Post on 18-Dec-2015
Outline
• Queueing System Introduction• Queueing System Elements
• Arrivals, Queue, Service• Performance Measures
• Waiting time, Queue Length, Standards/Service Levels
• Single Server Queues (M/M/1, M/G/1)• Multiple Server Queues (M/M/s)• Priority Queues• Economic Analysis• Chapter 11
1
Outline
• Queueing System Introduction• Queueing System Elements
• Arrivals, Queue, Service• Performance Measures
• Waiting time, Queue Length, Standards/Service Levels
• Single Server Queues (M/M/1, M/G/1)• Multiple Server Queues (M/M/s)• Priority Queues• Economic Analysis• Chapter 11
2
Queueing Models: Introduction
• A basic queueing system
http://www.youtube.com/watch?v=N5TAWW_LIsw
3
CustomersQueue
Served Customers
Queueing System
Service facility
SSSS
CCCC
C C C C C C C
Served Customers
Queueing Models: Introduction
• Customers• People waiting to be served• Machines waiting to be repaired• Jobs waiting to be completed• Airplanes waiting to takeoff• Trucks waiting to be loaded/unloaded….
• Servers• People serving the customers• A machine processing a job• Forklifts for unloading….
4
Queueing Models: Examples
• Some examples: Commercial service systems
5
Type of System Customers Server(s)
Barber shop People Barber
Bank teller services People Teller
ATM machine service People ATM machine
Checkout at a store People Checkout clerk
Plumbing services Clogged pipes Plumber
Ticket window at a movie theater People Cashier
Check-in counter at an airport People Airline agent
Brokerage service People Stock broker
Gas station Cars Pump
Call center for ordering goods People Telephone agent
Call center for technical assistance People Technical representative
Travel agency People Travel agent
Automobile repair shop Car owners Mechanic
Vending services People Vending machine
Dental services People Dentist
Roofing Services Roofs Roofer
Queueing Models : Examples
• Some examples: Internal service systems
6
Type of System Customers Server(s)
Secretarial services Employees Secretary
Copying services Employees Copy machine
Computer programming services Employees Programmer
Mainframe computer Employees Computer
First-aid center Employees Nurse
Faxing services Employees Fax machine
Materials-handling system Loads Materials-handling unit
Maintenance system Machines Repair crew
Inspection station Items Inspector
Production system Jobs Machine
Semiautomatic machines Machines Operator
Tool crib Machine operators Clerk
Queueing Models : Examples
• Some examples: Transportation service systems
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Type of System Customers Server(s)
Highway tollbooth Cars Cashier
Truck loading dock Trucks Loading crew
Port unloading area Ships Unloading crew
Airplanes waiting to take off Airplanes Runway
Airplanes waiting to land Airplanes Runway
Airline service People Airplane
Taxicab service People Taxicab
Elevator service People Elevator
Fire department Fires Fire truck
Parking lot Cars Parking space
Ambulance service People Ambulance
Queueing Models: Introduction
• Herr Cutter’s Barber Shop• Herr Cutter opens his shop at 8:00 A.M.• The table shows his queueing system in action over a typical morning.
8
CustomerTime ofArrival
HaicutBegins
Durationof Haircut
HaircutEnds
1 8:03 8:03 17 minutes 8:20
2 8:15 8:20 21 minutes 8:41
3 8:25 8:41 19 minutes 9:00
4 8:30 9:00 15 minutes 9:15
5 9:05 9:15 20 minutes 9:35
6 9:43 — — —
Queueing Models: Introduction
• Evolution of the Number of Customers
9
20 40 60 80
1
2
3
4
Number of Customers in the System
0
Time (in minutes)100
Queueing Models: Elements
• Arrivals• The time between consecutive arrivals to a queueing system are called the
interarrival times• The expected number of arrivals per unit time is referred to as the mean
arrival rate.• The symbol used for the mean arrival rate is
l = Mean arrival rate for customers coming to the queueing system (l lambda)
• The mean of the probability distribution of interarrival times is1 / l = Expected interarrival time
10
Queueing Models: Elements
• Herr Cutter’s Barber Shop• After gathering more data, Herr Cutter finds that 300 customers have arrived
over a period of 100 hours• Mean arrival rate
• Expected interarrival time
• Most queueing models assume that the form of the probability distribution of interarrival times is an exponential distribution
11
Queueing Models: Elements
• The Exponential Distribution for Interarrival Times• The most commonly used queuing models are based on the assumption of
exponentially distributed service times and interarrival times• A random variable Texp( ), i.e., is exponentially distributed
with parameter , if its density function is:
• The mean = E[T] = 1/• The Variance = Var[T] = 1/ 2
12
0twhen0
0twhene)t(f
t
Tt
T e1)t(F
Queueing Models: Elements
13
Time between arrivalsMean= E[T]=1/
Pro
bab
ility
den
sity
t
fT(t)
• Probability density function is decreasing
• Memoryless property:P(T>t+t | T>t) = P(T >t)
Queueing Models: Elements
• Properties of the Exponential Distribution• There is a high likelihood of small interarrival times, but a small chance of a
very large interarrival time. This is characteristic of interarrival times in practice.
• For most queueing systems, the servers have no control over when customers will arrive. Customers generally arrive randomly.
• Having random arrivals means that interarrival times are completely unpredictable, in the sense that the chance of an arrival in the next minute is always just the same.
• The only probability distribution with this property of random arrivals is the exponential distribution.
• The fact that the probability of an arrival in the next minute is completely uninfluenced by when the last arrival occurred is called the lack-of-memory property (memoryless property like my fish!!!).
14
Queueing Models: Elements
• The number of customers in the queue (or queue size) is the number of customers waiting for service to begin.
• The number of customers in the system is the number in the queue plus the number currently being served.
• The queue capacity is the maximum number of customers that can be held in the queue.
• An infinite queue is one in which, for all practical purposes, an unlimited number of customers can be held there.
• When the capacity is small enough that it needs to be taken into account, then the queue is called a finite queue.
• The queue discipline refers to the order in which members of the queue are selected to begin service.
• The most common is first-come, first-served (FCFS).• Other possibilities include random selection, some priority procedure, or even last-
come, first-served.
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Queueing Models: Elements
• When a customer enters service, the elapsed time from the beginning to the end of the service is referred to as the service time.
• Basic queueing models assume that the service time has a particular probability distribution.
• The symbol used for the mean of the service time distribution is1 / m = Expected service time (m mu)
• The interpretation of m itself is the mean service rate.m = Expected service completions per unit time for a single
busy server
16
Queueing Models: Elements
• Some Service-Time Distributions• Exponential Distribution
• The most popular choice.• Much easier to analyze than any other.• Although it provides a good fit for interarrival times, this is much less true for service
times.• Provides a better fit when the service provided is random than if it involves a fixed set of
tasks.• Standard deviation: s = Mean
• Constant Service Times• A better fit for systems that involve a fixed set of tasks.• Standard deviation: s = 0.
17
Queueing Models: Notation
• Kendall’s Notation
• M = Exponential• Ek = Erlang-k• U = Uniform• G = General• D = Deterministic (constant times)
18
Interarrival Time
Distribution
Service TimeDistribution
# ofServers
SystemCapacity
Dropped if infinite
Queueing Models: Notation
• Single Server Queueing Models M/G/1 M/D/1 M/Ek/1 M/M/1
• Multiple Server Queueing Models M/G/s M/D/s M/Ek/s M/M/s
• Finite Capacity Queueing Models M/M/s/k M/M/1/k M/M/s/s
• Priority Queues
19
Queueing Models: Assumptions
• Assumptions:• Interarrival times are independent and identically distributed according to a
specified probability distribution.• All arriving customers enter the queueing system and remain there until
service has been completed.• The queueing system has a single infinite queue, so that the queue will hold
an unlimited number of customers (for all practical purposes).• The queue discipline is first-come, first-served.• The queueing system has a specified number of servers, where each server is
capable of serving any of the customers.• Each customer is served individually by any one of the servers.• Service times are independent and identically distributed according to a
specified probability distribution.
20
Queueing Models: Performance
• Choosing a performance measure• Managers who oversee queueing systems are mainly concerned with
two measures of performance:• How many customers typically are waiting in the queueing system?• How long do these customers typically have to wait?
• When customers are internal to the organization, the first measure tends to be more important.
• Having such customers wait causes lost productivity.• Commercial service systems tend to place greater importance on the
second measure.• Outside customers are typically more concerned with how long they
have to wait than with how many customers are there.
21
Queueing Models: Performance
• Measures of performance• L = Expected number of customers in the system, including
those being served (the symbol L comes from Line Length).• Lq= Expected number of customers in the queue, which excludes
customers being served.• W = Expected waiting time in the system (including service time) for
an individual customer (the symbol W comes from Waiting time).• Wq = Expected waiting time in the queue (excludes service time) for
an individual customer.
These definitions assume that the queueing system is in a steady-state condition
22Not start-up, not
temporary rush hour
Queueing Models: Performance
• Relationship between L, W, Lq, and Wq
• Little’s formula states that
L = lW and Lq = lWq
23
• Since L is the expected number customers in the queueing system at any time, a customer looking back at the system after completing service should see L customers on average
• Under FCFS, all L customers would have arrived during this customer’s waiting time in the queueing system, this waiting time is W on average
• Since l is the expected number of arrivals per unit time
L = lW
Queueing Models: Performance
• Relationship between L, W, Lq, and Wq• Since 1/m is the expected service time
• Using Little’s law
• These are important!!! • Once we know one value, we can determine the others
24
W = Wq + 1/m
L = lW = (l Wq + 1/ )m = (l Lq / l+ 1/ )=m Lq + l /m
Lq = lWq
L = Lq + l/m
Queueing Models: Performance
• Relationship between L, W, Lq, and Wq• l = 3 customers per hour arrive on average• m = 4 customers per hour served (leave) on average• Wq = ¾ hour waiting in the queue on average
• 1/ m = ¼ hour service time on average• W=Wq + 1/m = ¾ + ¼ = 1 hour
• 1 hour waiting in the queueing system on average• L=lW =3 customers/hour * 1 hour/customer = 3 customers
• 3 customers in the queueing system on average• L = Lq + l/ m 3 = Lq + ¾ Lq = 9/4 customers
• 9/4 customers in the queue on average
25
Queueing Models: Performance
• In addition to knowing what happens on the average, we may also be interested in worst-case scenarios.
• What will be the maximum number of customers in the system? (Exceeded no more than, say, 5% of the time.)
• What will be the maximum waiting time of customers in the system? (Exceeded no more than, say, 5% of the time.)
• Statistics that are helpful to answer these types of questions are available for some queueing systems:
• Pn = Steady-state probability of having exactly n customers in the system.• P(W ≤ t) = Probability the time spent in the system will be no more than t.• P(Wq ≤ t) = Probability the wait time will be no more than t.
• Examples of common goals:• No more than three customers 95% of the time: P0 + P1 + P2 + P3 ≥ 0.95• No more than 5% of customers wait more than 2 hours: P(W ≤ 2 hours) ≥ 0.95
26
Queueing Models: A Case Study
• The Dupit Corporation is a longtime leader in the office photocopier marketplace.
• Dupit’s service division is responsible for providing support to the customers by promptly repairing the machines when needed. This is done by the company’s service technical representatives, or tech reps.
• Current policy: Each tech rep’s territory is assigned enough machines so that the tech rep will be active repairing machines (or traveling to the site) 75% of the time.
• A repair call averages 2 hours, so this corresponds to 3 repair calls per day.• Machines average 50 workdays between repairs, so assign 150 machines per rep.
• Proposed New Service Standard: The average waiting time before a tech rep begins the trip to the customer site should not exceed two hours.
27
Queueing Models: A Case Study
• Alternative Approaches• Approach Suggested by John Phixitt: Modify the current policy by
decreasing the percentage of time that tech reps are expected to be repairing machines.
• Approach Suggested by the Vice President for Engineering: Provide new equipment to tech reps that would reduce the time required for repairs.
• Approach Suggested by the Chief Financial Officer: Replace the current one-person tech rep territories by larger territories served by multiple tech reps.
• Approach Suggested by the Vice President for Marketing: Give owners of the new printer-copier priority for receiving repairs over the company’s other customers.
28
Queueing Models: A Case Study
• Queueing System:• The customers: The machines needing repair.• Customer arrivals: The calls to the tech rep requesting repairs.• The queue: The machines waiting for repair to begin at their sites.• The server: The tech rep.• Service time: The total time the tech rep is tied up with a machine,
either traveling to the machine site or repairing the machine. (Thus, a machine is viewed as leaving the queue and entering service when the tech rep begins the trip to the machine site.)
29
Outline
• Queueing System Introduction• Queueing System Elements
• Arrivals, Queue, Service• Performance Measures
• Waiting time, Queue Length, Standards/Service Levels
• Single Server Queues (M/M/1, M/G/1)• Multiple Server Queues (M/M/s)• Priority Queues• Economic Analysis• Chapter 11
30
Single Server Queues
• l = Mean arrival rate for customers= Expected number of arrivals per unit time
1/l = expected interarrival time
• m = Mean service rate (for a continuously busy server)= Expected number of service completions per unit time
1/m = expected service time
• r = the utilization factor= the average fraction of time that a server is busy serving
customers= l / m
31
M/M/1 Queue
• Simplest model
• That is a Markov Chain• Each state (the yellow nodes) is a possible number of people in your
queueing system• Since infinitely possible states, we have infinite Markov Chain
32
0 1 n-1 n n+1
M/M/1 Queue
• Transition rates
• Transition rate is the rate which you leave from a state and the rate which you enter a state
l m
33
0 1 n-1 n n+1
M/M/1 Queue
• Continuous time Markov Chain• Conversation of Flow• Rate in = Rate out
• System State:• n = # of jobs in system• pn = P(n jobs is the system)
• Then
lpn-1 + mpn+1 = pn(l + m)
34
M/M/1 Queue
• Rate in = Rate out
lpn-1 + mpn+1 = pn(l + m)
• Recursively, it can be shown that:
pn = rn(1-r) for r < 1
where r = l/m = utilization
35
M/M/1 Queue
Recursive calculation using lpn-1 + mpn+1 = pn(l + m)• l P0= m P1 P1 = / l m P0 P1 = rP0
• l P0 + m P2 = P1(l + m) P2 = r2P0
• l P1 + m P3 = P2(l + m) P3 = r3P0
• …………Pn = rnP0
36
=1- pn = rn(1-r)
M/M/1 Queue
• Sum of probabilities is equal to 1• P0+ P1+ P2+ P3+ P4+…..
37
Geometric sequence and <1
1/(1- )
M/M/1 Queue
• Recall the performance measures• L = Expected number of customers in the system, including
those being served (the symbol L comes from Line Length).• Lq= Expected number of customers in the queue, which excludes
customers being served.• W = Expected waiting time in the system (including service time) for
an individual customer (the symbol W comes from Waiting time).• Wq = Expected waiting time in the queue (excludes service time) for
an individual customer
38
M/M/1 Queue
• L = E[# in system]
39
1)1(
1)1(
)1()1(
)1(
2
1
1
0
00
n
n
n
n
n
n
nn
nn
nnp
1L
M/M/1 Queue
• Using Little’s Law L=lW and LQ=lWQ• W = E[Waiting Time in System]
• Lq = E[# in the queue]
• W = E[Waiting Time in System]
40
1
W
1)(
22
LLQ
)(1
WWQ
M/M/1
• Performance measures
41
1L
1
W
1)(
22
LLQ
)(1
WWQ
M/M/1
• Utilization law
42100%Utilization
L and W
Effect of High-Utilization Factors
43
345
B C D E G HData Results
0.5 (mean arrival rate) L = 1 1 (mean service rate) Lq = 0.5
9
10
1112
13
141516171819202122232425
A B C D EData Table Demonstrating the Effect ofIncreasing on Lq and L for M/M/1
Lq L
1 0.5 1
0 0.01 0.0001 0.01010 0.25 0.0833 0.33330 0.5 0.5 10 0.6 0.9 1.50 0.7 1.6333 2.33330 0.75 2.25 30 0.8 3.2 40 0.85 4.8167 5.66670 0.9 8.1 90 0.95 18.05 190 0.99 98.01 990 0.999 998.001 999
0
20
40
60
80
100
0 0.2 0.4 0.6 0.8 1
System Utilization (r)
Ave
rag
e L
ine
Le
ng
th (
L)
M/M/1
• The probability of having exactly n customers in the system is
Pn = (1 – r)rn
Thus,P0 = 1 – rP1 = (1 – r)rP2 = (1 – r)r2
::
• The probability that the waiting time in the system exceeds t is
P(W > t) = e–m(1–r)t for t ≥ 0
• The probability that the waiting time in the queue exceeds t is
P(Wq > t) = re–m(1–r)t for t ≥ 0
44
M/M/1
• Recall our example…• The Dupit Corporation is a longtime leader in the office photocopier
marketplace.• Dupit’s service division is responsible for providing support to the customers by
promptly repairing the machines when needed. This is done by the company’s service technical representatives, or tech reps.
• Current policy: Each tech rep’s territory is assigned enough machines so that the tech rep will be active repairing machines (or traveling to the site) 75% of the time.
• A repair call averages 2 hours, so this corresponds to 3 repair calls per day.• Machines average 50 workdays between repairs, so assign 150 machines per rep.
• Proposed New Service Standard: The average waiting time before a tech rep begins the trip to the customer site should not exceed two hours.
45
M/M/1
• Alternative Approaches• Approach Suggested by John Phixitt: Modify the current policy by
decreasing the percentage of time that tech reps are expected to be repairing machines. M/M/1
• Approach Suggested by the Vice President for Engineering: Provide new equipment to tech reps that would reduce the time required for repairs.
• Approach Suggested by the Chief Financial Officer: Replace the current one-person tech rep territories by larger territories served by multiple tech reps.
• Approach Suggested by the Vice President for Marketing: Give owners of the new printer-copier priority for receiving repairs over the company’s other customers.
46
M/M/1
• M/M/1 Queueing Model for the Dupit’s Current Policy• Current policy: Each tech rep’s territory is assigned enough machines so that
the tech rep will be active repairing machines (or traveling to the site) 75% of the time.
• A repair call averages 2 hours, so this corresponds to 3 repair calls per day.• Service rate 4 machines per day
• Machines average 50 workdays between repairs, so assign 150 machines per rep.• Arrival rate 3 machines per day
• ¾ =0.75 daily utilization currently
47
M/M/1
• Speadsheet for Dupit
48
Data Results 3 (mean arrival rate) L = 3 4 (mean service rate) Lq = 2.25
s = 1 (# servers)W = 1
Pr(W > t) = 0.368 Wq = 0.75
when t = 1 0.75
Prob(Wq > t) = 0.276
when t = 1 n Pn
0 0.251 0.18752 0.14063 0.10554 0.07915 0.05936 0.04457 0.03348 0.02509 0.0188
10 0.0141
M/M/1
• Proposed New Service Standard: The average waiting time before a tech rep begins the trip to the customer site should not exceed two hours.
• The proposed new service standard is that the average waiting time before service begins <= two hours (i.e., Wq ≤ 1/4 day).
• John Phixitt’s suggested approach is to lower the tech rep’s utilization factor sufficiently to meet the new service requirement.
Lower r = l / m, until Wq ≤ 1/4 day,where
l = (Number of machines assigned to tech rep) / 50.
49
M/M/1
• What can we control?• The number of machines assigned to a tech rep• Let’s say we make it 100 machines
• Machines average 50 workdays between repairs, we assign 100 machines per tech rep.• Arrival rate = 100/50 = 2 machines per day, =2l
50
M/M/1
• Under new policy
51
Data Results 2 (mean arrival rate) L = 1 4 (mean service rate) Lq = 0.5
s = 1 (# servers)W = 0.5
Pr(W > t) = 0.135 Wq = 0.25
when t = 1 0.5
Prob(Wq > t) = 0.068
when t = 1 n Pn
0 0.51 0.252 0.12503 0.06254 0.03135 0.01566 0.00787 0.00398 0.00209 0.0010
10 0.0005
We have it
Cost?
We will need to increase the number of tech reps
M/M/1
• Mathematically…
• l / ( - )<=0.25 m m l =4m• l / 4(4- )<=0.25l• l <=0.25*4*(4- )l• l <= 4- l• l <=2
• Number of machines assigned/50 <=2• Number of machines assigned<=100
52
)(1
WWQ
M/G/1
• Alternative Approaches• Approach Suggested by John Phixitt: Modify the current policy by
decreasing the percentage of time that tech reps are expected to be repairing machines.
• Approach Suggested by the Vice President for Engineering: Provide new equipment to tech reps that would reduce the time required for repairs.
• Approach Suggested by the Chief Financial Officer: Replace the current one-person tech rep territories by larger territories served by multiple tech reps.
• Approach Suggested by the Vice President for Marketing: Give owners of the new printer-copier priority for receiving repairs over the company’s other customers.
53
M/G/1
• The Vice President for Engineering has suggested providing tech reps with new state-of-the-art equipment that would reduce the time required for the longer repairs.
• After gathering more information, they estimate the new equipment would have the following effect on the service-time distribution:
• mean from 1/4 day to 1/5 day• standard deviation from 1/4 day to 1/10 day.
(in exponential, mean=standard deviation)
• No longer M/M/1
54
M/G/1
• Assumptions:• Interarrival times have an exponential distribution with a mean of 1/l• Service times can have any probability distribution. You only need the mean
(1/m) and standard deviation (s)• The queueing system has one server
55
M/G/1
• The probability of zero customers in the system isP0 = 1 – r
• The expected number of customers in the queue isLq = [l2s2 + r2] / [2(1 – r)]
• The expected number of customers in the system isL = Lq + r
• The expected waiting time in the queue isWq = Lq / l
• The expected waiting time in the system isW = Wq + 1/m
56
M/G/1
Service Distribution Model s
Deterministic M/D/1 0
Erlang-k M/Ek/1
Exponential M/M/1
57
)1(
2
QL
12
1 2
k
kLQ
)1(2
2
QL
)1(2
222
QL
k1
1
M/G/1
• The proposed new service standard is that the average waiting time before service begins be two hours (i.e., Wq ≤ 1/4 day).
• The Vice President for Engineering has suggested providing tech reps with new state-of-the-art equipment that would reduce the time required for the longer repairs.
• After gathering more information, they estimate the new equipment would have the following effect on the service-time distribution:
• Decrease the mean from 1/4 day to 1/5 day.• Decrease the standard deviation from 1/4 day to 1/10 day.
58
M/G/1
• Approach of the Vice President for Engineering
59
3456789
10
1112
B C D E F GData Results
3 (mean arrival rate) L = 1.1631 0.2 (expected service time) Lq = 0.563 0.1 (standard deviation)s = 1 (# servers) W = 0.388
Wq = 0.188
0.6
P0 = 0.4
Outline
• Queueing System Introduction• Queueing System Elements
• Arrivals, Queue, Service• Performance Measures
• Waiting time, Queue Length, Standards/Service Levels
• Single Server Queues (M/M/1, M/G/1)• Multiple Server Queues (M/M/s)• Priority Queues• Economic Analysis• Chapter 11
60
Multiple Server Queues
l = Mean arrival rate for customers= Expected number of arrivals per unit time
m = Mean service rate (for a continuously busy server)= Expected number of service completions per unit time
• Utilization factor• s servers
r = l/sm
61
Multiple Server Queues
• M/G/s – no useful analytical results• M/D/s – limited analytical results• M/Ek/s – limited analytical results• M/M/s – analytical results
• Mathematical derivations are complex!!!• We will use Excel• Utilization
62
s
Multiple Server Queues
• Alternative Approaches• Approach Suggested by John Phixitt: Modify the current policy by
decreasing the percentage of time that tech reps are expected to be repairing machines.
• Approach Suggested by the Vice President for Engineering: Provide new equipment to tech reps that would reduce the time required for repairs.
• Approach Suggested by the Chief Financial Officer: Replace the current one-person tech rep territories by larger territories served by multiple tech reps.
• Approach Suggested by the Vice President for Marketing: Give owners of the new printer-copier priority for receiving repairs over the company’s other customers.
63
M/M/s
• Assumptions• Interarrival times have an exponential distribution with a mean of 1/l.• Service times have an exponential distribution with a mean of 1/ .m• Any number of servers (denoted by s).
• With multiple servers, the formula for the utilization factor becomesr = l / sm
but still represents the average fraction of time that individual servers are busy.
64
M/M/s
• The proposed new service standard is that the average waiting time before service begins be two hours (i.e., Wq ≤ 1/4 day).
• The Chief Financial Officer has suggested combining the current one-person tech rep territories into larger territories that would be served jointly by multiple tech reps.
• A territory with two tech reps:• Number of machines = 300 (versus 150 before)• Mean arrival rate = l = 6 (versus l = 3 before)• Mean service rate = m = 4 (as before)• Number of servers = s = 2 (versus s = 1 before)• Utilization factor = r = l/sm = 0.75 (as before)
65
M/M/s
66
3456789
1011121314151617181920212223
B C D E G HData Results
6 (mean arrival rate) L = 3.4286 4 (mean service rate) Lq = 1.9286s = 2 (# servers)
W = 0.5714Pr(W > t) = 0.169 Wq = 0.3214
when t = 1 0.75
Prob(W q > t) = 0.087when t = 1 n Pn
0 0.14291 0.21432 0.16073 0.12054 0.09045 0.06786 0.05097 0.03818 0.02869 0.0215
10 0.0161
Still not what we want
M/M/s
• The Chief Financial Officer has suggested combining the current one-person tech rep territories into larger territories that would be served jointly by multiple tech reps.
• A territory with three tech reps:• Number of machines = 450 (versus 150 before)• Mean arrival rate = l = 9 (versus l = 3 before)• Mean service rate = m = 4 (as before)• Number of servers = s = 3 (versus s = 1 before)• Utilization factor = r = l/sm = 0.75 (as before)
67
M/M/s
68
3456789
1011121314151617181920212223
B C D E G HData Results
9 (mean arrival rate) L = 3.9533 4 (mean service rate) Lq = 1.7033s = 3 (# servers)
W = 0.4393Pr(W > t) = 0.090 Wq = 0.1893
when t = 1 0.75
Prob(W q > t) = 0.028when t = 1 n Pn
0 0.07481 0.16822 0.18933 0.14194 0.10655 0.07986 0.05997 0.04498 0.03379 0.0253
10 0.0189
M/M/s
• Comparison of s=2 and s=3
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Number ofTech Reps
Number ofMachines l m s r Wq
1 150 3 4 1 0.75 0.75 workday (6 hours)
2 300 6 4 2 0.75 0.321 workday (2.57 hours)
3 450 9 4 3 0.75 0.189 workday (1.51 hours)
Insights
• When designing a single-server queueing system, beware that giving a relatively high utilization factor (workload) to the server provides surprisingly poor performance for the system
• Decreasing the variability of service times (without any change in the mean) substantially improves the performance of a queueing system.
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Insights
• Multiple-server queueing systems can perform satisfactorily with somewhat higher utilization factors than can single-server queueing systems. For example, pooling servers by combining separate single-server queueing systems into one multiple-server queueing system greatly improves the measures of performance.
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Insights
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. . .
. . .
. . .
Customers ServiceCenters
. . .
Customers
ServiceCenters
Impact of Pooling Servers:• Suppose you have n identical M/M/1• Suppose you combine the servers so you have a single M/M/n
Wq(for combined system) <
Wq(for each single-server system)/n
Outline
• Queueing System Introduction• Queueing System Elements
• Arrivals, Queue, Service• Performance Measures
• Waiting time, Queue Length, Standards/Service Levels
• Single Server Queues (M/M/1, M/G/1)• Multiple Server Queues (M/M/s)• Priority Queues• Economic Analysis• Chapter 11
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Queueing Models: A Case Study
• Alternative Approaches• Approach Suggested by John Phixitt: Modify the current policy by
decreasing the percentage of time that tech reps are expected to be repairing machines.
• Approach Suggested by the Vice President for Engineering: Provide new equipment to tech reps that would reduce the time required for repairs.
• Approach Suggested by the Chief Financial Officer: Replace the current one-person tech rep territories by larger territories served by multiple tech reps.
• Approach Suggested by the Vice President for Marketing: Give owners of the new printer-copier priority for receiving repairs over the company’s other customers.
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Priority Queueing Models
• General Assumptions:• There are two or more categories of customers. Each category is assigned to a
priority class. Customers in priority class 1 are given priority over customers in priority class 2. Priority class 2 has priority over priority class 3, etc.
• After deferring to higher priority customers, the customers within each priority class are served on a first-come-fist-served basis.
• Two types of priorities• Nonpreemptive priorities: Once a server has begun serving a customer, the
service must be completed (even if a higher priority customer arrives). However, once service is completed, priorities are applied to select the next one to begin service.
• Preemptive priorities: The lowest priority customer being served is preempted (ejected back into the queue) whenever a higher priority customer enters the queueing system.
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Preemptive Priorities Queues
• Additional Assumptions1. Preemptive priorities are used as previously described.2. For priority class i (i = 1, 2, … , n), the interarrival times of the customers
in that class have an exponential distribution with a mean of 1/li.3. All service times have an exponential distribution with a mean of 1/m,
regardless of the priority class involved.4. The queueing system has a single server.
• The utilization factor for the server is
r = (l1 + l2 + … + ln) / m
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Non-Preemptive Priorities Queues
• Additional Assumptions1. Nonpreemptive priorities are used as previously described.2. For priority class i (i = 1, 2, … , n), the interarrival times of the customers
in that class have an exponential distribution with a mean of 1/li.3. All service times have an exponential distribution with a mean of 1/m,
regardless of the priority class involved.4. The queueing system can have any number of servers.
• The utilization factor for the servers is
r = (l1 + l2 + … + ln) / sm
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Priority Queueing Models
• VP of Marketing Approach (Priority for New Copiers)• The proposed new service standard is that the average waiting time before
service begins be two hours (i.e., Wq ≤ 1/4 day).• The Vice President of Marketing has proposed giving the printer-copiers
priority over other machines for receiving service. The rationale for this proposal is that the printer-copier performs so many vital functions that its owners cannot tolerate being without it as long as other machines.
• The mean arrival rates for the two classes of copiers are• l1 = 1 customer (printer-copier) per workday (now)• l2 = 2 customers (other machines) per workday (now)
• The proportion of printer-copiers is expected to increase, so in a couple years• l1 = 1.5 customers (printer-copiers) per workday (later)• l2 = 1.5 customers (other machines) per workday (later)
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Priority Queueing Models
• VP of Marketing Approach (Priority for New Copiers)• Nonpreemptive Priorities Model for VP of Marketing’s Approach (Current Arrival
Rates)
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1011121314151617
B C D E F GData
n = 2 (# of priority classes) 4 (mean service rate)s = 1 (# servers)
i L Lq W Wq
Priority Class 1 1 0.5 0.25 0.5 0.25Priority Class 2 2 2.5 2 1.25 1Priority Class 3 1 #DIV/0! #DIV/0! #DIV/0! #DIV/0!Priority Class 4 1 #DIV/0! #DIV/0! #DIV/0! #DIV/0!Priority Class 5 1 1.75 1.5 1.75 1.5
3 0.75
Results
Priority Queueing Models
• VP of Marketing Approach (Priority for New Copiers)• Nonpreemptive Priorities Model for VP of Marketing’s Approach (Future Arrival
Rates)
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9
1011121314151617
B C D E F GData
n = 2 (# of priority classes) 4 (mean service rate)s = 1 (# servers)
i L Lq W Wq
Priority Class 1 1.5 0.825 0.45 0.55 0.3Priority Class 2 1.5 2.175 1.8 1.45 1.2Priority Class 3 1 #DIV/0! #DIV/0! #DIV/0! #DIV/0!Priority Class 4 1 #DIV/0! #DIV/0! #DIV/0! #DIV/0!Priority Class 5 1 1.75 1.5 1.75 1.5
3 0.75
Results
Priority Queueing Models
• Expected Waiting Times with Nonpreemptive Priorities
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s When l1 l2 m r Wq for Printer Copiers Wq for Other Machines
1 Now 1 2 4 0.75 0.25 workday (2 hrs.) 1 workday (8 hrs.)
1 Later 1.5 1.5 4 0.75 0.3 workday (2.4 hrs.) 1.2 workday (9.6 hrs.)
2 Now 2 4 4 0.75 0.107 workday (0.86 hr.) 0.439 workday (3.43 hrs.)
2 Later 3 3 4 0.75 0.129 workday (1.03 hrs.) 0.514 workday (4.11 hrs.)
3 Now 3 6 4 0.75 0.063 workday (0.50 hr.) 0.252 workday (2.02 hrs.)
3 Later 4.5 4.5 4 0.75 0.076 workday (0.61 hr.) 0.303 workday (2.42 hrs.)
Outline
• Queueing System Introduction• Queueing System Elements
• Arrivals, Queue, Service• Performance Measures
• Waiting time, Queue Length, Standards/Service Levels
• Single Server Queues (M/M/1, M/G/1)• Multiple Server Queues (M/M/s)• Priority Queues• Economic Analysis• Chapter 11
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Economic Analysis
• In many cases, the consequences of making customers wait can be expressed as a waiting cost.
• The manager is interested in minimizing the total cost.TC = Expected total cost per unit timeSC = Expected service cost per unit timeWC = Expected waiting cost per unit time
The objective is then to choose the number of servers so as toMinimize TC = SC + WC
• When each server costs the same (Cs = cost of server per unit time),SC = Cs s
• When the waiting cost is proportional to the amount of waiting (Cw = waiting cost per unit time for each customer),
WC = Cw L
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Economic Analysis
Acme Machine Shop• The Acme Machine Shop has a tool crib for storing tool required by shop
mechanics.
• Two clerks run the tool crib.
• The estimates of the mean arrival rate l and the mean service rate (per server) m are
l = 120 customers per hourm = 80 customers per hour
• The total cost to the company of each tool crib clerk is $20/hour, so Cs = $20.
• While mechanics are busy, their value to Acme is $48/hour, so Cw = $48.
• Choose s so as to Minimize TC = $20s + $48L.
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Economic Analysis
Acme Machine Shop
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1011121314151617181920
B C D E F GData Results
120 (mean arrival rate) L = 1.736842105 80 (mean service rate) Lq = 0.236842105s = 3 (# servers)
W = 0.014473684Pr(W > t) = 0.02581732 Wq = 0.001973684
when t = 0.05 0.5
Prob(W q > t) = 0.00058707when t = 0.05 n Pn
0 0.2105263161 0.315789474
Cs = $20.00 (cost / server / unit time) 2 0.236842105Cw = $48.00 (waiting cost / unit time) 3 0.118421053
4 0.059210526Cost of Service $60.00 5 0.029605263Cost of Waiting $83.37 6 0.014802632
Total Cost $143.37 7 0.007401316
Economic Analysis:
Economic Analysis
Acme Machine Shop
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H I J K L M N
Data Table for Expected Total Cost of Alternatives
Cost of Cost of Totals r L Service Waiting Cost
0.50 1.74 $60.00 $83.37 $143.371 1.50 #N/A $20.00 #N/A #N/A2 0.75 3.43 $40.00 $164.57 $204.573 0.50 1.74 $60.00 $83.37 $143.374 0.38 1.54 $80.00 $74.15 $154.15
5 0.30 1.51 $100.00 $72.41 $172.41
$0
$50
$100
$150
$200
$250
0 1 2 3 4 5
Number of Servers (s)
Cos
t ($/
hour
)
Cost ofService
Cost ofWaiting
Total Cost
Further Study
• Read Chapter 11• Practice problems
• 11.6, 11.7, 11.8, 11.15, 11.16, 11.23, 11.27
• The following problems are in Homework 3:• 11.9, 11.13
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