Queuing theory

Queuing theory

1. KULDEEP MATHUR
M.B.A. JIWAJI UNIVERSITY
GWALIOR
OPERATIONS RESEARCH
Queuing Theory

2. 1-2
Queuing theory can be applied to a variety of operational situations
where it is not possible to predict accurately the arrival rate (or time)
of customers and service rate (or time) of service facility or facilities.
Thus it is needed to determine the level of service (either the service
rate or the number of service facilities) that balances the following two
conflicting costs:
Queuing Theory
i)cost of offering the service
ii)cost incurred due to delay in offering service
The first cost is associated with the service facilities and their
operation, and the second represents the cost of customer’s waiting
time.

3. 1-3
Figure illustrates both types of costs as a function of level of service. The
optimum service level is one that minimizes the sum of the two costs.
Since cost of waiting is difficult to estimate, it is usually measured in terms
of loss of sales or goodwill when the customer is a human being who has
no sympathy with the service. But, if the customer is a machine waiting for
repair, then cost of waiting is measured in terms of the cost of lost
production.
. . . queuing theory

4. 1-4
Queuing Costs vs Level of Service

5. 1-5
Situation Customers Service Facilities
Petrol pumps (stations)Automobiles Pumps/Passionel
Hospital Patients Doctors/Nurses/Rooms
Airport Aircraft Runways
Post office Letters Sorting system
Job interviews Applicants Interviewers
Examples of Queues

6. 1-6
The Structure of a Queuing System
A queuing system is composed of the following components (or parts):
Calling population (or input source)
Queuing process
Queue discipline
Service process (or mechanism)
Major Components of a Queuing System

7. Calling population characteristics
1-7
The calling population need not be homogeneous and may consist of
several subpopulations. For example, patients arriving at OPD of a
hospital are usually of three categories: walk-in patients, patients
with appointments and emergency patients. Each patient class
places different demands on service facility, but the waiting
expectations of each category differ significantly.

8. Size of calling population The size of calling population whether it is
homogeneous or consist of several subpopulations is considered to be
either finite (limited) or infinite (unlimited). If customer’s arrival depends
on the number of customers already in the system (in service plus in
queue), the calling population is called limited or finite.
1-8
A factory having only four machines which often require
repair/service and two of them (say) are in working condition is an
example of finite population. Then at any point in time, there are
only two machines which could possibly require service.
Alternately, if new customer’s arrival is independent of the number
of customers already in the system, the calling population is called
unlimited or infinite. For example customers arriving at a bank or
super market, students arriving to get admission at the university,
cars arriving at a highway petrol pump, etc.

9. 1-9
Behaviour of arrivals
Balking: Customers do not join the queue either by seeing the number
of customers already in service system or estimating the excessive
waiting time for desired service.
Reneging: Customers after joining the queue, wait for sometime in the
queue but leave before being served on account of certain reasons.
Jockeying: Customers move from one queue to another hoping to
receive service more quickly (a common scene an railway booking
window).

10. 1-10
Pattern of arrivals at the system Customers may arrive in batches (such as
the arrival of a family at a restaurant) or individually (such as the arrival of a
train at a platform). These customers may arrive at a service facility either on
scheduled time (by prior information) or on unscheduled time (without
information).
The arrival process (or pattern) of customers to the service system is classified
into two categories: static and dynamic.
In static arrival pattern, the control depends on the nature of arrival rate
(random or constant): In random (or unscheduled) arrivals the times are
random variable, and hence requires to understand the average and
frequency distribution of the times. In both the cases, the arrival process
can be described either by the average arrival rate (average number of
arrivals per unit of time) or by the average inter-arrival time (average time
between two consecutive arrivals).

11. 1-11
The dynamic arrival process is controlled by both service facility and
customers. The service facility adjusts its capacity to match changes
in the service intensity, by either varying the staffing levels at different
timings of service, varying service charges (such as telephone call
charges at different hours of the day or week) at different timings, or
allowing entry with appointments.
The variation in the service intensity also affects the customer’s
behaviour. They either balk or renege from the service system when
confronted with a long or slow moving waiting line.

12. 1-12
The arrival time distribution
Poisson distribution it is a discrete probability distribution that
provides probabilities for the number of customers that may arrive
in any specific interval to time, whereas exponential distribution
provides probabilities for times gap between two consecutive
arrivals.
Let n customers arrive during a time interval 0 to t. If l is the
expected (or average) number of arrivals per unit time, then
expected number of arrivals during a time interval t will be lt. Then
Poisson probability distribution function is given by
P(x = n | P
n
= lt) =
(lt )
n
e
-
lt
n!
, n = 0, 1, 2, …

13. 1-13
Let T be the time between successive arrivals. Since a customer can arrive
at any time, T must be a continuous random variable. The probability of no
arrival in the time interval from 0 to t will be equal to the probability that T
exceeds t, so we have
The cumulative probability that the time T between two successive arrivals is
t or less is given by
PTtPx P te
n
t
() ( | )>= = = =
-
0 l
l
PTt PTt e t
t
( ) ( ) ;£=- >=- ³
-
1 1 0
l
This probability is also called the cumulative probability distribution function,
F(t) of T. Also the distribution of the random variable T is referred to as the
exponential distribution. Its probability distribution function (pdf) is given by
; 0
( )
0 ; 0
t
e t
f t
t
-l
ìl ³ï
=í
<ïî

14. 1-14
The inter-arrival time in many situations is approximated by an exponential
distribution (sometimes called the negative exponential distribution). The
shape of the typical exponential probability distribution function and
cumulative exponential probability distribution function is:
Exponential Probability Distribution Cumulative Exponential Probability Distribution

15. 1-15
The mean of exponential distribution is the expected or average time E(T)
between arrivals. Thus, with l arrivals per unit of time, E(T) = 1/l.
P(T £ 0.4) = 1 – e
–4(0.4)
= 1 – e
–1.6
= 1 – 0.202 = 0.798
Poisson distribution of arrivals with arrival rate l is equal to the
negative exponential distribution of inter-arrival times with average
inter-arrival time 1/l. For example, if l=4 customers per minute, then
the inter-arrival time between any two customers is t = 0.40 minute (or
24 seconds) or less is

16. 1-16
If customers arrive at an average rate of l = 24 customers per hour, and a
customer has already arrived, then probability of a customer arriving in the
next 5 minutes (i.e. t = 1/12 hr) is
P(T < 1/12) = 1 – e
–24(1/12)
= 1 – e
–2
= 1 – 0.1353 = 0.8647
If l = 24 customers per hour, t = 1/12 hr and lt = 24 × (1/12) = 2, the
probability of n = 2 customers arriving within the next 10 minutes is:
P(x = 2) = 0.27
e
-
=
2412 2
2412
2
(1/)
(/)
!

17. 1-17
Queuing Process
The queuing process refers to the number of queues, and their respective
lengths. The number of queues single, multiple or priority queues depend
upon the layout of a service system. The length (or size) of the queue
depends upon the operational situation such as physical space, legal
restrictions, and attitude of the customers.

18. 1-18
If a service system is unable to accommodate more than the required
number of customers at a time. No further customers are allowed to
enter until space becomes available to accommodate new customers.
Such type of situations are referred to as finite (or limited) source
queue. Examples of finite source queues are cinema halls,
restaurants, etc.
If a service system is able to accommodate any number of customers
at a time, then it is referred to as infinite (or unlimited) source queue.
For example, in a sales department where the customer orders are
received, there is no restriction on the number of orders that can come
in so that a queue of any size can form.

19. 1-19
If arriving customers found long queue(s) in front of a service facility,
they often do not enter the service system even though additional
waiting space is available. The queue length in such cases depends
upon the attitude of the customers. For example, when a motorist finds
that there are many vehicles waiting at the petrol station, in most of the
cases he does not stop at this station and seeks service elsewhere.

20. 1-20
Queue Discipline
The queue discipline is the order or manner in which customers from the queue
are selected for service.
Static queue disciplines are based on the individual customer’s status in
the queue. Few of such disciplines are:
If customers are served in the order of their arrival, such discipline is
known as: first-come, first-served (FCFS) service discipline. Prepaid
taxi queue at airports where a taxi is engaged on a ‘first-come, first-
served’ basis is an example of this discipline.
Last-come, first-served (LCFS) is another discipline practised in most
cargo handling situations where the last item loaded is removed first
because it reduces handling and transportation cost, the last ones
being easier to reach closer.

21. 1-21
Similarly, in the production process items arrive at a workplace and
are stacked one on top of the other. Item on the top of the stack is
taken first for processing which is the last one to have arrived for
service.
Dynamic queue disciplines are based on the individual customer
attributes in the queue. Few of such disciplines are:
Service in Random Order (SIRO), where customers are selected for
service at random irrespective of their arrivals in the service system.
Priority Service where customers are grouped in priority classes on
the basis of some attributes such as service time or urgency, and
FCFS rule is used within each class to provide service. The
payment of telephone or electricity bills by cheque or cash are
examples of this discipline.

22. 1-22
pre-emptive Priority (or Emergency), where an important customer
is allowed to enter into the service immediately after entering into
the system even if a customer with lower priority is already in
service. That is, lower priority customer’s service is interrupted (pre-
empted) to start service for such a customer. This interrupted
service is resumed again after the priority customer is served.
Non-pre-emptive Priority where an important customer goes ahead
in the queue, but service is started immediately on completion of
the current service.

23. 1-23
Service Process (or Mechanism)
The service mechanism is concerned with the manner in which customers
are serviced and leave the service system. It is characterized by
The arrangement (or capacity) of service facilities
The distribution of service times
Server’s behaviour
Management policies
Arrangement of
Service Facilities
in Series

24. 1-24
… service process
Arrangement of Service Facilities in Parallel
Single Queue, Multiple Service Facilities in Parallel and in Series

25. 1-25
Distribution of Service Time
a)Average service rate It measures the service capacity of service facility in
terms of customers served per unit of time. If µ is the average service rate,
then the expected number of customers served during time interval 0 to t
will be µt. If the service time is exponentially distributed, then the service
rate can be shown to be Poisson distributed. If time at the start of service,
is considered as zero probability that service is not completed by time t is
given by
P(x = 0 | P
n = µt) = e
–µt
If the random variable T represents the service time, then the probability of
service completion within time t is given by
P(T £ t) = 1 – e
–µt
,  t ³ 0
b)Average length of service time A fluctuating service time may follow
negative exponential probability distribution. The average service time is
denoted by E(T) = 1/µ.

26. 1-26
Performance Measures of a Queuing System
1. Time-related questions for the customers
a)What is the average (or expected) time an arriving customer has to wait
in the queue (denoted by W
q
) before being served.
b)What is the average (or expected) time an arriving customer spends in
the system (denoted by W
s
) including waiting and service. This data can
be used to make economic comparison of alternative queuing systems.
2. Quantitative questions related to the number of customers
a)Expected number of customers who are in the queue (queue length) for
service, and is denoted by L
q
.
b)Expected number of customers who are in the system either waiting in
the queue or being serviced (denoted by L
s
). The data can be used for
finding the mean customer time spent in the system.

27. 1-27
3. Questions involving value of time both for customers and servers
a)What is the probability that an arriving customer has to wait before
being served (denoted by P
w
)? It is also called blocking probability.
b)What is the probability that a server is busy at any particular point in
time (denoted by r)? It is the proportion of the time that a server actually
spends with the customer, i.e. the fraction of the time a server is busy.
c)What is the probability of n customers being in the queuing system
when it is in steady state condition? It is denoted by P
n
, n = 0, 1 . . ..
d)What is the probability of service denial when an arriving customer
cannot enter the system because the queue is full? It is denoted by P
d
.
4. Cost-related questions
a)What is the average cost needed to operate the system per unit of
time?
b)How many servers (service centres) are needed to achieve cost
effectiveness?

28. 1-28
Transient-State and Steady-State
When a service system is started it progresses through a number of changes.
and attains stability after sometime. Before the start of the service operations it
gets influenced by the initial conditions specially by the number of customers in
the system and the elapsed time. This period of transition is termed as transient-
state. However, as time passes, the system becomes independent of the initial
conditions and of the elapsed time (except under very special conditions) and
enters a steady- state condition.
Let P
n
(t) be the probability of n customers in the system at time t. In the case of
steady-state, we have
lim()
t
n nPtP
®¥
=
lim()
t
n
Pt
®¥
¢=0
If arrival rate of customers is more than the service rate, then a steady-state cannot be
reached regardless of the length of the elapsed time.

29. 1-29
Line length (or queue size)  The total number of customers in the system who
are actually waiting in the line and not being serviced.
Queue length  It is the line length plus number of customers being served.
Relationships among Performance Measures
a) Expected number of customers in the queue and system are given by

and
L
s
= L
q
+ Expected number of customers in service
=
L nP
s
n
n
=
=
¥
S
0
L nsP
q
ns
n
= -
=
¥
S()
b)Expected number of customers in the system = Expected number of
customers in queue plus in service.
L
q
+
l
m

30. 1-30
c)Expected waiting time of the customer in the system = Average waiting
time in queue plus the expected service time
W
s
= W
q
+
d)Probability of being in the system (waiting and being served) longer than
time t is
P(T > t)= e
–(m + l)tc
and

P(T £ t) = 1 – P(T > t)
whereT = time spent in the system,
t = specified time period, e = 2.718
e)Probability of only waiting for service longer than time t is
P(T > t) = e
–(m+l)t
f)Probability of exactly n customers in the system is
P
n
=
1
m
1
æ öæ öl l
-
ç ÷ç ÷
m mè øè ø
n
l
m

31. 1-31
g)Probability that the number of customers in the system, n exceeds a given
number, r is
P(n > r) =
h)Expected number of customers in the system = Average number of arrivals
per unit of time multiplied by the average time spent by the customer in the
system
L
s
=
L
q
= L
s
– =
1+
æ öl
ç ÷
mè ø
r
1 1
orl = + =
m l
s s q s
W W W L
l
m
1 1
orl = - =
m l
q q s q
W W W L

32. 1-32
Different models in queuing theory are classified by using notations described
initially by D.G. Kendall in 1953 in the form (a/b/c). Later A.M. Lee in 1966
added the symbols d and c to the Kendall’s notation. The standard format
used to describe the main characteristics of parallel queues is
{(a/b/c) : (d/e)}
wherea = arrivals distribution
b = service time (or departures) distribution
c = number of service channels (servers)
d = maximum number of customers allowed in the system
(in queue plus in service)
e = queue (or service) discipline.
Classification of Queuing Models

33. 1-33
Certain descriptive notations used to replace notation a and b for the arrival
and service times distribution are:
M = Markovian (or Exponential) inter-arrival time or service-time distribution.
D = Deterministic (or constant) inter-arrival time or service time.
G = General distribution of service time (departures), i.e. no assumption
is made about the type of distribution with mean and variance.
GI = General probability distribution – normal, uniform or any empirical
distribution, for inter-arrival time
E
k
= Erlang-k distribution for inter-arrival or service time with parameter k
. . . queuing models

34. 1-34
Model I  {(M/M/1) : (¥/FCFS)} Exponential Service – Unlimited Queue
Assumptions for the queuing system:
Single-Server Queuing Models
Arrivals are described by Poisson probability distribution and come from
an infinite population.
Single waiting line and each arrival waits to be served regardless of the
length of the queue (i.e. infinite capacity) and no balking or reneging.
Queue discipline is ‘first-come, first-served’.
Single server and service times follow exponential distribution.
Customers arrival is independent but the arrival rate (average number of
arrivals) does not change over time.
The average service rate is more than the average arrival rate.

35. 1-35
The following events (possibilities) may occur during a small interval of time,
Dt just before time t.
. . . queuing models
The system is in state n (number of customers) and no arrival and no
departure, leaving the total to n customers.
Ths system is in state n + 1 (number of customers) and no arrival and
one departure, reducing the total to n customers.
The system is in state n – 1 (number of customers) and one arrival and
no departure, bringing the total to n customers.

36. 1-36
Single server queuing system states are shown below to illustrate the
process of determining P
n
by considering each possible number of customers
either waiting or receiving service at each state which may be entered by the
arrival of a new customer or left by the completion of the leading customer’s
service.
. . . queuing models
One Arrival

37. 1-37
Based on single server queuing system, the probability of no customer and
n customers in the system at time t are given by
P
0
= 1 – = 1 – r
P
w
= Probability that an arriving customer has to wait
= 1 – P
0
= l/m.
. . . queuing models
l
m
l
m
r rr- = - <=
n
n n1 1 1012();,,,,...())(P
n
=
l
m

38. 1-38
(a) Expected number of customers in the system (customers in the line
plus the customer being served)
L
s
= 0 < r < 1
=
(b) Expected number of customers waiting in the queue (i.e. queue length)
L
q
=
= ;
=
Performance Measures for Model  I
S S
n
n
n
n
nP n
=
¥
=
¥
= -
0 0
1(),rr
r
r
l
ml
r
l
m1-
=
-
=;
å- =å -å
=
¥
=
¥
=
¥
n
n
n
n
n
n
nP nP P
1 1 1
1()
L P
s
-- =
-
-( )1
0
l
lm
l
m
1
0-=P
l
m
l
mml
2
( )-

39. 1-39
2. a) Expected waiting time for a customer in the queue
W
q
=
=
= =
b) Expected waiting time for a customer in the system (waiting and service)
W
s
= Expected waiting time in queue + Expected service time
=
=
t e dt
t
×-
- -
¥
l r
m r
( )
( )
1
1
0
l
l
mml
1
1
2
-
-( )
()
l
mml()-
W
q
+=
-
+
1 1
m
l
mml m()
1
ml l-
=
L
s
L
q
l

40. 1-40
3. The variance (fluctuation) of queue length
Var (n)=
=
4. Probability that the queue is non-empty
P(n > 1) = 1 – P
0
– p
1
=
nP nP
n
n
n
n
2
1 1
2
=
¥
=
¥
å -å()
r
r
lm
ml() ( )1
2 2
-
=
-
11 1
2
---- =
l
m
l
m
l
m
l
m
()()()()

41. 1-41
5. Probability that the number of customers, n in the system exceeds
a given number k.
P (n ³ k) = =
= ; P (n > k) =
P
k
nk=
¥
å ()1-å
=
¥
rr
k
nk
l
m
k
()
l
m
+k1
()
6. Expected length of non-empty queue
L =
= =
Expected length of waiting line
Prob (>1)n
L
Pn
q
()
/( )
(/)>
=
-
1
2
2
lmml
lm
m
ml-

42. 1-42
7. Probability of an arrival during the service time when system
contains r customers
P (n = r)=
=
()
!
l
m
l
mt
e
r
edt
r t
t
-
-
¥
×
0
l
lm
m
lm+ +
r
()()

43. 1-43
Example Arrivals at telephone booth are considered to be Poisson with
an average time of 10 minutes between one arrival and the next. The
length of phone call is assumed to be distributed exponentially, with
mean 3 minutes.
What is the probability that a person arriving at the booth will have to
wait?
The telephone department will install a second booth when
convinced that an arrival would expect waiting for at least 3 minutes
for a phone call. By how much should the flow of arrivals increase in
order to justify a second booth?
What is the average length of the queue that forms from time to
time?
What is the probability that it will take him more than 10 minutes
altogether to wait for the phone and complete his call?

44. 1-44
Solution
l = 1/10 = 0.10 person per minute and m = 1/3 = 0.33 person per minute
a)Probability that a person has to wait at the booth.
=
b)The installation of second booth will be justified only if the arrival
rate is more than the waiting time. Let l be the increased arrival
rate. Then expected waiting time in the queue will be
W
q
=
3 = or
Hence, the increase in the arrival rate is 0.16 – 0.10 = 0.06 arrivals per
minute.
Pn()>0=-=1
0
Plm/ 01003303./. .=
l
mml
¢
-¢( )
¢
-¢
l
l033033.(. )
¢=l016.

45. 1-45
(c)Average length of non-empty queue
customers (approx.)
(d)Probability of waiting for 10 minutes or more is given by
P ( t ³ 10)

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.
0.33
2
0.23
L
m
= = =
m-l
= -
¥
-
-
10
l
m
ml
ml
( )
( )
e dt
t
=
¥
-
10
023
03023(.)(.)
.
e dt
t
=
-
=
-
¥
0069
023
003
023
10
.
.
.
.
e
t

46. 1-46
Example A road transport company has one reservation clerk on duty at a
time. He handles information of bus schedules and makes reservations.
Customers arrive at a rate of 8 per hour and the clerk can service 12
customers on an average per hour. After stating your assumptions, answer
the following:
What is the average number of customers waiting for the service of the
clerk?
What is the average time a customer has to wait before getting service?
The management is contemplating to install a computer system to
handle the information and reservations. This is expected to reduce the
service time from 5 to 3 minutes. The additional cost of having the new
system works out to Rs 50 per day. If the cost of goodwill of having to
wait is estimated to be 12 paise per minute spent waiting before being
served, should the company install the computer system? Assume 8
hours working day.

47. 1-47
Solution Given that l = 8 per hour; m = 12 per hour
a)The average number of customers waiting for the service in the
system
b)The average time spent by a customer in the system
The average waiting time in the queue for a customer
Ls=
-
=
-
=
l
ml
8
128
2customers
Ws=
-
=
-
= =
1 1
128
1
4
15
ml
hour minutes.
Wq=
-
=
-
= =
l
mml() ( )
8
12128
1
6
10hour minutes.

48. 1-48
c)Comparing additional cost of Rs 50 per day with the goodwill cost of
customers with existing system before installing computer system.
The average cost for waiting time in the system,
W
s
= 15 minutes is 0.12 × 15 = Rs 1.80.
Since 8 customers arrived in 1 hour or in 8 hours 64 (=8×8) customers
request service at a total goodwill cost of 1.80 × 64 = Rs 115.20.
The installation of computer system, will increase service rate up to l = 20
customers per hour (3 customers per minute). So, the average time spent
by a customer in the system would be
and the average daily customer queuing (or goodwill) cost is reduced to: 64
× (0.12 × 5) = Rs 38.40.
Ws=
-
=
-
= =
1 1
208
1
12
5
ml
hourminutes

49. 1-49
Operational cost of computer is Rs 50 per day. So, the average total daily
cost would be
TC = Computer cost + Goodwill cost = 50 + 38.40 = Rs 88.40
This cost is less than the existing goodwill loss cost and net saving is =
(115.20 – 88.40) = Rs 26.80. Hence, company may install a computer.
Model III {(M/M/1) : (N/FCFS)} Exponential Service – Finite (or Limited)
Queue
Suppose, no more than N customers can be accommodated at any time in
the system due to certain reasons. Then any additional customer arriving
can not enter into the system and is lost.
Emergency room in a hospital; one-man barber shop with certain number of
chairs for waiting customers, etc. are the examples of finite queue.

50. 1-50
;
In such a case the steady-state solution exists even for r > 1. Because
limited capacity of the system controls the arrivals by the queue length
(=N–1) not by the relative rates (l ) of arrival and departure (µ). If l < µ and
N          , then P
n
= ( 1 – l/µ)

(l/m)
n
, which is same as in Model I.
P
N
0
1
1
1
=
-
-
+
r
r
r r rlm¹ < =1 1and ; /
P
nN
N
n
N
n
=
-
-
£ ¹ ¹
+
= =
+
1
1
1
1
1
1
1
r
r
r rlm
rlm
; ; ( )
; ( )
( )
®¥
The steady-state probability distribution of n customers in the system is:

51. 1-51
Performance measures for model III
1. Expected number of customers in the system
L
s
=
=
2. Expected queue length or expected number of customers waiting in the
system
L
q
=
å =å
-
-= =
+
n
N
n
n
N
N
n
nP n
1 1
1
1
1
r
r
r
( )
r
r
r
r
r lm
r lm
1
1
1
1
2
1
1
1-
-
+
-
¹ ¹
= =
+
+
( )
; ( )
; ( )
N
N
N
N
L L
P
s s
N
- -
-l
m
l
m
=
( )1

52. 1-52
. . . model III
3. Expected waiting time of a customer in the system (waiting + service)
W
s
=
4. Expected waiting time of a customer in the queue
W
q
=
5. Fraction of potential customers lost ( = fraction of time system is full)
P
N
= P
0
r
N
Effective arrival rate, l
eff
= l(1 – P
N
)
Effective traffic intensity, r
eff
= l
e 
/ µ.
L
P
s
N
l( )1-
W
L
P
s
q
N
-
-
1
1m l
or
( )

53. 1-53
. . . model III
Example Consider a single server queuing 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 hour and the maximum
permissible 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.
Solution From the data of the problem, we have
 l = 3 units per hour; µ = 4 units per hour, and N = 2
Then traffic intensity, rlm= = =/ / .34075

54. 1-54
. . . model III
The steady-state probability distribution of n customers in the system is:
P
n
=
=
andP
0
=
The expected number of calling units in the system is given by
L
s
=
=
()
;
1
1
1
1
-
-
¹
+
rr
r
r
n
N
(.)(.)
(.)
(.)(.)
1075075
1075
043075
21
-
-
=
+
n
n
() .
(.)
1
1
1075
1075
1 21
-
-
=
-
-
+ +
r
r
N
=
-
=
025
1075
0431
3
.
(.)
.
å =å
= =n
N
n
n
n
nP n
1 1
2
043075(.)(.)
043 075
1
2
. (.)å
=n
n
n = + =0430752075 081
2
.{(.)(.)}.