Queueing Theory
Presented by
Dr. B.B. Upadhyay
Assistant Professor,
Department of Mathematics.
Indian Institute of Technology Patna
Patna, Bihar-801106
Contents I
1
Introduction
2
Queueing system
3
Elements of queueing system
4
Operating characteristics of queueing system
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Introduction
Introduction
A flow of customers from infinite/finite population towards the service
facility forms a queue (waiting line) on account of lack of capability
to serve them all at a time.
For example
1
Persons waiting at a doctor’s clinic.
2
Persons waiting at railway booking office.
3
Machines waiting to be repaired.
4
Ships in the harbour waiting to be unloaded.
5
Letters arriving at a typist’s desk.
In the absence of a perfect balance between the service facilities and
the customers, waiting is required either of the service facilities or for
the
customer’s arrival.
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Introduction
Note.
1 By the term ’customer’ we mean the arriving unit that requires some
service to be performed. The customer may be of persons, machines,
vehicles, parts, etc.
2
Queues (waiting line) stands for a number of customers waiting to
be serviced. The queue does not include the customer being serviced.
3
The process or system that performs the services to the customer is
termed by service channel or service facility.
4
The subject of queueing is not directly concerned with
optimization (maximization or minimization). Rather, it attempts
to explore, understand, and compare various queueing situations and
thus indirectly achieves optimization approximately.
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Queueing system
Queueing system
The mechanism of a queueing process is very simple.
Customers arrive at a service counter and are attended to by one or
more of the servers. As soon as a customer is served, it departs from
the system.
Thus a queueing system can be described as consisting of customers
arriving for service, waiting for service if it is not immediate, and
leaving the system after being served.
The general framework of a queueing system is shown in following
figure:
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Queueing system
Continued.
Queueing system
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Elements of queueing system
Elements of queueing system
The basic elements of a queueing system are as follows:
1. Input (or Arrival) Process
This element of queueing system is concerned with the pattern in which the
customers arrive for service.
Input source can be described by following three factors:
(a) Size of the queue.
If the total number of potential customers requiring service are only few,
then size of the input source is said to be finite.
On the other hand, if potential customers requiring service are sufficiently
large in number, then the input source is considered to be infinite.
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Elements of queueing system
Continued.
The customers may arrive at the service facility in batches of fixed
size or of variable size or one by one.
In the case when more than one arrival is allowed to enter the system
simultaneously (entering the system does not necessarily mean
entering into service), the input is said to occur in bulk or in batches.
Ships discharging cargo at a dock, families visiting restaurants, etc.
are the examples of bulk arrivals.
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Elements of queueing system
(b) Pattern of arrivals.
Customers may arrive in the system at known (regular or otherwise)
times, or they may arrive in a random way.
In case the arrival times are known with certainty, the queueing
problems are categorized as deterministic models.
If the time between successive arrivals (inter-arrival times) is
uncertain, the arrival pattern is measured by either mean arrival rate
or inter-arrival time.
These are characterised by the probability distribution associated with
this random process.
The most common stochastic queueing models assume that arrival
rate follow a Poisson distribution and/or the inter-arrival times follow
an exponential distribution.
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Elements of queueing system
(c) Customers’ behaviour.
It is also necessary to know the reaction of a customer upon entering
the system.
A customer may decide to wait no matter how long the queue
becomes (patient customer), or if the queue is too long to suit him,
may decide not to enter it (impatient customer).
For impatient customers,
1
If a customer decides not to enter the queue because of its length, he is
said to have balked.
2
If a customers enters the queue, but after some time loses patience and
decides to leave, then he is said to have reneged.
If a customer moves from one queue to another (providing
similar/different services) for his personal economic gains, then he is
said
to have jockeyed forQueueing
position.
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3
Elements of queueing system
The final factor to be considered regarding the input process is the
manner in which the arrival pattern changes with time.
The input process which does not change with time is called a
stationary input process.
If it is time dependent then the process is termed as transient.
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Elements of queueing system
2. Queue Discipline
It is a rule according to which customers are selected for service when
a queue has been formed.
The most common queue discipline is the ”first come, first served”
(FCFS), or the “first in, first out” (FIFO) rule under which the
customers are serviced in the strict order of their arrivals.
Other queue discipline include: “last in, first out” (LIFO) rule
according to which the last arrival in the system is serviced first.
Under priority discipline, the service is of two types:
a
Pre-emptive priority.
Under this rule, the customers of high priority are given service over the
low priority customers.
That is, lower priority customer’s service is interrupted (pre-empted )
to start service for a priority customer.
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The interrupted service is resumed again as soon as the highest
priority customer has been served.
b
Non pre-emptive priority: In this case the highest priority customer
goes ahead in the queue, but his service is started only after the
completion of the service of the currently being served customer.
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3. Service Mechanism.
The service mechanism is concerned with service time and service
facilities.
Service time is the time interval from the commencement of service
to the completion of service.
If there are infinite number of servers then all the customers are
served instantaneously on arrival and there will be no queue.
If the number of servers is finite, then the customers are served
according to a specific order.
Further, the customers may be served in batches of fixed size or of
variable size rather than individually by the same server, such as a
computer with parallel processing or people boarding a bus.
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Elements of queueing system
Continued.
In the case of parallel channels ”fastest server rule” (FSR) is adopted.
For its discussion we suppose that the customers arrive before parallel
service channels.
If only one service channel is free, then incoming customer is assigned
to free service channel.
But it will be more efficient to assume that an incoming customer is
to be assigned a server of largest service rate among the free ones.
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Elements of queueing system
Service facilities can be of the following types:
a
Single queue-one server, i.e., one queue -one service channel,
wherein the customer waits till the service point is ready to take him
in for servicing.
b
Single queue-several servers wherein the customers wait in a single
queue until one of the service channels is ready to take them in for
servicing.
c
Several queues-one server wherein there are several queues and
the customer may join any one of these but there is only one service
channel.
d
Several servers. When there are several service channels available to
provide service, much depends upon their arrangements. They may be
arranged in parallel or in series or a more complex combination of
both, depending on the design of the system’s service mechanism.
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Elements of queueing system
Note
i
By parallel channels, we mean a number of channels providing
identical service facilities.
ii
Further, customers may wait in a single queue until one of the service
channels is ready to serve, as in a barber shop where many chairs are
considered as different service channels; or customers may form
separate queues in front of each service channel as in the case of
super markets.
iii
For series channels, a customer must pass through all the service
channels in sequence before service is completed. The situations may
be seen in public offices where parts of the service are done at
different service counters.
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Elements of queueing system
4. Capacity of the System
The source from which customers are generated may be finite or
infinite.
A finite source limits the customers arriving for service. i.e., there is a
finite limit to the maximum queue size.
The queue can also be viewed as one with forced balking where a
customer is forced to balk if he arrives at a time when queue size is at
its limit.
Alternatively, an infinite source is forever ”abundant” as in the case of
telephone calls arriving at a telephone exchange.
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Operating characteristics of queueing system
Operating characteristics of queueing system
Some of the operational characteristics of a queueing system, that are
of general interest for the evaluation of the performance of an existing
queueing system and to design a new system are as follows:
1
Expected number of customers in the system denoted by E (n) or L is
the average number of customers in the system, both waiting and in
service. Here, n stands for the number of customers in the queueing
system.
2
Expected number of customers in the queue denoted by E (m) or Lq is
the average number of customers waiting in the queue. Here
m = n − 1, i.e., excluding the customer being served.
3
Expected waiting time in the system denoted by E (v ) or W is the
average total time spent by a customer in the system. It is generally
taken to be the waiting time plus servicing time.
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Continued.
4 Expected waiting time in queue denoted by E (w ) or Wq is the
average time spent by a customer in the queue before the
commencement of his service.
5
The server utilization factor (or busy period) denoted by P(= µλ ) is
the proportion of time that a server actually spends with the
customers. Here, λ stands for the average number of customers
arriving per unit of time and µ stands for the average number of
customers completing service per unit of time. The server utilization
factor is also known as traffic intensity or the clearing ratio.
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DETERMINISTIC QUEUEING SYSTEM
A queueing system wherein the customers arrive at regular intervals
and the service time for each customer is known and constant, is
known as a deterministic queueing system.
Let the customers come at the teller counter of a bank for withdrawl
every 3 minutes. Thus the interval between the arrival of any two
successive customers, that is the inter-arrival time, is exactly 3
minutes. Further, suppose that the incharge of that particular teller
takes exactly 3 minutes to serve a customer. This implies that the
arrival and service rates are both equal to 20 customers per hour. In
this situation there shall never be a queue and the incharge of the
teller shall always be busy with servicing of customers.
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Now suppose instead, that the incharge of the teller can serve 30
customers per hour. i.e., he takes 2 minutes to serve a customer and
then has to wait for one minute for the next customer to come for
service. Here also, there would be no queue, but the teller is not
always busy.
Further, suppose that the incharge of the teller can serve only 15
customers per hour, i.e., he takes 4 minutes to serve a customer.
Clearly, in this situation he would be always busy and the queue length
will increase continuously without limit with the passage of time.
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This implies that when the service rate is less than the arrival rate,
the service facility cannot cope with all the arrivals and eventually the
system leads to an explosive situation.
In such situations, the problem can be resolved by providing
additional service facilities, like opening parallel counters.
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We can summarize the above as follows:
Let the arrival rate be λ customers per unit time and the service rate
be µ customers per unit time.
Then,
i
if λ > µ the waiting line (queue) shall be formed and will increase
indefinitely; the service facility would always be busy and the service
system will eventually fail.
ii
(ii) if λ ≤ µ, there shall be no queue and hence no waiting time; the
proportion of time the service facility would be idle is 1 − µλ .
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However, it is easy to visualize that the condition of uniform arrival
and uniform service rates has a very limited practicability.
Generally, the arrivals and servicing time are both variable and
uncertain.
Thus, variable arrival rates and servicing times are the more realistic
assumptions.
The probabilistic queueing mode are based on these assumptions.
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PROBABILITY DISTRIBUTIONS IN QUEUEING SYSTEMS
It is assumed that customers joining the queueing system arrive in a
random manner and follow a Poisson distribution or equivalently the
inter-arrival times obey exponential distribution.
In most of the cases, service times are also assumed to be
exponentially distributed.
It implies that the probability of service completion in any short time
period is constant and independent of the length of time t the service
has been in progress.
Now, the arrival and service distributions for Poisson queues are
derived. The basic assumptions (axioms) governing this type of
queues are stated below:
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Axiom 1. The number of arrivals in non-overlapping intervals are
statistically independent, that is the process has independent increments.
Axiom 2. The probability of more than one arrival between time t and
time t + ∆t is o(∆t) ; that is, the probability of two or more arrivals
during the small time interval ∆t is negligible. Thus
P0 (∆t) + P1 (∆t) + o(∆t) = 1
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Axiom 3 The probability that an arrival occurs between time t and time
t + ∆t is equal to λ∆t + o(∆t). Thus
P1 (∆t) = λ∆t + o(∆t),
where λ is a constant and is independent of the total number of arrivals
upto time t, ∆t is an incremental element and o(∆t) represents the terms
such that
lim
∆t→0
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σ(∆t)
∆t
= 0.
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Distribution of Arrivals ( Pure Birth Process)
The model in which only arrivals are counted and no departure takes
place are called pure birth models.
Stated in terms of queueing, birth-death processes usually arise when
an additional customer increases the arrival (referred as birth) in the
system and decreases by departure (referred as death) of a serviced
customer from the system.
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Let Pn (t) denote the probability of n arrivals in a time interval of
length t (both waiting and in service) where n ≥ 0 is an integer.
Then Pn (t + ∆t) being the probability of n arrivals in a time interval
of length t + ∆t (making use of axiom 1) is as follows:
Pn (t + ∆t) = P{n arrivals in time t and one arrival in time ∆t}
+ P{(n − 1) arrivals in time t and one arrival in time ∆t}
+ P{(n − 2) arrivals in time t and two arriva1s in time∆t}
+ . . . + P{ no arrival in time t and n arrivals in time ∆t}
for n ≥ 1.
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Making use of axiom 2 and axiom 3, this difference equation reduces to
Pn (t + ∆t) = Pn (t)P0 (∆t) + Pn−1 (t)P1 (∆t)P1 (∆t) + o(∆t)
Pn (t)[1 − λ∆t − o(∆t)] + Pn−1 (t){λ∆t + o(∆t)} + o(∆t)
=
where the last term, o(∆t), represents the terms
Pn [(n − k) arrivals in time t and k arrivals in time ∆t ]
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Continued
The above equation can be re-written as
Pn (t + ∆t) − Pn (t) = −λ∆tPn (t) + λ∆t.Pn−1 (t) + o(∆t)
Dividing it by ∆t on both sides and then taking the limit as ∆t → 0, the
equation reduces to
d
Pn (t) = −λPn (t) + λPn−1 (t), n ≥ 1
dt
(1)
For the case when n = 0,
P0 (t + ∆t) = P0 (t)P0 (∆t) = P0 (t)[1 − λt∆t − o(∆t)]
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Rearranging the terms and then dividing on both sides by ∆t, taking the
limit as ∆t → 0, we have
d
Po (t) = −λP0 (t)
dt
(2)
To solve the n + 1 differential-difference equations given in (1) and (2), we
make use of the generating function
ϕ(z, t) =
∞
X
Pn (t)z n
n=0
in the unit circle |z| ≤ 1.
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Now multiplying the differential-difference equations given in (2) and (1)
by z 0 , z 1 , z 2 , ..., z n respectively and taking summation 0 to ∞, we get
∞
X
d
dt
n=0
Pn (t)z n = −λϕ(z, t) + λzϕ(z, t).
This can also be written as
d
ϕ(z, t) = λ(z − 1)ϕ(z, t).
dt
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An obvious solution of this differential equation is
ϕ(z, t) = Ce λ(z−1)t
where C is an arbitrary constant.
To determine the value of C, we use the initial condition that there is no
arrival by time t=0 and this gives
ϕ(z, 0) = P0 (0) +
∞
X
Pn (0)z n = 1
n=1
Now put Pn (0) = 0 for n ≥ 1. therefore C = 1 Hence
ϕ(z, t) = e λ(z−1)t
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Now,
d
ϕ(z, t)|z=0 = P1 (t),
dz
d2
ϕ(z, t)|z=0 = 2!P2 (t)
dz 2
..
.
dn
ϕ(z, t)|z=0 = n!Pn (t)
dz n
Using the value of ϕ(z, t) as given in equation (3), we get
P0 (t) = e −λt
P2 (t) =
Dr. B.B. Upadhyay
P1 (t) = (λt)e −λt
1
−λt
2! (λt)e
Pn (t) =
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n −λt
n! (λt) e
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The general formula, therefore, is
Pn (t) =
1
(λt)n e −λt , n ≥ 0
n!
which is the well-known Poisson probability law with mean λt. Thus, the
random variable defined as the number of arrivals to a system in time t,
has the Poisson distribution with a mean of λt arrivals, or a mean arrival
rate of λ.
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Example
For a Poisson arrival process with a mean 2 per unit time, if an arrival
occurred at time t = 30, what is the probability that an arrival will occur
by t = 40?
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2
Distribution of Inter-arrival Times (Exponential Process)
Inter-arrival times are defined as the time intervals between two successive
arrivals. Here, we shall show that if the arrival process follows the Poisson
distribution, an associated random variable defined as the time between
successive arrivals (inter-arrival time) follows the exponential distribution
f (t) = λe −λt and vice-versa.
Let the random variable T be the time between successive arrivals, then
P(T > t) = P(no arrival in time t) = P0 (t) = e −λt .
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The cumulative distribution function of T denoted by F (t) is given by
F (t) = P(T ≤ t) = 1 − P(t > t)
= 1 − P0 (t) = 1 − e −λt , t > 0.
The density function f (t) for inter-arrival times, therefore is
f (t) =
d
F (t) = λe −λt , t > 0.
dt
The expected (or mean) inter-arrival time is given by
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Z ∞
E (t) =
Z ∞
t.f (t)dt =
0
0
λte −λt dt =
1
.
λ
where λ is the mean arrival rate.
Thus, T has the exponential distribution with mean λ1 . We would
intuitively expect that, if the mean arrival rate is λ, then the mean time
between arrivals is λ1 .
Conversely, we can also show that if the inter-arrival times are independent
and have the same exponential distribution then the arrival rate follows
the Poisson distribution.
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3
Distribution of Departures (Pure Death Process)
The model in which only departures are counted and no other arrivals
allowed are called pure death models. The queueing system starts with N
customers at time t = 0, where N ≥ 1. Departures occur at the rate of µ
customers per unit time. To develop the differential-difference equations
for the probability of n customers remaining after ’t’ time units, Pn (t), we
make use of similar assumptions as was done for arrivals.
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Let the three axioms, given at the beginning of this section, be changed by
using the word service instead of arrival and condition the probability
statements by requiring the system to be non-empty.
Let us define:
µ∆t = probability that a customer in service at time t will complete
service during time ∆t.
For small time interval ∆t > 0, µ∆t gives probability of one departure
during ∆t. Using the same arguments as in pure birth process case, the
differential-difference equations for this can easily be obtained:
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Pn (t + ∆t) =
Pn (t){1 − µ∆t + o(∆t)}+
Pn+1 (t).{µδt + o(∆t)}, 1 ≤ n ≤ N − 1
P0 (t + ∆t) =
P0 (t) + P1 (t){µ∆t + o(∆t)}, n = 0
PN (t + ∆t) = PN (t).{1 − µ∆t + o(∆t)}, n = N
Re-arranging the above equations, dividing them by ∆t on both sides and
then taking the limits as ∆t → 0, we get
d
dt Pn (t)
= −µPn (t) + µPn+1 (t); 0 ≤ n ≤ N − 1, t > 0
d
dt P0 (t)
= µP1 (t); n = 0, t ≥ 0
d
dt PN (t)
= −µPn (t); n = N, t ≥ 0
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The solution of these equations with initial condition:
(
Pn (0) =
1; n = N ̸= 0;
0; n =
̸ N
can easily be obtained as earlier. The general solution to the above
equation so obtained is
Pn (t) =
N
X
(µt)N−n e −µt
; 1 ≤ n ≤ N and P0 (t) = 1 −
Pn (t)
(N − n)!
i=1
which is known as a truncated Poisson law.
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4
Distribution of Service Times
Making similar assumption as done above, for arrivals, one could utilize
the same type of process to describe the service pattern.
Let the three, axioms be changed by using the word service instead of
arrival and condition the probability statement by requiring the system to
be non-empty. Then we can easily show that, the time t to complete the
service on a customer follows the exponential distribution:
(
s(t) =
µe −µt ;
0;
t>0
t<0
where µ is the mean service rate for a particular service channel. This
shows that service times follows exponential distribution with mean µ1 .
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The number, n, of potential services in time T will follow the Poisson
distribution given by
ϕ(n) = P[ n service in time T, if servicing is going on throughout ]
=
(µT )N −µT
e
n!
Consequently, we can also show that
P[ no service in ∆t] = 1 − µ∆t + o(∆t)
and
Dr. B.B. Upadhyay
P [one service in ∆t] = µ∆t + o(∆t).
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CLASSIFICATION OF QUEUEING MODELS
Generally queueing model may be completely specified in the following
symbolic form:
(a/b/c) : (d/e).
The first and second symbols denote the type of distributions of
inter-arrival times and of inter-service times, respectively. Third symbol
specifies the number of servers, whereas fourth symbol stands for the
capacity of the system and the last symbol denotes the queue discipline. If
we specify the following letters as:
M = Poisson arrival or departure distribution,
Ek = Erlangian or Gamma inter-arrival for service time distribution,
GI = General input distribution,
G = General service time distribution,
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Operating characteristics of queueing system
then (M/Ek /1) : (∞/FIFO) defines a queueing system in which arrivals
follow Poisson distribution, service times are Erlangian, single server,
infinite capacity and ”first in, first out” queue discipline.
DEFINITION OF TRANSIENT AND STEADY STATES
A queueing system is said to be in transient state when its operating
characteristic (like input, output, mean queue length, etc.) are dependent
upon time.
If the characteristic of the queueing system becomes independent of time,
then the steady-state condition is said to prevail.
If Pn (t) denotes the probability that there are n customers in the system
at time t, then in the steady-state case, we have
lim Pn (t) = Pn (independent of t)
t→∞
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September 2, 2023
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Operating characteristics of queueing system
Due to practical viewpoint of the steady-state behaviour of the systems,
the present chapter is amply focused on studying queueing systems under
the existence of steady-state conditions.
However, the differntial- difference equations which can be used for
deriving transient solutions will be presented.
Dr. B.B. Upadhyay
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September 2, 2023
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Operating characteristics of queueing system
POISSON QUEUEING SYSTEMS
Queues that follow the Poisson arrivals (exponential inter-arrival time) and
Poisson services (exponential service time) are called Poisson queues. In
this section, we shall study a number of Poisson queues with different
characteristics.
Model I {(M/M/1) : (∞/FIFO)}.
This model deals with a queueing system having single service channel.
Poisson input, Exponential service and there is no limit on the system
capacity while the customers are served on a ”first in, first out” basis.
The solution procedure of this queueing model can be summarized in the
following three steps:
Dr. B.B. Upadhyay
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Operating characteristics of queueing system
Step 1:- Construction of Differential-Difference Equations. Let Pn (t) be
the probability that there are n customers in the system at time t.
The probability that the system has n customers at time (t + ∆t) can be
expressed as the sum of the joint probabilities of the four mutually
exclusive and collectively exhaustive events as follows:
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Operating characteristics of queueing system
Pn (t + ∆t)
= Pn (t).P[no arrival in ∆t].P[no service completion in ∆t]
+Pn (t).P[one arrival in ∆t].P[one service completed in ∆t]
+Pn+1 (1).P[no arrival in ∆t].P[one service completed in ∆t]
+Pn+1 (t).P[one arrival in ∆t].P[no service completion in ∆t]
This is re-written as:
Pn (t + ∆t)
= Pn (t)[1 − λ∆t + o(∆t)][1 − µ∆t + o(∆t)]
+Pn (t)[λ∆t][µ∆t]
+Pn+1 (t)[1 − λ∆t + o(∆t)][µ∆t + o(∆t)]
+Pn−1 (t)[λ∆t + o(∆t)][1 − µ∆t + o(∆t)]
or
Pn (t +∆t)−Pn (t) = −(λ+µ)∆tPn (t)+µ∆tPn+1 (t)+λ∆tPn−1 (t)+o(∆t)
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Operating characteristics of queueing system
Since ∆t is very small, terms involving (∆t)2 can be neglected. Dividing
the above equation by ∆t on both sides and then taking limit as ∆t → 0,
we get
d
Pn (t) = −(λ + µ)Pn (t) + µPn+1 (t) + λPn−1 (t); n ≥ 1
dt
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Operating characteristics of queueing system
Similarly, if there is no customer in the system at time (t + ∆t), there will
be no service. completion during ∆t. Thus for n = 0 and t ≥ 0, we have
only two probabilities instead of four. The resulting equation is
P0 (t+∆t) = P0 (t){1−λ∆t+o(∆t)}+P1 (t){µ∆t+o(t)}{1−λ∆t+o(∆t)}
or
P0 (t + ∆t) − P0 (t) = −λ∆tP0 (t) + µ∆tP1 (t) + o(∆t).
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Operating characteristics of queueing system
Dividing both sides of this equation by ∆t and then taking limit as
∆t → 0, we get
d
P0 (t) = −λP0 (t) + µP1 (t); n = 0.
dt
Step 2 Deriving the Steady-State Difference Equations. In the
steady-state, Pn (t) is independent, of time t and λ < µ when t → ∞.
Thus Pn (t) → Pn and
d
Pn (t) → 0 as t → ∞.
dt
Consequently the differential-difference equations obtained in Step 1
reduce to
0 = −(λ + µ)Pn + µPn+1 + λPn−1 ; n ≥ 1
and
0 = −λPn + µP1 ; n = 0
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Operating characteristics of queueing system
These constitute the steady-state difference equations.
Step 3. Solution of the Steady-State Difference Equations. For the
solution of the above difference equations there exist three methods,
namely, the iterative method, use of generating functions and the use of
linear operators. Out of these three the first one is the most
straightforward and therefore the solution of the above equations will be
obtained here by using the iterative method. Using iteratively, the
difference-equations yield
λ
P1 = P0 ,
µ
λ 2
λ+µ
λ
P2 =
P1 − P0 =
P0
µ
µ
µ
λ 3
λ
λ+µ
P2 − P1 =
P0
P3 =
µ
µ
µ
and in general
λ n
Pn =
P0 .
µ
Dr. B.B. Upadhyay
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Operating characteristics of queueing system
Now,
λ n+1
λ + µ λ n
λ λ n−1
P0 −
P0 =
P0 .
µ
µ
µ µ
µ
Thus, by the principle of mathematical induction, the general formulae for
Pn is valid for n ≥ 0.
Pn+1 =
To obtain the value of P0 , we make use of the boundary condition
∞
X
Pn = 1
n=0
Therefore
1 =
P∞ λ n
P ∞ λ n
P
=
P
;
0
0
n=0 µ
n=0 µ
= P0 1−1 λ ,
since Pn =
since
µ
λ
µ
n
λ
µ
P0
< 1,
This gives
P0 = 1 −
Dr. B.B. Upadhyay
λ
µ
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.
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Operating characteristics of queueing system
Hence, steady -state solution is
Pn =
λ
1−
µ
λ n µ
= ρn (1 − ρ);
ρ=
λ
µ
< 1 and n ≥ 0.
This expression gives us the probability distribution of queue length.
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Operating characteristics of queueing system
Characteristics of Model I
1
Probability of queue size being greater than or equal to k, the number
of custmors is given by
P(n ≥ k) =
P∞
=
P∞
k=n
Pk
k=n (1
− ρ)ρk
= (1 − ρ)ρn
=
P∞
k=n
ρk−n
(1−ρ)ρn
1−ρ
= ρn
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Operating characteristics of queueing system
2
Average number of customers in the system is given by
P
E (n) = ∞
n=0 nPn
=
P∞
n=0 n(1
− ρ)ρn
= ρ(1 − ρ)
P∞
= ρ(1 − ρ)
P∞
n−1
n=0 nρ
d
= ρ(1 − ρ) dρ
d n
n=0 dρ ρ
P∞
n
n=0 ρ
(since ρ < 1)
1
= ρ(1 − ρ) (1−ρ)
2
=
=
Dr. B.B. Upadhyay
ρ
1−ρ
λ
µ−λ .
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Operating characteristics of queueing system
3
Average queue length is given by
P
E (m) = ∞
m=0 mPn , where m = n − 1
being the number of customers in the queue. excluding the customer
which is in service. Therefore
P
E (m) = ∞
n=1 (n − 1)Pn
Dr. B.B. Upadhyay
−
P∞
n=0 nPn −
hP
=
P∞
=
P∞
=
ρ
1−ρ
− {1 − (1 − ρ)}
=
ρ
1−ρ
−ρ
=
ρ2
1−ρ
=
λ2
µ(µ−λ) .
n=1 nPn
Queueing Theory
n=1 Pn
∞
n=0 Pn
− P0
i
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Operating characteristics of queueing system
4
Average length of non-empty queue is given by
E (m|m > 0) =
E (m)
P(m>0)
=
λ2
µ(µ−λ)
=
µ
µ−λ
×
1
λ 2
)
(µ
since
P(m > 0) = P(n > 1) =
∞
X
Pn − P0 − P1 =
n=0
5
λ 2
µ
The fluctuation (variance ) of queue length is given by
V (n) =
∞
X
[n − E (n)]2 Pn =
n=0
Dr. B.B. Upadhyay
∞
X
n2 Pn − [E (n)]2 .
n=0
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Operating characteristics of queueing system
Using some algebraic transformation and the value of Pn , the result
reduces to
V (n) = (1 − ρ)
Dr. B.B. Upadhyay
h ρ i2
ρ + ρ2
ρ
λµ
=
−
=
.
3
2
(1 − ρ)
1−ρ
(1 − ρ)
(µ − λ)2
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Operating characteristics of queueing system
Waiting Time Distribution for Model I
The waiting time of a customer in the system is, for the most part, a
continuous random variable except that there is a non-zero probability that
the delay will be zero, that is a customer entering service immediately
upon arrival.
Therefore, if we denote the time spent in the queue by w and if Ψw (t)
denotes its cumulative probability distribution then from the complete
randomness of the Poisson distribution, we have
Ψw (0) = P(w = 0) (no customer in the system upon arrival)
= P0 = (1 − ρ).
It is now required to find Ψw (t) for t > 0. Let there be n customers in the
system upon arrival, then in order for a customer to go into service at a
time between 0 and t, all the n customers must have been served by time t.
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Operating characteristics of queueing system
Let s1 , s2 , . . . , sn denote service times of n customers respectively. Then
w=
n
X
si , (n ≥ 1) and w = 0 (n = 0).
i=1
The distribution function of waiting time, w, for a customer who has to
wait is given by
P(w ≤ t) = P
n
hX
i
si ≤ t ; n ≥ 1 and t > 0.
i=1
Since, the service time for each customer is independent and identically
distributed, therefore its probability density function is given by
µe −µt (t > 0), where µ is the mean service rate. Thus
Ψn (t) =
P∞
n=1 Pn × P(n − 1 custmer served at time t)
×P(1 custmor is servedin time ∆t)
=
Dr. B.B. Upadhyay
P∞ n=1
1−
λ
µ
n
λ
µ
(µt)n−1 e −µt
(n−1)! .µ∆t.
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Operating characteristics of queueing system
The expression for Ψw (t), therefore, can be written as
Ψw (t) = P(w ≤ t)
Dr. B.B. Upadhyay
Rt
n=1 Pn 0
=
P∞
=
P∞
n=1 (1
Ψn (t)dt
− ρ)ρn
= (1 − ρ)ρ
Rt
= (1 − ρ)ρ
Rt
0
0
R t (µt)n−1 −µt
.µdt
0 (n−1)! e
µe −µt
(µtρ)n−1
n=1 (n−1)! .dt
P∞
µe −µt(1−ρ) dt.
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Operating characteristics of queueing system
Hence, the waiting time of a customer who has to wait is give by
ψ(w ) =
λ −(µ−λ)t
d
[Ψw (t)] = ρ(1 − ρ).µe −µt(1−ρ) = λ 1 −
e
.
dt
µ
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Operating characteristics of queueing system
Characteristic of waiting Time Distribution
Average waiting time of a customer (in the queue) is given by
R
E (w ) = 0∞ tΨ(w )dt
=
R∞
0
=ρ
Dr. B.B. Upadhyay
t.ρµ(1 − ρ)e −µt(1−ρ) dt
R ∞ xe −x
0 µ(1−ρ) dx , for µ(1 − ρ)t = x
=
ρ
µ(1−ρ)
=
λ
µ(µ−λ) .
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Operating characteristics of queueing system
Average waiting time of an arrival who has to wait is given by
n
E (w |w > 0) =
E (w )
=
P(w > 0)
λ
µ(µ−λ)
λ
µ
o
=
1
.
µ−λ
[ Here P(w > 0) = 1 − P(w = 0) = 1 − (1 − ρ) = ρ.]
Dr. B.B. Upadhyay
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September 2, 2023
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Operating characteristics of queueing system
For the busy period distribution, let the random variable v denote the
total time that a customer has to spend in the system including
service. Then the probability density of its cumulative density
function is given by
Ψ(w |w > 0) =
Ψ(w )
P(w >0)
h =
λ
λ 1− µ
e −(µ−λ)t
i
λ
µ
= (µ − λ)e −(µ−λ)t , t > 0.
Dr. B.B. Upadhyay
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September 2, 2023
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Operating characteristics of queueing system
Average waiting time that a customer spends in the system including
service is given by
R
E (v ) = 0∞ t.Ψ(w |w > 0)dt
Dr. B.B. Upadhyay
=
R∞
=
1 R∞
−x dx ,
µ−λ 0 xe
=
1
µ−λ
0
t.(µ − λ)e −(µ−λ)t dt
Queueing Theory
for (µ − λ)t = x
September 2, 2023
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Operating characteristics of queueing system
Relationships among Operating Characteristics
We have deriving the following important characteristics of an M/M/1
queueing system:
E (n) =
λ
λ2
λ2
1
, E (m) =
, e(W ) =
and E (v ) =
.
µ−λ
µ(µ − λ)
µ(µ − λ)
µ−λ
Using these expressions, we observe some general relationships between
the average system characteristics as follows:
Expected number of customers in the system is equal to the expected
number of customers in the queue plus a customer currently in
service, i.e.,
λ
E (n) = E (m) +
µ
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September 2, 2023
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Operating characteristics of queueing system
Expected waiting time of a customer in the system is equal to the
expected waiting time in the queue plus the expected service time of
a customer in service, i.e.,
E (v ) = E (w ) +
Dr. B.B. Upadhyay
Queueing Theory
1
.
µ
September 2, 2023
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Operating characteristics of queueing system
Expected number of customers in the system is equal to the average
number of arrivals per unit of time multiplied by the average time
spent by the customer in the system, i.e.
E (n) = λE (v ).
Expected number of customers in the queue is equal to the average
number of arrivals per unit time multiplied by the average time spent
by a customer in the queue, i.e.,
E (m) = λE (w ).
Note. Relations between Average Queue Length and Average waiting
Time are known as Little’s Formulae.
Dr. B.B. Upadhyay
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September 2, 2023
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Operating characteristics of queueing system
Problem
A TV repairman finds that the time spent on his jobs has an Exponential
distribution with mean 30 minutes. If he repairs sets in the order in which
they come in, and if the arrival of sets is approximately Poisson with an
average rate of 10 per 8-hour day, what is repairman’s expected idle time
each day ? How many jobs are ahead of the average set just bought in ?
Dr. B.B. Upadhyay
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Operating characteristics of queueing system
Solution
We are given,
λ = 10 = set per day
µ = 16 sets per day
Therefore,
ρ=
Dr. B.B. Upadhyay
λ
10
=
= 0.625
µ
16
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Operating characteristics of queueing system
The probability for the repairman to be idle is
P0 = 1 − ρ = 1 − 0.625 = 0.375
Expected idle time per day = 8 × 0.375 = 3 hours.
Expected (or average) number of T.V. sets in the system
E (n) =
Dr. B.B. Upadhyay
ρ
1−ρ
=
0.625
1−0.625
=
5
3
= 2(approx ) T.V. sets.
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Operating characteristics of queueing system
Problem
In a railway marshalling yard, goods trains arrive at a rate of 30 trains per
day. Assuming that the inter-arrival time follows an exponential
distribution and the service time distribution is also exponential with an
average 36 minutes. Calculate the following.
1
the mean queue size (line length ), and
2
the probability that the queue size exceeds 10.
If the input of trains increases to an average 33 per day, what will be
the change in (i ) and (ii )?
Dr. B.B. Upadhyay
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September 2, 2023
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Operating characteristics of queueing system
Solution
Here, we have
λ=
30
60×24
=
1
48
and µ =
1
36
trains per minute.
Therefore,
ρ=
i
E (m) =
ρ
1−ρ
=
0.75
1−0.75
36
λ
=
= 0.75
µ
48
= 3 trains.
ii
P(≥ 10) = ρ10 = (0.75)10 = 0.06.
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Operating characteristics of queueing system
When the input increases to 33 trains per day, we have
λ=
33
60×24
=
11
480
and µ =
1
36
trains per minute.
Therefore,
ρ=
λ
µ
=
11
480
× 36 = 0.83
Then, we get
i
E (n) =
ρ
0.83
=
= 4.9 or 5 trains (approx .)
1−ρ
1 − 0.83
ii
P(≥ 10) = ρ10 = (0.83)10 = 0.2 (approx )
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September 2, 2023
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Operating characteristics of queueing system
Problem
The rate of arrival of customers at a public telephone booth follows
Poisson distribution, with an average time of 10 minutes between one
customer and the next. The duration of a phone call is assumed to follow
exponential distribution, with mean time of 3 minutes.
i
What is the probability that a person arriving at the booth will have
to wait?
ii
What is the average length of the non-empty queues that form from
time to time ?
iii
The Mahanagar Telephone Nigam Ltd. will install a second booth,
when it is convinced that the customers would expect waiting for at
least 3 minutes /or their turn to make a call. By how much time
should the flow of customers increase in order to justify a second
booth?
iv
Estimate the fraction of a day that the phone will be in use.
v
What is the probability that it will take him more than 10 minutes
altogether to wait for phone and complete his call ?
Dr. B.B. Upadhyay
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September 2, 2023
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Operating characteristics of queueing system
Solution
Here, we are given:
λ=
i
1
10
× 60 or 6 per hours and µ =
1
3
Probability that a person arriving at the booth will have to wait
P(w > 0) = 1 − P0 = 1 − 1 −
ii
λ
6
=
or 0.3.
µ
20
Average length of non-empty queues
E (m|m > 0) =
iii
× 60 or 20 per hour.
µ
20
=
= 1.43.
µ−λ
20 − 6
The installation of a second booth will be justified, if the arrival rate
is greater than the waiting time. Now, if λ′ denotes the increased
arrival rate, expected waiting time is:
E (w ) =
Dr. B.B. Upadhyay
λ′
3
λ′
=⇒
=
or λ′ = 10.
µ(µ − λ′ )
60
20(20 − λ′ )
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Operating characteristics of queueing system
Hence the arrival rate should become 10 customers per hour to justify the
second booth.
4
The fraction of a day that the phone will be busy = traffic intensity
ρ = µλ = 0.3.
5
Consider,
P(w ≥ 0) =
=
R∞ 10
R∞
10
λ 1−
λ
µ
e −(µ−λ)t dt
(0.30)(0.23)e −0.23t dt,
where λ = 0.10 per minute, and µ = 0.33 per minute. Therefore,
P(w ≥ 10) = (0.069)
e −0.23t
−0.23
∞
10
= 0.03.
This shows that 3 per cent of the arrivals on an average will have to wait
for 10 minutes or more before they can use the phone.
Dr. B.B. Upadhyay
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September 2, 2023
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Operating characteristics of queueing system
Exercise
Problem
A warehouse has only one loading dock manned by a three-person crew.
Trucks arrive at the loading dock at an average rate of 4 trucks per tour
and the arrival rate is Poisson distributed. The loading of a truck takes 10
minutes on an average and can be assumed to be exponentially
distributed. The operating cost of a truck is Rs. 20 per hour and the
members of the loading crew are paid Rs. 6 each per hour. Would you
advise the truck owner to add another crew of three persons?
Dr. B.B. Upadhyay
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Operating characteristics of queueing system
Problem
At a certain petrol pump, customers arrive in a Poisson process with an
average time of 5 minutes between arrivals. The time interval between
servers at the petrol pump follows an exponential distribution and the
mean time taken to service a unit is 2 minutes. Find the following:
1
Expected average queue length.
2
Average number of customers in the system.
3
Average time a customer has to wait in the queue.
4
Average time a customer has to spend in the system.
5
By how much time the flow of the customer be increased to justify
the opening of another service point, where the customer has to wait
for 5 minutes for the service?
Dr. B.B. Upadhyay
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September 2, 2023
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Operating characteristics of queueing system
Problem
Delhi medical store has only one cashier who handles all the payments. On
an average, the cashier can serve 5 customers per 15 minutes, who arrive
at his counter randomly on an average rate of 18 per hour. The
Management of the store noticed that the cashier was idle many times and
some of the customers complained against long waits number of times. lt
was, therefore, decided (by making suitable assumptions) to investigate
the following:
1
What portion of the time the cashier is likely to be idle?
2
What is the average length of the waiting line to be expected under
the existing servicing conditions ?
3
How many customers would be expected to be in the service area?
4
Suppose that the management of the store will employ another
cashier if they are convinced that a Customer has to wait for at least
five minutes for making payments. What arrival rate would justify the
employment of second cashier?
Dr. B.B. Upadhyay
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September 2, 2023
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Operating characteristics of queueing system
Model II {(M/M/1) : (∞/SIRO)}
This model is essentially the same as Model I, except that the service
discipline follows the SIRO-rule (service in random order) instead of
FIFO-rule.
As the derivation of Pn for Model I does not depend on any specific queue
discipline, it may be concluded that for the SIRO-rule case, we must have
Pn = (1 − ρ)ρn , n ≥ 0.
Consequently, whether the queue discipline follows the SIRO-rule or
FIFO-rule the average number of customers in the system E (n), will
remain the same.
Dr. B.B. Upadhyay
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Operating characteristics of queueing system
In fact E (n) will remain the same for any queue discipline provided, of
course, Pn remains unchanged. Thus,
E (v ) =
1
E (n)
λ
under the SIRO-rule is the same as under the FIFO-rule.
This result can be extended to any queue discipline as long as Pn remains
unchanged. Specifically, the result applies to the three most common
disciplines, namely, FIFO, LIFO and SIR0.
The three queue disciplines differ only in the distribution of waiting time
where the probabilities of long and short waiting times change depending
upon the discipline used. Thus we can use the symbol GD (general
discipline) to represent the disciplines FIFO, LIFO and SIRO, when the
waiting time distribution is not required.
Dr. B.B. Upadhyay
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September 2, 2023
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Operating characteristics of queueing system
Model III {(M/M/1) : (N/FIFO)}
This model differs from that of Model 1 in the sense that the maximum
number of customers in the system is limited to N. Therefore, the
difference equations of Model 1 are valid for this model as long as n < N.
The additional difference equation for n = N, is
PN (t + ∆t) = PN (t)[1 − µ∆t] + PN−1 (t)[λ∆t][1 − µ∆t] + o(∆t).
This gives, after simplification, the differential-difference equation
d
PN (t) = −µPN (t) + λPN−1 (t)
dt
from which the resultant steady-state difference equation is
0 = −µPN + λPN−1 .
Dr. B.B. Upadhyay
Queueing Theory
September 2, 2023
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Operating characteristics of queueing system
This complete set of steady-state difference equation for this model,
therefore, can be written as
µP1 = λP0
µPn+1 = (λ + µ)Pn − λPN−1 , 1 ≤ n ≤ N − 1,
µPN = λPN−1 .
Using the iterative procedure (as in Model I), the first two difference
equation gives
λ n
Pn =
P0 , n ≤ N − 1
µ
Also, we see that for this value of Pn , the third (last) difference equation
holds for n = N.
Therefore we have
Pn =
Dr. B.B. Upadhyay
λ n
µ
P0 = ρn P0 , n ≤ N and
Queueing Theory
λ
µ
=ρ
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Operating characteristics of queueing system
For obtaining the value of P0 we make use of the boundary conditions,
n=0 Pn = 1. Therefore,
PN
1 = P0
N
X
n=0
Thus,
n
ρ =

P 1−ρN+1 ,
P (N + 1),
0
(ρ = 1)
0
1−ρ

 1−ρ ,
N+1
P0 = 1−ρ
1

,
N+1
Dr. B.B. Upadhyay
Queueing Theory
ρ ̸= 1
ρ ̸= 1
ρ = 1.
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Operating characteristics of queueing system
Hence,
Pn =

 (1−ρ)ρn ;
ρ ̸= 1; 0 ≤ n ≤ N
 1 ,
N+1
(ρ = 1)
1−ρN+1
Remark. The steady-state solution exists even ρ ≥ 1. Intuitively this
makes since the maximum limit prevents the process from “blowing up”. If
N → ∞, then the steady-state solution is
Pn = (1 − ρ)ρn ;
n < ∞,
This result is in complete agreement with that of Model I.
Dr. B.B. Upadhyay
Queueing Theory
September 2, 2023
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Operating characteristics of queueing system
Characteristics of Model III
1
Average number of customers in the system is a given by
E (n) =
PN
n=0 nPn
= P0
PN
n
n=0 nρ
= P0 ρ
or
d
E (n) = P0 ρ dρ
PN
d
n
n=0 ρ P0 ρ dρ
n
PN
h
d n
n=0 dρ ρ
1−ρN+1
1−ρ
i
N+1 ]
+Nρ
= P0 ρ[1−(N+1)ρ
(1−ρ)2
=
Dr. B.B. Upadhyay
ρ[1−(N+1)ρN +NρN+1 ]
(1−ρ)(1−ρN+1 )
since P0 =
Queueing Theory
1−ρ
;
1−ρN+1
ρ ̸= 1
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Operating characteristics of queueing system
2
Average queue length is given by
E (m) =
PN
n=1 (n
= E (n) −
− 1)Pn
PN
n=1 Pn
= E (n) − (1 − P0 )
= E (n) −
=
Dr. B.B. Upadhyay
ρ(1−ρn )
,
1−ρN+1
since P0 =
1−ρ
;
1−ρN+1
ρ ̸= 1
ρ2 [1−NρN−1 +(N−1)ρN ]
(1−ρ)(1−ρN+1 )
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Operating characteristics of queueing system
3
The average waiting time in the system can be obtained by using
Little’s formulae, that is,
E (v ) =
E (n)
,
λ′
where λ′ is the mean rate of costumers entering the system and is
equal to λ(1 − PN ). The average waiting time in the queue can be
obtained by using the relations
E (w ) = E (v ) −
Dr. B.B. Upadhyay
1
E (m)
or E (w ) =
.
µ
λ′
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Operating characteristics of queueing system
Problem
At a railway station, only one train is handled at a time. The railway yard
is sufficient only for two trains to wait while other is given signal to leave
the station. Trains arrive at the station at an average rate of 6 per hour
and the railway station can handle them on an average of 12 per hour.
Assuming Poisson arrivals and exponential service distribution, find the
steady-state probabilities for the various number of trains in the system.
Also find the average waiting time of a new, train coming into the yard.
Dr. B.B. Upadhyay
Queueing Theory
September 2, 2023
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Operating characteristics of queueing system
Solution
Here, λ = 6 and µ = 12 so that
ρ=
6
= 0.5.
12
The maximum queue length is 2; i.e., the maximum number of trains in
the system is 3 (= N).
The probability that there is no train in the system (both waiting and in
service ) is given by
P0 =
1−ρ
1 − 0.5
=
= 0.53
1 − ρN+1
1 − (0.5)3+1
Now since
Pn = P0 ρn ,
Dr. B.B. Upadhyay
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Operating characteristics of queueing system
therefore
P1 = (0.53)(0.5) = 0.27,
P2 = (0.53)(0.5)2 = 0.13,
P3 = (0.53)(0.5)3 = 0.07.
Hence we get
E (n) = 1(0.27) + 2(0.13) + 3(0.07) = 0.74.
Thus the average number of trains in the system is 0.74 and each train
1
takes on an average 12
(= 0.08) hours for getting service.
As the arrival of new train expects to find an average 0.74 trains in the
system before it.
E (w ) = (0.74)(0.08) hours = 0.0592 hours or 3.5 minute
Dr. B.B. Upadhyay
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September 2, 2023
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Operating characteristics of queueing system
Problem
Assume that the goods trains are coming in a yard at the rate of 30 trains
per day and suppose that the inter-arrival times follow an exponential
distribution. The service time for each train is assumed to he exponential
with an average of 36 minutes. If the yard can admit 9 trains at a time
(there being 10 lines, one of which is reserved for shunting purposes ),
calculate the probability that the yard is empty and find the average queue
length.
Dr. B.B. Upadhyay
Queueing Theory
September 2, 2023
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Operating characteristics of queueing system
Solution
We have,
λ=
30
60×24
1
48
=
and µ =
1
16
trains per minute.
Therefore
ρ=
36
λ
=
= 0.75.
µ
48
The probability that the yard is empty is given by
P0 =
=
Dr. B.B. Upadhyay
1−ρ
1−ρN+1
0.25
0.90
=
1−0.75
,
1−(0.75)10
since N = 9
= 0.28.
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Operating characteristics of queueing system
Average queue length is given by
E (m) =
=
ρ2 [1−NρN−1 +(N−1)ρN ]
(1−ρ)(1−ρN+1 )
(0.75)2 [1−9(0.75)8 +8(0.75)9 ]
0.25[(0.75)10 ]
(1−0.303)
= (2.22) (1−0.005)
= (2.22)(0.70)
= 1.55.
Dr. B.B. Upadhyay
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September 2, 2023
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Operating characteristics of queueing system
Exercise
Problem 1
Consider a single server queueing system with Poisson input, exponential
service times. Suppose the mean arrival rate is 3 calling units per hour,
the expected service time is 0.25 hours and the maximum permissible
number of calling units in the system is two. Derive the steady-state
probability distribution of the number of calling units in the system, and
then calculate the expected number in the system.
Dr. B.B. Upadhyay
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September 2, 2023
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Operating characteristics of queueing system
Problem 2
Patients arrive at a clinic according to a Poisson distribution at a rate of
30 patients per hour. The waiting room dons not accommodate more than
14 patients. Examination time per patient is exponential with mean rate
20 per hour.
i
Find the effective arrival rate at the clinic.
ii
What is the probability that an arriving patient will not wait?
iii
What is the expected waiting time until a patient is discharged form
the clinic?
Dr. B.B. Upadhyay
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September 2, 2023
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Operating characteristics of queueing system
Model IV ( Generalized Model: Birth-Death Process )
This model deals with a queueing system having single service channel,
Poisson input with no limit on the system capacity. Arrivals can be
considered as birth to the system, whereas a departure can be looked upon
as a death. Let
n
λn
µn
Pn
= number of customers in the system,
= arrival rate of customers given n customers in the system,
= departure rate of customers given it customers in the system,
= steady − state probability of n customers in the system.
Dr. B.B. Upadhyay
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September 2, 2023
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Operating characteristics of queueing system
The model determines the values of Pn in terms of λn and µn . Now from
the Poisson process, we observe that an arrival during the small time
interval ∆t is negligible.
This implies that for n > 0, state n can change only to two possible states:
State n-1 when a departure occurs at the rate µn and state n + 1 when an
arrival occurs at rate λn .
State 0 can only change to state 1 when an arrival occurs at the rate λ0 .
Since no departure is possible when the system is empty, µ0 is undefined.
Dr. B.B. Upadhyay
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September 2, 2023
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Operating characteristics of queueing system
Under steady-state conditions, for n > 0, the rate of flow into and out of
state n must be equal. Thus is illustrated in the transition-rate diagram
given below:
The balance equation is:
λn−1 Pn−1 + µn+1 Pn+1 = λn Pn + µn Pn , n ≥ 1
Dr. B.B. Upadhyay
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September 2, 2023
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Operating characteristics of queueing system
and
µ1 P1 = λ0 P0 ,
n = 0.
Using the iterative procedure (as in Model 1), we have
λ0
P0 ,
µ1
λ1 + µ 1
P2 =
P1 −
µ2
λ2 + µ 2
P2 −
P3 =
µ3
P1 =
Dr. B.B. Upadhyay
λ0
λ1 λ0
P0 =
P0 ,
µ2
µ 2 µ1
λ1
λ2 λ1 λ0
P1 =
P0 .
µ3
µ 3 µ2 µ 1
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Operating characteristics of queueing system
In general, we can write the following formula
Pn =
n−1
Y λi
λn−1 λn−2 ...λ0
P0 , or Pn = P0
n ≥ 1.
µn µn−1 ...µ1
µ
i=0 i+1
Now,
Pn+1 =
n
Y
λn−1
λi
λ n + µn
Pn −
Pn−1 = P0
µn+1
µn+1
µ
i=0 i+1
Thus, by mathematical induction the general value of Pn holds for all n.
To obtain the value of P0 , we use the boundary condition
∞
X
Pn = 1
or P0 +
n=0
∞
X
Pn = 1,
n=1
to get
P0 = 1 +
∞ n−1
X
Y λi −1
n=1 i=1
Dr. B.B. Upadhyay
Queueing Theory
µi+1
September 2, 2023
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Operating characteristics of queueing system
Remark. P0 = 0 if R.H.S is divergent series. In case R.H.S is convergent,
the value of P0 will depend on λ′i s and µ′i s.
Special Cases
Case 1)When λn = λ for n ≥ 0 ; and µn = µ for n > 1
"
P0 = 1 +
#
∞ n −1
X
λ
n=1
µ
= 1 − ρ.
In this case, therefore
Pn = ρn (1 − ρ), for n ≥ 0.
This result is exactly the same as that of Model 1.
Dr. B.B. Upadhyay
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September 2, 2023
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Operating characteristics of queueing system
Case 2)When λn =
λ
n+1
for n ≥ 0 and µn = µ for n > 1.
h
P0 = 1 +
h
λn
n=1 n!µn
P∞
= 1+ρ+
1 2
2! ρ
i−1
+ ... +
1 n
n! ρ
i−1
+ ...
= e −ρ
∴ Pn =
1 n −p
λ
ρ e
for n ≥ 0 and ρ = .
n!
µ
which is Poisson distribution with mean E (n) = ρ.
Dr. B.B. Upadhyay
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September 2, 2023
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Operating characteristics of queueing system
Case 3)- When λn = λ for n ≥ 0 and µn = nµ for n > 1,
"
P0 = 1 +
∞
X
λn
n=1
h
= 1+ρ+
#−1
n!µn
i−1
1 2
1
ρ + . . . + ρn + . . .
2!
n!
= e −ρ .
Therefore,
∴ Pn =
λ
1 n −p
ρ e , for n ≥ 0 and ρ = .
n!
µ
which is again a Poisson with mean E (n) = ρ and E (m) = 0, E (w ) = 0.
In this case the service rate increases with increase in queue length and
hence is known as a queueing problem with infinite number of channels,
i.e., (M/M/∞) : (∞/FIFO). This model is known as a Self-service
Model.
Dr. B.B. Upadhyay
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September 2, 2023
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Operating characteristics of queueing system
Problem
Problems arrive at a computing centre in Poisson fashion at an average
rate of five per day. The rules of the computing centre are that any man
waiting to get his problem solved must aid the man whose problem is being
solved. If the time to solve a problem with one man has an exponential
distribution with mean time of 31 a day, and if the average solving time is
inversely proportional to the number of people working on the problem,
approximate the expected time in the centre for a person entering the line.
Dr. B.B. Upadhyay
Queueing Theory
September 2, 2023
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Operating characteristics of queueing system
Solution
Here λ = 5 problems per day, and µ = 3 problem per day.
It is given that the service rate increases with increase in the number of
persons. Therefore,
∴ µn = nµ when there are n problems and Pn =
E (n) =
∞
X
n=0
Dr. B.B. Upadhyay
nPn =
∞
X
n=0
n
1 n −p
n! ρ e
1 n −ρ
5
ρ e = e −ρ .ρ.e ρ = ρ =
or 1.67
n!
3
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Operating characteristics of queueing system
Now the average solving time, which is inversely proportional to the
number of people working on the problem, is given by 15 day per problem.
∴ Expected time for a person entering the line is given by
1
1 5
1
E (n) = × days = day or 8 hours.
5
5 3
3
Dr. B.B. Upadhyay
Queueing Theory
September 2, 2023
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Operating characteristics of queueing system
Exercise
Problem
A shipping company has a single unloading berth with ships arriving in a
Poisson fashion at an average rate of three per day. The unloading time
distribution for a ship with n unloading crews is found to be exponential
with average unloading time n2 days. The company has a large labor
supply without regular working hours, and to avoid long waiting line the
company has a policy of using as many unloading crews on a ship as there
are ships getting in line or being unloaded.
a
Under these conditions what will be the average number of unloading
crews working at any time?
b
What is the probability that more than 4 crews will be needed?
Dr. B.B. Upadhyay
Queueing Theory
September 2, 2023
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Operating characteristics of queueing system
Problem
An international airport operates with three check-out counters. The sign
by the check-out area advises the customers that an additional counter will
be opened any time the number of customers in the queue exceeds three.
This means that for fewer than four customers, only one counter will be in
operation. For four to six customers, two counters will be open. For more
than six customers, all three counters will be open. The customers arrive
at the counters area according to a Poisson distribution with a mean of 10
customers per hour. The average check-out time per customer is
exponential with mean 12 minutes. Determine the following:
a
Steady-state probability Pn of n customers in the check-out area.
b
Average number of busy counters.
c
Average number of idle counters.
Dr. B.B. Upadhyay
Queueing Theory
September 2, 2023
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Operating characteristics of queueing system
Problem
For a General queue system (discuss in Model IV), find the steady-state
probability when
a
(
λn =
λ for n = 0, 1, . . . , C − 1
0 for n = C , C + 1, . . .
and µn = nµ for n = 1, . . . , C .
b
(
λn =
C − nλ
0
for n = 0, 1, . . . , k
for n = k + 1, k + 2, . . .
and µn = nµ for n > 1.
Dr. B.B. Upadhyay
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September 2, 2023
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Operating characteristics of queueing system
Model V {(M/M/C ) : (∞/FIFO)}
This model is a special case of Model IV in the sense that here we consider
C parallel service channels. The arrival rate is λ and the service rate is µ.
The effect of using C parallel service channels is a proportionate increase
in the service rate of the faculty to nµ if n ≤ C and C µ if n > C . Thus, in
term of generalized model (Model IV), λn and µn are defined as
λn = λ, n ≥ 0
and
µn = nµ if 1 ≤ n ≤ C
Dr. B.B. Upadhyay
and Cµ , if n ≥ C .
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Operating characteristics of queueing system
Utilizing the value of λn and µn , the steady-state probabilities of Model IV
becomes
Pn =

λn P 0


nµ(n−1)µ...(1)µ ;




1 ≤ n ≤ C,
λn P 0



(C µ)(C µ)...(C µ)(C −1)µ(C −2)µ...(1)µ ;



n>C
That is
Pn =

1 n
1 λn

 n! µn P0 = n! ρ P0




Dr. B.B. Upadhyay
λn P0
C n−C C !µn
=
1
ρ n P0
C n−C C !
Queueing Theory
if 1 ≤ n ≤ C
if n > C .
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Operating characteristics of queueing system
To find the value of P0 , we use the boundary condition
Thus, we have
C
−1
X
∞
X
Pn +
n=0
P∞
n=1 Pn
= 1.
=1
n=C
or
∞
−1
h CX
1 n X
ρ +
n!
n=0
n=C
1
i
C n−C C !
ρn P0 = 1
0r
P0 =
hP
=
hP
Dr. B.B. Upadhyay
C −1 1 n
n=0 n! ρ
C −1 1
n=0 n!
+ ρC
n
λ
µ
+
P∞
1
n=C C !
1
C!
C
Queueing Theory
λ
µ
n−C i−1
ρ
C
µ
. C Cµ−λ
i−1
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Operating characteristics of queueing system
Remark
1
The result obtain above is valid only if Cλµ < 1 ; that is, the mean
arrival rate must be less than the mean maximum potential service
rate of the system.
2
If C = 1, then the value of P0 is in complete agreement with the
value of P0 for Model I.
Dr. B.B. Upadhyay
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September 2, 2023
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Operating characteristics of queueing system
Characteristics of Model V
1
P(n ≥ C ) = Probability that an arrival has to wait
∞
X
=
n=C
2
Pn =
∞
X
n=C
C
λ n
1
C !C n−C µ
P0 =
λ
µ
Cµ
C !(C µ − λ)
P0
Probability that an arrival enters the service without wait
C
= 1 − P(n ≥ C ) or 1 −
Dr. B.B. Upadhyay
Queueing Theory
C
λ
µ
C !(C − µλ )
P0 .
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Operating characteristics of queueing system
3
Average queue length is given by
E (m) =
P∞
=
P∞
=
P∞
n=C (n
− C )Pn
for x = n − C
x =0 xPx +C
1
x =0 x . C !C x
C +x
λ
µ
P0
or
E (m) =
1
C
C ! (λ/µ)
=
1
C!
=
1
C!
C
λ
µ
C
λ
µ
=
Dr. B.B. Upadhyay
λ
µ
x =0 x .(λ/C µ)
P0
xP
0
P∞ d x .y , where y = Cλµ
x =0 dy y
d
P0 y dy
C
λµ
P∞
1
1−y
P0
(C −1)!(C µ−λ)2
.
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Operating characteristics of queueing system
4
Average number of customers in the system is given by
E (n) = E (m) +
λµ
=
5
λ
µ
C
λ
µ
P0
(C − 1)!(C µ −
λ)2
+
λ
.
µ
Average waiting time of an arrival is given by
E (w ) =
1
E (m)
λ
µ
=
Dr. B.B. Upadhyay
C
λ
µ
P0
(C − 1)!(C µ − λ)2
Queueing Theory
.
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Operating characteristics of queueing system
7
Average waiting time an arrival spends in the system is given by
1
µ
E (v ) = E (w ) +
µ
=
C
λ
µ
P0
(C − 1)!(C µ − λ)2
+
1
µ
or
E (n)
.
λ
Average number of idle servers is equal to
E (v ) =
8
C − Average number of customers served.
Dr. B.B. Upadhyay
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September 2, 2023
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Operating characteristics of queueing system
Problem
A supermarket has two girls serving at the counters. The customers arrive
in a Poisson fashion at the rate of 12 per hour. The service time for each
customer is exponential with the mean 6 minute. Find
1
the probability that an arriving customer has to wait for service,
2
the average number of customers in the system, and
3
the average time spent by a customer in the super-market.
Dr. B.B. Upadhyay
Queueing Theory
September 2, 2023
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Operating characteristics of queueing system
Solution
We are give λ = 12 customers per hour, µ = 10 per hour, and C = 2 girls.
1
∴ P0 =
hP
n
=
hP
=
1
4
C −1 1
n=0 n!
2−1 1
n=0 n!
λ
µ
12
10
n
+
1
C!
+
1
2!
C
λ
µ
12
10
µ
. C Cµ−λ
2
2×10
20−12
i−1
i−1
(or 0.25).
Probability of having to wait for service
P(w > 0) =
=
1
C!
1
2
C
λ
µ
12
10
2
Cµ
C µ−λ P0
20
20−12
×
1
4
= 0.45.
Dr. B.B. Upadhyay
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Operating characteristics of queueing system
2
Average queue length is
C
λµ
∴ E (m) =
λ
µ
P0
(C −1)!(C µ−λ)2
=
12×10×(1.2)2 ×0.25
(2−1)!(20−12)2
=
27
40
Average number of customers in the system
E (n) = E (m) +
3
λ
27 12
=
+
= 1.87 (or 2 customers) approx .
µ
40 10
Average time spent by customer in super market
E (v ) =
Dr. B.B. Upadhyay
E (n)
1.87
=
= 0.156 hours or 9.3 minutes.
λ
12
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Operating characteristics of queueing system
Problem
A company currently has two tools cribs, each having a single clerk, in its
manufacturing area. One tool cribs handles only the tools for the heavy
machinery, while the second one handles all others tools. It is observed
that for each tool crib the arrival follows a Poisson distribution with a
mean of 20 per hour and the service time distribution is negative
exponential with mean of 2 minute.
The tool manager feels that, if tool cribs are combined such a way that
either clerk can handle any kind of tool as demands arises, would be more
efficient and the waiting problem could be reduced to some extent. It is
believed that the mean arrival rate at the two cribs will be 40 per hours;
while the service time will remain unchanged.
Compare in status of queue and the proposal with respect to the total
expected number of mechanics at the tool crib(s), the expected waiting
time including service time for each mechanics and the probability that he
has to wait more than five minute.
Dr. B.B. Upadhyay
Queueing Theory
September 2, 2023
130 / 207
Operating characteristics of queueing system
Solution
For each tool crib, we have λ = 20 machine per hour, and µ = 30 machine
per hour.
20
λ
=
= 2 mechanics
∴ E (n) =
µ−λ
30 − 20
and
E (m) =
λ2
(20)2
4
=
= or 1.33
µ(µ − λ)
30 × (30 − 20)
3
Expected waiting time in the system is given by
E (v ) =
Dr. B.B. Upadhyay
1
1
1
=
=
hour or 6 minutes.
µ−λ
30 − 20
10
Queueing Theory
September 2, 2023
131 / 207
Operating characteristics of queueing system
For the proposed system (when both the tool cribs are combined ), we
have λ = 40 mechanics per hour, µ = 30 mechanics per hour, C = 2.
∴ P0 =
hP
n
=
hP
n
C −1 1
n=0 n!
2−1 1
n=0 n!
h
40
30
1
5
or 0.2
= 1+
=
Dr. B.B. Upadhyay
+
λ
µ
λ
µ
1
2!
+
1
C!
+
1
2!
40
30
2
Queueing Theory
C
λ
µ
2
×
λ
µ
µ
. C Cµ−λ
2µ
. 2µ−λ
2×30
2×30−40
i−1
i−1
i−1
September 2, 2023
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Operating characteristics of queueing system
C
λµ
E (n) =
λ
µ
2
λµ
=
P0
(C −1)!(C µ−λ)2
λ
µ
+
P0
(2−1)!(2µ−λ)2
+
λ
µ
λ
µ
= 2.4
E (v ) =
=
E (n)
λ
2.4
40
= 0.06 or 3.6 minutes.
Dr. B.B. Upadhyay
Queueing Theory
September 2, 2023
133 / 207
Operating characteristics of queueing system
From above, it is clear that combining the two tool cribs and asking both
servers to attend the customers leads to a more efficient handling of the
demand. This is not surprising result.
By separating the functions, certain advantages occurs but by combining
the two, the net waiting time is reduced because in the former system it is
possible that one tool crib is idle, whereas the second tool crib may have a
long queue.
Dr. B.B. Upadhyay
Queueing Theory
September 2, 2023
134 / 207
Operating characteristics of queueing system
Exercise
Problem 1
A petrol pump station has two pumps. The service time follows the
exponential distribution with a mean of 4 minutes and cars arrive for the
service in a Poisson process at the rate of 10 cars per hour. Find the
probability that the customer has to wait for service. What proportion of
the time the pump remain idle?
Dr. B.B. Upadhyay
Queueing Theory
September 2, 2023
135 / 207
Operating characteristics of queueing system
Problem 2
Given an average arrival rate of 20 per hour, is it better for a customer to
get service at a single channel with mean service rate of 22 customer or at
one of the two channels in parallel, with mean service rate of 11 customers
for each of the two channels? Assume both queues are of M/M/S type.
Dr. B.B. Upadhyay
Queueing Theory
September 2, 2023
136 / 207
Operating characteristics of queueing system
Problem 3
A telephone company is planning to install telephone booths in a new
airport. It has established the policy that a person should not have to wait
more than 10 per cent of the times he tries to use a phone. The demand
for use is estimated to be Poisson with an average of 30 per hour. The
average phone call has an exponential distribution with the mean time of 5
minutes. How many phones booth should be installed?
Dr. B.B. Upadhyay
Queueing Theory
September 2, 2023
137 / 207
Operating characteristics of queueing system
Model VI {(M/M/C ) : (N/FIFO)}
This model is essentially the same as Model V except that the maximum
number in the system is limited to N where N ≥ C . Therefore, utilizing
the steady-state probabilities of Model IV, with
λn = λ
if 0 ≤ n ≤ N; and 0 otherwise
µn = nµ if 0 ≤ n < C ; and Cµ if C ≤ n ≤ N
We get

 1 λ n P ;
0
Pn = n! 1µ
λ n
 n−C
P0 ;
C
C! µ
0≤n<C
C ≤n≤N
where
Dr. B.B. Upadhyay
Queueing Theory
September 2, 2023
138 / 207
Operating characteristics of queueing system
P0 =
−1
n CX
1 λ n
n=0
=
n! µ
+
N
X
n=C
λ n o−1
C n−C C ! µ
1
h
C
N−C +1
i−1
PC −1 1 λ n

Cµ
1 λ
λ

+
{1
−
+
}
;

n=0

n! µ
C! µ
Cµ
C µ−λ

λ
Cµ
̸= 1

hP
n
C
i−1



C −1 1 λ
1 λ

+
(N
−
C
+
1)
;
n=0 n! µ
C! µ
λ
Cµ
= 1.
Remark If we take N → ∞ and consider Cλµ < 1, then the result
corresponds to that of Model V. Also, if we take C = 1 then the result
reduced corresponds to that of Model 1.
Dr. B.B. Upadhyay
Queueing Theory
September 2, 2023
139 / 207
Operating characteristics of queueing system
Characteristics of Model VI
1
Average queue length is given by
E (m) =
PN
=
PN
=
P
1 λ C
P0 .ρ N−C
x =0
C! µ
=
(C ρ)C ρP0
C!
=
(C ρ)C ρP0 d 1−ρN−C +1
. dρ
C!
1−ρ
=
P0 (C ρ)C ρ
[1
C !(1−ρ)2
Dr. B.B. Upadhyay
n=C (n
n=C (n
− C )Pn
λ
µ
n
− C ) C !C n−C P0
x .ρx −1 , for ρ =
λ
Cµ
PN−C d x
x =0 dρ ρ
h
i
− ρN−C +1 − (1 − ρ)(N − C + 1)ρN−C ].
Queueing Theory
September 2, 2023
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Operating characteristics of queueing system
2
Now since,
E (m) =
N
X
N
X
(n − C )Pn =
n=C
nPn − C
n=C
N
X
Pn
n=C
therefore, he average number of customers in the system is given by
E (n) =
PN
n=0 nPn
=
PC −1
nPn +
=
PC −1
nPn + E (m) + C
n=0
n=0
= E (m) + C
PN
n=C
hP
N
n=0 Pn
= E (m) + C − P0
Dr. B.B. Upadhyay
nPn
−
PN
Pn
PC −1
Pn +
n=C
n=0
i
PC −1 (C −n)(ρC )n
n=0
Queueing Theory
n!
PC −1
n=0
nPn
.
September 2, 2023
141 / 207
Operating characteristics of queueing system
3
Average waiting time in the system can be obtained by using Little’s
formula, that is
E (n)
E (v ) =
λ′
where λ′ = λ(1 − PN ) is effective arrival rate.
Average waiting time in the queue can be obtained by using
E (w ) = E (v ) −
or
E (w ) =
Dr. B.B. Upadhyay
1
µ
[E (m)]
.
λ′
Queueing Theory
September 2, 2023
142 / 207
Operating characteristics of queueing system
Problem
A car servicing station has 3 stall where service can be offered
simultaneously. The cars wait in such a way that when a stall becomes
vacant, the car at the head of the line pulls up to it. The station can
accommodate at most four cars-waiting (seven in the system) at one time.
The arrival pattern is Poisson with the mean of one car per minute during
the peak hours. The service time is exponential with mean 6 minutes.
Find the average number of cars in the service station during the peak
hours, the average waiting time and the average number of cars per hours
that cannot enter the station because of full capacity.
Dr. B.B. Upadhyay
Queueing Theory
September 2, 2023
143 / 207
Operating characteristics of queueing system
Solution
Here λ = 1 car per minute, µ =
ρ = µλ = 6 and
P0 =
h 3−1
X 1
n=0
1
n!
1
6
car per minute, C = 3, N = 7 and
n
(6) +
7
X
1 × 6n i
n=3
3n−3 3!
=
1
.
1141
Expected number of cars in the queue is
E (m) =
=
P0 (C ρ)C ρ
[1
C !(1−ρ)2
− ρN−C +1 − (1 − ρ)(N − C + 1)ρN−C ]
(3×6)3 ×6 1
.
(1
3!×(−5)2 1141
− (6)5 − (−5)(5)(6)4 )
= 3.09 cars.
Dr. B.B. Upadhyay
Queueing Theory
September 2, 2023
144 / 207
Operating characteristics of queueing system
2
Expected number of cars in the service station
E (n) = 3.09 + 3 − P0
2
X
(3 − n)
n=0
3
n!
(6)n = 6.06 cars.
Expected waiting time a car spends in the system
E (v ) =
Since, Pn =
Dr. B.B. Upadhyay
0.06
6.06
=
7
(6)
1(1 − P7 )
1 − 3!34 ×
1
C !C n−C
n
λ
µ
1
1141
= 12.3 minutes
P0 , for c ≤ n ≤ N.
Queueing Theory
September 2, 2023
145 / 207
Operating characteristics of queueing system
4
Expected number of cars per hour that cannot enter the station is
60λPN = 60 × 1 × P7
(6)7
1
×
4
3!3
1141
= 30.3 cars per hour .
= 60 ×
Dr. B.B. Upadhyay
Queueing Theory
September 2, 2023
146 / 207
Operating characteristics of queueing system
Exercise
Problem 1
A barber shop has two barber and three chair for waiting customers.
Assume that customers arrive in a Poisson fashion at the rate of 5 per
hour that each barber services according to an exponential distribution
with the mean of 15 minute. Further, if a customer arrives and there are
no empty chairs in the shop he will leave. Find the steady-state
probabilities. What is the probability that the shop is empty ? What is the
expected number of customers in the shop?
Dr. B.B. Upadhyay
Queueing Theory
September 2, 2023
147 / 207
Operating characteristics of queueing system
Problem 2
A car servicing station has two bays where service can be offered
simultaneously. Due to space limitation, only four cars are accepted for
servicing. The arrival pattern is Poisson with 12 cars per day. The service
time in both the bays is exponentially distributed with µ = 8 cars per day
per bay. Find the average number of cars waiting to be serviced and the
average time a car spends in the system.
Dr. B.B. Upadhyay
Queueing Theory
September 2, 2023
148 / 207
Operating characteristics of queueing system
Model VII {(M/M/C ) : (C /FIFO)}
This model is essentially the same as Model VI except that here N = C .
Therefore, we consider the situation where no waiting queue is allowed to
form. This gives rise to a stationary distribution known as Erlang’s first
formula and can be easily obtained by using result of Model VI with N=C.
Thus, we have
 n
1 λ P
0 if 0 ≤ n ≤ C
Pn = n! µ
0
otherwise
where
P0 =
C
hX
1 λ n i−1
n=0
n! µ
The resultant formulae for P0 is itself called Erlang’s Loss Formula.
Dr. B.B. Upadhyay
Queueing Theory
September 2, 2023
149 / 207
Operating characteristics of queueing system
Model VIII {(M/M/R) : (K /GDK > R)}
This model describes a queuing system in which there are R service
channels and the maximum limit imposed on incoming customers, say
K, is fixed.
A practical situation corresponding to this model is that of machine
servicing where the calling population is that of machines and an
arrival corresponds to a machine breakdown.
The repairmen, of course, are the servers.
Whereas a machine breaks down, it will result in loss in production
until it is repaired.
Dr. B.B. Upadhyay
Queueing Theory
September 2, 2023
150 / 207
Operating characteristics of queueing system
Consequently, a broken machine cannot generate new calls while in
service. The model is generally known as ”Machine-Repairman”
model.
Let there be R servers and the service times be identical exponential
random variable with mean µ1 .
Using the balance equations for the generalized model (Model IV)
with
(
(K − n)λ, if 0 ≤ n ≤ K
λn =
0,
if n ≥ K .
µn =



nµ,
Rµ,


0,
Dr. B.B. Upadhyay
0≤n<R
R≤n≤K
n>K
Queueing Theory
September 2, 2023
151 / 207
Operating characteristics of queueing system
we have
Pn =
 K λ.(K −1)λ...(K −n+1)λ

P0 ,

µ.2µ...nµ




=
 K! n

(K −n)!
λ

P0 ,


n!
µ






=
K λ.(K −1)λ...(K −n+1)λ
(Rµ).(Rµ)...(Rµ)(R−1)µ.(R−2)µ...(1) µ,
K!
(K −n)!
R n−R R!
n
λ
µ
P0
0≤n<R
R≤n≤K
if 0 ≤ n < R;
if R ≤ n ≤ K
!


K λ n


P0


µ


 n
if 0 ≤ n < R

!


n

K

n!
λ


P0
n−R

n R R! µ
if R ≤ n ≤ K .
Dr. B.B. Upadhyay
Queueing Theory
September 2, 2023
152 / 207
Operating characteristics of queueing system
To obtain the value of P0 , we use the boundary condition
K
X
Pn = 1.
n=0
Therefore,
P0 =
h R−1
X K
n=0
Dr. B.B. Upadhyay
n
!
λ n
µ
+
K
X
n=R
K
n
Queueing Theory
!
n!
λ n i−1
R n−R R! µ
.
September 2, 2023
153 / 207
Operating characteristics of queueing system
Characteristics of Model VIII
1
Average number of customers in the system is given by
E (n) =
=
PK
n=0 nPn
PR−1
= P0
2
nPn +
n=0
PK
hP
R−1
n=0 n
n=R
K
n
!
Pn
λ n
µ
+
1
R!
K
n=R n.
n
!
PK
n!
R n−R
λ n
µ
i
Average queue length is given by
E (m) =
PK
=
PK
n=R (n
n=R
nPn − R.
= E (n) −
Dr. B.B. Upadhyay
− R)Pn
PR−1
n=0
PR
n=K
Pn
nPn − R 1 −
PR−1
= E (n) −Queueing
R + Theory
n=0 (R − n).
PR−1
n=0
K
!
Pn
λ n
September
µ 2, 2023
154 / 207
Operating characteristics of queueing system
3
= E (n) − R +
PR−1
= E (n) − R +
PR−1
n=0 (R
− n)Pn
K
n=0 (R − n). n
!
λ n
µ .
To find the average waiting, we first of all find out the mean rate of
arrival into the system. Now, let there be n units in the system, which
mean arrival rate λ. Therefore, the mean arrival rate to the system in
(K − n)λ. Thus by taking summation over all n and after weighting
each term by its probability Pn , we find that
λeff =
X
(K − n)λPn = λK
X
Pn −
X
λnPn = λ[K − E (n)].
Hence by using Little’s formulae,
E (n)
[E (n)]
=
E (v ) =
λeff
λ[K − E (n)]
and
E (m)
E (m)
E (w ) =
=
.
λeff
λ[K − E (n)] September 2, 2023
Dr. B.B. Upadhyay
Queueing Theory
155 / 207
Operating characteristics of queueing system
Problem
We have 5 machine, each of which when running, suffers breakdown at an
average rate of 2 per hour. There are 2 servicemen and only one man can
work on machine at a time. If n machines are out off order when n > 2
then (n − 2) of them wait untill serviceman is free. Once a serviceman
starts work on a machine the time to complete the repair has an
exponential distribution with mean 5 minutes. Find the distribution of the
number of machines out of action at a given time. Find also the average
time an out-of-action has to spend waiting for the repairs to start.
Dr. B.B. Upadhyay
Queueing Theory
September 2, 2023
156 / 207
Operating characteristics of queueing system
Solution
We are given
K = total number of machines in the system =5
R = number of serviceman =2, λ = 2, µ = 60
5 .
Let n = number of machines out of order, Then
Pn
 !

5 2 n



P0

60
 n
5
!
=
n


n!
2
 5

n−2

 n 2 2! 605 P0
if 0 ≤ n < 2,
if 2 ≤ n ≤ 5.
and
P0
=
hP
2−1
n=0
!
n P
5
5
+ 5n=2
Pn 602
n
n
5
!
n!
2n−2 2!
Dr. B.B. Upadhyay
Queueing
Theoryis given by
∴ Average
time of machine out of
action
2
n i−1
60
5
September 2, 2023
157 / 207
Operating characteristics of queueing system
= P3 + 2P4 + 3P5
=
165
1493
= 0.110 hours.
Average time an out-of-action machine has to spend waiting for the repairs
start is,
E (w ) =
But
λ′ =
P5
n=0 (5
E (m)
.
λ′
− n)Pn
= λ[5P0 + 4P1 + 3P2 + 2P3 + P4 ]
= λ + 0.434 × 9.189
= 8 machines (approx )
Dr. B.B. Upadhyay
Queueing Theory
September 2, 2023
158 / 207
Operating characteristics of queueing system
Problem
1 Two repairman are attending five machine in workshop. Each
machine breaks down according to a Poisson distribution with mean 3
per hour. The repair time per machine is exponential with mean 15
minutes.
a
b
2
Find the probability that the two repairman are idle, that one
repairman is idle.
What is the expected number of idle machine not being serviced ?
A machine services four machines. For each machine, the mean time
between service requirement is 10 hours and is assumed to follow
exponential distribution. The repair time tends to follow the same
distribution with mean of two hours. When a machine is down for
repairs, the time lost has a value of Rs. 20 per hour. The mechanics
costs Rs. 50 per day.
a
b
c
What is the expected number of machine in operation?
What is the expected down-time cost per day?
Would it be desirable to provide two mechanics each to service only
two machines?
Dr. B.B. Upadhyay
Queueing Theory
September 2, 2023
159 / 207
Operating characteristics of queueing system
Problem
1 A group of engineer has two terminals available to aid in their
calculation. The average computing job requires 20 minutes of
terminal time, each engineer requires some computation about once
every 0.5 hours, i.e., the mean time between a call for service is 0.5
hours. Assume these are distributed according to an exponential
distribution. If there are six engineers in the group, find:
a
b
2
The expected number of engineers waiting to use one of the terminals.
The total lost time per day.
A repairman is to be hired to repair machines which break down at an
average rate of 6 per hour. The breakdown follows a Poisson
distribution. The non-production time of machine is considered to
cost Rs. 20 per hour. Two repairmen, Mr. X and Mr. Y, have been
interviewed for this purpose. Mr. X charges Rs. 10 per hour and he
services breakdown machines at the rate of 8 per hour. Mr. Y
demands Rs. 14 per hour and he services at an average rate of 12 per
hour. Which repairman should be hired ? (Assume 8-hours shift per
day.)
Dr. B.B. Upadhyay
Queueing Theory
September 2, 2023
160 / 207
Operating characteristics of queueing system
Model IX (Power Supply model)
This model describes a situation where an electric circuit supplies power of
’a’ consumer. The requirement of consumers are assumed to follow
Poisson distribution with parameter λ, and the supply schedule also follows
Poisson distribution with parameter µ.
Using the balance equation for the generalized model (Model IV) with
λn = (a − n)λ, 0 ≤ n ≤ a and µn = nµ.
We have,
Pn =
1
n! a(a
− 1)...(a − n + 1)
a! λ a
a! µ P0 .
P
condition ∞
n=0 Pn
λ n
µ P0
Pa =
Using the boundary
h
P0 1 +
Dr. B.B. Upadhyay
= 1, we get
i
λ a(a − 1)
+
(λ/µ) + ... + (λ/µ)a = 1
µ
2!
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September 2, 2023
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Operating characteristics of queueing system
i.e.,
P0 [1 + λ/µ]a or P0 =
µ a
µ+λ
Therefore,
Pn =
=
1
n! a(a
− 1)...(a − n + 1))(λ/µ)n
µ
a!
n
µ+λ
(a−n)!n! (λ/µ)
µ
µ+λ
a
a
!
=
a λ n µ a−n
λ+µ
λ+µ
n
which is a binomial distribution.
Dr. B.B. Upadhyay
Queueing Theory
September 2, 2023
162 / 207
Operating characteristics of queueing system
Non-Poisson Queueing System
Queues in which arrival and/or departure may not follow the Poisson
axiom, are called non-Poisson queue. The development of these queuing
system is more complicated, mainly because the Poisson axioms no longer
hold good.
Following are the techniques which are usually employed in studying the
non-Poisson queues:
a Phase technique,
b The technique of Imbibed Markov Chain
c
The supplementary Variable technique
The phase technique is used when an arrival demands phases of services,
say k in number. The technique by which non-Markovian queues are
reduced to Markovian is termed as Imbedded Markov Chain Technique.
When one or more random variables are added to convert a non-Markovian
process into the Markovian one, the technique involved in this conversion
Dr. B.B. Upadhyay
Queueing Theory
September 2, 2023
163 / 207
Operating characteristics of queueing system
1
GI/G/C
2
M/G/I
3
GI/M/S
4
GI/Ek /1
5
D/Ek /1.
The study of these non-Poisson queues in not presented here initially.
However, we shall introduce the queuing system M/Ek /1 and the
steady-state result of M/G/1.
Dr. B.B. Upadhyay
Queueing Theory
September 2, 2023
164 / 207
Operating characteristics of queueing system
Erlangian Service Time Distribution with k-phases
When one unit has to pass through k stages for its service, then these k
stages of servicing are called k-phases.
The distribution of servicing time in each of these phases will be an
independent variable and the distribution of the total servicing time of a
unit in the system will be some combined distribution of time in all these
phases.
In the previous section we have assumed that the servicing time of a unit
follows the exponential distribution given by µe −µt with µ1 as the average
servicing time. Since the exponential distribution involves only one
parameter (the parameter µ ), it is known as one parameter distribution.
A two-parameter generalization of the exponential distribution is called
Erlangian (Gamma ) service time distribution. The p.d.f. of this
Erlangian service time distribution is defined by
Dr. B.B. Upadhyay
−kµt
g(t; µ; k) = CkQueueing
t k−1 eTheory
, k = 1, 2, ...
September 2, 2023
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Operating characteristics of queueing system
The value of constant Ck is given by
Z ∞
Z ∞
g(t; µ; k)dt or
0
or
Thus Ck =
0
Ck
(k − µ)k
(kµ)k
(k−1)!
Z ∞
Ck t k−1 e −kµt dt = 1
y k−1 e −y dy = 1 for kµt = y
0
and therefore g(t, µ, k) =
(kµ)k t k−1 e −kµt
;
(k−1)!
o ≤ t < ∞.
where,
µ = expected number of customers completing service per unit of time,
and
k = a positive constant.
The expected value of the total service time and variance of Erlang
distribution are k(1/kµ), i.e., 1/µ and k(1/kµ)2 ,i.e., 1/kµ2 respectively.
The following figure shows how the shape of the Erlang distribution
changes for various values of k. When k = 1, it reduces to the exponential
Dr. B.B. Upadhyay
Theory to a constant
September
2, 2023
166 / 207
distribution,
and for 1 < k < ∞,Queueing
it reduces
distribution
for
Operating characteristics of queueing system
Erlang Distribution
Model I {(M/Ek /1) : (∞/FIFO)}
This model consist of a single service channel queuing system in which
there are n phases in the system (waiting or in service ). It has been
assumed that a new arrival creates k − phase of service and departure of
one customer reduces k − phases of service. Let
n = number of customers in the system,
λn = λ, constant arrival rate per unit time,
µn = kµ, k phases of service pet unit time.
when Pn denotes the steady-state probability of n phases in the system,
the transition-rate
diagram of theQueueing
model
under consideration
is:
Dr. B.B. Upadhyay
Theory
September 2, 2023
167 / 207
Operating characteristics of queueing system
The balance equation, therefore, are:
λPn−k + kµPn+1 = λPn + kµPn
kµP1 = λP0
n≥1
n=0
Letting (λ/kµ) = ρ, these equations are:
(1 + ρ)Pn = ρPn−k + Pn+1
P1 = ρP0
n≥1
n=0
The solution of these difference equation is beyond the scope of this book.
However, we discuss the characteristics of this model using these difference
equations.
Dr. B.B. Upadhyay
Queueing Theory
September 2, 2023
168 / 207
Operating characteristics of queueing system
Characteristics of Model I
1
Average number of phases in the system E (np ) is obtained as follows:
Multiplying by n2 on both the side and then taking summation, the
above difference equation gives
(1 + ρ)
P∞
n=1 n
2P
n
=ρ
P∞
=ρ
P∞
n=k
n2 Pn−k +
x =0 (x
P∞
n=1 n
+ k)2 Px +
2P
n+1
P∞
y =2 (y
− 1)2 Py
(where n − k = x and n + 1 = y )
=ρ
P∞
y =0 (x
since
Dr. B.B. Upadhyay
2
+ 2xk + k 2 )Px +
P∞
y =2 (y
Queueing Theory
− 1)2 Py =
P∞
P∞
y =1 (y
y =1 (y
2
− 2y + 1
− 1)2 Py
September 2, 2023
169 / 207
Operating characteristics of queueing system
or
(1 − ρ)
∞
X
n2 Pn = (1 + ρ)
n=1
∞
X
n 2 Pn + ρ
n=1
∞
X
(2nk + k 2 )Pn +
n=0
∞
X
(−2n + 1)Pn
n=1
or
h
0 = ρ 2k
∞
X
nPn + k
n=0
2
∞
X
i
Pn +
n=0
∞
X
Pn − 2
n=1
∞
X
nPn
n=1
or
2(1−kρ)
∞
X
nPn = ρk 2 −P0 +1 since
n=0
∴ E (np ) =
∞
X
Pn = 1 and
n=0
∞
X
nPn =
n=1
∞
X
nPn .
n=0
k(k + 1)ρ
ρk 2 + 1 − 1(1 − kρ)
=
where P0 = (1 − kρ)
2(1 − kρ)
2(1 − kρ)
i.e., E (np ) =
Dr. B.B. Upadhyay
k(k + 1)
λ/kµ
k +1 λ
.
=
2
1 − kλ/kµ
2 µ−λ
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Operating characteristics of queueing system
2
Average waiting time of the phase in the system is given by
E (wp ) =
3
E (np )
k +1 λ
=
.
µ
2µ µ − λ
Average waiting time of an arrival is given by
E (w ) =
E (wp )
k +1
λ
=
.
k
2k µ(µ − λ)
4
Average time an arrival spends in the system is given by
1
k +1
λ
1
E (v ) = E (w ) + =
+ .
µ
2k µ(µ − λ) µ
5
Average number of units in the system is given by
k +1
λ2
λ
+ .
2k µ(µ − λ) µ
E (n) = λE (v ) =
6
Average queue length is given by
E (m) = E (n) −
Dr. B.B. Upadhyay
λ
k +1
λ2
=
.
µ
2k µ(µ − λ)
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Operating characteristics of queueing system
Model II {(M/Ek /1) : (1/FIFO)}
This model differs from Model I in the sense that in this case the capacity
of the system is unity, i.e., there is no queue and the system contains only
one customer.
Let the customer be in nth phase of service, where 1¡n¡k. Like Model, it
can be easily seen that the following set of steady-state difference
equations govern this model:
kµ = kµPn ,
1≤n<k
λP0 = kµPk
n=k
kµP1 = λP0 , n = 0
These equation reduce to
Pn =
λ
kµ P0 ;
This gives
us
Dr. B.B. Upadhyay
n = 1, 2, ..., k and Pn+1 = Pn ; n = 1, 2, ..., k − 1.
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September 2, 2023
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Operating characteristics of queueing system
Now, since
Pk
j=0 Pj
= 1, we have
P0 +
Pk
j=1 Pj
= 1 or P0 +
=⇒ P0 = 1 +
Hence
Pn =
Dr. B.B. Upadhyay
Pk
λ
j=1 kµ P0
=1
k
1 −1 λX
λ −1
= 1+
µ j=1 k
µ
1 λ
.
; n = 1, 2, ..., k
k λ+µ
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Operating characteristics of queueing system
Problem
A hospital clinic has doctor examining every patient brought in for a
general check-up. The doctor spends 4 minute on each phase of the
check-up although the distribution of time spent on each phase is
approximately exponential.
If each patient goes through four phases in the check-up and if the arrival
of the patients to the doctor’s office are approximately Poisson at the
average rate of three per hour, what is the average time spent by a patient
waiting in the doctor’s office ? What is the average time spent in the
check-up ? What is the most probable time spent in the check-up?
Dr. B.B. Upadhyay
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September 2, 2023
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Operating characteristics of queueing system
Solution
We are given
k = 4 and Mean arrival rate = 3 patients per hour, i.e., λ = 3 per hour.
Service time per phase =1/4µ = 4 minute .
∴µ=
E (w ) =
1
1
=
patient per minue
4×4
16
4+1
4×4
3
15
4
15
4
−3
= 40 minutes.
Average time spent in the examination 1/µ = 16 minutes.
Most probable time spent in the examination
=
Dr. B.B. Upadhyay
k −1
4−1
3
=
=
= 12 minutes.
1
kµ
1/4
4 × 16
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Operating characteristics of queueing system
Problem
At a certain airport it takes exactly 5 minutes to land an aeroplane, once it
is given the signal to land. Although incoming planes have scheduled
arrival times wide variability in arrival time produces an effect which makes
the incoming planes appear to arrive in a Poisson fashion at an average
rate of 6 hour. This produces occasional stockups at the airport which can
be dangerous and costly. Under these circumstances, how much time will
a pilot expect to spend circuling the field waiting to land?
Dr. B.B. Upadhyay
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September 2, 2023
176 / 207
Operating characteristics of queueing system
Solution
From the data of the problem, we have
λ = 6 per hour pr 1/10 per minute;
µ = 1/5 per minute
k = ∞, as servicing time is constant.
Hence, the average time which a pilot expects to spend circling the field
waiting to land is given by
k+1
λ
k→∞ 2k µ(µ−λ)
E (w ) = lim
1
k→∞ 2
= lim
1+
= 21 (1 + 0)
Dr. B.B. Upadhyay
1
k
λ
µ(µ−λ)
1/10 1
5
Queueing Theory
1
1
− 10
5
September 2, 2023
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Operating characteristics of queueing system
Exercise
1
A barber with one-man takes exactly 25 minutes to complete one
haircut. If customers arrive in a Poisson fashion at an average rate of
one every 40 minutes, how long on the average must a customer wait
for service? Also find the average time a customer spends in the
barber shop.
2
A tailoring shop with one man takes exactly one day to stitch a suit.
Customer’s arrival follows Poisson pattern with mean rate of arrival
one in every two day. How long, on an average, a customer is
expected to wait in such a situation?
3
A barber runs his own saloon. It takes him exactly 25 minutes to
complete one haircut. Customers arrive in Poisson fashion at an
average rate of one every 35 minutes.
a
b
For what percent of time would the barber be idle?
What is the average time of customers spent in the shop?
Dr. B.B. Upadhyay
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September 2, 2023
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Operating characteristics of queueing system
Model III {(M/G/1) : (∞/GD)}.
This model consists of single server, Poisson arrivals and general service
time distribution. In this case, the probability that a customers leaves the
system in (t, t + ∆t) is quantity dependent on when the service of that
particular customer begins, i.e., they would involve not merely t, but also
another time variable, say t ′ measured from the beginning of the current
service.
Let q0 denote the queue length when the service of customer n terminates,
and q1 be the queue length when the service of customer (n + 1)
terminates.
Further, let k(k = 1, 0, 2, ..) denotes the number of customers who arrive
during the service of the customer (n + 1). Then we can have the
following relation between q0 , q1 and k:


k;





q1 = Dr. B.B. Upadhyay
if q0 = 0 , i.e., the queue length is zero after serviing
customer
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Operating characteristics of queueing system
The expected value of the above relation is obtained as,
E (q1 ) = E (q0 ) − E (δ) + E (k)
Since E (q1 ) = E (q0 ) in the steady state, it follows that
E (δ) = E (k).
Further,
q12 = (q0 − δ + k)2
= (q02 + δ 2 k 2 + 2kq0 − 2kδ − 2δq0 ).
But by definition, δ 2 = δ and δq0 = q0
∴ q12
= q02 + k 2 + 2kq0 + δ − 2kδ − 2δq0
or E (q12 ) = E (q02 ) + E (k 2 ) + 2E (kq0 ) + E (δ) − 2E (kδ) − 2E (q0 )
Again, since E (q12 ) = E (q02 ) in the steady-state, it follows that
2E (q0 ) − 2E (kq0 ) = E (k 2 ) + E (δ) − 2E (kδ)
Dr. B.B. Upadhyay
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September 2, 2023
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Operating characteristics of queueing system
E (q0 ) =
or
+ E (k) − 2E 2 (k)
2{1 − E (k)}
E (k 2 )
since E (kδ) = E (k)E (δ), E (δ) = E (k) and q0 , k and δ are independent.
If λ is the mean arrival rate, the number of arrivals in a service time of
length t has a Poisson distribution with mean λt. The probability of k
arrivals in time t, therefore, is given by
(λt)k e −λt
, k ≥ 1,
k!
and the probability of k arrivals during the service time of a customer is
Z ∞
(λt)k e −λt
0
k!
p(t)dt,
where p(t) is the probability density function of service time.
Dr. B.B. Upadhyay
Queueing Theory
September 2, 2023
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Operating characteristics of queueing system
The mean number of arrivals in the service time of a customer is given by
E (k) = λE (t) = λ(1/µ);
where µ is the mean service rate.
The variance of arrival in the service time of a customer is
E (k 2 ) = λE (t) + λ2 E (t 2 )
= λ × [mean service time] + λ2 × [variance of service time + squar
mean service time]
= (λ/µ) + [λ2 (σ 2 + 1/µ2 )]
= ρ + λ2 σ 2 + ρ2 .
Substituting the value of E (k) and E (k 2 ) in the relation for E (q0 )
obtained above, we get
2 2
Dr. B.B. Upadhyay
2
2
(ρ+λ σ +ρ )+ρ−2ρ
E (q0 ) =Queueing
2(1−ρ)
Theory
September 2, 2023
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Operating characteristics of queueing system
Further, if the customer (n+1) has to wait for time ω before being taken
into service, the length of the queue must be q1 during the time ω + t.
∴ E (q1 ) = λE (ω + r ) = λE (ω) + λE (r )
i.e.,
λE (ω) = Average waiting time = E (q1 ) − (λ/µ),
where E (r ) = 1/µ,
or
E (ω) =
1h
λ
ρ+
λ2 σ 2
i
+ ρ2
λ2 σ 2 + ρ2
−ρ =
,
2(1 − ρ)
2λ(1 − ρ)
Since E (q1 ) = E (q0 ) in the steady-state.
Dr. B.B. Upadhyay
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September 2, 2023
183 / 207
Operating characteristics of queueing system
We now summarize the various formulae for (M/G/1) : (∞/GD) as
follows:
1
Average number of customers in the system =
λ2 σ 2 + ρ2
+ ρ,
2(1 − ρ)
2
Average queue length =
h λ2 σ 2 + ρ2
2(1 − ρ)
3
i
+ρ −ρ=
λ2 σ 2 + ρ2
,
2(1 − ρ)
Average waiting time of a customer in the queue=
λ2 σ 2 + ρ2
.
2λ(1 − ρ)
4
Average waiting time that a customer spends in the system =
λ2 σ 2 + ρ2
1
+ .
2λ(1 − ρ)
µ
Dr. B.B. Upadhyay
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September 2, 2023
184 / 207
Operating characteristics of queueing system
Remark. The formulae for the average queue length and the average
number of customers in the system are known as
Pollaczek − Khintchine (or P − K ) formulae.
Problem
In a heavy machine shop, the overhead crane is 75 per cent utilized. Time
study observation gave the average slinging time as 10.5 minute with a
standard deviation of 8.8 minutes.
What is the average calling rate for the services of the crane, and what is
the average delay in getting service?
If the average service time is cut to 8.0 minutes, with standard deviation
of 6.0 minutes, how much reduction will occur, on average, in the delay of
getting served?
Dr. B.B. Upadhyay
Queueing Theory
September 2, 2023
185 / 207
Operating characteristics of queueing system
Solution
This is a (M/G/1) : (∞/FCFS) process.
The average delay in getting service is given by
E (w ) =
ρ(1 + µ2 σ 2 )
2µ(1 − ρ)
Initial solution:
ρ = 0.75.
µ=
60
10.5
= 5.71 per hour,
λ = ρ × µ = 0.75 × 5.71 = 4.29 per hour.
∴ E (w ) =
=
Dr. B.B. Upadhyay
0.75
2(1−0.75)
0.75
0.5
h
1 + (5.71)2
× 1.70 ×
8.8 2
60
i
×
60
5.71
60
5.71
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Operating characteristics of queueing system
If service time is cut to 8 minute, then
µ=
60
4.29
= 7.5 per hour , and ρ =
= 0.571
8
7.5
or utilisation of the crane reduced to 57.1 present. Then
E (w ) =
=
0.571
2(1−0.571)
0.571
2×0.429
h
1 + (7.5)2 ×
6.0 2
60
i
×
60
7.5
× 1.562 × 8
= 8.3 minutes,
a reduction of 18.5 minutes or approximately 70 per cent.
Dr. B.B. Upadhyay
Queueing Theory
September 2, 2023
187 / 207
Operating characteristics of queueing system
Exercise
1
Consider a queuing system where arrivals are according to Poisson
distribution with mean 5. Find expected waiting time in the system if
the service time distribution is
i
ii
2
Uniform from t=5 to t=15 min,
Normal with mean 3 min. and variance 4 min2 .
At certain health-care center, patients arrive at a mean rate of 4 per
hour and they are checked by doctor at a mean rate of 5 per hour.
The center feels that service time have some unspecified positive
skewed unimodel two-tailed distribution with a standard deviation of
0.05 hours.
i
ii
Determine the queuing characteristics for the health care center.
How much the assumption of exponential service time would distort
these values. Discuss.
Dr. B.B. Upadhyay
Queueing Theory
September 2, 2023
188 / 207
Operating characteristics of queueing system
COST MODELS IN QUEUING
The service level in a queuing facility is a function of the service rate µ,
that balances the following two conflicting costs:
1
Cost of offering the service, and
2
Cost incurred due to delay in offering the service (customer waiting
time ).
The two types of costs conflict because an increase in the existing service
facility would reduce the customer’s waiting time (i.e., cost of waiting).
On the other hand, a decrease in the level of service would increase the
cost of waiting.
Following figure illustrate both the type of costs as a function of level of
service. The optimum service level is one that minimizes the sum of two
costs.
Dr. B.B. Upadhyay
Queueing Theory
September 2, 2023
189 / 207
Operating characteristics of queueing system
The cost model can be expressed as
T = K µ + CE (n),
where
K=cost of service rate per unit time,
C=cost of waiting customers per unit time, and
E(n)=average number of customers in the system.
Dr. B.B. Upadhyay
Queueing Theory
September 2, 2023
190 / 207
Operating characteristics of queueing system
Optimum service rate. The procedure for determining the optimum
service rate, µ, is based on Model I of Poisson Queuing system.
Now, in the case of an (M/M/1) : (∞/FIFO) model,
E (n) =
Cλ
λ
; and therefore T = K µ +
.
µ−µ
µ−λ
Since the service rate µ is continuous, its optimum value can be obtained
using the technique of maxima and minima. Thus, we get
d
Cλ
d2
2λC
(T ) = K −
and
(T ) =
> 0.
2
2
dt
(µ − λ)
dµ
(µ − λ)3
Hence, the optimum value of µ is µ0 = λ +
p
λC /K .
This result shows that the optimum value of µ is not only dependent on K
and C, but on the arrival rate λ also.
Dr. B.B. Upadhyay
Queueing Theory
September 2, 2023
191 / 207
Operating characteristics of queueing system
Remark
In the case of (M/M/1) : (N/FIFO), the above cost model may be
modified to reflect that the larger the value of N, the smaller will be the
number of lost customers. This modified cost function is:
T = K µ + CE (n) + D1 N + λPN D2
Where
D1 = cost of servicing each additional customers per unit time,
D2 = cost per lost customer, and
λPN = number of lost customers per unit time.
Dr. B.B. Upadhyay
Queueing Theory
September 2, 2023
192 / 207
Operating characteristics of queueing system
Optimum number of Servers
For determining the optimum number of service channel, we consider the
operational characteristics of queuing model (M/M/S) : (∞/FIFO). In
this case the cost function can be written as:
T = KS + CE (n)
where
K = cost per service unit time,
C = cost of waiting per customers per unit time,
E (n) = average number of customers in the system when there are S
servers.
As the number of servers S can be measured in discrete units only,
differentiation is not applicable to this case. Instead, one can apply the
method of finite difference. The function T will have a local minimum at
Dr. B.B. Upadhyay
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September 2, 2023
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Operating characteristics of queueing system
∆T (S0 − 1) < 0, i.e., T (S0 ) − T (S0 − 1) < 0,
We have
[K × S0 + C × L(S0 )] < K (S0 − 1) + C × L(S0 − 1)
or
L(S0 − 1) − L(S0 ) > K /C .
Similarly, considering ∆T (S0 ) > 0, we get
L(S0 ) − L(S0 + 1) < K /C .
Both these results together yield
L(S0 ) − L(S0 + 1) <
K
< L(S0 − 1) − L(S0 ).
C
The value K/C indicates where the search for optimum S should start.
Dr. B.B. Upadhyay
Queueing Theory
September 2, 2023
194 / 207
Operating characteristics of queueing system
Other Queuing Models
We have discussed the model that are rather simple to analysis. However,
it is easy to conceive several other models by changing the queue discipline
or by considering different pattern of arrival and service rate other than
Poisson and exponential.
We may have finite or infinite queues, circular queues, queues in series,
queue with balking and reneging. We list some of them:
1
Queues in series where input of a queue is derived from the output of
the another queue.
2
Circular queue where a customer after being served joins the queue
again,
3
Multi-channel queues with different service rate at each channel,
4
Queues with priority or pre-emptive priority, etc.
Dr. B.B. Upadhyay
Queueing Theory
September 2, 2023
195 / 207
Operating characteristics of queueing system
Queuing Control
The origin of Queuing Theory dates back to attempts of determining the,
now, well- known queuing characteristics, such as, mean queue length,
variance, etc.
During the past 30 year and so, there has been a rapidly interest in the
study of designing and controlling of the queuing system behavior
(prescriptive model as opposed to descriptive models that make up the
majority of queuing literature to date).
Prescriptive models were viewed as statics optimization models. The
statics optimization models are those in which we set up steady-state
conditions for the system and determine some long-run average criterion
such as cost and/or profit.
In statics models, the system configuration is set once for all. When the
queuing system are allowed to depend upon time and are controlled, then
these system are known as dynamic control systems.
Dr. B.B. Upadhyay
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September 2, 2023
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Operating characteristics of queueing system
A lot of work has been done in dynamic control of queuing systems.
Following are the objective of dynamics control:
Arrival Process Control
a
b
c
d
e
To accept or reject control.
To adjust mean arrival rate.
Customer exercised control.
Self versus social optimization.
Projection times.
Service Control System
1 Varying a number of servers.
2 Varying the service time.
Apart from the two aspect, namely, Statics (design) and Dynamics
(control) Optimization, there is one more aspect of optimization, known as
”Control of Queue Discipline”-Priority Models, Scheduling Models and
Dr. B.B. of
Upadhyay
Queueing Theory
September 2, 2023
197 / 207
allocation
customers to multi-servers
fall in this category.
Operating characteristics of queueing system
Queuing Theory And Inventory Control
We now give example suggesting a queuing theory approach to an
inventory problem.
In inventory technique we have commodity input, its storage and finally
the sales demands. We can imagine the queue to be consisting of demand
orders waiting to be satisfied, but the real queue is often that of waiting
for replenishment.
Consider the manufacturer of a very expensive item who uses the following
inventory control procedure:
Let the manufacturer keep a buffer stock of Q units on hand. The
customer demand us described by Poisson distribution with mean λ.
Whenever a demand for a unit comes (the customer arrive ) the
manufacturers places an order on the factory to manufacture another unit.
Let the amount of time required to manufacture a unit be exponential
Dr. B.B. Upadhyay
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September 2, 2023
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Operating characteristics of queueing system
Further, let C1 be the carrying cost for inventory storage cost per unit
time. Whenever, the buffer stock depletes to zero, Let C2 be the shortage
cost per unit time.
It is also assumed that in the case of shortage, the customer will wait until
the stock is replenished by order due.
To look at it from another view point, we may consider the shortage cost
C2 as the discount allowed to the waiting customers for his not moving to
some other firm.
The problem now is that of determining the optimum value of Q so as to
minimize the total expected cost
TEC (Q) = C1
Q
X
r =1
rp(r ) + C2 λ
0
X
p(r ),
r =−∞
where r is on-hand inventory level. The positive range for r corresponds to
the items in stock whereas the negative range for r corresponds to items in
back Dr.
order
; and p(r) is its pointQueueing
probability
function. September 2, 2023
B.B. Upadhyay
Theory
199 / 207
Operating characteristics of queueing system
We observe that
Q
X
p(r )
r =1
is the average value of the safety stock and
λ
0
X
p(r )
r =−∞
is the expected number of breakdown per unit time.
This is so because
0
X
p(r )
r =−∞
is the percentage of time during which there is no-stock safety stock and λ
is the average demand rate.
If one is able to determine p(r ), TEC (Q) could easily be optimized with
respect to Q.
Dr. B.B. Upadhyay
Queueing Theory
September 2, 2023
200 / 207
Operating characteristics of queueing system
Review Question
1
List the components of queueing system.
2
What is queueing theory? In what types of problem situations it can
be applied successfully? Discuss giving examples.
3
What are the limitations of Queueing Theory? What information can
be obtained by analysing a queueing system?
4
What do you understand by queue discipline and service process?
5
Briefly explain the important characteristics of queueing system.
6
In what kind of situations is queueing analysis most appropriate?
7
If the number of arrivals in some time interval follows a Poisson
distribution, obtain the distribution of the time interval between two
consecutive arrivals.
Dr. B.B. Upadhyay
Queueing Theory
September 2, 2023
201 / 207
Operating characteristics of queueing system
9
With respect to the queue system, explain; the following:
i
ii
iii
iv
v
Input process,
Queue discipline,
Reneging,
Jockeying, and
Balking
10
Explain the single channel and multi-channel queueing models.
11
What do you understand by:
a
b
i
ii
iii
i
ii
queue length
traffic intensity
the service channels
steady and transient state
utilization factor
12
What is service discipline? Describe some forms of common service
discipline and illustrate with examples.
13
For a single server, Poisson arrival and Exponential service time
queuing system, obtain the steady state equations satisfied by the
Dr. B.B. UpadhyayPn of n customers
Queueing
Theory
September
2, 2023 Pn . 202 / 207
probability
in the
system and hence
obtain
Operating characteristics of queueing system
14
What are the operating characteristic s of a single channel waiting
line with Poisson arrivals and exponential service times?
15
Show that for a single service station with Poisson arrival and
exponential service time, the probability that exactly n calling units
are in the queue system is
Pn = (1 − ρ)ρn , n ≥ 0.
where ρ is the traffic intensity. Also, find the expected number of
units in the system.
16
Show that for {(M/M/1) : (∞/FCFS)}, the distribution of waiting
time in the queue is
(
P(w ) =
Dr. B.B. Upadhyay
1−ρ
µρ(1 − ρ)e −(µ−λ)w , w > 0 and ρ = µλ .
Queueing Theory
September 2, 2023
203 / 207
Operating characteristics of queueing system
17
For {(M/M/1) : (∞/FIFO)} queueing model in the steady-state
case, show that:
1
The expected number of units in the system and in the queue are given
by
λ2
λ
and E (m) =
E (n) =
µ−λ
µ(µ − λ)
2
Expected waiting time the customer spends in the system (including
1
service) is µ−λ
.
18
Explain (M/M/1) : (N/FCFS) system and solve it under steady-state
conditions.
19
For {(M/M/1) : (N/FIFO)} queueing model:
i
ii
20
find the expression for E (n).
derive the formula for Pn and E (n), where ρ = 1.
State and explain the condition for the existence of steady state in
case of MIMIC queueing system.
Dr. B.B. Upadhyay
Queueing Theory
September 2, 2023
204 / 207
Operating characteristics of queueing system
21
For {(M/M/C : (∞/FIFO)} queueing model, obtain in a steady
state, an expression for
i
ii
22
the probability that there are k(≤ C ) customers in the system,
proportion of times the counter remains idle.
For a {(M/M/C ) : (∞/FCFS)} queueing model, derive the
expressions for:
i
ii
iii
the steady state equation.
probability that a customer will not have to wait.
expected number of customers in the queue.
23
Obtain the steady-state solution for the model
{(M/M/C ) : (GD/N/∞)}.
24
Obtain the steady-state solution for the model
{(M/M/C ) : (GD/∞/∞)}.
25
Explain Erlangian service time distribution and show when it can be
approximately assumed in a queueing problem.
Dr. B.B. Upadhyay
Queueing Theory
September 2, 2023
205 / 207
Operating characteristics of queueing system
27
Define an M/G/l queueing system and explain briefly the various
characteristics of an M/G/1 queueing system.
28
Basic problem in queueing theory is to determine an optimal service
level.” Elucidate this statement.
29
Queueing Theory can be used effectively in determining optimal
service level. Elucidate this statement with the help of an example.
30
Discuss the costs associated with queueing system. Also explain the
concepts of optimum servicing rate and optimum cost.
31
Why do waiting lines form even though a service system is
underloaded? In a multiple-channel system, what is the ration ale for
having customers wait in a single line, as is now being done in many
banks and post offices, rather than multiple lines?
32
What are the components of total relevant cost in case of single
channel
queueing model? Queueing Theory
Dr. B.B. Upadhyay
September 2, 2023
206 / 207
Operating characteristics of queueing system
Thank You!
Dr. B.B. Upadhyay
Queueing Theory
September 2, 2023
207 / 207