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UNIT – IV
QUEUEING MODELS
Introduction:
Waiting for service is part of our daily life. We wait at hotels, we queue up at the railway
reservation counter and line-up for service in banks and the waiting phenomenon is not an experience
limited to human beings alone jobs wait to be processed on a machine Aeroplanes circle in air before
given permission to land at an airport. Cars wait at the traffic signals and so on.
Queues are formed, if the demand for service is more than the capacity to provide the service.
Queueing system:
A Queueing system can be described as customers arriving for service, waiting if service is not
available immediately and leaving the system after having been served.
Characteristics of a queueing system:
1. Arrival (or) input pattern of customers
2. Service pattern (or) service mechanism of server.
3. Queue discipline
4. System capacity.
1. Arrival pattern:
Since the customers arrive in a random fashion, therefore their arrival refers or inter-arrival time
can be described interms of probabilities.
Mean arrival rate Poisson distribution, Mean =
Mean inter-arrival rate Exponential distribution, mean = 1
2. Service Pattern:
The pattern according to which the customers are served i.e., Distribution of the time to service a
customer.
Service time Poisson distribution, Mean=
Inter service time Exponential distribution mean= 1
.
(i) Single Server:
(ii) Multi servers:
3. Queue Discipline:
It is the rule according to which customers are selected for service when a queue has been formed.
The common queue disciplines are
(i) FIFO – First In First Out (Or) FCFS – First Come First Served.
(ii) LIFO – Last In First Out (Or) LCFS – Last Come First Served.
(iii) SIRO – Service In Random Order (Or) Selected In Random Order.
(iv) GD – General Service Discipline
There are also priority schemes for selection.
4. System Capacity:
In some queueing process, there is limited waiting space, So that when the queue reaches a certain
length, further customers are not allowed to join the queue, until the space becomes available after service
completion. Thus there is a finite limit to the maximum system size. If any number of customers are
allowed to join the queue, we may say that the capacity is infinite.
Customers Behaviour:
A customer generally behaves in four ways.
(i) Balking – A customer may leave the queue, if there is no waiting space.
(ii) Reneging – This occurs when the waiting customers leave the queue due to inpatients.
(iii) Priorities – In certain applications some customers are served before others, regardless of their order
of arrival.
(iv) Jockeying – The customer may jump one waiting line to another.
Transcient state:
A system is said to be transcient state then its operating characteristics are dependent on time.
Steady State:
The system is said to be steady state when the behaviour of the system is independent of time.
Let ( )nP t denotes the probability that there are n units in the system at time t. Then the steady
state is lim '( ) 0nn
P t
Kendall’s Notation:
The assumptions made for the simples queueing model is (a|b|c): (d|e) where
a = Probability distribution of the inter-arrival time.
b = Probability distribution of the service time.
c = Number of services in the system.
d = Maximum number of customers allowed in the system.
e = Queue discipline.
Notations and Symbols:
The following notations and symbols will be used in connection with the queueing systems.
n = Total number of customers in the system, both waiting and in service.
= Average number of customers arriving per unit time.
= Average number of customers being served per unit time.
C = Number of parallel service channels (servers).
sL (or) E(n) = Expected or average number of customers in the system, both waiting and in service
[ ( )]s sL E N
qL (or) E(M) = Average or expected number of customer in the queue (excluding those who are
receiving service) [ ( )]q qL E N
sW (Or) E ( sW ) = Average or expected waiting time of a customer in the system both waiting and in
service (Including the service time).
qW (Or) E ( qW ) = Average or expected waiting time of a customer in the queue. (Excluding service time)
( )nP t = Probability that there are n customers in the system at any time t (both waiting and in service)
assuming that the system has started its operation at time zero.
= Traffic intensity or utilization factor, which represents the proportion of time the servers are busy
nP = Steady state probability of having n customers in the system.
MARKOVIAN MODELS:
Birth and Death Queueing Models:
(i) (M|M|I): ( | FIFO)
(ii) (M|M|C): ( | FIFO)
(iii) (M|M|I): (K|FIFO)
(iv) (M|M|S): (K|FIFO)
Where M stands for Markovian Model
MODEL – I
Single Server Poisson Queue Model – I:
[M|M|I] : [ | FIFO]:
Let 0P and nP be the values of the steady state probabilities for Poisson queue systems.
0 1P
Where 0P is the probability of system being idle. (ie). No customers in the queue.
0
n
nP p
(i) Average number sL of customers in the system:
Let N denotes the number of customers in the queueing system. Then the values of N are 0,1,2,3,…..
0
( )s n
n
L E N np
= 0
0
n
n
n p
= 0
1
n
n
n
= 0
1
n
n
n
=
2 3
1 0 2 3 .....
=
2
1 1 2 3 .....
=
2
1 1
=
1
1
=
( )
(1 )
= .
sL
(ii) The average number qL of customers in the queue or average length of the queue:
Let N denotes the number of customers in the queueing system. Then the number of customers
in the queue is (N-1).
1
( 1) ( 1)q n
n
L E N n p
= 1
( 1) 1
n
n
n
= 1
1 ( 1) 1n
n
=
2 3
1 0 2 .....
=
2 2
1 1 2 3 .....
=
2 2
1 1
=
2 1
1
=
2( )
(1 )
=
2 2
2.
( )
2
( )qL
(iii) Average number wL of customers in the non-empty queues:
Since the queue is non-empty, the number of customer in the system must be atleast 2, one
in the queue and one being served.
wL = E {(N-1) |P (N-1) > 0}
1 1
11 0
E N E N
P NP N
2
2
1
1 ( )
Q
n
n
L
P NP
2
2
1
1
n
n
2
2
1
1
n
n
2
2 3
1
( )1 .....
=
2
2 2
1
( )1 1 .....
2
2 1
1
( )1 1
2 2 2
2 2
1
( ) ( )
wL
(iv) The probability that the number of customers in the system exceeds K:
1 1
( ) 1
n
n
n k n k
P N K P
1
1
n
n k
1 2 3
1 .....
k k k
1 2
1 1 .....
k
=
1 1
1 1
k
P (N>K) =
1k
Little’s formula for (M|M|I):[ | FIFO) Model:
(i) )( ( ) [ ( ) ]S S S SE N E W E N L
(ii)
2
( ) ( ) [ ( ) ]( )
q q q qE N E W E N L
(iii) 1
( ) ( )S qE W E W
(iv) ( ) ( )S qE N E W
FORMULAS
1 = Mean arrival rate Average number of customers
arrivals per unit time
2 = Mean service rate per channel
1
[ ]E T
3 = Traffic intensity or utilization factor =
4 sL = Expected number of customers in the system
1
sL
5 qL = Expected number of customer in the queue
2
q sL L
6 sW = Expected time a customer spends in the system
1 1
s sW L
7 qW = Expected waiting time per customer in the queue
1
( )q qW L
8 0P = Probability of having zero customers in the system 0 1P
9 nP = Steady state probability of having n customers in the
system
1 , 0,1,2,....n
nP n
0
n
nP P
10 N=Maximum number of customer permitted in the system N-1 = Number of customer in the
queue.
11 (i) [ ]P N n = Probability of queue length being greater
than or equal to n.
(ii) [ ]P N n = Probability of queue length being greater
than n.
[ ] nP N n
1[ ] nP N n
12 P(w) = The probability density function of the waiting time
in the system.
( )( ) ( ) wP W e
13 ( )sP W t = The probability that the waiting time of a
customer in the system exceeds t.
( )( ) t
sP W t e
14 wL = Average length of the non-empty queue. Queue formed
from time to time.
1
1 1wL
15 ( )qP W t = The probability that the waiting time of a
customer in the queue exceeds t.
( )( )qP W t e
16 n = Total number of customer in the system both waiting and
service.
Problems based on (M/M/1): ( / )FIFO Model:
1. Customers arrive at one-man barber shop according to a Poisson process with a mean inter arrival
of 12 min. Customers spend an average of 10 min. in the Barber’s chair.
(i) What is the expected number of customers in the barber shop and in the queue?
(ii) Calculate the percentage (%) of time an arrival can walk straight into the barber’s chair
without having to wait.
(iii) How much time can customer expect to spend in the barbers shop?
(iv) Management will provide another chair and here another barber, when a customer’s
waiting time in the shop exceeds 1.25h. How much must the average rate of arrivals
increase to warrant a second barber.
(v) What are the average time customers spends in the queue?
(vi) What is the probability that the waiting time in the system is greater than 30 min?
(vii) Calculate the percentage (%) of customers who have to wait prior to getting into the
barber’s chair.
(viii) What is the Probability that more than 3 customers are in the system?
Solution:
Given: One man barber shop Single server.
Customers infinite capacity.
The given problem is (M/M/1): ( / )FIFO Model.
Given: Mean = 1
= 12 minutes.
Arrival rate = = 1
12 minutes.
And service rate = = 1
10 minutes.
(i)(a) To find the expected number of customers in the barber shop:
sL
1
12
1 110 12
=1
12
2120
=1 120
*12 2
5 sL Customers
(b) To find the expected number of customers in the queue:
2
( )qL
=
21
12
1 1 110 10 12
=
1144
1 210 120
= 1
144
21200
= 1 1200
*144 2
4.17 4 qL Customers
(ii) To Find P[A customer straightly go to the barber’s chair]:
P [No customer in the system] = 0 1P
1
120 1
10
1P
1 5 1
1 *10 112 6 6
0
1
6P
The Percentage of time an arrival need not wait = 1
*100 16.676
(iii)To find a customer expected time to spend in the barber’s shop:
1
sW
= 1 1 2
10 12 120
1 1 120
2
60 ( ) 1 .sW Min or hr
(iv) To find a customer’s waiting time in the shop exceeds 1.25 hrs:
1.25sW hrs
75 1.25 75minsW Min hrs
1
75
1
75
1
75
1
75
1 1 1
75 10 75
13
150
Hence to warrant a second barber, the average arrival rate must increase by
13 1 1min.
150 12 300per
(v) To find the average time customers spend in the queue:
( )
qW
112
1 1 110 10 12
112
1 210 120
50 minqW utes
(vi) To find the probability that the waiting time in the system is greater than 30 minutes
.sP W t
W.K.T ( )t
sP W t e
1 110 12
30 0.530sP W e e
30 0.6065sP W
(vii) To Find P[a customer has to wait]:
0 1 0P W P W
0 0
1 11 1
6 6P P
5
06
P W
The Percentage of customers who have to wait = 5
*100 83.336
(vii) To find P [More than 3 customers in the system] ie. 3 :P N
W.K.T 1n
P N n
3 1 41
2
110
13 *10
12P N
45
6
3P N 0.4823.
2. Automatic car wash facility operates with only one bay cars arrive according to a Poisson
process, with mean of 4 cars per hour and may wait in the facility’s parking lot if the bay is busy.
If the service time for all cars is contact and equal to 10 minutes. Determine , ,s q s qL L W and W
Solution:
Given: one bay Single server
Cars infinite capacity.
The given problem is (M/M/1): ( / )FIFO Model.
Given: Mean arrival rate = = 4 cars per hour.
Mean service rate = = 6 cars per hour.
1
10 *60 610
Min hrs
(i)To find sL :
sL
4 4
6 4 2
2sL cars
(ii) To find qL :
2
( )qL
2(4) 16 4
6 6 4 6*2 3
1.333 .qL cars
(iii) To find sW :
1 1 1
6 4 2
1.
2
s
s
W
W hours
(iv) To find qW :
4 4 1
( ) 6(6 4) 6*2 3qW
1
.3
qW hours
Hence,
(i) 2sL cars
(ii) 1.333 .qL cars
(iii)1
2sW hours
(iv)1
3qW hours
3. Customers arrive at the first class ticket counter of a theatre at a rate of 12 per hour. There is one
clerk servicing the customer’s at the rate of 30 per hour.
(i) What is the probability that there is no customer at the counter?
(ii) What is the probability that there are more than 2 customers at the counter?
(iii) What is the probability that there is no customer waiting to be served?
(iv) What is the probability that a customer is being served and nobody is waiting?
(v) Probability that a customer has to wait for at most 4 minutes in the queue?
Solution:
Given: one clerk Single server
Customers infinite capacity.
The given problem is (M/M/1): ( / )FIFO Model.
Given, Mean arrival rate = = 12 per hour.
Mean service rate = = 30 per hour.
(i)To find P [There is no customer at the counter]: 0P
0
0
12 181 1
30 30
0.6
P
P
(ii) To find P[There are more than 2 customers at the counter]: 2P N
W.K.T 1n
P N n
2 1 3
12 122
30 30P N
0.064
2P N 0.064.
(iii) To find P {There is no customer waiting to be served]:
P[There is at the most one customer in the counter] = 1P n
= P(n=0) + P(n=1) 0nP P
= 0 1
0 0 0 0.6P P P
= (0.6)(1) + (0.6)12
30
= 0.6 + 0.24
1P n = 0.84
(iv) P[A customer is being served and nobody is waiting]:
P[There is exactly one customer in the system who is being served]
1
1 0
1
120.6 0.4 0.6
30
0.24
P P
P
(v) To find P[n customer has to wait for at most 4 minutes in the queue]:
14
15q qP W Min P W
1
115
qP W
1
15( )
1 e
1
15(30 12)12
130
e
18
151 (0.4)e
= 1 – 0.1205
4qP W Min 0.8795
4. Customers arriving at a watch repair shop according to Poisson process at a rate of one per every
10 minutes and the service time is an exponential random variable with mean 8 minutes.
(i) Find the average number of customers sL in the shop.
(ii) Find the average time a customer spends in the shop sW
(iii) Find the average number of customer in the queue qL
(iv) What is the probability that the server is idle?
Solution:
Given: Watch repair shop Single server
Customer Infinite Capacity.
The given problem is (M|M|1) :( | FIFO) Model:
Given: Mean = 1
= 10 min
Arrival rate = 1
10
And service rate = 1
8
(i)To Find the average number of customers in the shop SL :
SL
1
10
1 18 10
1 80
10 2
SL = 4 customers.
(ii) To find the average time a customer spends in the shop SW :
1
SW
1 18 10
1 8040
2
SW = 40 minutes.
(iii) To find the average number of customer in the queue qL :
2
( )qL
21 1
10 100
1 1 1 1 28 8 10 8 80
1 808
( ) 100 2
16
5
qL = 3.2 3 customers
(iv) To Find the Probability that the server is idle: oP
1
10
18
1 41 1 1 8 1
10 5oP
1
5oP
5. Arrival at a telephone both are considered to be Poisson with an average time of 12 min between
one arrival and the next. The length of a Phone call is assumed to be distributed exponentially
with Mean 4 min.
(i) Find the average number of Persons waiting in the system.
(ii) What is the Probability that a person arriving at the booth will have to wait.
(iii) What is the Probability that it will take him more than 10 min altogether to wait for the
Phone and complete his call?
(iv) Estimate the fraction of the day when the Phone will be in use.
(v)The telephone department will install a second booth, when convinced that an arrival has to
wait on the average for atleast 3 min for Phone. By how much the flow of arrivals should increase
in order to justify a second booth?
(vi) What is the average length of the Non-empty queue that Forms from time to time?
Solution:
Given: Telephone booth Single server.
Customer s Infinite Capacity.
The given problem is (M|M|1) :( | FIFO) Model:
Given: Mean = 1
= 12 min
Arrival rate = 1
12
And service rate = 1
4
(i)To Find the average number of Persons in the system: SL
SL
1
12
1 14 12
1 48 1
12 8 2
1
2SL
(ii) To Find P[The Person arriving in the booth has to wait]: P[N > 0]
P[N > 0] = 1 – P[n = 0]
= 1 – P[no customer in the system]
= 1 - 0P
1
12
14
1 1 1 1
P[N > 0] = 1
3
(iii) To Find A Person takes more than 10 min to wait and complete his call ie., P[w > 10]:
W.K.T. ( )tP w t e
1 1 54 12 3
1010P w e e
[ 10] 0.1889P w
(iv) To Find P [The Phone will be in use]:
P[Phone in use] = P[Phone in busy]
= 1 – P[Phone in idle]
= 1 - 0P 1
12
14
1 1 1 1
P[Phone in use] = 1
3
(v) To Find the second Phone will be installed, if an arrival has to wait on the average for at least 3 min
for Phone: [ qW >3].
qW > 3
3( )
3 ( )
23 3
23 3
21 1
3 34 4
3 1
34 16
7 3
4 16
3
74
3
28
Hence the arrival rate should increase by,
3 1 1
min28 12 42
Per
(vi) To Find the average length of the Non-empty queue: wL
wL
14
1 14 12
1 48
4 8
3
2wL
6. 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
(i) The Mean size queue.
(ii) The Probability that the queue size exceeds 10.
(iii) If the input of trains increases to an average of 33 Per day what will be change in the above
quantities?
Solution:
Given: A railway Marshalling yard Single server
good trains Infinite Capacity.
The given problem is (M|M|1):( | FIFO) Model:
Given: Mean arrival rate = = 30 trains per day.
Mean service time = 36 minutes.
Mean service rate = (24)(60)
4036
trains per day
(i)To find the Mean queue size: qL
2
( )qL
230 900 900
40 40 30 40 10 400
9
4
qL = 2.25 2 trains.
(ii) To Find P[The queue size exceeds 10] (or) 10P N :
W.K.T. n
P N n
10 10
30 310
40 4P N
10 0.056P N
(iii) To Find if the input of trains increases to an average of 33 per day.
Here, 33, 40.
(a)To Find Mean queue size:
2
( )qL
233 1089 900
40 40 33 40 7 280
3.889
qL = 4 trains.
Change in queue size = 3.889 – 2.25 = 1.64
2 trains.
(b) To Find P[The queue size exceeds 10]
ie., 10P N
10
3310
40P N
10P N = 0.146
Change in Probability = 0.146 – 0.056 = 0.09.
MODEL – II
Multi server Poisson Queue Model – II [M|M|1]:[ | FIFO]
(i)The value of 0 nP and P :
The Probability of zero customers in the system 0P
01
0
1
1 1
! !
n cc
n
Pc
n c c
Where 0P is idle.
The Probability of Poisson Queue system nP is,
0
0
10
!
1
!
n
n n
n c
P If n cn
P
P If n cc c
(ii) The average number of customers in the queue: qL
1
02
1 1
!1
c
qL Pc c
c
Where C is Multi server or Number of server
(iii) The Average number of customers in the system SL
S qL L
1
02
1 1
!1
c
sL Pc c
c
(iv) The Average time a customer has to spend in the system sW :
1
s sW L
1
02
1 1 1
!1
c
sW Pc c
c
(v) The average time a customer has to spend in the queue qW :
1
q qW L
1
02
1 1 1
!1
c
qW Pc c
c
(vi) The Probability that an arrival has to wait (or) The Probability that there are C or more customers in
the system.
0( ) ( 0) ( )
! 1
c
sP N c P P W P N c
cc
(vii) ) The Probability that an arrival has to get the service without waiting:
P(Getting the service without waiting) = P(System is idle)
= 1 – P(Arrival has to wait)
01
! 1
c
P
cc
(viii) The Probability that someone will be waiting:
1
0( )
! 1
c
P N c P
c cc
(ix) The Mean waiting time in the queue for those who actually wait | 0q sE W W
1
| 0q sE W Wc
(x) The Average number of customers (in non-empty queues) who have to actually wait:
wLc
(xi) Traffic intensity (or) utilization factor = c
.
Problem based on (M|M|C) :( | FIFO) Model:
1. There are 3 typists in an office. Each typist can type an average of 6 letters per hour. If letters arrive for
being typed at the rate of 15 letters per hour.
(i) What fraction of the time all the typists will be busy?
(ii) What is the average number of letters waiting to be typed?
(iii) What is the average time a letter has to spend for waiting and for being typed?
(iv) What is the Probability that a letter will take longer than 20 min waiting to be typed and being
Typed?
Solution:
Given: 3 Typists Multi server
Number of letters Infinite Capacity.
The given problem is (M|M|C) :( | FIFO) Model:
Given: Arrival rate = = 15 per hour
Service rate = = 6 per hour
Number of server = C = 3.
To Find 0P :
01
0
1
1 1
! !
n cc
n
Pc
n c c
33 1
0
1
1 5 1 5 3 6
! 2 3! 2 3 6 15
n
n n
2
0
1
1 5 1 125 18
! 2 6 8 18 15
n
n n
2
0
1
1 5 1 125 18
! 2 6 8 3
n
n n
2
0
1
1 5 125
! 2 8
n
n n
0 1 2
1
1 5 1 5 1 5 125
0! 2 1! 2 2! 2 8
1 1
5 25 125 8 20 25 1251
2 8 8 8
8
178
0 0.0449P
(i)To Find P[All typists are busy]
ie., 3P N :
W.K.T. 0( )
! 1
c
P N c P
cc
315
6( 3) (0.0499)
153! 1
3 6
P N
125
8(0.0499)
156 1
18
125
8(0.0499)
36
18
3 0.7016P N
(ii) The average number of letters waiting to be typed:
1
02
1 1
!1
c
qL Pc c
c
3 1
2
1 15 1(0.0499)
3!3 6 151
3 6
4
2
1 15 1(0.0499)
6 3 6 18 15
18
1 625 324
(0.0499)18 16 9
9092.25
2592
3.5078qL
(iii) To find the average time a letter has to spend for waiting: sW
1
s sW L
W.K.T. S qL L
15
3.50786
6.0078sL
1 1
(6.0078)s sW L
1
(6.0078) 0.4005 .15
hours
0.4005 60minutes
24.03sW 24 minutes.
(iv) To find the probability that a letter will take longer than 20 min. waiting to be typed:
W.K.T
( 1 / )
0
1
1
! 1 1
/
c
t c
t
eu
e P
c cc
P w t
1 3 2 ( 0.5)6
3(2.5) (1 )
1 0.04492
20 1
.56 1 0.
3
3
560
P w P we
e
10.4616
3P w
2. A Supermarket has two girls attending to sales at the counters. If the service time for each customer is
exponential with Mean 4 minutes and if people arrive in a Poisson fashion at the rate of 10 per hour
(i) What is the probability of having to wait for service?
(ii) What is the expected percentage of idle time for each girl?
(iii) Find the average queue length and the average number of units in the system?
(iv) If a customer has to wait, what is the expected length of his waiting time?
Solution:
Given : Two girls multi server
Customer Infinite capacity
The given problem is (M|M|C): ( | FIFO) model:
Given: Arrival rate = = 10 per hour
Service time =1
= 4 minutes
Service rate = = 60
4 = 15 per hour
Number of server = C = 2.
To Find 0P :
01
0
1
1 1
! !
n Cc
n
PC
n C C
22 1
0
1
1 10 1 10 2 15
! 15 2! 15 2(15) 10
n
n n
21
0
1
1 10 1 10 30
! 15 2 15 20
n
n n
0 1 2
1
1 2 1 2 1 2 3
0! 3 1! 3 2 3 2
1
2 1 4 31
3 2 9 2
=1 1
2 1 3 2 11
3 3 3
1 3 1
6 / 3 6 2
0
1
2P
(i) To find P[A customer having to wait]:
P[N 2]:
W.K.T P[NC] 0
! 1
c
P
CC
2 210 1 2 1
15 2 3 2
10 202! 1 2
2 15 30
=4 1 3
9 4 2
1
6P N C
(ii) To find expected percentage of idle time for each girl:
Fraction of time that a girl is busy = C
10 1
2 5 3
1
3
Fraction of time when a girl is idle = 0
11 1
3P
0
2
3P
Percentage of idle time for each girl = 2
100 67%3
(iii) (a) To find average queue length in the system:
qL
1
02
1 1
!1
c
qL PC C
C
2 1
2
1 10 1 1
2!2 15 2101
2 15
3
2
1 2 1 1
4 3 220
30
=1 2 1 1 1 1
4 3 2 4 3 12
1
12Lq
(b) To find average number of units in the system: Ls
s qL L
=1 10 15 120 135
12 15 12 15 12 15
3
4sL
(iv) To find the expected length of his waiting time: ( | 0) :q sE W W
1
( | 0)q sE W WC
1 1
2 15 10 20hour
=1
6020
Minutes
( | 0)q sE W W = 3 minutes
3. A petrol pump station has 4 pumps. The service time follows the exponential distribution with a mean
of 6 min and cars arrive for service in a Poisson process at the rate of 30 cars per hour.
(i) What is the probability that an arrival would have to wait in line?
(ii) Find the average waiting time in the queue, average time spends in the system and the
average number of cars in the system.
(iii) For what percentage of time would a pump be, idle on an average?
Solution:
Arrival rate = = 30 per hour
Service time =1
= 6 minutes
Service rate = = 60
6 = 10 per hour
Number of server = C = 4.
To find 0P :
01
0
1
1 1
! !
n Cc
n
PC
n C C
44 1
0
1
1 30 1 30 4 10
! 10 4! 10 4 10 30
n
n n
3
4
0
1
1 1 403 3
! 4! 10
n
n n
0 1 2 3 4
1
1 1 1 1 13 3 3 3 3 4
0! 1! 2! 3! 24
=3 4
1
9 1 11 3 (3) (3)
2 6 6
0
2
53P
(i) To find P[an arrival would have to wait in line]
0
! 1
c
P N C P
CC
430
210
30 534! 1
10 4
43 2
10 5324
40
=
3(3)
53
= 0.5094.
(ii) (a) To find the average waiting time in the queue: :qW
1
02
1 1 1
!1
c
qW PC C
C
4 1
2
1 1 30 1 2
4! 30 4 10 53301
4(10)
5
2
1 1 1 2(3)
24 120 531
4
qW = 0.0509 hours
3.05minqW utes
(b) To find average time spends in the system: sW
1
02
1 1 1 1
!1
c
sW PC C
C
5
2
1 1 30 1 2 1( )
30 4!4 10 53 10301
4(10)
5
2
1 1 1 2 13 ( )
30 24(4) 53 101
4
=0.1509 hours
9.05sW Minutes
c) To find the average number of cars in the system: sL
1
02
1 1
!1
c
sL PC C
C
5
2
1 30 1 2 30
4!4 10 53 10301
4(10)
5
2
1 1 2 303 ( )
24(4) 53 101
4
= 4.5283 cars
sL = 4.5283 cars
(iii) To find the percentage of time a pump is idle:
P(A pump is idle) = 30
1 14(10)C
=3 1
14 4
Percentage of time a pump is idle
1
100 25%4
Model-III
Single server, Finite capacity Poisson Queue Model-III:
(M|M|l) : (K/FIFO)
(i) The value of 0 nP and P :
The probability of Zero customer in the system 0P is,
1
0
1
1
1
1
kIf
P
Ifk
Where 0P is idle.
The probability of Poisson queue system nP is
1
1
1
1
1
n
k
n
If
P
Ifk
(ii) The effective arrival rate :
0' (1 )P
(iii) P[A customer turned away]: kP
0
k
kP P
(iv) The Average number of customers in the system: sL
1
1
1
1 1
2
k
k
S
k
If
L
kIf
(v) The Average number of customers in the queue : qL
'
q sL L
(vi) The average time a customer has to spend in the system : sW
'
ss
LW
(vii) The average time a customer has to spend in the queue : qW
'
q
q
LW
(viii) Traffic intensity (or) Utilization factor:
Problems based on (M|M|l) : (K|FIFO) Model:
1. Patients arrive at a clinic according to Poisson distribution at a rate of 30 patients per hour. The
waiting room does not accommodate more than 14 patients. Examination time per patient is
exponential with mean rate of 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 from the clinic?
Solution:
Given : A clinic Single server.
Patient finite capacity
The given problem is (M|M|l) : (K|FIFO) Model:
Given : Arrival rate = = 30 per hour
Service rate = = 20 per hour
Finite capacity = k = capacity of the system
= [14 waiting patients + 1patient in service]
= 14 + 1
15k
To find 0P :
0 1
1
1
kP
1/615 1
301
1 3 / 220
33011
220
=0.5
0.00076655.84
0 0.00076P
(i) To find the effective arrival rate at the clinic:
The effective arrival rate = 0' (1 )P
= 20(1-0.00076)
=20(0.99924)
=19.98
' 20 Per hour
(ii) To find P(A patient will not wait):
0 0.00076P
(iii) To find the expected waiting time until a patient is discharged: sW
'
ss
LW
To find sL :
W.K.T .
1
1
( 1)
1 1
k
s k
k
L
15 1
15 1
3030 (15 1)2020
30 301 120 20
16
16
3 316
2 23 16.02
1 31
2 2
13.02sL
Expected waiting time '
ss
LW
13.02
20
= 0.651 hours
= 0.651 x 60 minutes
= 39.06 Minutes
39sW Minutes
2. At a railway station only one train is handled at a time. The railway yard is sufficient only for 2
trains to wait, while the 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 6 per hour.
Assuming Poisson arrivals and Exponential service distribution find the probabilities for the
numbers of trains in the system. Also find the average waiting time of a new train coming into the
yard. If the handling rate is doubled, how will the above results are modified?
Solution:
Given: A railway station Single server
Train Finite capacity
The given problem is (M|M|l) : (K|FIFO) model :
Given : Arrival rate = = 6 per hour [ ]
Service rate = = 6 per hour
Finite capacity = k = 3 ( 2 trains + 1 train = 3 trains)
To find 0P :
0
1[ ]
1P
K
0
1
4P
0
1 11,2,3..
1 4P for n
K
(i) To find P [The numbers of trains in the system] sL
[ ]2
s
KL
31.5
2sL
(ii) To find the average waiting time in the station of a new train coming in to the yard:
'
ss
LW
To find ' :
0
1 3' (1 ) 6 1 6
4 4P
' 4.5
'
1.5 1 1*60min
4.5 3 3
ss
LW hours utes
20minsW utes
(iii) To find if the handling rate is doubled:
i.e., Service rate is doubled = 2 6 12
To find 0P :
0 1
1
( )
1
KP
0 3 1
61
1 212
15 1661
12
P
0
8
15P
0
6 8
12 15nP P
1 8, 1,2,3..
2 15
n
nP n
For n=1,
1
1
1 8 4
2 15 15P
For n=2,
2
2
1 8 2
2 15 15P
For n=3,
3
3
1 8 1
2 15 15P
(a) To find sL :
1
1
1
1 1
K
s K
KL
3 1
3 1
6 12 3 1 6 121 0.2667
1 6 12 1 6 12sL
0.7333 .sL train
(b) To find sW :
'
ss
LW
To find ' :
0' (1 )P
=128
115
' 5.6 per hour
'
0.73330.131
5.6
ss
LW hour
=0.131x60 minutes
=7.86 minutes
7.9minutessW
MODEL-IV
Multiserver, Finite capacity Poisson queue model-IV
(M|M|C): (K|FIFO)
(i) The values of 0P and P n :
The probability of zero customers in the system 0P is
01
0
1
1 1
! !
n C n CC K
n n C
P
n c C
The Probability of Poisson Queue system P n is:
0
0
1
!
1
!
0
n
n
n n C
P for n Cn
P P for C n KC C
for n K
(ii) The average queue length or average number of customers in the queue:
0 21 1
! 1
C K C K C
q
CL P K C
C C CC
C
(iii) The average number of customers in the system:
'
s qL L
(iv) The average number of waiting time in the system:
'
1s sW L
(v)The average number of waiting time in the queue:
'
1q qW L
(vi)The effective arrival rate: '
1'
0
0
[ ( ) ]C
n
C C n P
(vii) Traffic intensity (or) Utilization factor:
= C
Problems based on (M|M|C) : (K|FIFO) model:
1. A 2 person barber shop has 5 chairs to accommodate waiting customers. Potential customers, who
arrive when all 5 chairs are full leave without entering barber shop. Customers arrive at the
average rate of 4 per hour and spend on average of 12min in the barber’s chair compute
0 1 7, , , q sP P P L and W
Solution:
Given: Two person barber shop Multiserver
Five chair finite capacity
The given problem is (M|M|C): (K|FIFO) model:
Given: Arrival rate= 4 per hour
Inter-service rate=1
=12 min
1 1 112
60 5x
Service rate= =5 per hour
Finite capacity =K=Capacity of the system
5 2K
K=7
Number of servers=C=2
2C
(i) To find 0P
01
0
1
1 1
! !
n C n CC K
n n C
P
n c C
=2 22 1 7
0 2
1
1 4 1 4 4
! 5 2! 5 2 5
n n
n nn
=21 7
0 2
1
1 4 1 16 4
! 5 2! 25 10
n n
n nn
=0 1 0 1 2 3 4 5
1
1 4 1 4 8 2 2 2 2 2 2
0! 5 1! 5 25 5 5 5 5 5 5
=1
4 8 2 4 8 16 321 1
5 25 5 25 125 625 3125
=1
9 8(1.65984)
5 25
=1
9 13.27872
5 25
=1
45 13.27872
25
=25
0.42958.27872
0 0.429P
(ii) To find 1P :
W.K.T nP 0
1
!
n
P for n Cn
Put n = 1, 1P
11 4
(0.429) [ 1 2]1! 5
=4
(0.429)5
1 0.3432P
(iii) To find 7P :
W.K.T nP 0
1
!
n
n CP for C n K
C C
Put n=7, 7P
7
7 2
1 4(0.429)
2!2 5
=
7
5
1 4(0.429)
2(2 ) 5
7 0.0014P
(iv) To find :qL
W.K.T 0 21 1
! 1
C K C K C
q
CL P K C
C C CC
C
2 7 2 7 2
2
4 4 5 2 4 4 4(0.429) 1 7 2 1
5 5 2 5 2 5 242! 1
5 2
5 5
2
16 4 10 2 2 2(0.429) 1 5 1
25 5 5 542 1
10
16 2 5 32 480(0.429) 1
925 3125 156252
25
= (0.429)(0.64)(0.5556)[1-0.0102-0.030]
=0.1463
0.15qL customer
(v) To find :sW
'
1s sW L
To find ' :
1'
0
[ ( ) ]C
n
n
C C n P
2 1
0
5[2 (2 ) ]n
n
n P
1
0
5[2 (2 ) ]n
n
n P
5[2 (2(0.429) (0.3432)]
' 3.994
To find :sL
'
s qL L
3.9940.15 0.9488
5
0.95sL customer
'
1 10.95
3.994s sW L
=0.2376 hours
=0.2376 60minutes
14.256 minsW utes
2. A car servicing station has 2 bays where service can be offered simultaneously. Because of space
limitations. Only 4 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 8 cars per day. Find
the average number of cars in the service station, the average number of cars-waiting for service
and the average time a car spends in the system.
Solution:
Given: Two bays Multiserver
Four cars finite capacity
The given problem is (M|M|C): (K|FIFO) model:
Given: Arrival rate= 12 per day
Service rate= =8 per day
Finite capacity =K=Capacity of the system
K=4
Number of servers=C=2
2C
(vi) To find 0P
01
0
1
1 1
! !
n C n CC K
n n C
P
n c C
=2 22 1 4
0 2
1
1 12 1 12 12
! 8 2! 8 8 2
n n
n nn
=21 4
0 2
1
1 3 1 9 3
! 2 2 4 4
n n
n nn
=0 1 0 1 2
1
1 3 1 3 9 3 3 3
0! 2 1! 2 8 4 4 4
=1
3 9 3 91 1
2 8 4 16
=1
5 9 37
2 8 16
=1 1
2.5 1.125(2.3125) 5.10156
0 0.1960P
(ii) To find the average number of cars waiting for service in the queue :qL
W.K.T 0 21 1
! 1
C K C K C
q
CL P K C
C C CC
C
2 4 2 4 2
2
12 12 8 2 12 12 12(0.1960) 1 4 2 1
8 8 2 8 2 8 2122! 1
8 2
2 2 2
2
3 3 4 3 1 3(0.1960) 1 2
2 4 4 412
4
9 3 4 9 9(0.1960) 1
14 16 322
16
= (0.1960) (2.25)(6)[1-0.5625-0.28125]
= 2.646 0.15625
= 0.4134
0.4134qL
(iii) To find the average number of cars in the service station: sL :
'
s qL L
To find ' :
1'
0
0
[ ( ) ]C
n
C C n P
2 1
0
8[2 (2 ) ]n
n
n P
1
0
8[2 (2 ) ]n
n
n P
=8[2-(2 0P +1. 1P )]
8[2 (2 0.1960 0.294)]
=8(1.314).
' 10.512
'
s qL L
10.512
0.4134 0.4134 1.3148
1.7274 1.73sL
(iv) To find the average time a car spend in the system :sW
'
1s sW L
1.72740.1643
10.512day
=0.1643 24hours
3.94sW hours