Unit V - Queuing Theory

Syllabus
• Introduction
• Queuing System
• Classification of Queuing models
• Distribution of Arrivals (The Poisson Process): Pure Birth Process
• Distribution of Inner-arrival Time
• Distribution of Departures (Pure Death Process)
• Distribution of Service Time
• Solution of Queuing models
• Model 1 (M/M/1) : (∞/FCFS): Birth and Death Model

Introduction
• In 1903, A.K. Erlang, a Swedish engineer started theoretical analysis of
waiting line problem in telephone calls.
• Later, in 1951, D.G. Kendall provided a systematic and mathematical
approach to waiting line problem.
• Queue is a waiting line or act of joining a line.
• Example: Ration shop, airport runways, traffic signal, etc.
• Waiting has become an integral part of out day-to-day life.
• The objective is to formulate the system in such a manner that the
average waiting time of the customers in queue is minimized.
• The utilization of server is maximized.

Queuing System
• The input or arrival pattern (customers)
• The service mechanism (service pattern)
• The ‘queue discipline’
• Customer’s behavior

Arrivals
Exit
Queues Service mechanism
Figure 1

Input process or Arrival pattern
• This process is usually called arrival process and arrivals are called
customers.
• No more than one arrival can occur at any given time. If more than
one arrival can occur at any given time, we call it as bulk arrival.
• In general, customers arrive in a more or less random fashion. The
arrival pattern can be described in terms of probabilities and
consequently the probability distribution for inter-arrival time.
• There may be many arrival patterns but Poisson, Exponential and
Erlang distributions are most commonly considered.

The service mechanism
• It includes the distribution of time to service a customer, number of
servers is more than one, example is parallel counters for providing
service and arrangement of servers.
• Queues in tandem service is to be provided in multistage in
sequential order.
• Service time is a random variable with the same distributions for all
arrivals.
• It follows ‘negative exponential distribution’

The queue discipline
• It refers to the order or manner in which customers join queues.
• It describes the method used to determine the order in which
customers are served.
• First In First Out (FIFO) or First Come First Served (FCFS): The
customers are served in the order in which they join in the queue.
• Example: cinema ticket windows, bank counters etc.
• Last in First Out (LIFO) or Last Come First Served (LCFS): In this
system, the element arrived last will have a chance of getting service
first.

• Service in Random Order (SIRO): A service in a random order
irrespective of their arrival order in the system.
• Example: Government offices, etc.
• Service by Priority or Priority Selection: Priority disciplines are those
where any arrival is chosen for service ahead of some other
customers already in queue.
• Example: in a doctor’s chamber, doctor may ask the patient with
stomach pain to wait for sometime and give attention to the heart
patient.

Customer’s behavior
• It is assumed that the arrivals in the system are one by one.
• Patient customer: If a customer, on arriving at the service system,
waits in the queue until served and does not switch between waiting
lines.
• Impatient customer: Machines arrived at the maintenance shop are
examples of patient customers.
• The customer who waits for a certain time in the queue and leaves
the service system without getting service due to certain reasons.
• Bulk arrival: Customers may arrive in groups.
• This is observed when groups of ladies arrive at shopping malls during
afternoon.

• Balking: A customer may not like to join the queue due to long
waiting line and has no time to wait or there is not sufficient space for
waiting.
• Reneging: A customer may leave the queue after waiting for
sometime due to impatience.
• Priorities: In some circumstances, some customers are served before
others regardless of their order of arrival. These customers have
priority over others.
• Jockeying: When there are a number of queues, a customer may
move from one queue to another with the hope of receiving the
service quickly.
• Collusion: Several customers may cooperate and only one of them
may stand in the queue.

Queue model needs a question
• What probability distribution is followed by arrival and service
mechanism?
• What is the busy period distribution?
• How much time is spent by a customer in the queue before his
service starts and total time spent by him in system in terms of
waiting time and service time?

• Busy period: Server is free when customer arrives, latter will be
served immediately during this time some more customers arrive,
they will be served in their turn. It continues until no customer is left
unserved and server becomes free.
• Idle period: No customer is present in the system.
• Busy cycle: Comprises a busy period and idle period following it.
• A system is said to ‘transient state’ its operating characteristics are
dependent on time.
• The system is said to be ‘steady or prevail state’ when its behavior
becomes independent on time.

• Let Pn(t) denote the probability that there are n units in the system at
time t and P’n(t) be change in Pn(t) with respect to time t.
• Pn(t)  Pn as t  ∞ and P’n(t)  0 as t  ∞
• Explosive state: If arrival rate of system is greater than its service rate,
a steady-state may not be reached regardless of long run-time and
queue length will increase with time.

Only Steady-state queuing system. The mathematical
models are developed by using this notations.
• N Number of customers in the system
• C Number of servers in the system
• Pn(t) Transient state probability that exactly n-customers of units are
in the system at time t
• Pn Steady state probability of having n-customers of units in the
system
• P0 Probability of having no customers in the system
• Lq Average number of customers waiting in the queue

• Ls Average number of customers waiting in the system
• Wq Average waiting time of customers in the queue
• Ws Average waiting time of customers in the system
• λ Average arrival rate of customers
• μ Average service rate of server
• ρ Traffic intensity or utilization factor of the server defined as the
ration of λ and μ (λ < μ)
• M Poisson arrival
• N Maximum number of customers allowed in the system
• GD General discipline for services

Classification of Queuing models
• Using kendall’s notation, the queuing model can be defined by
(a/b/c) : (d/e)
• where
• A = arrival rate distribution
• B = service rate distribution
• C = number of servers
• D = capacity of the system
• E = service discipline
• Example: (M/M/1) : (∞/FCFS) means arrival follow poisson distribution.
• Poisson service rate, single server, system can accommodate infinite number
of customers and service discipline is FCFS.

Solution of Queuing models
• Step 1: To derive the system of steady-state equations governing the
queue.
• Step 2: To solve these equations for finding the probability distribution
of the queue length.
• Step 3: To obtain probability density function for waiting time
distribution.
• Step 4: To find busy period distribution.
• Step 5: To derive formula for Ls, Lq, Ws, Wq, and var{n} or v{n}.
• Step 6: To obtain the probability of arrival during service time of any
customer.

Distribution of Arrivals (The poisson process):
Pure Birth Process
• Pure Birth Model: The arrival pattern of customers varies from one
system to another and it is random too, mathematically, have a
poisson distribution and only arrivals are counted and also no
departure takes place.

1. Assume that there are n-units in the system at time t, and the
probability that exactly one arrival occur during small interval ∆t be
given by λ∆t + 0(∆t), where λ is the arrival rate independent of time
t and 0(∆t) contains the terms of higher powers of ∆t.
2. Under the assumption that ∆t is very small, the probability of more
than one arrival in time ∆t is negligible.
3. The number of arrivals in non-overlapping intervals is mutually
independent.
The probability of n-arrivals in time t. Denote it by Pn(t), (n > 0).
The difference-differential equations governing the process in two
different situations.
Order to derive the arrival distribution in queues

• Case 1: For n > 0, there are two mutually exhaustive events of having
n-units at time (t+ ∆t) in system.
• There are n-units in system at time t and no arrival takes place during
time interval ∆t.
• So at time (t + ∆t), there will be n-units in system.
n-units n-units
t t + ∆t
no-arrivals
Figure 2

• The probability of these two combined events will be
= Probability of n-units at time t x Probability of no arrivals during ∆t
= Pn(t) (1 - λ∆t) ______________ (1)
• There are (n – 1) units in the system at time r and one arrival takes place
during time interval ∆t.
• So at time (t + ∆t), there will be n-units in system.
(n-1)-units n-units
t t + ∆t
One arrivals
Figure 3

• The probability of these two combined events will be
• = Probability of (n-1)-units at time t x Probability of no arrivals during
∆t
= Pn-1(t) (1 - λ∆t) ______________ (2)
• Adding (1) and (2) two probabilities, we get probability of n-arrivals at
time t + ∆t as
Pn(t) (1 - λ∆t) = Pn(t) (1 - λ∆t) + Pn-1(t) λ∆t _________ (3)

• Case 2: When n=0, that is, there is no customer in the system.
• P0(t + ∆t) = Probability of no units at time t x Probability of no arrivals
during ∆t
= P0(t) (1 - λ∆t) ___________ (4)
Pn(t + ∆t) – Pn(t) = Pn(t)(1 - λ∆t) + Pn-1(t) λ∆t, n > 0 ________ (5)
Pn(t + ∆t) – Pn(t) = Pn(t)(1 - λ∆t), n = 0 ________ (6)
• The above two equations constitute the system of differential-
difference equations.
• Equation 6 can be written as
•
𝑃′
0
(𝑡)
𝑃0
(𝑡)
= - λ

• Integrating both sides with respect to t,
• log P0(t) = - λt + A ___________ (7)
• where A is constant of integration. Its value can be computed using
the boundary conditions
• ___________ (8)
• t = 0, P0(0) = 1 and hence A = 0.
• P0(t) = 𝑒−𝑖𝜆𝑡 ___________ (9)

• Putting n = 1, in equation (5)
• P′1(t) = -λP1(t) + λP0(t) = -λP1(t) + λ𝑒−𝜆𝑡
• which is linear differential equation of first order.
• ________ (10)
• Equation (8), B =0. Thus, Equation (10) can be written as
• P1(t) = λt𝑒−𝜆𝑡
• Arguing we get
• P2(t) =
(𝜆𝑡)2
2!
𝑒−𝜆𝑡 ……. Pn(t) =
(𝜆𝑡)𝑛
𝑛!
𝑒−𝜆𝑡

Distribution of Inter-Arrival Time
• The time T between two consecutive arrivals is called inter-arrival
time.
• Mathematical development is given to show that T follows negative
exponential law.
• Let f(T) be the probability density function of arrivals in time T.
• f(T) = λ𝑒−𝜆𝑡
________ (11)
• where λ is an arrival rate in time t.

• Proof: Let t be the instant of an arrival.
• Since there is no arrival during (t, t + T) and (t + T, t + T, T + ∆t), arrival be at
t + T + ∆t.
• Putting n = 0 and t = T + ∆t in Equation (10)
• P0(T + ∆t) = 𝑒−𝜆(𝑇+∆t) = 𝑒−𝜆𝑇[1 −𝜆∆t+О ∆t ]
• But P0(T) = 𝑒−𝜆𝑇
• So P0(T + ∆t) – P0(T) = P0(T)[- 𝜆 ∆t + О ∆t ]
• Dividing both sides by ∆t and taking time ∆t  0,
P’0(T) = -𝜆P0(T) ___________ (12)
• Left Hand Side (LHS) of equation (11) is probability density function of T,
say f(T).
f(T) = 𝜆𝑒−𝜆𝑇
___________ (13)
• which is negative exponential law of probability for T.

Markovian property of inter-arrival time
• It states that at any instant of the time until the next arrival occurs is
independent of the time that has elapsed since the occurrence of the
last arrival, i.e., Prob. (T > t1/T > to) = Prob. (0 < T < t1 – to).
0 to t1
Figure 4

• Proof: Using formula of conditional probability,
• Prob. (T > t1/T > to) =
𝑃𝑟𝑜𝑏.[ 𝑇≥𝑡1 𝑎𝑛𝑑 𝑇 ≥𝑡𝑜 ]
𝑃𝑟𝑜𝑏.(𝑇 ≥𝑡𝑜)
• 𝑡0
𝑡1
= 𝜆𝑒−𝜆𝑇𝑑𝑡
𝑡0
∝
= 𝜆𝑒−𝜆𝑇𝑑𝑡
• 1 − 𝑒−𝜆(𝑡1 −𝑡𝑜) = Prob. (0 < T < t1 – to)
• Hence, proved.

Distribution of departures (Pure Death Process)
• The model in which only departures occur and no arrival takes place is
called pure death process.
• There are N-customers in the system at time t = 0, no arrivals occur in the
system, and departures occur at a rate 𝝁 𝒑𝒆𝒓 𝒖𝒏𝒊𝒕 𝒕𝒊𝒎𝒆.
• The distribution of departures from the system on the basis of three
axioms.
1. The probability of exactly one departure during small interval ∆t be given
by 𝜇 ∆t + 0(∆t ).
2. The term ∆t is so small that the probability of more than one departure
in time ∆t is negligible.
3. The number of departures in non-overlapping intervals is mutually
independent.

• Case 1: When 0 < n < N (1 < n < N – 1)
• The probability will be
Pn (t + ∆t) = Pn(t)(1 - 𝜇 ∆t) + Pn+1(t) 𝜇 ∆t ___________ (14)
• Case 2: When n = N, that is, there are N-customers in the system.
PN (t + ∆t) = PN(t)(1 - 𝜇 ∆t) ___________ (15)
• Case 3: When n = 0, that is, there is no customer in the system.
P0 (t + ∆t) = P0(t) + P1(t) 𝜇 ∆t ___________ (16)
• Thus,
P0 (t + ∆t) = P0(t) + P1(t) 𝜇 ∆t, n = 0
Pn (t + ∆t) = Pn(t)(1 - 𝜇 ∆t) + Pn+1(t) 𝜇 ∆t, 0 < n < N
PN (t + ∆t) = PN(t)(1 - 𝜇 ∆t), n = N

• Rearranging the above equations, dividing by ∆t and taking limit
∆t0,
P’0(t) = P1(t) 𝜇, n =0 _________ (17)
P’n(t) = -Pn(t) 𝜇 + Pn+1(t) 𝜇, 0 < n < N _________ (18)
P’N(t) = - PN(t) 𝜇, n = N _________ (19)
• The above three equations are the required system of differential-
difference equations for pure death process.
• The solution of equation (19) is
log PN(T) = - 𝜇t + A
• where A is constant of integration.

• Its value can be computed using the boundary conditions
• Pn(0) = 0,
1, 𝑛 = 𝑁 ≠ 0
𝑛 ≠ 𝑁
gives A = 0. Therefore, putting n = N – 1 in
equation 18.
P’N-1(t) + 𝜇PN-1(t) = 𝜇𝑒−𝜇𝑡
• which is linear differential equation of first order. Its solution is
• PN-1(t) = 𝜇𝑡𝑒−𝜇𝑡
• Putting n = N – 2, N- 3, …… N – n, in equation (18)
Pn-k(t) =
(𝜇𝑡)𝑘
𝑘!
𝑒−𝑖𝜇𝑡 In general, Pn(t) =
(𝜇𝑡)𝑁 − 𝑛
(𝑁 − 𝑛)!
𝑒−𝑖𝜇𝑡
• which is a poisson distribution.

Distribution of Service Time
• Let T be the random variable denoting the service time and t be the
possible value of T.
• Let S(t) and s(t) be the cumulative density function and the probability
density function of T respectively.
• To find s(t) for the poisson departure case, it can be observed that the
probability of no service during time (0, t), which means the probability
of having no departures during the same period.
• Prob. (service time T > t) = Prob. (no departure during t) = PN(t)
• where there are N-units in the system and no arrival is allowed after N.
• PN(t) = 𝑒−𝜇𝑡
.

• S(t) = Prob. (service time T < t) = 1 – Prob. (service time T > t) = 1 - 𝑒−𝑖𝜇𝑡
• Differentiating both sides with respect to ‘t’
• S(t) = 0,
𝜇𝑒−𝜇𝑡, 𝑡 ≥ 0
𝑡 < 0
• This shows that the service time distribution is exponential with mean
service time 1/𝜇 and variance 1/𝜇2.

Model 1(M/M/1): (M/M/1) : (∞/FCFS): Birth and Death Model
• Birth and death model deals with a queuing system having a single
server, Poisson arrival, exponential service and there is no limit on the
system’s capacity while the customers are served on a “FCFS” basis.
• Step 1: Formulation of difference-differential equations
• Let Pn(t) denote the probability of n-customers in the system at time t.
• Then the probability that the system has n-customers at time (t + ∆t)
may be expressed as the sum of the combined probability of following
four mutually exclusive and exhaustive events.

• Pn(t + ∆t) = Pn(t) x P(no arrival in ∆t) x P(no service completion in ∆t) +
Pn(t) x P(one arrival in ∆t) x P(one service completion in ∆t) +
Pn+1(t) x P(no arrival in ∆t) x P(one service completion in ∆t) +
Pn-1(t) x P(one arrival in ∆t) x P(no service completion in ∆t)
= Pn(t)(1 - 𝜆∆t) (1 - 𝜇∆t) + Pn(t) 𝜆∆t 𝜇∆t + Pn+1(t) )(1 - 𝜆∆t)
𝜇∆t + Pn−1(t) 𝜆∆t (1 - 𝜇∆t)
or
= Pn(t + ∆t) - Pn(t) = -(𝜆 + 𝜇)Pn(t) ∆t + Pn+1(t) 𝜇∆t + Pn−1(t)𝜆∆t
• Dividing by ∆t and taking limit ∆t  0,
• P’n(t) = -(𝜆 + 𝜇)Pn(t) +𝜇Pn+1(t) +𝜆Pn−1(t), n > 1
• If there is no customer in the system at time (t + ∆t), there will be no service
during ∆t. Then for n = 0,

• P0(t + ∆t) = P0(t) x P(no arrival in ∆t) + P1(t) x P(no arrival in ∆t) x
P(one service completion in ∆t)
= P0(t) (1- 𝜆∆t) + P1(t) (1- 𝜆∆t) 𝜇∆t
P0(1 + ∆t) – P0(t) = - 𝜆 P0(t) ∆t + P1(t) (1- 𝜆∆t) 𝜇∆t
• Dividing by ∆t and taking limit ∆t → 0,
• P’0(t) = - 𝜆 P0(t) + 𝜇 P1(t), n = 0
• Thus, the required difference-differential equations are
• P’n(t) = -(𝜆 + 𝜇)Pn(t) +Pn+1(t)𝜇 + Pn−1(t)𝜆, n > 1 ___________ (20)
• P’0(t) = - 𝜆 P0(t) + 𝜇 P1(t), n = 0 ___________ (21)

• Step 2: Deriving the steady-state difference-differential equations
• Using equation (20), the steady-state difference-differential equations
are
• 0 = -(𝜆 + 𝜇)Pn + 𝜇Pn+1 + 𝜆Pn−1, n > 1 _________ (22)
• 0 = - 𝜆P0 + 𝜇P1, n = 0 _________ (23)
• Step 3: To solve above equation
• Take P0. Then equation 23 gives P1 = (𝜆/𝜇) P0.
• For n = 1, equation 22 gives P2 = (𝜆/𝜇) P1=(𝜆/𝜇)2 P0.
• In general, Pn=(𝜆/𝜇)n P0 = pnP0.
• To obtain the value of P0, we proceed as follows

• 1 = 𝑛=0
∞
𝑃𝑛 = 𝑛=0
∞
𝑝𝑛𝑃0 = P0 𝑛=0
∞
𝑃𝑛 =
𝑃0
1−𝑝
• or P0 = 1 – p _____________ (24)
• and Pn = pn(1 – p) _____________ (25)
• Step 4: Characteristics of model
1. Probability of queue size being greater than or equal to n = pn
• = 𝑘=𝑛
∞
𝑃𝑘 = 𝑘=𝑛
∞
1 − 𝑝 𝑝𝑘 = pn ______________ (26)
2. Average number of customers in the system
• Ls = 𝑛=0
∞
𝑛𝑃𝑛 = 𝑛=0
∞
𝑛(1 − 𝑝)𝑃𝑛
• = p(1 – p) 𝑛=1
∞
𝑛𝑝𝑛_1

• = p(1 – p) 𝑛=1
∞ 𝑑
𝑑𝑝
𝑝𝑛
• = p(1 – p)
𝑑
𝑑𝑝 𝑛=1
∞
𝑝𝑛 = p(1 – p)
𝑑
𝑑𝑝
1
1 −𝑝
• =
𝑝
1 −𝑝
=
𝜆
𝜇−𝜆
__________ (27)
3. Average queue length
• Lq = 𝑛=1
∞
(𝑛 − 1)𝑃𝑛 = 𝑛=1
∞
𝑛𝑃𝑛 - = 𝑛=1
∞
𝑃𝑛
• = 𝑛=1
∞
𝑛𝑃𝑛 - 𝑛=0
∞
𝑃𝑛 −𝑃0
• =
𝑝
1 −𝑝
- [1 – (1 – p)] (Using equation 26 and 24)
• =
𝑝2
1 −𝑝
=
𝜆2
𝜇(𝜇−𝜆)
__________ (27)

4. Average length of non-empty queue
• =
𝐿𝑞
𝑃(𝑛> 1)
• P(n > 1) = 𝑛=0
∞
𝑃𝑛- P0 – P1 = 1 – P0 – pP0 = 1 – (1 + p)(1 – p) = p2
• Therefore, the average length of the non-empty queue
•
𝑝2
1 −𝑝
1
𝑝2 =
1
1 −𝑝
=
𝜇
𝜇−𝜆
__________ (28)
5. Variance of queue length
• V[n] = 𝑛=0
∞
𝑛2𝑃𝑛 - 𝑛=0
∞
𝑛𝑃𝑛
2
• = 𝑛=0
∞
𝑛2 1 − 𝑝 𝑝𝑛 -
𝑝
1 −𝑝
2
• = (1 – p) 𝑛=0
∞
𝑛2𝑝𝑛 -
𝑝
1 −𝑝
2
• = (1 – p) pX -
𝑝
1 −𝑝
2

• where X = 𝑛=1
∞
𝑛2𝑃𝑛_1 , Integrating both sides by p,
• 0
𝑝
𝑋 𝑑 𝑝 = 𝑛=1
∞ 𝑛2
𝑃𝑛
𝑛
=
𝑝
1 −𝑝
2
• Now differentiating both sides with respect to p, we get,
• X = (1 + p)/(1 – p)3
• V[n] =
𝑝(1 −𝑝)(1+𝑝)
(1 – p)3
-
𝑝
1 −𝑝
2 =
𝑝
1 −𝑝 2 ________ (29)

Waiting time distribution
• The waiting time of a customer in a system is that time a customer
entering for service immediately upon arrival.
• Let w be the time spent in the queue and 𝜓𝑤(𝑡) be its cumulative
probability distribution.
• 𝜓𝑤 0 = 𝑃 𝑤 − 0 = 𝑃 𝑛𝑜 𝑐𝑢𝑠𝑡𝑜𝑚𝑒𝑟 𝑖𝑛 𝑡ℎ𝑒 𝑠𝑦𝑠𝑡𝑒𝑚 𝑢𝑝𝑜𝑛 𝑎𝑟𝑟𝑖𝑣𝑎𝑙
• = P0(1 – p) __________ (30)
• We want to find 𝜓𝑤 𝑡 .
• Let there be m-customers in the system upon arrival.
• Then for a customer, in order to go into service at a time between (0, t),
all n-customers must have been served by time t.

• Let s1, s2, …..sn denote the service time of n-customers respectively.
• w 𝑖=1
𝑛
𝑠𝑖,
0,
𝑛 ≥1
𝑛=0
_____________ (31)
• Then the probability distribution function of waiting time, w, for a
customer who has to wait is given by
• P(w < t) = P 𝑖=1
𝑛
𝑠𝑖 ≤ 𝑡 , n > 1 ___________ (32)
• Since the service time for each customer is independent, its
probability distribution function is 𝜇𝑒−𝜇𝑡(t > 0) where 𝜇 is the mean
service rate.

• 𝜓𝑤(𝑡) = 𝑛=1
∞
𝑃𝑛 𝑥 𝑝[ 𝑛 − 1 − 𝑐𝑢𝑠𝑡𝑜𝑚𝑒𝑟𝑠 𝑎𝑟𝑒 𝑠𝑒𝑟𝑣𝑒𝑑 𝑖𝑛 𝑡𝑖𝑚𝑒 𝑡]
• x P (one customer is served in time ∆t)
• = 𝑛=1
∞
(1 − 𝑝)𝑃𝑛 𝜇𝑡 𝑛
_
1
𝑛−1 !
𝑒−𝜇𝑡
𝜇 ∆t ___________ (33)
• Hence, the waiting time of a customer who has to wait is given by
• 𝜓 𝑤 =
𝑑
𝑑𝑡
𝜓𝑤(t) = p(1 – p) 𝜇𝑒−𝜇𝑡(1 −𝑝) = 𝜆
𝜇 −𝜆
𝜇
𝑒− 𝜇 −𝜆 𝑡 __ (34)

Characteristics of waiting time distribution
1. Average waiting time of a customer in the queue
• 𝜓 𝑤 = 0
∞
𝑡 𝜓 𝑤 dt = 0
∞
𝑡p(1 – p) 𝜇𝑒−𝜇𝑡(1 −𝑝)dt
•
𝑝
𝜇(1 −𝑝)
=
𝜆
𝜇(𝜇−𝜆)
________ (35)
2. Average waiting time of an arrival that has to wait in the system
• Ws =
𝐿𝑞
𝑃(𝑤>0)
P w > 0 = 1 − P w = 0 = 1 − P += 1 − 1 − p = p
• Therefore, Ws =
𝑝
𝜇 1 −𝑝 𝑝
=
1
𝜇−𝜆
_________ (36)
3. The busy period distribution is
• 𝜓
𝑤
𝑤
> 0 =
𝜓 𝑤
𝑃(𝑤>0)
= (𝜇 - 𝜆) 𝑒− 𝜇−𝜆 𝑡 __________ (37)

Example 12.1:
• The arrival rate of a customer at a service window of a cinema hall
follows a probability distribution with a mean rate of 45 per hour. The
service rate of the clerk follows poisson distribution with a mean of
60 per hour.
• (a) What is the probability of having no customer in the system?
• (b) What is the probability of having five customers in the system?
• (c) Find Ls, Lq, Ws and Wq

• Arrival rate, 𝜆 = 45 per hour and service rate, 𝜇 = 60 per hour.
• p =
𝜆
𝜇
=
45
60
= 0.75
• (a) No customer = P0 = 1- p = 1 – 0.75 = 0.25
• (b) Five customer = P5 = (1- p)p5 = (0.25) x (0.75)5 = 0.0593
• (c) Ls =
𝑝
1 −𝑝
=
0.75
1 −0.75
=
0.75
0.25
= 3 customers
• Lq =
𝑝2
1 −𝑝
=
0.75 2
1 −0.75
=
0.5625
0.25
= 2.25 customers
• Ws =
1
𝜇−𝜆
=
1
60 −45
=
1
15
= 0.067 hour
• Wq =
𝑝
𝜇−𝜆
=
0.75
60 −45
=
0.75
15
= 0.05 hour

Example 12.2:
• An arrival rate at a telephone booth is considered to be poisson with
an average time of 10 minutes and exponential call lengths averaging
3 minutes.
• (a) Find the fraction of a day that the telephone will be busy
• (b) What is the probability that an arrival at the booth will have to
wait?
• (c) What is the probability that an arrival will have to wait more than
10 minutes before the phone is free?
• (d) What is the probability that it will take him more than 10 minutes
altogether to wait for phone and complete his call?

• Arrival rate, 𝜆 = 1/10 per minute
• Service rate, 𝜇 = 1/3 per minute
(a) The fraction of the day that phone will be busy,
• p =
𝜆
𝜇
=
1
10
1
3
=
1
10
×
3
1
=
1
3
𝑝𝑒𝑟 𝑚𝑖𝑛𝑢𝑡𝑒.
(b) Probability that an arrival at the booth will have to wait
• = 1 – P0 = 1 – (1 – p) = 1 – (1 – 0.3) = 1 – 0.7 = 0.3
(c) Probability that an arrival will have to wait for more than 10 minutes
before the phone is free
• = 10
∞
1 −
𝜆
𝜇
𝜆𝑒− 𝜇 −𝜆 𝑡
dt = 1 −
1
10
1
3
1
10
𝑒
− (
1
3
) −(
1
10
) 𝑡
• = 0.3𝑒− 0.23 ×𝑡= 0.3𝑒−2.3 = 0.03

• (d) Probability of an arrival waiting in the system is greater than or
equal to 10
• = 10
∞
𝜇 − 𝜆 𝜆𝑒− 𝜇 −𝜆 𝑡 dt
• = 10
∞ 1
3
−
1
10
𝑒
− (
1
3
) −(
1
10
)
𝑡 𝑑𝑡
• = e-2.3 = 0.1

Example 12.3:
• Vehicles pass through a toll gate a rate of 90 per hour. The average
time to pass through the gate is 36 seconds. The arrival rate and the
service rate follow Poisson distribution. There is a complaint that the
vehicles wait for long duration. The authority is willing to install one
more gate to reduce the average time to pass through the toll gate to
30 seconds if the idle time of the toll gate is less than 10% and the
average queue length at the gate is more than 5 vehicles. Discuss
whether the installation of the second gate is justified or not.

• Arrival rate of vehicles at toll gate, 𝜆 = 90 per hour
• Departure rate of vehicles through the gate, 𝜇 = 36 seconds =
3600
36
= 100 vehicles
per hour
• P =
𝜆
𝜇
=
90
100
= 0.9
• (a) Waiting number of vehicles in the queue
• Lq =
𝑝2
1 −𝑝
=
0.9 2
1 −0.9
=
0.81
0.1
= 8.1 vehicles
• (b) Expected time taken to pass through the gate = 30 seconds.
• Service rate, 𝜇 = 30 seconds =
3600
30
= 120 vehicles per hour
• So, P =
90
120
= 0.75
• Percent of the idle time of the gate = 1 – p = 1 – 0.75 = 0.25 after percentage = 25%
• Thus, the average waiting number of vehicles in the queue is more than 5 but the
idle time of toll gate is not less than 10%. Hence, the installation of another gate is
not justified.

Example 12.3:
• At what average rate must a clerk at a super market work in order to
ensure a probability of 0.90 that the customer will not wait longer
than 12 minutes? It is assumed that there is only one counter, at
which customers arrive in a Poisson fashion at an average rate of 15
per hour. The length of service by the clerk has an exponential
distribution.

• Arrival rate , 𝜆 =
15
60
= 0.25 customers per minute
• Departure rate of customers be, 𝜇
• Probability that the customer will not have to wait more than 12
minutes.
• = 1 – 0.9 = 0.1
• Therefore, 0.1 = 12
∞ 𝜆
𝜇
(𝜇 - 𝜆) 𝑒−(𝜇 − 𝜆)dt
• 0.1 =
𝜆
𝜇
𝑒−12(𝜇 − 𝜆)
• 0.4 𝜇 = 𝑒−12(𝜇 − 𝜆) = 𝑒−12𝜇 +12𝜆
=𝑒−12𝜇 +12 𝑥 3
= 𝑒−12𝜇 +3
• 0.4 𝜇 = 𝑒3 −12𝜇 gives
1
𝜇
= 2.48 minutes/customer

Example 12.5:
• In a railway Marshall 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 distribution is also an exponential with an
average 36 minutes, calculate
• (a) The mean size queue and
• (b) The probability that the queue size exceeds 10
• If the input of trains increases to an average 33 per day, what will be
the changes in (a) and (b)?

30
60 𝑥 24
=
30
1440
=
1
48
trains per minute
• Service rate, 𝜇 =
1
36
= trains per minute
• So, P =
𝜆
𝜇
=
1/48
1/36
=
36
48
= 0.75
• (a) The mean queue size, c. Ls =
𝑝
1 −𝑝
=
0.75
1 −0.75
=
0.75
0.25
= 3 trains
• (b) Probability (queue size > 10 trains) = p10 = 0.5410 = 0.056
• When the input increases to 33 trains per day i.e.,
• 𝜆 =
33
60 𝑥 24
=
33
1440
=
11
480
trains per minute
• p =
𝜆
𝜇
=
11/480
1/36
=
11
480
x
36
1
=
396
480
= 0.83

• (a) The mean queue size, c. Ls =
𝑝
1 −𝑝
=
0.83
1 −0.83
=
0.83
0.17
= 4.88 = 5 trains
• (b) Probability (queue size > 10 trains) = p10 = 0.8310 = 0.155

Example 12.6:
• In a maintenance shop, the inter-arrival times at tool crib are
exponential with an average time of 10 minutes. The length of the
service (i.e. the amount of time taken by the tool crib operator to
meet the needs of the maintenance man) time is assumed to be
exponentially distributed with a mean 6 minutes.
• (a) The probability that a person arriving at the booth will have to
wait
• (b) The average length of the queue that forms and the average time
that an operator spends in the queuing system
• (c) The probability that an arrival will have to wait for more than 12
minutes for service and to obtain his tools
• (d) The estimate of the fraction of the day that the tool crib operator
will be idle.

• (e) The probability that there will be six or more operators waiting for
the service
• (f) The manager of the shop will install a second booth when an
arrival would expect to wait 10 minutes or more for the service. By
how much must the rate of arrival be increased in order to justify a
second booth>

60
10
= 6 per hour
• Departure rate, 𝜇 =
60
6
= 10 per hour
• (a) Probability that the arrival will have to wait = p =
𝜆
𝜇
=
6
10
= 0.6
• (b) Average number of arrivals waiting time in the queue,
• Lq =
𝑝2
1 −𝑝
=
0.6 2
1 −0.6
=
0.36
0.4
= 0.9 and
• Average waiting time in system, Ws =
1
𝜇−𝜆
=
1
10−6
=
1
4
= 0.25 hour
• (c) Probability that the arrival will have to wait more than 12 minutes
• = 12
∞ 𝜆
𝜇
(𝜇 − 𝜆) 𝑒−(𝜇 − 𝜆)dt = 0.6𝑒−(4/5)= 0.6𝑒−0.8 = 0.6 x 0.44 = 0.27

• (d) Probability of the tool crib operator to be idle = P0 = 1- p = 1 – 0.6 = 0.4
• Therefore, for 40% of the time the tool crib operator will remain free.
• (e) Probability of six or more operators waiting for service
• = p6 = (o.6)6 = 0.05
• (f) Average waiting time of a customer in the queue
• Wq =
𝜆
𝜇(𝜇−𝜆)
=
6
10(4)
=
6
40
=
3
20
hours = 3 x
60
20
= 9 minutes
• The installation of the second booth will be justified if the arrival rate,
𝜆′
𝑠𝑎𝑦 is greater than the waiting time.
• Wq = 10 minutes =
1
6
hours
•
1
6
=
𝜆′
𝜇(𝜇−𝜆′)
= gives 𝜆′ = 6.25
• Therefore, if the arrival rate exceeds 6.25 per hour, the second booth will be
justified.

Unit V - Queuing Theory

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Unit V - Queuing Theory