Lecture 11 Queueing Models. 2 Queueing System Queueing System: A system in which items (or...
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Transcript of Lecture 11 Queueing Models. 2 Queueing System Queueing System: A system in which items (or...
Lecture 11Lecture 11Lecture 11Lecture 11
Queueing ModelsQueueing ModelsQueueing ModelsQueueing Models
2
Queueing SystemQueueing 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
Characteristics of Characteristics of Queueing SystemQueueing 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
Characteristics of Characteristics of Queueing System Queueing System (cont’)(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
Queueing SystemQueueing System
CustomerPopulatio
n
ArrivalProcess
SystemSystem
.
.
.
WaitingCustomers
QueueingProcess
Depart
Network of queuesNetwork of queues
Arrival Depart
One-stepqueueing system
One-stepqueueing system
ArrivalDepart
Components of a Queueing System:
Departure Processes:
Servers
ServiceProcess
6
Arrival ProcessArrival 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
Arrival Process Arrival Process (Example)(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
tetf )(
8
Queueing ProcessQueueing 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
Classifications of Classifications of Queueing ModelsQueueing 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
Classifications of Classifications of Queueing Models Queueing Models (cont’)(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
Classifications of Classifications of Queueing Models Queueing Models (cont’)(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
Performance MeasuresPerformance Measures
Performance Measures for Evaluating a Queueing System
Performance Measures for Evaluating a Queueing System
Transient Phase Steady State Phase
Customer number
(in order of arrival)
13
Common performance Common performance MeasuresMeasures
Arrival rate – Service rate of one server – Server utilization– Mean number in the system – L Mean queue length – Lq Average time in the system – W Average waiting time – Wq
Steady-state probability distribution – Pn, n = 0,1,2,…
Arrival rate – Service rate of one server – Server utilization– Mean number in the system – L Mean queue length – Lq Average time in the system – W Average waiting time – Wq
Steady-state probability distribution – Pn, n = 0,1,2,…
14
Lq = W qLq = W q
L = WL = W
W = Wq + (1 / )W = Wq + (1 / )
Relationships among Relationships among Performance MeasuresPerformance Measures
Average timein the system
Average timein the system = Average
waiting time
Averagewaiting time + Average
service time
Averageservice time
Average no. of customersin the system
Average no. of customersin the system = Average no. of arrivals
per unit of time
Average no. of arrivalsper unit of time
Average timein the system
Average timein the system
Average no. of customersin the queue
Average no. of customersin the queue = Average no. of arrivals
per unit of time
Average no. of arrivalsper unit of time
Average timein the queue
Average timein the queue
15
Relationships among Relationships among Performance Measures Performance Measures (Example)(Example)
Suppose = 12, = 4, Lq = 3 Suppose = 12, = 4, Lq = 3
4
1
12
3
q
q
LW
2
1
4
1
4
11
qWW
62
112 WL
16
MM / / MM / 1 Queueing / 1 Queueing SystemSystem
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
Formulas (Formulas (MM / / MM / 1) / 1)
Performance Measure General Formula
Utilization
Average number in line
Average waiting time in queue
Average waiting time in system
Performance Measure General Formula
Utilization
Average number in line
Average waiting time in queue
Average waiting time in system
1
2
qL
)1(
q
q
LW
)1(
11
qWW
18
Formulas (Formulas (MM / / MM / 1) / 1) (cont’)(cont’)
Performance Measure General 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 Measure General 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
01 PPw
)1(0 nnn PP
10P
1
WL
19
ComputationComputation
Utilization (): = / = 60 / 66 = 0.9091
Prob.(no customers are in system) (P0): P0 = 1 - = 1 – 0.9091 = 0.0909
Average no. in line (Lq): Lq = 2 / (1 - ) = 9.0909
Average waiting time in the queue (Wq): Wq = Lq / = 9.0909 / 60 = 0.152
Average waiting time in the system (W): W = Wq + 1 / = 0.1667
Average no. in the system (L): L = W = 60 0.1667 = 10
Prob.(arriving customer having to wait) (Pw): Pw = 1 - P0 = = 0.9091
Prob.(n customers in the system) (Pn): Pn = n P0
Utilization (): = / = 60 / 66 = 0.9091
Prob.(no customers are in system) (P0): P0 = 1 - = 1 – 0.9091 = 0.0909
Average no. in line (Lq): Lq = 2 / (1 - ) = 9.0909
Average waiting time in the queue (Wq): Wq = Lq / = 9.0909 / 60 = 0.152
Average waiting time in the system (W): W = Wq + 1 / = 0.1667
Average no. in the system (L): L = W = 60 0.1667 = 10
Prob.(arriving customer having to wait) (Pw): Pw = 1 - P0 = = 0.9091
Prob.(n customers in the system) (Pn): Pn = n P0
n 0 1 2 3 …
pn 0.0909 0.0826 0.0751 0.0683 …
20
M M / / M M / / cc Queueing Queueing SystemSystem
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 scaleWeigh Station
21
Formulas (Formulas (MM / / MM / / cc))
Performance Measure General Formula
Server utilization
Probability that no customers are in system
Average number in line
Performance Measure General Formula
Server utilization
Probability that no customers are in system
Average number in line
11
!1
!)(
11
0
0
cc
nc
Pc
c
n
n
20
1
)1(!
)(
cc
PcL
c
q
c
22
Formulas (Formulas (MM / / MM / / cc) ) (cont’)(cont’)
Performance Measure General Formula
Average waiting time in queue
Average waiting time in system
Average number in system
Performance Measure General Formula
Average waiting time in queue
Average waiting time in system
Average number in system
1
qWW
q
q
LW
20
1
)1(!
)(
cc
PccL
c
W
23
ComputationComputation
Utilization (): = / c = 70 / (2 40) = 0.875
Prob.(no customers are in system) (P0):
Utilization (): = / c = 70 / (2 40) = 0.875
Prob.(no customers are in system) (P0):
75.275.11!1
75.1
!0
75.1
!)1(
)(
!1
)(
!0
)(
!
)( 101101
0
c
ccc
n
c cc
n
n
25.12853125.1875.01
1
!2
75.1
1
1
!
)( 2
c
c c
06667.015
1
25.1275.2
10
P
11
!1
!)(
11
0
0
cc
nc
Pc
c
n
n
24
Computation (cont’)Computation (cont’)
Average no. in line (Lq):
Average waiting time in the queue (Wq):
Average waiting time in the system (W):
Average no. in the system (L):
Average no. in line (Lq):
Average waiting time in the queue (Wq):
Average waiting time in the system (W):
Average no. in the system (L):
7168.5
06667.0643398.1
06667.0875.01
1
!22
75.12
3
08167.070/7168.5/ qq LW
10667.040/108167.0)/1( qWW
4669.710667.070 WL
20
1
)1(!
)(
cc
PcL
c
q
25
Economic Analysis of Economic Analysis of Queueing SystemQueueing 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
Economic Analysis of Economic Analysis of Queueing System Queueing System (cont’)(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
Analyzing the Current Analyzing the Current SystemSystem
M / M / c Queue StatisticsNumber of identical servers 7
Mean arrival rate 25.0000
Mean service rate per server 4.0000
Mean server utilization (%) 89.2857
Expected number of customers in queue 5.8473
Expected number of customers in system 12.0973
Probability that a customer must wait 0.7017
Expected time in the queue 0.2339
Expected time in the system 0.4839
M / M / c Queue StatisticsNumber of identical servers 7
Mean arrival rate 25.0000
Mean service rate per server 4.0000
Mean server utilization (%) 89.2857
Expected number of customers in queue 5.8473
Expected number of customers in system 12.0973
Probability that a customer must wait 0.7017
Expected time in the queue 0.2339
Expected time in the system 0.4839
28
Cost AnalysisCost Analysis
cs = cost per server per unit of time = $50/hour
cw = 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
= cs c + cw L
= 50 7 + 100 12.0973
= $1559.73 per hour
cs = cost per server per unit of time = $50/hour
cw = 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
= cs c + cw L
= 50 7 + 100 12.0973
= $1559.73 per hour
29
Cost Analysis (cont’)Cost Analysis (cont’)
Crew size Expected no. in the system Hourly cost ($)
7 12.0973 50 7 + 100 12.0973 = 1559.73
8 7.7436 50 8 + 100 7.7436 = 1174.36
9 6.7863 50 9 + 100 6.7863 = 1128.63
10 6.7594 50 10 + 100 6.4594 = 1145.94
11 6.3330 50 11 + 100 6.3330 = 1183.30
0
200
400
600
800
1000
1200
1400
1600
1800
7 8 9 10 11
Service Cost
Waiting Cost
Total Cost
0
200
400
600
800
1000
1200
1400
1600
1800
7 8 9 10 11
Service Cost
Waiting Cost
Total Cost