueing Systems:  Analysis Problems  Need to know if a given system is performing satisfactorily.  Design Problems  Want to design the features of a system to accomplish an overall objective.  Queueing System:  A system in which items (or customers) arrive at a station, wait in a line (or queue), obtain some kind of service, an then leave the system.  Managerial Problem Related to Queueing Systems:  Analysis Problems  Need to know if a given system is performing satisfactorily.  Design Problems  Want to design the features of a system to accomplish an overall objective.

3. 3 Characteristics of Queueing System  A customer population, which is the collection of all possible customers.  An arrival process, which is how the customers from this population arrive.  A queueing process, which consists of  the way in which the customers wait for service and  the queueing discipline, which is how they are then selected for service.  A customer population, which is the collection of all possible customers.  An arrival process, which is how the customers from this population arrive.  A queueing process, which consists of  the way in which the customers wait for service and  the queueing discipline, which is how they are then selected for service.

4. 4 Characteristics of Queueing System (cont’)  A service process, which is the way and rate at which customers are served.  A departure process, which is one of the following:  Items leave the system completely after being served, resulting in a one-step queueing system.  Items, once finished at a station, proceed to some other work station to receive further service, resulting in a network of queues.  A service process, which is the way and rate at which customers are served.  A departure process, which is one of the following:  Items leave the system completely after being served, resulting in a one-step queueing system.  Items, once finished at a station, proceed to some other work station to receive further service, resulting in a network of queues.

5. 5 Queueing System Customer Population Arrival Process System...... Waiting Customers Queueing Process Depart Network of queues ArrivalDepart One-step queueing system One-step queueing system Arrival Depart Components of a Queueing System: Departure Processes: Servers Service Process

6. 6 Arrival Process  Classes of Interarrival Times:  Deterministic, in which each successive customer arrives after the same fixed and known amount of time. A classic example is an assembly line, where the jobs arrive at a station at unvarying time intervals (called the cycle time).  Probabilistic, in which the time between successive arrivals is uncertain and varies. Probabilistic interarrival times are described by a probability distribution.  Classes of Interarrival Times:  Deterministic, in which each successive customer arrives after the same fixed and known amount of time. A classic example is an assembly line, where the jobs arrive at a station at unvarying time intervals (called the cycle time).  Probabilistic, in which the time between successive arrivals is uncertain and varies. Probabilistic interarrival times are described by a probability distribution.

7. 7 Arrival Process (Example)  Exponential Distribution where is the average number of arrivals per unit of time. Probability of next customer arriving within T units of previous arrival: P (interarrival time  T) = 1- e - T For example: P (interarrival time  1/6 hour)= 1- e -20(1/6) = 0.964  Exponential Distribution where is the average number of arrivals per unit of time. Probability of next customer arriving within T units of previous arrival: P (interarrival time  T) = 1- e - T For example: P (interarrival time  1/6 hour)= 1- e -20(1/6) = 0.964

8. 8 Queueing Process  Single-line Queueing System:  A queueing system in which the customers waiting in a single line for the next available server.  Multiple-line Queueing System:  A queueing system in which arriving customers may select one of several lines in which to wait for service.  Single-line Queueing System:  A queueing system in which the customers waiting in a single line for the next available server.  Multiple-line Queueing System:  A queueing system in which arriving customers may select one of several lines in which to wait for service.

9. 9 Classifications of Queueing Models  Queueing Disciplines  First-In-First-Out (FIFO)  Last-In-First-Out (LIFO)  Priority Selection  Notations (Kendall) A / B / c / K / L where A: interarrival time distribution B: service time distribution c: no. of services K: the system capacity L: the size of the calling population  Queueing Disciplines  First-In-First-Out (FIFO)  Last-In-First-Out (LIFO)  Priority Selection  Notations (Kendall) A / B / c / K / L where A: interarrival time distribution B: service time distribution c: no. of services K: the system capacity L: the size of the calling population

10. 10 Classifications of Queueing Models (cont’)  The arrival process: Symbol A to describe interarrival distribution:  D = deterministic interarrival time  M = probabilistic interarrival time with exponential distribution  G = probabilistic interarrival time with general distribution other than exponential.  The service process: Symbol B to describe the service time distribution:  D = deterministic service time  M = probabilistic service time with exponential distribution  G = probabilistic service time with general distribution other than exponential.  The arrival process: Symbol A to describe interarrival distribution:  D = deterministic interarrival time  M = probabilistic interarrival time with exponential distribution  G = probabilistic interarrival time with general distribution other than exponential.  The service process: Symbol B to describe the service time distribution:  D = deterministic service time  M = probabilistic service time with exponential distribution  G = probabilistic service time with general distribution other than exponential.

11. 11 Classifications of Queueing Models (cont’)  The queueing process: this number c denotes how many parallel stations or channels are in the system (Servers are assumed to be identical in rate of service).  When the waiting space and/or customer population size is finite:  K = Maximum number of customers that can be in the system at any one time = number of parallel stations plus total number of customers that can wait for service  L = Total number of customers in the population  Example:  M/M/3//10 = a system that has room for an infinite number of customer (K has been left off) but only 10 possible customers exists.  The queueing process: this number c denotes how many parallel stations or channels are in the system (Servers are assumed to be identical in rate of service).  When the waiting space and/or customer population size is finite:  K = Maximum number of customers that can be in the system at any one time = number of parallel stations plus total number of customers that can wait for service  L = Total number of customers in the population  Example:  M/M/3//10 = a system that has room for an infinite number of customer (K has been left off) but only 10 possible customers exists.

12. 12 Performance Measures  Performance Measures for Evaluating a Queueing System Transient PhaseSteady State Phase Customer number (in order of arrival)

13. 13 Common performance Measures  Arrival rate –  Service rate of one server –   Server utilization–   Mean number in the system – L  Mean queue length – L q  Average time in the system – W  Average waiting time – W q  Steady-state probability distribution – P n, n = 0,1,2,…  Arrival rate –  Service rate of one server –   Server utilization–   Mean number in the system – L  Mean queue length – L q  Average time in the system – W  Average waiting time – W q  Steady-state probability distribution – P n, n = 0,1,2,…

14. 14 L q =  W q L =  W W = W q + (1 /  ) Relationships among Performance Measures Average time in the system Average time in the system = Average waiting time Average waiting time + Average service time Average service time Average no. of customers in the system Average no. of customers in the system = Average no. of arrivals per unit of time Average no. of arrivals per unit of time  Average time in the system Average time in the system Average no. of customers in the queue Average no. of customers in the queue = Average no. of arrivals per unit of time Average no. of arrivals per unit of time  Average time in the queue Average time in the queue

15. 15 Relationships among Performance Measures (Example)  Suppose = 12,  = 4, L q = 3

16. 16 M / M / 1 Queueing System  Analyzing a Single-Line Single-Channel Queueing System with Exponential Arrival and Service Processes (M / M / 1)  Queueing System for Weigh Station = the average number of trucks arriving per hour = 60  =the average number of trucks that can be weighed per hour = 66  Analyzing a Single-Line Single-Channel Queueing System with Exponential Arrival and Service Processes (M / M / 1)  Queueing System for Weigh Station = the average number of trucks arriving per hour = 60  =the average number of trucks that can be weighed per hour = 66 Weigh Station

17. 17 Formulas (M / M / 1) Performance MeasureGeneral Formula Utilization Average number in line Average waiting time in queue Average waiting time in system Performance MeasureGeneral Formula Utilization Average number in line Average waiting time in queue Average waiting time in system

18. 18 Formulas (M / M / 1) (cont’) Performance MeasureGeneral Formula Average number in system Probability that no customers are in system Probability that an arriving customer has to wait Probability of n customers in the system Performance MeasureGeneral Formula Average number in system Probability that no customers are in system Probability that an arriving customer has to wait Probability of n customers in the system

19. 19Computation Utilization (  ):  = /  = 60 / 66 = 0.9091 Prob.(no customers are in system) (P 0 ): P 0 = 1 -  = 1 – 0.9091 = 0.0909 Average no. in line (L q ): L q =  2 / (1 -  ) = 9.0909 Average waiting time in the queue (W q ): W q = L q / = 9.0909 / 60 = 0.152 Average waiting time in the system (W): W = W q + 1 /  = 0.1667 Average no. in the system (L): L =  W = 60  0.1667 = 10 Prob.(arriving customer having to wait) (P w ): P w = 1 - P 0 =  = 0.9091 Prob.(n customers in the system) (P n ): P n =  n  P 0 Utilization (  ):  = /  = 60 / 66 = 0.9091 Prob.(no customers are in system) (P 0 ): P 0 = 1 -  = 1 – 0.9091 = 0.0909 Average no. in line (L q ): L q =  2 / (1 -  ) = 9.0909 Average waiting time in the queue (W q ): W q = L q / = 9.0909 / 60 = 0.152 Average waiting time in the system (W): W = W q + 1 /  = 0.1667 Average no. in the system (L): L =  W = 60  0.1667 = 10 Prob.(arriving customer having to wait) (P w ): P w = 1 - P 0 =  = 0.9091 Prob.(n customers in the system) (P n ): P n =  n  P 0 n0123… pnpn 0.09090.08260.07510.0683…

20. 20 M / M / c Queueing System  Analyzing a Single-Line Multiple-Channel Queueing System with Exponential Arrival and Service Processes (M / M / c)  Queueing System for Weigh Station with Two Scales c= 2 servers = 70 trucks per hour  =40 trucks per hour on each scale  Analyzing a Single-Line Multiple-Channel Queueing System with Exponential Arrival and Service Processes (M / M / c)  Queueing System for Weigh Station with Two Scales c= 2 servers = 70 trucks per hour  =40 trucks per hour on each scale Weigh Station

21. 21 Formulas (M / M / c) Performance MeasureGeneral Formula Server utilization Probability that no customers are in system Average number in line Performance MeasureGeneral Formula Server utilization Probability that no customers are in system Average number in line

22. 22 Formulas (M / M / c) (cont’) Performance MeasureGeneral Formula Average waiting time in queue Average waiting time in system Average number in system Performance MeasureGeneral Formula Average waiting time in queue Average waiting time in system Average number in system

23. 23Computation Utilization (  ):  = / c  = 70 / (2  40) = 0.875 Prob.(no customers are in system) (P 0 ): Utilization (  ):  = / c  = 70 / (2  40) = 0.875 Prob.(no customers are in system) (P 0 ):

24. 24 Computation (cont’) Average no. in line (L q ): Average waiting time in the queue (W q ): Average waiting time in the system (W): Average no. in the system (L): Average no. in line (L q ): Average waiting time in the queue (W q ): Average waiting time in the system (W): Average no. in the system (L):

25. 25 Economic Analysis of Queueing System  American Weavers, Inc.  The plant has a large number of machines that jam frequently.  Machines are repaired on a first-come-first-serve basis by one of seven available repair persons.  An average of 10 to 12 machines are out of operation at any one time due to jams.  Hiring more repair people will reduce the number of jammed machines. How many should be hired?  American Weavers, Inc.  The plant has a large number of machines that jam frequently.  Machines are repaired on a first-come-first-serve basis by one of seven available repair persons.  An average of 10 to 12 machines are out of operation at any one time due to jams.  Hiring more repair people will reduce the number of jammed machines. How many should be hired?

26. 26 Economic Analysis of Queueing System (cont’)  Modeling Current System:  Current number of repair persons (c = 7).  The occurrence of jammed machines can be approximated by a Poisson arrival process with an average rate of 25 per hour ( = 25).  Each jammed machine requires a random amount of time for repair that can be approximated by an exponential distribution with an average service time of 15 minutes, which is an average rate per server of four machines per hour (  = 4).  Modeling Current System:  Current number of repair persons (c = 7).  The occurrence of jammed machines can be approximated by a Poisson arrival process with an average rate of 25 per hour ( = 25).  Each jammed machine requires a random amount of time for repair that can be approximated by an exponential distribution with an average service time of 15 minutes, which is an average rate per server of four machines per hour (  = 4).

27. 27 Analyzing the Current System M / M / c Queue Statistics Number of identical servers7 Mean arrival rate25.0000 Mean service rate per server4.0000 Mean server utilization (%)89.2857 Expected number of customers in queue5.8473 Expected number of customers in system12.0973 Probability that a customer must wait0.7017 Expected time in the queue0.2339 Expected time in the system0.4839 M / M / c Queue Statistics Number of identical servers7 Mean arrival rate25.0000 Mean service rate per server4.0000 Mean server utilization (%)89.2857 Expected number of customers in queue5.8473 Expected number of customers in system12.0973 Probability that a customer must wait0.7017 Expected time in the queue0.2339 Expected time in the system0.4839

28. 28 Cost Analysis c s = cost per server per unit of time = $50/hour c w = cost per unit of time for a customer waiting in the system = $100/hour for lost production, etc. L= average number of customers in the system Total Cost = cost of crew + cost of waiting = c s  c + c w  L = 50  7 + 100  12.0973 = $1559.73 per hour c s = cost per server per unit of time = $50/hour c w = cost per unit of time for a customer waiting in the system = $100/hour for lost production, etc. L= average number of customers in the system Total Cost = cost of crew + cost of waiting = c s  c + c w  L = 50  7 + 100  12.0973 = $1559.73 per hour

29. 29 Cost Analysis (cont’) Crew sizeExpected no. in the systemHourly cost ($) 712.0973 50  7 + 100  12.0973 = 1559.73 87.7436 50  8 + 100  7.7436 = 1174.36 96.7863 50  9 + 100  6.7863 = 1128.63 106.7594 50  10 + 100  6.4594 = 1145.94 116.3330 50  11 + 100  6.3330 = 1183.30