Study Material by M. Shadab Khan
Queuing Theory
Introduction
A group of items waiting to receive service are known as queue. Queues or waiting lines
are in every day life. Some of the situations where queues may be seen are; patients
waiting for doctor at hospital outdoor, customers waiting at barber shop for hair cut,
passengers waiting at railway booking counter, students waiting at college fee counter,
vehicles lines up at petrol pumps, machines waiting to be repaired, trucks in line to be
unloaded or airplane lined up on a run way waiting for permission to take off.
Queues are formed to get service at that time when demand for a service is more than
capacity of service facility. In a queue, customers arrive at a greater rate than the service
rendered. In case queue start building up and a time will come when queue will become
so large that customers may leave the queue and the new customers may not join it. This
will cause losses to the business and also customers waste time standing in a queue,
known as waiting time cost. But if there are no queues, the service facility may remain
idle. The waiting phenomenon is the direct result of randomness in the operation of
service facility. In general, otherwise the operation of the customers arrival otherwise
operations of the service time are not known in advance; for facilities could be scheduled
in a manner that would eliminate waiting completely. Hence, the objective to study
Queuing Theory is to maintain a balance between waiting time cost and service cost so
that these two are minimum.
Features of Queuing Theory
Most important features of queuing model are as follows:
1.
length of the queue. It indicates the average number of customers waiting in
the line. Large queues indicate poor service and small queues imply too much
service capacity.
2.
System. System means maximum capacity of the queue i.e. system = customers
waiting + customers being served.
3.
Waiting Time. This is the average time that a customer has to wait to get
service. If waiting time is too much long, it may result in potential loss of revenue
and set back to the goodwill of the business.
4.
Total Time. Total time means time taken from entry in the queue to the
completion of the service.
5.
Idle Time. The relative frequency for which the service system remains idle. Idle
time leads to increase in the related cost.
6.
Arrival Pattern. The pattern of arrival of customers at service station is known
as arrival pattern of the queue. The arrival pattern can be regular or it can be
irregular at random interval of time. A regular pattern of arrival is not very common
in the case of customers coming for service, generally it is completely at random. In
case the numbers of customers are very large, the probability of an arrival in next
interval of time does not depend upon the number of customers already in the
system. Rather where the number of customers are large the arrival completely at
random and it follows Poisson distribution with mean equal to average number of
arrivals per unit of time. A simple test to verify that the collected data follow
Poisson distribution i.e. are truly random, is to substitute the mean of one
distribution in the following equation and obtain the mean of the other distribution.
Mean of Exponential = 1/ Mean of Poisson. For example, mean arrival per hour is 3, the
mean interval between arrivals will be 1/3 or 20 minutes i.e. if the mean of Poisson
distribution is 3 then the mean of exponential distribution will be 20 minutes. As the
arrival time can be expressed either in Poisson or in exponential terms, the arrival pattern
is proved to be truly random
7.
Service Pattern. Although arrival pattern is random in most situations and can be
satisfactorily modeled using the Poisson distribution, service pattern exhibits no
obvious or consistent pattern. However for the sake of simplicity and in order to
reduce the necessity of complex mathematical models, simple queue theory assumes
that the service pattern can be represented by treating them as being exponential.
8.
Service Management. For producing service to the incoming customers at the
service station certain service points are established. The number of these service
points mostly dependent upon the number of customers, rate of arrivals time taken
for providing service too a single customer, availability of persons /resources for
producing service etc. Depending upon these variables a customer service channel
can be either a single channel or a multi channel.
Service System
By structure of the service system we mean how the service facilities exist and what is
the service time and arrival pattern including the queue discipline.
1.
A Single Channel, Single Phase System. A library counter is an example
of this system
Arrival
Queue
XXX
Departure
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Service Centre
2.
Service Facility in a Series. In this case when a customer enters he gets one kind
of service and then moves to the next station, gets some service and then again moves
to the next station and so on and ultimately leaves the system. For example, machining
of a certain steel item may consist of cutting, turning, knurling, drilling, grinding and
packing operations each of which is performed by a single server in a series.
Arrival
Departure
Queue
XXXX
3.
Service Centre I
XXX
Service centre II
centre
Multiple, Parallel Facilities with Single Queue. In this model, there are
more than one server each server provides the same facility e.g. washing of a car on
shop having more than one washers or cutting of hair on barbers shop.
Departure
Service Centre I
Queue
Arrival
XXXX
Service Centre II
Service Centre III
4.
Multiple Parallel Facilities with Multiple Queues. Reservation of
railway tickets at the railways station counters are example of this type of model.
Arrival
5.
Queue
Departure
XXXX
Service Centre I
XXXX
Service Centre II
XXXX
Service Centre III
XXXX
Service Centre IV
Service Time and Arrival Pattern. The time taken in providing service to a
customer depends upon the nature of the service to be provided as well as the
requirement of the customers. Accordingly, the service time can be constant or vary
with the customers. However, for the sake of simplicity of queue model, it is
assumed that the time requirement for service is constant for all customers.
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Moreover, since the arrival pattern is assumed to be random, the service time is also
taken as random where the server does not distinguish between different arrivals
and does not change deliberately the duration of service on the basis of time taken
to serve the previous arrival. Under these conditions, the service time follows
exponential distribution with mean equal to reciprocal of the mean rate of service as
explained earlier. In cases where assumption of an exponential distribution for
service time is not valid, Erlang distribution is applied. Service time also decides
the expected waiting time for the customers.
6.
Queue Discipline. The order in which the customers are selected from the
queue for service is known as queue discipline. This can be of one of the
following type:
(a)
First in First Served (FIFS). Under FIFS arrangement selection of
customers for service is done in the order in which they arrive and form the
queue.
(b)
Last in First Served (LIFS). Under LIFS arrangement selection of
customers for service is done in the reverse order in which they arrive and the
person entering last is the first to be selected for service. It is not an appropriate
method, such practice is common in the case of issue of stores, as the
storekeeper finds it more convenient to remove the item in LIFS order while
issuing for job.
(c)
Service in Random Order (SIRO). In case of service in random order
customers are selected at random out of those present at the service state at that
particular time
(d)
Service in Priority (SIP). Under such an arrangement certain customers
are given priority over others in selection for service. The priority can be of
two types viz. non-pre-emptive priority, where the customer already getting
served is allowed to continue with service till it is completed even when a
priority customer arrives mid way during the service, and pre-emptive priority,
where service to non-priority customer is stopped as soon as priority customer
arrives. This is in the case of when some critical machine breaks down in a
factory. Repair of these critical machines is given top priority even by stopping
the repair of less important machines
7.
Customer Behaviour. Various customers while in queue behave differently.
(a)
Balking. Some customers show reluctance for waiting in the queue. They do not
join the queue at their correct position and attempt to jump the queue and reach the
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service centre by passing other ahead of them. This is known as balking. In simple
words it is a situation, when customers decide not to join the waiting line.
(b)
Collusion. When one customer represents a group of customers is called as collusion.
(c)
Jockeying. When a customer who has already joined a queue leaves it and joins
another queue is called jockeying.
(d)
Reneging. Some customers loose their patience after some time and leave the
queue without getting service.
Queuing Situation
Situation
Arriving
customers
Situations
1. Passage of customers through a
supermarket checkout
2. Floor of automobile traffic through a
road network
3. Transfer of electronic message
4. Banking Transaction
5. Flow of computer programmes
through a computer system
6. Sale of theater tickets
7. Arrival of trucks to carry fruits and
vegetables from a central market
8. Registration of Unemployed at
employment exchange
9. Occurrence of fire
10. Flow of ships to the seashore
Shoppers
Checkout counter
Automobiles
Road network
Electronic message
Bank patron
Computers
programmes
Theater- goers
Trucks
Transmission line
Bank tellers
Central processing unit
Unemployed
personnel
Fires
Ships
Ticket booking window
Loading crews &
facilities
Registration Assistants
Firemen & equipment
Harbor & docking
facilities
The waiting lines develop because the service to a customer may not be rendered
immediately as the customer reaches the service facility. Thus lack of adequate service
facility would be waiting line of customers to be framed. The only way that the service
demand can be met with ease is to increase the service capacity (and raising the
efficiency of the existing capacity if possible) to a higher level. The capacity might be
built to such high level as can always meet the peak demand with no queues. But adding
to may be a costly affair and uneconomic after a stage because then it shall remain idle to
verify degrees when there are no or few customers. A manager, therefore, has to decide
on an appropriate level of service which is neither too low nor too high. I efficient or poor
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service would cause excessive waiting which has a cost in terms of customer frustration,
loss of goods will in the long run, direct cost of idle employees etc.
Notations
1.
Customers. A unit coming for service to the service station is known as customer.
This can be a person, machine, telephone calls or demand for some commodity.
2.
Queue or Waiting Time. A line formed by customers waiting to get service is
known as queue or waiting line. The customers already getting the service shall not
be included in the queue.
3.
Service Channel. A system or channel consisting of service points which
provide service to the incoming customers is known as service channel.
4.
Arrival Rate. The rate at which the customers arrive at the service centre for service
per unit of time is known as arrival rate. It is denoted by lamda ( ) . The arrival is
calculated by dividing the total number of arrivals by total units of time. If the rate of
arrivals is in Poisson distribution then the distribution of time between two arrivals is
1
exponential. Thus, the arrival rate is then the mean of time intervals will be .

5.
Service Rate. The rate at which service is provided at the service centre per unit
of time is known as service rate. It is denoted by mew (  ) . This is calculated by
dividing total number of customers served by total units of time. The service rate
should be greater than arrival rate otherwise the queue will never terminate.
6.
Traffic Intensity or Utilization Rate. The rate at which the available
service facility is utilized is known as traffic intensity or utilization rate or average
service rate and denoted by 
7.
Idle Rate. The rate at which service facility remains utilized is known as idle rate
and it is denoted by P0 . Therefore, P0  1  P  1 
8.

.

Expected Number of Customers in the System ( En ) . The Expected
number of Customers in the system is equal to the customers in the queue plus those
getting the service. It is denoted by En , in simple terms the customers waiting and in
service, i.e. E n 

 
.
287
9.
Expected Number of Customers in a Queue or Waiting Line ( EL ).
The expected number of customers in a queue is equal to the expected number of
customers in the system minus the customers getting service. This is known as
queue length and is denoted by EL .
E L  E n  p or E L 
10.



  
.
       (   )
2
Expected Time Spent by Customers in the System ( ET ) . The expected
time spent by customer in the system is equal to the expected time spent while
waiting for service plus the service time. Since the rate of arrival is ( ) , the
expected number of customers in the system En will be equal to   ET

11.
En
(   )
1
.


 
Expected Waiting Time in a Queue ( Ew ). The expected time spent by
customers in the system is equal to the expected time spent while waiting for
1
service plus the service time or ET  E w  . By this relation the expected waiting
ET 



time in the queue is determined by the equation
 1  1

1
 
E w  ET 
or
.



         (   )
The last four formulas are given below in short for the convenience of the students
 

2

En 
or  E L    E L 
or  E n   
 
 (   )


Ew 
12.
1

or  ET   
 (   )

ET 
1
1
or  E w   
 

Probability of Zero Customers Waiting. The probability of zero
customers waiting means a customer arriving for service does not have to wait in a
queue or the idle rate of the system. According to the probability of zero customer
waiting is determined by the equation
 
P0  1  P  1  
 
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Similarly the probability of 1,2,3.......n persons waiting in a queue for service is determined
by the formulas given below:
2
 
   
p1  p 0    1   
   
 
3
 
    
p 3  p 0    1   

    

    
p 2  p 0    1   

    
3
n
2

    
p n  p 0    1   

    
n
Cost of Service to the Customers
(a)
Cost of waiting in a queue = E L  E w  cost of waiting per customer per unit of time.
(b)
Cost of Idle Time = E n  ET  cost of waiting per customer per unit of time
(c)
Total Cost = Cost of waiting in a queue + Cost of Idle Time
13.
S = Number of service channels, N = maximum number of customers allowed in the
system and Em = Average length of non-empty queue.
Queuing Models
The queuing models can be of the following three types:
Model
Probabilistic
Model-I
Model-II
Erlang
(M/M/1) : (  /FCFS)
Deterministic
(D/D/1) : (K-1/FCFS)
D stands Deterministic arrival
D Stands Deterministic service
Time distribution
General Erlang Model
(M/M/1) : (  /FCFS)
(Rate of arrival and service
Depends on the length)
Model -III
(M/M/1) : (N/FCFS)
Capacity is limited (finite)
Model -IV
(M/M/S) : (  /FCFS)
(Parallel service station)
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Mixed
(M/D/1) : (  /FCFS) D
stands Deterministic
service time
Where FCFS stands for first come first served
M : Markovian property of exponential process
M : Markovian property of Poisson distribution
Probabilistic Model
When arrival and service rate are unknown and assumed to be random variable
Poisson Distribution. A Poisson distribution is a discrete probability distribution
which predicts the number of arrivals in a given time. The Poisson distribution involves
the probability of occurrence of an arrival. Poisson assumption is quite restrictive in some
cases, it assumes that the arrivals are random and independent of all other operating
conditions. Poisson distribution assumes fixed time interval of continuous servicing
which is never sure in all services.
Exponential Distribution. The most common type of distribution used for service
time is exponential distribution which involves the probability of completion of a service.
The queuing theory provides a mathematical frame work to present these dimensions of a
queue problem in precise statistical terms and to develop a solution which can, avoiding
both the extremes, meet the requirements of customers as well as service unit. The
queuing theory then is to minimize the total cost of queue i.e. the cost of producing
service and the cost of waiting time both, with the help of suitable models. Various
constraints are taken into consideration in developing the queuing model. There is no
maximization and minimization of objective function. Various alternatives are considered
and evaluated through the queuing model and s final choice of appropriate alternative ad
appropriate model is made.
Single Server Queuing Model
Model I {(M/M/1) : (Y/FCFS)}
In this model the service is exponential and queue is unlimited.
Assumptions of this model are as under :
(1)
Exponential distribution of inter arrival time or Poisson distribution of arrivals.
(2)
Single waiting line no restriction on length of queue (i.e. infinite capacity) and no
balking or reneging.
(3)
Queue discipline is first come first served.
(4)
Single server with exponential distribution of service time.
290
(5)
The number of arrivals per unit of time id described by Poisson distribution. Here
the mean arrival rate is denoted by ( ) Lambda.
(6)
The service time has exponential distribution. The average service time is denoted
by  Mue.
(7)
Arrivals are from infinite population.
(8)
The queue discipline in on the basis first come first serve basis (FIFO).
(9)
Arrivals are independent of preceding arrivals but the average number of arrivals
(the arrival rate) does not change over time.
(10) There exists only a single service station.
(11) The mean arrival rate is less than the mean service rate.
(12) The waiting space available for customers in the queue is infinite.
(13) The customer arriving does not leave without getting the service.
(a)
(b)
(c)
Expected number of customers in the system E n 

 
2
Queue length (expected number of customers waiting in queue) E L 
 (   )

Expected waiting for a customer in the queue EW 
 (   )
(d)
Expected waiting time for a customer in the system (waiting and service)
1
ET 
 
(e)
Expected length of non-empty queue
Example 1. A television repairman finds that the time spent on his job has an
experimental distribution with a mean of 30 minutes. If he repairs sets in the order in
which they came in, and if the arrival of sets follows a Poisson’s distribution
approximately with an average rate of 10 per hour day, assuming 8 hours shift. What is
the repairmen’s expected idle time each day? How many jobs are ahead of the average set
just brought in.
Solution. From the above data we can drive the following
  10/8
sets per hour
  (1/30)60
= 2 sets per hour
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(i)
(ii)

Expected idle time of repairman each day = 8   8(5 / 8)  5 hours because traffic

intensity is 8 hour per day.
Idle time for a repairman in an 8 hours day will be = 8-5 =3 hours.
(iii) Expected (average) number of TV sets in the system E  

 

5/ 4
5
 TV
2  5/ 4 3
sets.
Model II {(M/M/1) : (  /SIRO)}}
This model is same as the Model I with a difference only in queue discipline. Since the
derivation of Pn is independent of any specific queue discipline, therefore in this model
we have
Pn  (1  p ) p n ; n = 1, 2, 3…
Hence, the results remain unchanged in any queue system as long as Pn remains unchanged.
Model III{(M/M/1) : (N/FCFS)}
Condition : Exponential service, finite or limited queue. As per as the capacity of the
system is concerned this model is different from Model I because any customer arriving
when the system already contains N customers does not enter the system and is gone.
This may happen an account of physical constraint such as one chair saloon, one chair
dentist with certain number of chairs for waiting customers etc. Hence most of the
equations derived in Model I removes the same as long as n < N. In case of Model I
length of the queue is unlimited but in this case is limited. Hence the service rate will be
less than the arrival rate    .
(i)
(ii)
Expected number of customers in the system
N
En 
 P  1(   ) or P  1(   )
2
Expected number of customers waiting in the system or expected queue length
E L  En  P
or
E L  En 


(iii) Expected waiting time of a customer in the system i.e. (waiting + service)
En
Ew 
 (1  Pn )
292
(iv) Expected waiting time of a customer in the queue
ET  EW 
1

Example 2. Consider a single server queuing system with Poisson’s input, exponential
service times. Suppose the mean arrival rate is 36 calling units per hours, the expected
service time is 0.25 hour and the maximum permissible calling units in the system in true.
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 above data
3
units per hour
 4
units per hour
Traffic intensity,  

= 3/4= .75

The steady state probability distribution of number of n customers (calling units) in the
system is
(1   )  n
Pn 
;  1
1   N 1

P0 
and
(1  0.75)(0.75) n
 (0.43)(0.75) n
1  (0.75) 2 1
(1   )
1  0.75
0.25


 0.431
N 1
2 1
1 
1  (0.75)
1  (o.75) 3
The expected number of calling units in the given system is given by
N
2
2
n 1
n 1
n 1
Ls   nPn   n(0.43)(0.75) n  0.43 n(0.75) n  0.43{(0.75)  2(0.75) 2 }  0.81
Multi – Server Queuing Models
Model IV {(M/M/s) : (  / FCFS)} Exponential service Unlimited Queue
This model is an extension of Model I. In this case instead of single service channel, there
are multiple servers in parallel equal to s. For this queuing system it is assumed that
customers arrive according to a Poisson processes at an average rate of  customers per
293
unit of time and are served on a first come, first served basis at any of the servers. These
servers are identical, each serving customers according to an exponential distribution
with an average of  customers per unit of time. It is further assumed that only one queue
is formed. The overall service rate of servers in obtained in two situations, when n
customers are in the system:
(i)
If n  s (number of customers in the system are less than the number of servers), then
there will be no queue. However, numbers of servers are not busy. The combined
service rate will then be
 n  n : n  s
(ii)
If n  s (number of customers in the system are more than or equal to the number of
servers), then all servers will be busy and the maximum number of customers in the
queue will be (n  s) . The combined service rate will be
 n  s : n  s
1.
Expected number of customers waiting with a queue
 1   n
 
 
EL  
 P0
2
 ( s  1)!    ( s   ) 
2.
Expected number of customers in the system
En  EL 
3.


Expected waiting time of a customer in the queue
 1   n


 
Ew 

 P0
  ( s  1)!    ( s   ) 2 
EL
4.
Expected waiting time a customer spends in the system
ET  Ew
1 EL EL
  

1

Thus the probability that the system shall be idle is
1
n
s
 n 1 ( sp) n 1 ( sp) s 
1
s 
  n 1 1   






P0  

,






s!    s   
s  n 0 n!   
 n 0 n!  s! 1   
294
1
Example 3. A bank has two tailors working on saving accounts. The first teller
handles withdrawn only and the second teller handle deposits only. It has been found that
the service time distribution for the deposits and withdrawals both are exponential with
mean service time 3 minutes per customer. Depositors are found to arrive in a Poisson
distribution throughout the day with a mean arrival rate of 14 per hour. What would be
the effect on the average waiting time for depositors and withdrawers if each teller could
handle both withdrawals and deposits? What would be the effect if this could only be
accomplished by increasing the service time to 3.5 minutes?
Solution. (I) Initially there are two independent queuing systems: withdrawers and
depositors where arrivals follow Poisson distribution and service time follows
exponential distribution.
(i)
For withdrawers it is given that  = 14/hour and  = 3/minutes or 20/ hour
Average waiting time in a queue E w 
(ii)
For depositors it is given that  = 14/hour and  = 3/minutes or 20/ hour
Average waiting time in a queue E w 
(II)

14

 7 / 60 hour or 7 minutes.
 (    ) 20(20  14)

16

 1 / 5 hour or 12 minutes.
 (    ) 20(20  16)
In combined case there will be a common queue with two servers (tellers). Thus, we
have  = 14+16 = 30/hour and  = 20/ hour, s = 2,  
(i)
 n 1 1    n 1    s s 
P0       

s!    s   
 n 0 n!   
 1 1  3  n 1  3  2 40 
      

2!  2  40  30 
 n 0 n!  2 
(ii)

=3/4.
s
1
1
 3 19 
 1    .4
 2 24 
1

1
7
Average waiting time of arrival in the queue
Ew 
2
 1   n


20
1
9
3

 

P

. 
hour or 3.85 minutes.


0
2
2
  ( s  1)!    ( s   ) 
 2  (40  30) 7 140


EL
(iii) Combined waiting time with increased service time when  = 30/hour and  =
60/3.5 or 120/7 per hour, we have
295
 1 1  21  n 1  21  s 2(120 / 7) 
P0       

2!  12  2(120 / 7)  30 
 n0 n!  12 
1

1
15
(iv) Average waiting time of arrivals in the queue
2
 1   n


120 / 7
1
7
 
Ew 

.  11.43 minutes.
 P0   
2
2
  ( s  1)!    ( s   ) 
 2  (120 / 7  30) 15
EL
Example 4. At a service centre customers arrive at the rate of 10 per hour and are
served at the rate of 15 per hour. Their arrival follows Poisson distribution and service
is exponentially distributed. Find the average length and average waiting time in the
system.
Solution. Average arrival
 =10 per hour
 = 15 per hour
Average service rate
Traffic Intensity or Utilization rate  

 10 / 15  .66

2
100
100


= 4/3 person
 (    ) 15(15  10) 75
(i)
Average queue length E L 
(ii)
Average waiting time in the system ET 
1
 1/5 hour or 12 minutes.
 
Example 5. Customers arrive at a sales counter managed by a single person according
to Poisson distribution with a mean rate of 20 per hour. The time required to serve a
customer has an exponential distribution with a mean of 100 seconds. Find the average
waiting time of a customer.
Solution. Mean arrival rate  =20 per hour. It takes 100 seconds to serve a customer,
hence number of customers served in 1 hour =  =36.
The average waiting time of a customer in the queue
EW 

20
5


hours
 (    ) 36(36  20) 36  4

5  3600
seconds=125 seconds
36  4
296
The average waiting time of a customer in the system
ET 
1
1

= 1/16
(    ) (36  20)
Example 6. An airline organization has one reservation clerk on duty in its local
branch at any given time. The clerk handles information regarding passenger’s
reservation and flight timings. Assume that the number of customers arriving during any
given period in Poisson distribution with an arrival rate of 8 per hour and that the
reservation clerk can serve a customer in 6 minutes on an average, with an exponentially
distributed time.
(i)
What is the probability that the system is busy.
(ii)
What is the average time a customer spends in the system.
(iii) What is the average length of the queue and what is the number of customers in the
system.
Solution. Average arrival rate  =8 per hour.
Average service rate  =60/6 = 10 per hour.
 8

 10
(i)
Traffic Intensity  
(ii)
Average time spend in the system ET 
1
1

= 0.5 hour or 30 minutes.
(    ) 10  8
(iii) Average length of queue and number of customer in the system
EL 
En 
2
64
64


 3.2 persons
 (    ) 10(10  8) 20

(   )

8
8
  4 persons = 1/16 hours = 225 seconds.
(10  8) 2
Example 7. A company distributes its product by trucks loaded at its only loading
station. Both, company’s trucks and contractor’s trucks are used for this purpose. It was
found that on an average every five minutes, one truck arrived and the average loading
time was three minutes. 50% of the trucks belong to the contractor. Find
(i) The probability that a truck has to wait
(ii) The waiting time of truck that waits.
297
(iii) The expected waiting time of contractor’s trucks per day, assuming 24 hours
shift.
Solution. Average arrival rate  =1 per five minute = 12 per hour.
Average loading rate  = 3 per minute = 20 per hour.

 12 / 20  0.6 0

(i)
Traffic Intensity  
(ii)
The waiting time of truck E w 

12
12


 3 / 40 per hour.
 (    ) 20(20  12) 160
We know, Expected waiting time of a truck = 3/40 p/h
Number of contractor’s truck per day = 6  24 = 144
(iii) Expected waiting time of contractor’s trucks per day
=
3
 (number of contractor’s trucks per day)
40

3
 144  108 hrs.
40
Example 8. Customers arrive at the first class ticket counter of a theater at a rate of 12
per hour. There is one clerk service for the customer at a rate of 30 per hour.
(i)
What is the probability that there is no customer at the counter (idle system).
(ii)
What s 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.
Solution.
Arrival rate  =12 per hour
Service rate  = 30 per hour.
Traffic Intensity P =

 12 / 30  2 / 5

(i)
p 0  (1  p)  (1  2 / 5)  0.6
(ii)
p (more than two customers at counter) = p (three or more customers in the queue)
298
= p 3  (2 / 3) 3  0.064
(iii) p (no customer is waiting to be served) = p ( at most one customer at counter)
= p0  p1  0.6  0.6(0.4)  0.6  0.24  0.84
(iv) p (a customer is being served and no body is waiting) = p1  0.6  0.4  0.24
Example 9. A bank plans to open a single sever drive in banking facility at a particular
centre. It is estimated that 28 customers will serve each hour on an average. If on an
average, it is required 2 minutes to process a customer’s transaction. Determine
(i) The proportion of time that the system will be idle.
(ii) On the average how long the customer will have to wait before reaching the server.
(iii) The length of the drive way required to accommodate all the arrivals on the average,
if 20 feet of drive way required for each car that is waiting for services.
Solution.
Average arrival rate  =28
Average service time  =60/2 = 30.
Traffic Intensity P =

 28 / 30  0.9333

(i)
The system will be idle P0  (1  P)  1  0.9333  0.067
(ii)
Customer waiting in queue E w 

28
28


 7/15 hours.
 (    ) 30(30  28) 30  2
2
282
764


 1307
(iii) Average number of customers waiting E L 
 (    ) 30(30  28) 30  2
Since 20 feet of drive way is required for each car, the length of drive way required to
accommodate all arrivals waiting for service is = 20  13 = 261.3 feet
Example 10. A Xerox machine in an office is operated by a person who does other
jobs also, the average service time for a job is 16 minutes per customers. On an average,
in every 12 minutes one customer arrives for Xeroxing. Find
(i)
The Xerox machine utilization.
(ii)
Percentage of time when an arrival has not to wait
(iii) Average time spent by a customer.
299
(iv) Average queue length.
(v)
The arrival rate if the management is willing to deploy the person exclusively for
Xeroxing when average time spent by the customer is 15 minutes.
Solution.
Arrival rate  =5 per hrs
Service time  = 10 per hrs.

 5 / 10  0.5

(i)
Utilization factor  
(ii)
Customer do not have to wait = 1 – 0.5 = 0.5
(iii) Average time spent by customer ET 
(iv) Average queue length E L 
(v)
1
1
 hours
(   ) 5
2
25

 25 / 50  0.5
 (    ) 10  5
Arrival rate ET  15 minutes i.e. 15/60 hours =1/4 hours
ET 
1
 
1
1

 6
4 10  
Example 11. In a bank every 15 minutes one customer arrives for cashing the cheque.
There is only one payment counter which takes 10 minutes for serving a customer on an
average. Find
(i)
The average queue length
(ii)
The average waiting time
(iii) Increase in the arrival rate in order to justify a second counter (when Ew  15
minutes, waiting time of customer is at least 15 minutes the management will
increase one more counter)
Solution. (i) Arrival rate   60 / 15  4 per hour.
Service rate   60 / 10  6 per hour
As  <  , using M/M/1/  queuing model, the average queue length is given by
300
EL 
(ii)
2
42
16


 1.33 customers.
 (    ) 6(6  4) 6  2
The average waiting time for the present system is
Ew 

4
4


 1/3 hours = 20 minutes.
 (    ) 6(6  4) 6  2
(iii) Whether a second counter will be opened if the waiting time is at least 15 minutes.
15
'

60 6(6   )
or
e.i.
60'  90(6  ' )  '  540 / 150  36
Example 12. A bank has one drive-in counter. It is estimated that cars arrive according
to Poisson distribution at the rate of 2 every 5 minutes and that there is enough space to
accommodate a line of 10 cars. Other arriving cars can wait outside this space, if
necessary. It takes 1.5 minutes on an average to serve a customer, but the service time
actually varies according to an exponential distribution. Find
(i)
The proportion of time the facility remains idle.
(ii)
Expected number of customers waiting but currently not being served at a particular
point of time.
(iii) Expected time a customer spends in the system.
(iv) Probability that the waiting line will exceed the capacity of the space leading to
drive in counter.
Solution. From the data given, we have
20  60
 24 per hour.
5
60
 40 per hour.
Mean arrival time  
1.5
Mean arrival rate  
(i)
The proportion of time, the facility remains the idle is given by
    24 
p0  1    1    1  0.6  0.4
    20 
Hence 40% of the time, the facility remains idle.
301
(ii)
Expected number of customers waiting but currently not being served at a particular
point of time
EL 
2
24 2
576


 0.9
 (    ) 40(40  24) 640
(iii) The expected time a customer spends in the system
ET 
1
1
1


hours = 3.75 minutes
(    ) (40  24) 16
(iv) Since there is enough space only to accommodate a line of 10 cars, the waiting line
will exceed the capacity of the space if there are 11 or more cars in the system. The
probability that the waiting line will exceed the capacity of the space (10 cars only)
leading to the drive in counter (i.e. the probability of having 11 or more cars in the
system) is given by
11
11

 24 
p(n  11)        0.0036
 40 

Hence, if the arrival rate of customers is greater 6 customers per hour, the average time
spent by a customer will exceed 15 minutes.
Example 13. A hospital is studying a proposal to reorganize its emergency service
facility. The present arrival rate at the emergency service is 1 call every 15 minutes and
the service rate is 1 call every 10 minutes. Current cost of service is Rs.100 per hour.
Each delay in service costs Rs. 125. If the proposal is accepted the service rate will
become 1 call every 6 minutes. Can the reorganization be justified on a strictly cost basis
if the acceptance of the proposal is to result in increase in cost of service by 50%.
Solution. Arrival rate   4 per hour and Service time   6 per hour.
Cost of service = Rs. 100 per hour
Opportunity cost = Rs. 125 for each delay in service
Utilization factor  
 4 2
 
 6 3
Average queue length E L 
2
42
16

  1.33 patient.
 (    ) 6(6  4) 12
Actual cost of present service system
302
En  E L 100  (2  4 / 3) 100  66.67
Opportunity cost of delay =
16
125  166.67
12
Hence Total cost of the system = 66.67 + 166.67 = 233.34
Now in case the proposal for reorganization is accepted
Actual cost of service ( En  E L )  C S ; opportunity cost of delay = E L  Ci
Arrival rate   4 per hour and Service time   10 per hour.
Cost of service = 100 + 50 = Rs. 150; opportunity cost = Rs. 125
Average queue length
En 

 

4
4
42
16
 

= 0.26 patients
10  4 6 10(10  4) 60
Actual cost of proposed service system
24
 4 16 
( En  EL )  150     150 
 150  Rs. 60
60
 6 10 
Opportunity cost of proposed service system =
16
125  Rs. 33
60
Total cost of proposed service system = 60 + 33 = 93
Total cost of proposed service system is less than the total cost of present service system,
therefore the proposal for reorganization of the facility should be accepted.
Example 14. A road transport company has one reservation clerk on duty at a time. He
handles information of bus schedule and makes reservations. Customers arrive at a rate of
8 per hour and the clerk can serve 12 customers on an average per hour. After stating
your assumption, answer the following :
(i)
What is the average number of customers waiting for service of the clerk in the
system and queue.
(ii)
What is the average time a customer has to wait before getting service in the system
and queue.
303
Solution. Arrival rate   8 per hour and Service time   12 per hour.
(i)
Average number of customers waiting for the service of the clerk ( in system)
En 

 

8
8
  2 customers.
12  8 2
Average number of customers waiting for the service of the clerk (in queue)
2
82
64
EL 


 1.33 customers.
 (    ) 12(12  8) 48
(ii) The average waiting time of customer before getting service (in system)
1
1
1
ET 

 hours or 15 minutes.
(    ) (12  8) 4
The average waiting time of customer before getting service (in queue)

8
1
EW 

 hours or 10 minutes.
 (    ) 12(12  8) 6
Example 15. A typist in an office, on the average 22 letters per day for typing. The
typist works 8 hours a day and it takes an average 20 minutes to type a letter. The office
in charge has determined that the cost of a letter waiting to be mailed is 80 paisa per hour
and the equipment operating cost plus the salary of the typist will be Rs. 40 per day.
(i)
What is the typist utility rate.
(ii)
What is the average number of letters waiting to be typed.
(iii) What is the average waiting time needed to have a letter typed.
(iv) What is the total daily cost of waiting letters to be mailed.
Solution. Average arrival
60  8

 24 per day.
20
rate
  22 per day and Average service rate

 22 / 24  0.917

(i)
Traffic Intensity  
(ii)
Average number of letters waiting to be typed
2
22 2
484
EL 


 10.08.
 (    ) 24(24  22) 48
304
(iii) Average waiting time, ET 
1
1
1

 day i.e. 4 hours.
(    ) (24  22) 2
(iv) Number of letters in system En 

 

22
 11 letters per day and the
2
opportunity cost = 0.80  8 = 6.4 per day
Total daily cost = Number of letters in system  opportunity cost + equipment operating
cost + salary of the typist = 11  6.4 + 40 = 70.4 + 40 = 110.40
Example 16. In a bank with a single server there are two chairs for waiting customers.
On an average one customers arrives 10 minutes and each customer takes 5 minutes for
getting served. Making the suitable assumptions, find
(i)
The probability that an arrival will get a chair to sit down.
(ii)
The probability that an arrival will have to stand.
(iii) Expected waiting time of a customer.
Solution. From the above date, we have
Arrival rate   10 minutes or 6 customers per hour.
Service rate   5 minutes or 12 customers per hour.
There are two chairs for waiting customers
(i) The probability that an arrival will get a chair to sit down is given by
    
P0  P1  P2  1    1     
    
2
 
1  
 
6 6
6 6

 1     1     
 12  12  12   12 
2
6

1  
 12 
2
1 1 1 1 1 7
  .    .   0.875
2 2 2 2 2 8
(ii) The probability that an arrival will have to stand in given by
1  ( P0  P1  P2 )  1 
7 1
  0.125
8 8
Alternatively, the probability that an arrival will have to stand in given by
305
3
3
  6 
6
p(n  3)         0.125
8
    12 
(iii) Expecting waiting time of a customer in the queue is given by
EW 

6
1 60



 5 minutes.
 (    ) 12(12  6) 12 12
Multi Channel Service System
A service system with queue served by parallel service channels, each server having an
independent and identical service time distribution with arrival process assumed to be
Poisson distribution is known as multichannel service system. The common type multi
channel service system is known as M/M/C/  /FIFS Model i.e. arrival rate in Poisson
distribution (M), service time in exponential distribution (M), multi service channels (c),
infinite customer (  ) and service in First in First Served order. The formulae as listed
from (i) to (xi) below are used for obtaining various results in multichannel models.
In case there is, instead of a single server channel, C service channels, the mean arrival rate
will be  n i.e.  for all n. Mean service rate in case (n > c) = c  as all the servers will be
busy. The mean service rate in case (n < c) = n  as only n servers out of c will be busy.
(i)
The probability of the service counters remaining idle i.e. no customers getting
service or waiting (n = 0)
 C 1 1    n   1    c  c  
P0           
 
 n0 n!      c!     c    
(ii)
1
The probability of 1, 2, 3,………..,n customers in the system
n
1
Pn    P0 if 0  n  c
n!   
2
1 
1
P1    P0 ; P2    P0
1!   
2!   
(iii) Probability that a customer arriving has to wait (a customer will have to wait if the
number of customers are more than the service counters)
306
c
Pnc

  


P0
(c  1)!(c   )
(iv) Expected number of customers waiting in queue
c
 
  

EL 
P0
(c  1)!(c   ) 2
(v)
Expected number of customers in the system
c

  


En 
P 
2 0

(c  1)!(c   )
or
 
En  E L   

(vi) Expected waiting time for a customer in queue
c

  

EW 
P0
(c  1)!(c   ) 2
(vii) Expected time for a customer in the system
c

  
1
1

ET 
P0 
or ET  EW 
(c  1)!(c   )


(viii) Expected number of idle service points = c – expected number of customers served
in the system.
(ix) Efficiency of the service system = Expected number of customers served / Total
number of customers.
Single Service Counter with Arrival from M Channels
(x) Probability of n units in the system
n

m!
Pn  P0  
where m ! permutations of m units and (m-n) ! = permutations (m   (m  n)!
n) units.
307
Probability of empty system i.e. no customer at the service station
P0 
1

   m     m 
   
  ..........
1   
   (m  1)!     (m  2)! 

(xi) Expected number of customers in the system
 1 P1  2  P2  3  P3  ..............
   m! 
 
  2 P0  
 P0  
   (m  1)! 

2
 m! 

  .............
 (m  1)! 
Example 17. A super market has two sales girls at its sales counters. If the service
time for each customer is exponential with mean 4 minutes and if the customers arrive in
Poisson’s distribution at the rate of 10 per hour. Find
(i)
The probability of having to wait for service.
(ii)
Expected idle time for each girl.
(iii) Expected number of customers in the super market at any point of time.
Solution. Arrival rate   1/6 per minute, service rate  =1/4 per minute for each
customer.
Number of service channel c = 2 girls at the sales counters.
Thus, utilization rate 

1/ 6

 1/ 3
c 2  1 / 4
Probability of empty system i.e. no customer at the sales counters
 C 1 1    n   1    c  c  
P0           
 
 n0 n!      c!     c    
1
1
1
 21 1  4  0   1  4  2  2  1 / 4  
1
 2 1
          
   1    
2
 3 3
 n0 0!  6    2!  6   2  1 / 4  1 / 6  
Probability of one customer at the sales counters
308
1  
1
 1/ 6 
P1    P0  1
 1 / 2 
1!   
3
 1/ 4 
(i)
Probability of having to wait for the service:
As there are two sales girls at the sales counters the customers will have to wait only if
the numbers of customers are more than two. So

Pn2   Pn
n 2
or
1  ( P0  P1 ) , since

P
n 0
n
1
 1 1 1
Pn2  1       16.67
 2 3 6
(ii)
Expected number of idle sales girls  (2  P0 )  (1  P1 )  (2  1 / 2)  (1  1 / 3)  4 / 3
Expected idle time for a sales girl = Expected number of idle sales girls / Total number
4
sales girls
=  2  4/6
3
(iii) Expected number of customers in the super market
c

 1 1  1 1  1
  
     

  6 4  6 4  2 1 / 6

En 
P0  

2
(c  1)!(c   ) 2

1/ 4
 1 1
(2  1)! 2   
 4 6
2
5 4 1
 
12
6 6  2  5  2  23

1
3 4 3 12
1
9
Example 18. A petrol service station has two petrol pumps. The service time follows
the exponential distribution with a mean of 4 minutes and the vehicles arrive for service
as per Poisson’s distribution at the rate of 10 per hour. Find the probability that a
customer has to wait for service. What is the exponential time of waiting of customer in
the idle.
Solution. Arrival rate   10 vehicles per hour, service rate  =1/4 vehicles per minute
or 5 vehicles per hour.
309
Number of service state c = 2
Combined service rate (c )  2  15  30 vehicles per hour.
Probability of no vehicle at the petrol service station
 C 1 1    n   1    c  c  
P0           
 
 n 1 n!      c!     c    
1
1
1
  10 1  1  2  10  2  30  
1
 2 1
 1         
   1    
2
 3 3
  15   2!   15   30  10  
Probability that a vehicle arriving has to wait:
As there are two petrol filling pumps, a vehicle will required to wait in case the numbers
of vehicles are more then two. So
c
PnC

 10 
  
15 
1
15  4 / 9  1 / 2 1
 15 


P0 
 

(c  1)!(c   )
(2  1)!(30  10) 2
1 20
6
2
Expected time of waits at the pumps
c
2

 10 
  
15  
1
15  4 / 9  1 / 2
1

 15 
EW 
P 
 

Hours.
2
2 0
(2  1)!(30  10) 2
1  400
120
(c  1)!(c   )
Example 19. A telephone exchange has two operators for long distance calls. The
telephone company finds that the peak load, long distance calls arrive in Poisson
distribution at an average rate of 15 per hours. The length of service time on these calls is
approximately exponentially distributed with mean time 5 minutes per calls. What is the
probability that a subscriber will have to wait for his long distance call during the peak
load hours of the day. If the subscribers wait and are serviced as per turn, find the
expected waiting time.
Solution. The arrival rate   15 calls per hour, service rate  =60/5 = 12 calls per hours.
Number of service points c = 2 operators
  
15
5
Utilization rate =   
 .
 c  1  12 8
310
Probability of no telephone call
 21 1    n   1    c  c  
P0           
 
 n0 n!      c!     c    
 1  15 1  1  15  2  2  12 
         

 0!  12   2!  21   2  12  15 
1
1

12
.
52
Probability of one telephone call
n
1
1 
1  15  12 15
P1    P0    
 .
n 
1  12  52 52
Since there are two operators for long distance calls a customer will have to wait if there
are two and more than two calls. So
Pn 2  1  ( P0  P1 )  1  (12 / 52  15 / 52)  25 / 52.
Expected waiting time for the customers
c

 15 
  
12 
12
 12 

EW 
P 

2 0
2
(c  1)!(c   )
(2  1)!(2  12  15) 52
2
25
16  12  25 hours or 3.2 minutes.

1  81 52 468
12 
Example 20. A insurance company has three claims adjusters at its branch office.
Policy holders are found to make claims against the company in a Poisson distribution at
an average of 20 per 8 hours day. The time claim adjuster spends with a claimant is found
to follow exponential distribution with mean service time 40 minutes per claim. Claims
are processed in the order of their submission. How many hours a claim adjuster is
expected to spend the claimants per week?
Solution. The arrival rate   20/8 or 5/2 claims per hour, service rate  =60/40 or 3/2
claims per hour.
Number of service points c = 3 claims adjusters.
311
Utilization rate 

5/ 2

 5/9 .
c 3  3 / 2
Probability of no policy holder being with the claim adjusters
 c 1 1    n   1    c  c  
P0           
 
n
!

c
!

c



n

0








 

 
1
  5  25   1  5  3  3  3 / 2 
 1         

  3  18   6  3   3  3 / 2  5 / 2 
1

24
.
139
Probability of one policy holder being with the claim adjusters
1
1
1 
40
 5 / 2  24
P1    P0  1  

.
 
n!   
 3 / 2  139 139
Probability of two policy holders being with the claim adjusters
2
2
1
24 100
 5/ 2 
P2    P0  1 / 2  

.
 
n!   
 3 / 2  139 417
Expected number of idle claim adjusters at any point of time = All three idle + any two
idle + any one idle
24  
40   100 

= 3P0  2 P1  1P2   3 
  2
  1 

 139   139   417 
= Expected number of idle claim adjusters / Total number of adjusters

4/3
 4/9 .
3
Probability of number of claim adjuster remaining idle  1 
4 5
 .
9 9
Time claim adjuster is expected to spend with the policy holders per week, assuming 5
working days of 8 hours each in a week = 5/9  40 hours = 22.2 hours.
Example 21. A bank has two tellers working for savings bank accounts. The first teller
handles withdrawals, while the second teller handles deposits. The service time
distribution for both deposits and withdrawals are exponential with mean service time 3
minutes per customer. Deposits are found to arrive in a Poisson distribution with mean
312
arrival rate of 14 per hours. Determine the expected waiting time for customers coming
for deposits and for withdrawals.
(i)
What would be the effect on the average waiting time for depositors and
withdrawers if each teller handles both withdrawals and deposits.
(ii)
What would be the effect if this is accompanied by increasing the service time to
3.5 minutes.
Solution. The arrival rate (depositors) 1  16 customers per hours
The arrival rate (withdrawers) 2  14 customers per hours
Service rate (both tellers)  =60/3 or 20 customers per hours.
(i) Existing expected waiting time for customers
For depositors E w1 
1
16
16


hours or 12 minutes.
 (   1 ) 20(20  16) 80
For withdrawers E w 2 
2
14
14


hours 7 minutes.
 (   2 ) 20(20  14) 120
Expected waiting time in case both the tellers do both the functions :
Combined arrival rate   16  14  30 customers per hours.
Number of service points c = 2 tellers.
Probability of no customer in the bank at a point of time
 c 1 1    n   1    c  c  
P0           
 
 n0 n!      c!     c    
 1  30   1  30  2  2  20 
         

 0!  20   2!  20   2  20  30 
1
1
1
1
1
 3 9
7
 1        .
7
 2 2
1
Expected waiting time for the depositors and withdrawers both combined
c
E(W 1&2 )

3
  
20 
1
9
2


P 
 
hours or 3.86 minutes.
2 0
2
(c  1)!(c   )
(2  1)!(40  30) 7 140
2
313
The waiting time for the depositors and the withdrawers both will be considerably
reduced. From 12 minutes to 3.86 minutes. Hence it is a better proportion.
(ii)
Expected waiting time in case the service time is increased to 3.5 minutes then the
1
new service rate is  =
 60  120 / 7 customers per hours. Probability of no
3 .5
customer in the bank
 c 1 1    n   1    c  c  
P0           
 
 n0 n!      c!     c    
1
 1  30   1  30  2  2  120 / 7 
  
   
 

 0!  120 / 7   2!  120 / 7   2  120 / 7  30 
Expected waiting time for the customers
1

1
.
15
c
E(W 1& 2)

  


P
2 0
(c  1)!(c   )
2
120 
7 
 30 

1
7 
120 

 .
120
2
(2  1)!(2 
 30) 5
7
343

hours or 11.43 minutes.
1800
The time for waiting will increase considerably, hence not desirable.
Erlang Queue Model
Erlang queue model are based on a situation where service is provided in a number of
exponentially distributed phases, one immediately following the other. It is assumed that
all these phases are completed before a new unit enters the service channel.
Main assumptions of this service distribution are
(i)
The arrival pattern is as per Poisson distribution
(ii)
One unit is allowed in the service channel at one time & when the service is
completed through all the phases another unit is admitted into the service channel.
Broad structure of the model is shown in the following diagram:
314
SERVICE PHASES
1
2
DEPARTURE
XXXX
SERVICE CHANNEL
The formulae as listed from (i) to (iv) below are used for obtaining various results in the
case of this model:
(i) Expected number of customers in the queue
K 1
2
EL 

2K  (   )
(ii) Expected number of customers in the system
K 1
2


En 


or E n  E L 
2 K  (   ) 

(iii) Expected waiting time in thee queue
K 1

EW 

2 K  (   )
(iv) Expected time in the system
K 1

1
1
ET 


or ET  EW 
2K  (   ) 

Example 22. An automobile workshop maintenance of vehicle is done in two stages.
Service time at each stage is one hour with exponential form. If one vehicle is brought for
maintenance every 4 hours at the workshop. Determine expected number of vehicles in
queue and in the system, expected waiting time in queue and in the system.
Solution. The arrival rate   1 / 4 vehicle per hour, service rate  =1 vehicle per hour at
each phase.
Number of phases in service K = 2 phases
(i)
Expected number of customers in the queue
2
1
 
2
K 1

2 1  4 
1
EL 



 vehicles per hours.
2 K  (    ) 2  2  1  16
11  
 4
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(ii)
Expected number of vehicles in the system
2
1
 
2
1/ 4 5
K 1

 2 1  4 
En 

 



vehicles per hours.
2 K  (   )  2  2  1 
1
16
11  
 4
(iii) Expected waiting time in the queue
1
 
2 1  4 
1
K 1

EW 



 hours.
2 K  (   ) 2  2  1  4
11  
 4
(iv) Expected time in the system
1
 
1 5
K 1

1 2 1  4 
ET 

 

  hours.
2K  (   )  2  2  1  4 4
11  
 4
Example 23. Repair of a certain machine which breaks down in the factory from time
to time requires five operations, which have to be performed in a sequence. The time
taken to perform each of these five operations is found to have an exponential distribution
with mean 5 minutes and is independent of other steps. If these machines break down in
Poisson distribution at an average rate of two per hour and if there is only one man for
repair, what is the average idle time for each machine break down.
Solution. Number of service phases K = 5 phases
Total service time for a machine = 5  5 = 25 minutes.
So, service rate  = 1/25 machines per minute.
Average inter arrival time is = 30 minute.
Utilization rate 

25

 0.167 %
K 30  5
Expected idle time for a machine or total time for repairs
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ET 
1 / 30
K 1

1 5 1

 

 25 =100 minutes.
2K  (   )  2  5 1  1 1 
  
25  15 30 
Example 24. Arrival of customers to payment counter (only one) in a bank follow
Poisson distribution with an average of 10 per hour. The service time follows negative
exponential distribution with an average of 4 minutes.
(i)
What is the average number of customers in the queue.
(ii)
The bank will open one more counter when the waiting time of a customer is at
least 10 minutes. By how much the flow of arrivals should increase in order to
justify the second counter.
Solution. Arrival rate   10 per hour, service rate  = 60/4 = 15 per hour.
(i)
2
10 2
4
Average number of customers in queue E L 

 .
 (    ) 15(15  10) 3
(ii)
Arrival rate for waiting time EW  10 minutes or 10/60 hour.
EW 

 (   )
10


   10.7
60 15(15   )
Arrival rate should increase by 10.7-10 = 0.7 persons or more per hour to justify the
second counter.
Example 25. A repairman is to be hired to repair machines which break down at the
average rate of 6 per hour. The break downs follow Poisson distribution. The productive
time of a 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
break down 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).
Solution. Arrival rate or break down rate   6 per hour.
Service rate of Mr. X 1  8 per hour and for Mr. Y  2  12 per hour.
There are two alternatives either to hire Mr. X or to hire Mr. Y.
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(i)
In case Mr. X is hired ;
Average break down time of machine
ET 
1
1

 1 / 2 Hours.
  86
Total cost of break down = Loss of productive time + cost of repair
 1/2 hrs  Rs. 20 + 1/8 hrs  Rs. 10 = Rs.11.25
(ii)
In case Mr. Y is hired :
Average break down time of machine ET 
1
1

 1 / 6 Hours
   12  6
.
Total cost of break down = Loss of productive time + cost of repair
 1/6
hrs  Rs. 20 + 1/12 hrs  Rs. 14 = Rs. 4.50.
Therefore repairman Y should be hired.
Review Questions
1.
What is the waiting line problem? What are the components in a waiting line
system.
2.
What are the assumptions underlying common queuing model?
3.
Why must the server rate be greater than the arrival rate in a single channel queuing
system?
4.
What is queue? Give an example and explain the basic elements of queue.
5.
Give some applications of queuing theory and explain the following terms (i)
Queue
(ii) Traffic Intensity (iii) service Channel (iv) Queue Discipline (v) Balking
6.
What do you understand by queuing structure? Explain (i) First Come First Serve
(ii) Last Come First Served (iii) service in Random Basis of Customer handling.
7.
Explain the role of queuing theory in decision making and discuss its applications.
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Exercise
1.
Customers arrive at a booking office window being manned by a single individual
at a rate of 25 per hours. The time required to serve a customer has exponential
distribution with a new man of 30 per hours. Discuss the different characteristics of
queuing system assuming that there is only one server.
2.
Customers arrive at a sale counter manned by a single person according to a
Poisson distribution with a mean rate of 20 per hours. The time required to serve a
customer has an exponential distribution with a mean of 100 seconds. Find the
average waiting time of a customer.
3.
In the insurance company on an average one customer arrive in every 12 minutes in
the company. Determine (i) Utilization rate of company (ii) Percentage of time that
on arrival has not to wait
(iii) Average time spent by a customers (iv) average queue length
4.
5.
The mean rate of arrival of planes at airport during the peak period in 20 per hours
and the actual number of arrivals in any hour follow a Poisson distribution. The
airport can land 60 planes per hour, on an average in good weather and 30 planes
per hour in a bad weather. But the actual number landed in any hour follows a
Poisson distribution with these respective averages when there is congestion. The
planes are forced to fly over the field in the stock awaiting the landing of other
planes that arrives earlier. Determine
(i)
How many planes would be flying over the field in stock on an average a
good weather and in bad weather?
(ii)
How long a plane would be in the stock and in the process landing in good
and bad weather?
In a service department manned by one server. On an average 8 customers arrive
every 5 minutes while the service can serve to customers in the same time.
Assuming Poisson distribution for arrival and exponential distribution for service
rate determine:
(i)
Average number of customers in system.
(ii)
Average number of customers in the queue.
(iii) Average time a customer spent in system.
(iv) Average time a customer waits before being served.
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6.
At Dr. Sharma clinic patients arrive at an average of 6 patients per hours. The clinic
is attended by Dr. Sharma himself. The doctor takes 6 minutes per patient to serve.
It can be assumed that arrival follows Poisson distribution and the doctor inspection
time follows an exponential distribution. Determine
(i)
The percent of time a patient can walk right inside the doctor’s cabin without
having to wait.
(ii)
The average number of patients in Dr. Sharma clinic.
(iii) The average number of patients waiting for their turn.
(iv) The average time a patient spends in the clinic.
7.
A self service store employees are cashier at its counter a customer arrive on an
average every 5 minute while the cashier can serve to customers in 5 minutes.
Assuming Poisson distribution for service rate. Determine
(i)
Average number of customer in the system.
(ii)
Average number of customer in the queue.
(iii) Average time a customer spends in the system.
(iv) Average time a customer waits before being served.
8.
In a bank, cheques are cashed at a single teller counter, customers arrive at the
counter in a Poisson distribution on an average rate of 30 customers pr hour. The
teller takes on an average a minute and a half to cash cheque. The service time has
been shown to be exponential distribution, calculate
(i)
The percentage of time the teller is busy.
(ii)
The average time a customer is expected to wait.
9.
Trucks arrive at a factory for collecting finished goods for transportation to distant
markets. As and when they come they are required to join a waiting time and are
served on first come first served basis truck arrive at a rate of 10 per hour. Where as
the loading rate is15 per hour. Transporters have complained that their trucks have
to wait for nearly 12 hours at the plant. Examine whether the complaint is justified
and also the probability that the loaders are idle in the above problem.
10.
In a tool crib manned by a single assistant, operators arrive at the tool crib at the
rate of 10 per hour. Each operator needs 3 minutes on an average to be served. Find
out the loss of production due to waiting of an operator on a shift of 8 hours of the
rate of production in 100 units per shift.
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11.
T.V. repairman find that the time short on the jobs has an exponential distribution
with mean 30 minutes. If the repair sets on the first come first service basis and if
the arrivals of the sets in approximately Poisson distribution with an average rate of
48 minutes in 8 hours day. What is repairmen’s expected idle time each day? Also
obtain the average number of units in the system.
12.
In a bank the present arrival rate of customers is 15 minutes while the present
service rate is 10 minutes. Current cost of providing service is Rs. 100 per hour.
Each delay in service cost Rs. 125, find out the total cost of the bank.
13.
A branch of Punjab National Bank has only one typist. Since the typing work varies
in length (number of pages to be typed), the typing rate is randomly distributed
approximately a Poisson distribution with mean service rate 8 letters per hour. The
letters arrive at a rate of 5 per hour during the entire 8 hour work day. If the type
writer is valued at Rs. 1.50 per hour, determine
(i)
Equipment utilization
(ii)
The percent time an arriving letters gas to wait.
(iii) Average system time
(iv) Average idle time cost of typewriter per day.
14.
The ABC Co’s quality control department in managed by a single clerk, who takes
on an average 5 minutes in checking parts of each of the machines coming for
inspection. The machines arrive once in every 8 minutes on an average. One hour of
the machine is valued at Rs. 15 and the clerk time is valued at Rs. 4 per hour. What
are the average hourly queuing system cost associated with the quality control
management.
15.
A load transport company is studying the proposals to reorganize its service facility.
The arrival of customers is 4 per hour while service rate is 10 minutes per customer.
Current cost of service is Rs. 100 per hour. Each delay in service cost is Rs. 125.
The total cost of the present system is calculated to be Rs. 233.34 if proposal is
accepted, the service rate will become 10 per hour. Can the reorganization be
justified on a strictly cost basis if acceptance of proposal is to result in increase in
cost of service by 50%.
16.
A firm has served machines and wants to install its own service facility for the
repair of its machines.. The average breakdown rate of the machines is 3 per day.
The repair time has exponential distribution. The loss incurred due to the last time
of an inoperative machine is Rs. 40 per day. There are two repair facilities
available. Facility A has an installation cost of Rs. 20.000 and the facility B requires
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cost of Rs. 40,000. with facility A, the total labour cost is Rs. 5000 per year and
with facility B, the total labour cost is Rs. 8000 per year. Facility A can repair 4.5
machines per day and facility B can repair 5 machines per day. Both facilities have
a life of 4 years , which facility should be installed.
17.
In the production shop of a company, the breakdown of the machines is found to be a
Poisson distribution with an average rate of 3 machines per hours. Breakdown time at
one machine costs Rs. 40 per hour to the company. There are 2 choices before the
company to hire the repairman. One of the repairmen is slow but cheap, but the other is
fast and expensive. The slow cheap repairman demands Rs. 20 per hour and will repair
the various breakdown machines exponentially at the rate of 4 per hour. The fast
expensive repairman demands Rs. 30 per hour and will repair machines exponentially
at an average rate of 6 per hour, which repairman should be hired.
18.
Self service at a university cafeteria, at an average rate of 7 minutes per customer, is
slower than attendant service, which has a rate of 6 minutes per customer. The
manager of the cafeteria wishes to calculate the average number of customers in the
cafeteria, the average time each customer spends and the average time each
customer spends waiting for service. Assume that customers arrive randomly at
each time at the rate of 5 per hour. Calculate the appropriate operating statistics for
this cafeteria.
19.
Arrival of cars at a petrol station are considered to be Poisson distribution with an
average time of 12 minutes between one arrival and the next. The length of filling
petrol is assumed to be distributed exponentially with mean of 3 minutes.
(i)
What is the probability that a person arriving at the petrol pump will have to wait.
(ii)
What is the average length of queue that forms from time to time.
(iii) The petrol station queue will install a second petrol pump when convinced that an
arrival would expect waiting for at least 3 minutes for service. By how much
should the flow of car arrivals increased in order to justify a second pump.
20.
On an average 96 patients per 24 hour day require the service of an emergency
clinic. Also on an average a patient requires 10 minutes of active attention. Assume
that the facility can handle only one emergency at a time. Suppose that it costs the
clinic Rs. 100 per patient treated to obtain an average serving time of 10 minutes
and that each decrease in this average time would cost Rs. 10 per patient. How
much would have to be budgeted by the clinic to decrease the average size of the
1
1
queue from 1 patients to patients.
3
2
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Answers
1. En  5 Customers,
E L  25 / 6 customers,
ET  1 / 5 hrs, EW  1 / 6 hrs) 2. EW  125 , ET  225 3.
ET  1 / 5 Hrs, E L  0.5 customers 4. EL  1 / 6 good,
4/3 bad, ET  1 / 40 good, 1/10 hours bad
5.
1
En  4 Customers, E L  3.2 customers, ET  Hrs, EW  2 hrs 6. En  3 / 2 Patients, E L  0.90 patients,
.4
ET  15 minutes)
7. En  9 Customers, E L  8.1 customers, ET  5 minutes, E E  4.5 minutes 8.
P  75% , ET  6 minutes 9. EW  8 Minutes, P0  33.33% 10. EW  1 / 20 Hrs, waiting time =2/5 Hrs,
loss of production = 5 units
11. En  5 / 3 units, Busy hours=5, Idle hours = 3
E L  2 / 3 , Total cost = Rs. 233.34 13.
12. En  2 Customers,
ET  20 min, Average cost = Rs. 4.50
14. ET  2 / 9 hrs,
Average cost = Rs. 25, Av. Hourly queuing cost = Rs. 25, Av. Hourly cost for clerk = 4
hrs, total cost = Rs. 34 per day 16. For A, En  2 days, TC = Rs. 39200: For B,
=
15. En  0.25
En  3 / 2 days,
TC
39900 17. Slow: En  3 machines, TC = Rs. 1402; Fast : En  1 machine, TC =Rs. 70, Fast should
employee 18.(i) self service line En  1.34 customers, ET  16.67 minutes, EW  9.68 minutes (ii) Attended
line En  .99 customers, ET  11.90 minutes, EW  591 minutes) 19. P =0.25,   10   10 per hour,
second pump should be installed
20. E L  4 / 3 Patients,  '  192 patients, Budget = Rs. 125
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