Reference books:
Anderson, Sweeney, and Williams, AN INTRODUCTION TO MANAGEMENT
SCIENCE, QUANTITATIVE APPROACHES TO DECISION MAKING, 7th edition,
West Publishing Company,1994
Hamdy A. Taha, OPERATIONS RESEARCH, AN INTRODUCTION, 5th edition,
Maxwell Macmillan International, 1992
Burley and O’Sullivan, OPERATIONAL RESEARCH, MacMillan, 1986
Lecture 1
Queuing Models
Imagine the following situations:
•
•
•
•
•
•
Shoppers waiting in front of a check-out stand in a supermarket;
Cars waiting at a red traffic light;
Patients waiting in a lounge of a medical centre;
Planes waiting for taking off in an airport;
Customers waiting for tables in Pizza Hut (especially at weekends);
……
What these situations have in common is the phenomenon of waiting. Waiting causes
inconvenience, but like it or not, it is part of our daily life. All we hope to achieve is
to reduce the inconvenience to bearable levels.
The waiting phenomenon is the direct result of randomness in the operation service
facilities. In general, the customer’s arrival and service time are not known in
advance. Our objective in studying the operation of a service facility under random
conditions is to secure some characteristics that measure the performance of the
system under study. Some of the characteristics that are of interest are the following:
1) The probability that there are no customers in the system
2) The average number of customers in the queue
3) The average number of customers in the system (the number in queue + the
number being served)
4) The average time a customer spends in the queue
5) The average time a customer spends in the system (the waiting time plus
the service time)
6) The probability that an arriving customer has to wait for service
7) The probability of n customers in the system
Managers who have the above information are better able to make decisions that
balance desirable service levels against the cost of providing the service.
1. The Structure of a Queue
1
As customers arrive at a facility, they join a queue. The server chooses a customer
from the queue to begin service. Upon the completion of a service, the process of
choosing a new waiting customer is repeated. This is how a queuing situation is
created. It is assumed that no time is lost between the completion of a service and the
admission of a new customer into the facility. A simple queuing system is illustrated
in Figure 1.
This diagram depicts a single channel queue, i.e., all customers entering the system
must pass through the one channel. When more customers arrives at the facility, It is
necessary the system have multiple channel queue to serve the customers.
System
Customer arrivals
Queue
Customer leaves
Server
Figure 1. A single channel queue system
Obviously, the main “actors” in a queuing system are the customers and the server.
The elements that are of most interest in the queuing system are the pattern in which
the customers arrive, the service time per customer, and the queue discipline.
The process of arrivals
Defining the arrival process for a queue involves determining the probability
distribution for the number of arrivals in a given period of time. In many queuing
situations the arrivals occur in a random fashion; that is, each arrival is independent
of other arrivals, and we cannot predict when an arrival will occur. In such cases, the
Poisson probability distribution is used to describe the arrival pattern.
Using Poisson probability function, the probability of x arrivals in a specific time
period is defined as follows:
P( x ) =
λx e −λ
x!
(for x = 0, 1, 2…)
where
x = the number of arrivals in the period
λ = the average or mean number of arrivals per period
e = 2.71828
Example:
2
(1)
The North-west Fried Chicken has analysed data on customer arrivals and has
concluded that the mean arrival rate is 45 customers per hour. For a 1-minute
period, the mean number of arrival would be λ = 45/60 = 0.75 arrivals per
minute.
We can use the following probability function to compute the probability of x
arrivals during a 1-minute period:
0.75 x e −0.75
P( x ) =
x!
The Poisson probability distribution during a 1-minute period is listed as
follows:
x
P(x)
0
1
2
3
4
≥5
0.4724
0.3543
0.1329
0.0332
0.0062
0.0010
The distribution of service times
The service time is the time the customer spends at the service facility once the
service has started. Service time normally varies according to the individual
situations. In the North-west Fried Chicken example, each customer’s order may be of
different sizes. It may take 1 minute to fill a small order, but 2 or 3 minutes to fill a
larger order.
It has been determined that the exponential probability distribution often provides a
good approximation of service times in queuing situations. If the probability
distribution for service times follows an exponential probability distribution, the
probability that the service time will be less than or equal to a time of length t is
given by
P(service time ≤ t) = 1 - e-μt
where
(2)
μ = the average or mean number of customers that can be served per time
period
Example:
Suppose that North-west Fried Chicken has studied the order-taking and
order- filling process and has found that the single server can process an
average of 60 customer orders per hour. On a 1-minute basis, the average or
mean service rate would be μ = 60/60 = 1 customer per minute. This makes
the equation (2) become
3
P(service time ≤ t) = 1 - e-t
for this case.
For example
P(service time ≤ 0.5 min.) = 1 - e-1(0.5) = 1 - 0.6065 = 0.3935
P(service time ≤ 1.0 min.) = 1 - e-1(1.0) = 1 - 0.3679 = 0.6321
P(service time ≤ 2.0 min.) = 1 - e-1(2.0) = 1 - 0.1353 = 0.8647
Queue discipline
In describing a queuing system, we must define the manner in which the queuing
customers are arranged for service. The possible queue disciplines are
1)
2)
3)
4)
First-in-first-out (FIFO) or first-come-first-served (FCFS)
Last-in-first-served (LIFO)
Service-in-random-order (SIRO)
Others
The FIFO or FCFS queue discipline applies to many situations.
Steady-state operation
A queuing system gradually builds up to a normal or steady state from its beginning.
The start-up period is known as the transient period. This period ends when the
system reaches the normal or steady-state operation. Queuing models describe the
steady-state operating characteristics of the queue.
Lecture 2
2. The Single-channel Queue Model with Poisson Arrivals and Exponential Service
Times
This is the simplest queuing model and it assumes the following:
1)
2)
3)
4)
The queue has a single channel
The pattern of arrivals follows a Poisson probability distribution
The service times follow an exponential probability distribution
The queue discipline is FCFS
The formulas that can be used to develop the steady-state operating characteristics in
this situation are given below, with
4
λ = the mean or average number of arrivals per time period (the mean arrival
rate)
μ = the mean or average number of services per time period (The mean service
rate)
• The probability that there are no customers in the system
P0 = 1 - λ/μ
(3)
• The average number of customers in the queue
Lq = λ2/[μ(μ-λ)]
(4)
• The average number of customers in the system (the number in queue + the
number being served)
L = Lq + λ/μ
(5)
• The average time a customer spends in the queue
Wq = Lq/λ
(6)
• The average time a customer spends in the system (the waiting time plus the
service time)
W = Wq + 1/μ
(7)
• The probability that an arriving customer has to wait for service
Pw = λ/μ
(8)
• The probability of n customers in the system
Pn = (λ/μ)n P0
(9)
Formula (8) provides the probability the service facility is busy, and thus the ratio λ/μ
is often referred to as the utilisation factor for the service facility.
The above formulas are applicable only when the mean service rate μ is greater than
the mean arrival rate λ, i.e., when λ/μ < 1. If this condition does not exist, the queue
will grow without limit since the service facility does not have sufficient ability to
handle the arriving customers. Therefore, to use these formulas, we must have μ > λ.
Example:
Recall that for the North-west Fried Chicken problem we had a mean arrival
rate of λ = 0.75 customers per minute and a mean service rate of μ = 1
5
customers per minute. Obviously the condition λ/μ < 1 is satisfied. Then we
can use the formulae to calculate the operation characteristics for North-west
Fried Chicken single queue:
P0 = 1 - λ/μ = 1 - 0.75/1 = 0.25
Lq = λ2/[μ(μ-λ)] = 0.752/[(1(1-0.75)] = 2.25 customers
L = Lq + λ/μ = 2.25 + 0.75/1 = 3 customers
Wq = Lq/λ = 2.25/0.75 = 3 minutes
W = Wq + 1/μ = 3 + 1/1 = 4 minutes
Pw = λ/μ = 0.75/1 = 0.75
Equation Pn = (λ/μ)n P0 can be used to determine the probability of any number of
customers in the system. Some computation results are listed in the following table.
Numbers of
Customers
Probability
0
1
2
3
4
5
6
≥7
0.2500
0.1875
0.1406
0.1055
0.0791
0.0593
0.0445
0.1335
Looking at the results of the single-channel queue for the North-west Fried Chicken,
we can learn several important things about the operation of the queuing system:
• Customers wait an average of 3 minutes before beginning to place an
order, which is somewhat long for a business based on fast service;
• the average number of customers waiting in line is 2.25 and 75% of the
arriving customers have to wait for service; this indicates something needs
to be done to improve the efficiency of the queue operation;
• There is a 0.1335 probability that seven or more customers are in the
queuing system at one time, which indicates a fairly high probability that
the North-west Fried Chicken will periodically experience some very long
queues if it continues to use the single channel operation.
If the operating characteristics are unsatisfactory in terms of meeting companies
standards for service, something will need to be done to improve the queue operation.
To reduce the waiting time mainly means to improve the services rate. The
service improvements can be made by either
6
• increasing the mean service rate μ by making a creative change or by
employing new technology; or
• adding parallel service channels so that more customers can be served at a
time.
Example:
Suppose that the North-west Fried Chicken management decides to increase
the mean service rate μ by employing an order filler who will assist the order
taker at the cash machine. By this change, the management estimates that the
mean service rate can be increased from the current 60 customers/hour to 75
customers/hour. The mean service rate for the revised system is μ = 75/60 =
1.25 customers per minute.
Then using λ = 0.75 and μ = 1.25, the characteristics of the revised system are
calculated as follows:
Operating Characteristics
Probability of no customers in the system
Average number of customers in the queue
Average number of customers in the system
Average time a customer spends in the queue
Average time a customer spends in the system
Probability of an arriving customer has to wait for service
Probability of 7 or more customers in the system
Original
0.2500
2.2500
3.000
3.000 min.
4.000 min.
0.75
0.1335
Revised
0.400
0.900
1.500
1.200 min.
2.000 min.
0.600
0.028
Comparison between the original system and the revised system shows the service of
the North-west Fried Chicken has been greatly improved due to the action taken.
Lecture 3
3. The Multiple-channel Queuing Model with Poisson Arrivals and Exponential
Service times
A multiple-channel queue consists of two or more channels or service locations that
are assumed to be identical in terms of service capability. In the multiple channel
system, arriving customers wait in a single queue and then move to the first available
channel to be served (such as the queue waiting to pass the security check in
Manchester airport). A 3-channel queuing system is depicted in Figure 2.
7
System
Server1
Customer leaves
Customer arrivals
Server2
Customer leaves
Server3
Customer leaves
Queue
Figure 2. A three channel queue system
Assumptions for the queue in the multiple-channel system:
• The queue has two or more channels
• The pattern of arrivals follows a Poisson probability distribution
• The service time for each channel follows an exponential probability
distribution
• The mean service time μ is the same for each channel
• The arrivals wait in the single queue and then move to the first open
channel for service
• The queue discipline is first-come, first-served (FCFS)
Notations:
λ = the mean arrival rate for the system
μ = the mean service rate for each channel
k = the number of the channels
Formulas:
1) The probability that there are no customers in the system
P0 =
1
k
⎛
⎞
∑ ( λ /nμ! ) + ( λ /kμ! ) ⎜⎝ kμkμ− λ ⎟⎠
n= 0
k −1
n
2) The average number of customers in the queue
8
(10)
( λ / μ ) k λμ
Lq =
P
( k − 1)!( kμ − λ ) 2 0
(11)
3) The average number of customers in the system
L = Lq + λ/μ
(12)
4) The average time a customer spends in the queue
Wq = Lq/λ
(13)
5) The average time a customer spends in the system
W = Wq + 1/μ
(14)
6) The probability that an arriving customer has to wait for service
1 ⎛λ⎞
Pw = ⎜ ⎟
k !⎝ μ ⎠
k
⎛ kμ ⎞
⎜
⎟P
⎝ kμ − λ ⎠ 0
(15)
7) The probability of n customers in the system
Pn =
( λ / μ )n
P0
n!
for n ≤ k
(16)
Pn =
( λ / μ )n
P
k ! k ( n−k ) 0
for n > k
(17)
While some of the formulas for the operating characteristics of multiple-channel
queues are more complex than their single-channel counterparts, the expressions
provide the same information. The following table is provided to simplify the use of
the formulas.
Values of P0 for multiple-channel queue with
Poisson arrivals and exponential service time
Ratio λ/μ
0.15
0.20
0.25
0.30
0.35
0.40
0.45
0.50
0.55
0.60
0.65
0.70
0.75
0.80
0.85
k=2
0.8605
0.8182
0.7778
0.7391
0.7021
0.6667
0.6327
0.6000
0.5686
0.5385
0.5094
0.4815
0.4545
0.4286
0.4035
k=3
0.8607
0.8187
0.7788
0.7407
0.7046
0.6701
0.6373
0.6061
0.5763
0.5479
0.5209
0.4952
0.4706
0.4472
0.4248
k=4
0.8607
0.8187
0.7788
0.7408
0.7047
0.6703
0.6376
0.6065
0.5769
0.5487
0.5219
0.4965
0.4722
0.4491
0.4271
9
k=5
0.8607
0.8187
0.7788
0.7408
0.7047
0.6703
0.6376
0.6065
0.5769
0.5488
0.5220
0.4966
0.4724
0.4493
0.4274
0.90
0.95
1.00
1.20
1.40
1.60
1.80
2.00
2.20
2.40
2.60
2.80
3.00
3.20
3.40
3.60
3.80
4.00
4.20
4.40
4.60
4.80
0.3793
0.3559
0.3333
0.2500
0.1765
0.1111
0.0526
0.4035
0.3831
0.3636
0.2941
0.2360
0.1872
0.1460
0.1111
0.0815
0.0562
0.0345
0.0160
0.4062
0.3963
0.3673
0.3002
0.2449
0.1993
0.1616
0.1304
0.1046
0.0831
0.0651
0.0521
0.0377
0.0273
0.0186
0.0113
0.0051
0.4065
0.3867
0.3678
0.3011
0.2463
0.2014
0.1646
0.1343
0.1094
0.0889
0.0721
0.0581
0.0466
0.0372
0.0293
0.0228
0.0174
0.0130
0.0093
0.0063
0.0038
0.0017
Example:
Consider the North-west Fried Chicken situation again. If the management
decided to add a parallel service channel to improve the service rather than to
increase the mean service rate μ, let’s evaluate the operating characteristics for
the two-channel system.
We know that k = 2, λ = 0.75, and μ = 1. So,
P0 = 0.4545 (from table)
Lq = 0.1227 customer
L = 0.8727 customer
Wq = 0.16 min.
W = 1.16 min.
Pw = 0.2045
Pn≥7 = 0.0017
Operating Characteristics
Probability of no customers in the system
Average number of customers in the queue
Average number of customers in the system
Average time a customer spends in the queue
Average time a customer spends in the system
Probability of an arriving customer has to wait for service
Probability of 7 or more customers in the system
Original
0.2500
2.2500
3.000
3.000 min.
4.000 min.
0.75
0.1335
Revised 1
0.400
0.900
1.500
1.200 min.
2.000 min.
0.600
0.028
Revised 2
0.4545
0.1227
0.8727
0.16 min.
1.16 min.
0.2045
0.0017
It is clear that the two-channel system will greatly improve the operating
characteristics of the queue.
Lecture 4
10
4. Some General Relationships for Queuing Models
In sections 2 and 3, we have come across a number of operating characteristics
including
Lq = the average number of customers in the queue
L = The average number of customers in the system
Wq = the average time a customer spends the queue
W = the average time a customer spends in the system
John D. C. Little showed that general relationships existing among these four
characteristics apply to a variety of different queues. Two of the relationships,
referred to as Little’s flow equation, are as follows:
L = λW
Lq = λWq
(18)
(19)
Another general relationship is the expression for the average time in the system:
W = Wq + 1/μ
The importance of these relations is that they apply to any queuing models regardless
the arrival patterns and the service time probability distribution.
5. Economic Analysis of Queues
For a queuing system, the manager has to decide at what level the system should be
run so that it is most economical. In other word, the manager has to make sure that the
total cost for operating the queuing system is minimised.
In queuing systems, the total cost comes from two directions, i.e. the waiting cost and
the service cost. Let’s use the following notations:
cw = the waiting cost per time period for each customer
L = the average number of customers in the system per time period
cs = the service cost per time period for each channel
k = the number of channels
TC = the total cost
Then the total cost is
TC = cwL + csk
(20)
A critical issue in conducting an economic analysis of a queue is to obtain reasonable
estimates of the waiting cost and service cost per period. Of these two, the waiting
cost per period is usually more difficult to evaluate. This is not a direct cost to a
company. However, if this cost is ignored and the queue is allowed to grow,
customers will ultimately take their business to elsewhere. In this way, the company
11
will experience lost sales, and incur a cost. The service cost is generally easier to
determine. This cost is the relevant cost associated with operating each service
channel.
Example:
Assume that in the case of Northwest Fried Chicken, the waiting cost rate is
£10 per hour for customer waiting time and the service cost rate is £7 per
hour. Calculate the total cost for single-channel and two-channel queuing
system:
Single-channel system:
We knew that the average number of customers in the system L = 3. So
TC = £10 (3) + £7 (1) = £37.00 per hour
Two-channel system:
L = 0.8727
TC = £10 (0.8727) + £7 (2) = £22.73 per hour
Thus, given the cost data provided by the North-west Fried Chicken, the two-channel
system provides the more economical operation.
The general shapes of the cost curves in the economic analysis of queues are show in
Figure 3.
Total
Cost per
Hour
Total cost
Service cost
Waiting cost
Figure 3
Number of Channels (k)
12
4. Other queuing Models
D G Kendall suggested a notation that is helpful in classifying the wide variety of
different queuing models that have been developed. The three-symbol notation
Kendall notation is as follows:
A/B/s
where
A denotes the probability distribution for the number of arrivals
B denotes the probability distribution for the service time
s denotes the number of channels
The values for A and B that are commonly used are as follows:
M
designates a Poisson probability distribution for the number of arrivals
or an exponential probability distribution for service time
D
designates that the number of arrivals or service time is deterministic
or constant
G
designates a general distribution with a known mean and variance for
the number of arrivals or service time
Using the Kendall notation, the single-channel queuing model with Poisson arrivals
and exponential service times is classified as an M/M/1 model. The two-channel
queuing model with Poisson arrivals and exponential service times would be
classified as an M/M/2 model.
We will not cover the other queuing models because of time limitation and the
similarity of the equations for the operating characteristics. These equations can be
found in the listed reference books.
13
Exercise 3
Queuing Models
1. The reference desk of a university library receives requests for assistance. Assume
that a Poisson probability distribution with a mean rate of 10 requests per hour can
be used to describe the arrival pattern and that service times follow the exponential
probability distribution with a mean service rate of 12 request per hour.
a.
b.
c.
d.
What is the probability that there are no requests for assistance in the system?
What is the average number of requests that will be waiting for service?
What is the average waiting time in minutes before the service begins?
What is the average time at the reference desk in minutes (waiting time +
service time)?
e. What is the probability that a new arrival has to wait for service?
2. Trucks using a single-channel loading dock arrive according to Poisson probability
distribution. The time required to load/unload follows the exponential probability
distribution. The mean arrival rate is 12 trucks per day, and the mean service rate is
18 trucks per day.
a.
b.
c.
d.
What is the probability that there are no trucks in the system?
What is the average number of trucks waiting for service?
What is the average time a truck waits for loading/unloading service to begin?
What is the probability that a new arrival will have to wait?
3. Marty’s Barber Shop has one barber. Customers arrive at the rate of 2.2 customers
per hour, and haircuts are given at the average rate of 5 per hour. Use the Poisson
arrivals and exponential service times model to answer the following questions:
a. What is the probability that there are no customers in the system?
b. What is the probability that 1 customer is receiving a haircut and no one is
waiting?
c. What is the probability that 1 customer is receiving a haircut and 1 customer is
waiting?
d. What is the probability that 1 customer is receiving a haircut and 2 customers
are waiting?
e. What is the probability that more than two customers are waiting?
f. What is the average time a customer waits for service?
4. Trosper Tire Company has decided to hire a new mechanic to handle all tire
changes for customers ordering a new set of tires. Two mechanics have applied for
the job. One mechanic has limited experience and can be hired for $14 per hour. It
is expected that this mechanic can serve an average of 3 customers per hour. The
other mechanic has several years of experience. This mechanic can service a
average of 4 customers per hour, but must be paid $20 per hour. Assume the
customers arrive at the Trosper garage at the rate of 2 customers per hour.
14
a. Compute the waiting line characteristics for each mechanic, assuming Poisson
arrivals and exponential service times.
b. If the company assigns a customer waiting cost of $30 per hour, which
mechanic provides the lowest operating cost?
5. Consider a two-channel waiting line with Poisson arrivals and exponential service
times. The mean arrival rate 14 units per hour, and the mean service rate is 10 units
per hour for each channel.
a.
b.
c.
d.
e.
What is the probability that there are no units in the system?
What is the average number of units in the system?
What is the average time a unit waits for service?
What is the average time a unit is in the system?
What is the probability of having to wait for service?
6. Refer to Problem 5. Assume that the system is expanded to a three-channel
operation.
a. Compute the operating characteristics for this waiting line system.
b. If the service goal is to provide sufficient capacity so that at most 25% of the
customers have to wait for service, is the two- or three-channel system
preferred?
7. Patients arrive at a dentist’s office at a mean rate of 2.8 patients per hour. The
dentist can service patients at the mean rate of 3 patient per hour. A study of
patient waiting times shows that, on average, a patient waits 30 minutes before
seeing the dentist.
a. What are the mean arrival rate and mean service rate in terms of patients per
minute?
b. What is the average number of patients in the waiting room?
c. If a patient arrives at 10:19 am. What is the expected time the patient will leave
the office?
8. The City Beverage Drive-Thru is constructing a two channel service system. Cars
arrive according to the Poisson probability distribution, with a mean arrival rate of
6 cars per hour. The service times have an exponential probability distribution,
with a mean service rate of 10 cars per hour for each channel.
a.
b.
c.
d.
e.
What is the probability that there are no cars in the system?
What is the average number of cars waiting for service?
What is the average time waiting for service?
What is the average time in the system?
What is the probability that an arrival will have to wait for service?
15