QUEUING THEORY

Introduction Queuing theory deals with problems that involve waiting (or queuing). It is quite common that instances of queue occurs everyday in our daily life. Examples of queues or long waiting lines might be     Waiting for service in bank and at reservation counter. Waiting for a train or bus. Waiting at barber saloon. Waiting at doctors clinic.

Whenever a customer arrives at a service facility, some of them usually have to wait before they receive the desired service. This form a queue or waiting line and customer feel discomfort either mentally or physically because of long waiting queue. We infer that queues from because the service facilities are inadequate. If service facilities are increased, then the question arise how much to increase? For example, how many buses would be needed to avoid queues? How many reservation counters would be needed to reduce the queue? Increase in number of buses and reservation counters requires additional resources. At the same time, cost due to customer dissatisfaction must also be considered. Symbols and notations: n = total number of customers in the system, both waiting and in service

= average number of customers being serviced per unit of time. = average number of customers arriving per unit of time. C = number of parallel service channels Ls or E(n) = average number of customers in the system, both waiting in the service. Lq or E(m) = average number of customers waiting in the queue

Ws or E(w) = average wating time of a customer in the system both waiting and in service Wq or E(w) = average waiting time of a customer in the queue Pn (t = probability that there are n customer in the queue

total cost of the system cost cost of service

cost of waiting

optical service level

level of service

Queuing system The customers arrive at service counter (single or in a group) and attended by one or more servers. A customer served leaves the system after getting the service. In general, a queuing system comprise with two components, the queue and the service facility. The queue is where the customers are waiting to be served. The service facility is customers being served and the individual service stations. SERVICE SYSTEM The service is provided by a service facility (or facilities). This may be a person (a bank teller, a barber, a machine (elevator, gasoline pump), or a space (airport runway, parking lot, hospital bed), to mention just a few. A service facility may include one person or several people operating as a team. There are two aspects of a service system(a) the configuration of the service system and (b) the speed of the service.

Configuration of the service system

The customers entry into the service system depends upon the queue conditions. If at the time of customers arrival, the server is idle, then the customer is served immediately. Otherwise the customer is asked to join the queue, which can have several configurations. By configuration of the service system we mean how the service facilities exist. Service systems are usually classified in terms of their number of channels, or numbers of servers. Single Server Single Queue The models that involve one queue one service station facility are called single server models where customer waits till the service point is ready to take him for servicing. Students arriving at a library counter is an example of a single server facility.

Several (Parallel) Servers Single Queue In this type of model there is more than one server and each server provides the same type of facility. The customers wait in a single queue until one of the service channels is ready to take them in for servicing

Several Servers Several Queues

This type of model consists of several servers where each of the servers has a different queue. Different cash counters in an electricity office where the customers can make payment in respect of their electricity bills provide an example of this type of model.

Service facilities in a series In this, a customer enters the first station and gets a portion of service and then moves on to the next station, gets some service and then again moves on to the next station. . and so on, and finally leaves the system, having received the complete service. For example, machining of a certain steel item may consist of cutting, turning, knurling, drilling, grinding, and packaging operations, each of which is performed by a single server in a series. Service Facility.

Characteristics of Queuing System

In designing a good queuing system, it is necessary to have a good Information about the model. The characteristic listed below would Provide sufficient information.

1. 2. 3. 4. 5.

The Arrival pattern. The service mechanism. The queue discipline. The number of service channels. Number of Service Stages

1. The Arrival pattern. Arrivals can be measured as the arrival rate or the interarrival time (time between arrivals). Interarrival time =1/ arrival rate

These quantities may be deterministic or stochastic (given by a propbability distribution). Arrivals may also come in batches of multiple customers, which is called batch or bulk arrivals. The batch size may be either deterministic or stochastic.

(i) Balking: The customer may decide not to enter the queue upon Arrival, perhaps because it is too long. (ii) Reneging: The customer may decide to leave the queue after Waiting a certain time in it. (iii) Jockeying: If there are multiple queues in parallel the customers May switch between them. (iv) Drop-os: Customers may be dropped from the queue for rea-

Sons outside of their control. (This can be viewed as a generalIsation of reneging.)

2. Service Pattern As with arrival patterns, service patterns may be deterministic or stochastic. There may also be batched services. The service rate may be state-dependent. (This is the analoge of impatience with arrivals.) Note that there is an important dierence between arrivals and services. Services do not occur when the queue is empty (i.e. in this case it is a no-op).

3. Queue Discipline This is the manner by which customers are selected for service. (i) (ii) (iii) First in First Out (FIFO). Last in First Out (LIFO), also called Service in Random Order (SIRO).

(iv) Priority Schemes. Priority schemes are either: Preemptive: A customer of higher priority immediately displaces any customers of lower priority already in service. The displaced customer's service may be either resumed from where it was left o, or started a new. Non-Preemptive: Customers with higher priority wait current service completes, before being served.

4.The number of service channels

5. Number of Service Stages Customers are served by multiple servers in series.

In general, a multistage queue may be a complex network with feedBack

Application of queuing theory: Queing theory has been applied to a great variety of business situations. Here we shall discuss a few problem s where the theory may be applied-

1) Waiting line theory can be applied to be determine the number of check out counters needed to secure smooth and economic operations of its stored at various time during the day of a super market or a departmental store . 2) Waiting line theory can be used to analyze the delays at the toll booths of bridges and tunnels. 3) Waiting line theory can be used to improve the customers service at restaurants,cafeteria ,gasoline service station , airline counters,hospitals etc, 4) Waiting line theory can be used to determine the proper determine the proper number docks to be constructed in the building of terminal facilities for trucks &ships. 5) Several manufacturing firms have attacked the problems of machine break down &repairs by utilizing this theory . Waiting line theory can be used to determine the number of personnal to be employed so that thee cost of the production loss from down time & the cost f repairman is minimized. 6) Queuing theory has been extended to study a wage incentive plan Queuing theory (Limitations)

1) Most of the queuing models are quite complex & cannot be easily
understood.

2) Many times

form of theoretical distribution applicable to given queuing situations is not known. is not in first in, first out, the study of queuing problems become more difficult.

3) If the queuing discipline

BASIC POINTS

Customer:> (Arrival) The arrival unit that requires some services to performed. Queue:>The number of Customer waiting to be served. Arrival Rate ( ):>The rate which customer arrive to the service station. Service rate () :> The rate at which the service unit can provide sevices the customer If Utilization Ratio Or Traffic intensity i.e / > 1 Queue is growing without end. / < 1 Length of Queue is go on diminishing. / = 1 Queue length remain constant. / to

When Arrival Rate ( ) is less than Service rate () the system is working . i.e < (system work)

Formulas =Service Rate = Arrival Rate

1. Traffic Intensity (P)=

2. Probability Of System Is Ideal (P0) =1-P P0 = 1- / 3. Expected Waiting Time In The System (Ws) = 1/ (- ) 4. Expected Waiting Time In Quie (Wq) = / (- ) 5. Expected Number Of Customer In The System (Ls)= / (- ) Ls=Length Of System 6. Expected Number Of Customers In The Quie (Lq)=
2

/ (- )

7. Expected Length Of Non-Empty Quie (Lneq)= / (- ) 8. What Is The Probability Track That That K Or More Than K Customers In The System. P >=K = ( /)K 9. What Is The Probability That More Than K Customers Are In The System ( P>K)= ( /)K+1 10. What Is The Probability That Atleast One Customer Is Standing In Quie. P=K=( /)2 11. What Is The Probability That Atleast Two Customer In The System P=K=( /)2 (P Is Greater Than Equal To K)

Solved Example. Question 1.People arrive at a cinema ticket booth in a poisson distributed arrival rate of 25per hour. Service rate is exponentially distributed with an average time of 2 per min.

Calculate the mean number in the waiting line, the mean waiting time , the mean number in the system , the mean time in the system and the utilization factor?

Solution: Arrival rate =25/hr Service rate = 2/min=30/hr Length of Queue (Lq)=
2

/ (- )

= 252/(30(30-25)) =4.17 persone Expected Waiting Time In Quie (Wq) = / (- ) =25/(30(30-25)) =1/6 hr= 10 min

Expected Waiting Time In The System (Ws) = 1/ (- ) =1/(30-25) =1/5hr= 12 min

Utilization Ratio =

/ =25/30

=0.8334 = 83.34%

Question 2. Assume that at a bank teller window the customer arrives at a average rate of 20 per hour according to poission distribution .Assume also that the bank teller spends an distributed customers who arrive from an infinite population are served on a first come first services basis and there is no limit to possible queue length. 1.what is the value of utilization factor? 2.What is the expected waiting time in the system per customer? 3.what is the probability of zero customer in the system?

Solution: Arrival rate =20 customer per hour

Service rate = 30 customer per hour 1.Utilization Ratio = /

= 20/30 = 2/3 2. Expected Waiting Time In The System (Ws) = 1/ (- ) =1/(30-20) =1/10 hour = 6 min 3. Probability of zero customers in the system P0 = 1 P =1- 2/3 = 1/3

Question 3 : Abc company has one hob regrinding machine. The hobs needing grinding are sent from companys tool crib to this machine which is operated one shift per day of 8 hours duration. It takes on the average half an hour to regrind a hob. The arrival of hobs is random with an average of 8 hobs per shift. 1. Calculate the present utilization of hob regrinding machine. 2. What is average time for the hob to be in the regrinding section?

3. The management is prepared to recruit another grinding operator when the utilization of the machine increases to 80%. What should the arrival rate of hobs then be? Solution: : Let us calculate arrival rate and service rate per shift of 8 hours. Arrival rate =8 shift Service rate =8x60/30=16 /shift 1. Percentage of the time the machine is busy Pb =arrival rate/service rate=8/16=0.50=50% 2. Average time for the hob to be in the grinding section. i.e., average time in the queue system=ws ws = 1/( - )=1/16-8=1/8 shift=1/8x8=1 hour 3. Let =arrival rate for which utilization of the machine will be 80%, Therefore, Pb = / i.e., = Pb . =0.80x16=12.8 per shift.

Question: 4 (a) calculate expected number of persons in the system if average waiting time pf a customer is 45 or more than 45 minutes . b)if service rate is same. c)if arrival rate is same. Solution:-(a)expected no. of persons in a system(Ls )= / =45/65-45 =9/4 =3/4=1/65=191/3

(b)Ws= 1/

=1/65-45

=1/20 x60/1=3 mins. (c)ws =1/ - =1/6-4 = 3/4=1/ -45 =3 -135=4 =3 =139 =46.33

Question: 5 In a factory, the machines break down and require service according to a poission Distribuation at the average of per day. What is the probability that exactly six Machines.

Solution :

Given = 4, n = 6, t = 2 p

P(n,t) = (6,4) when = 4 We know, p (n,t) = ( t)n e- t/ n!

p(6,2) = (42)6 e-42/ 6! =86 e-8/720 =0.1221

Question 6 On an average , 6 customer arrive in a coffee shop per hour. Determine the probability that Exactly 3 customers will reach in a 30 minute period, assuming that the arrivals follow poisson Distribution.

Solution: Given, = 6 customers / hour t = 30 minutes = 0.5 hour

n=2

we know, p(n,t) = ( t)n e- t/n! p(6,2) = (60.5)2 e-60.5/2! = 0.22404

Question 7 In a bank with a single sever, there are two chairs for waiting customers. On an average one customer arrives 12 minutes and each customer takes 6 minutes for getting served. Make suitable assumption, find (i) (ii) (iii) The probability that an arrival will get a chair to sit on, The probability that an arrival will have to stand, and Expected waiting time of a customer.

Solution following assumption are made for solving the given queuing problem : 1. The arrival rate is randomly distributed according to poission distribution. 2. The mean value of the arrival rate is . 3. The services time distribution approximated by an exponiential distribution and a nmean rate of services is . 4. The rate of services is greather than the rate of arrival ( > ) 5. The queue discipline id FIFO.

Arrival rate Services rate / = 5/10 =

= 12min or 5 customer / hr = 6 min or 10 customer/ hr

there are two chairs including services one. (i) The probality that an arrival get a chair to seat on is given by: Pn (n<=2) = 1- Pn(n>2) 1-( / )3 1-(1/2)3 = 7/8 (II) The probability that an arrival will have to stand is given by 1-(P0+p1+P2) = 1-(7/8)= 1/8 (III)Expected waiting time of a customer in the queue is given by Wq = / ( - ) =5/10(10-5) = 1/(2*5) hr = 6 min

Question 8 A television repairman finds that the time spent on his jobs has an expontial 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 follow a passion distribution approximately with an average rate of 10 per 8- hour day, what is the repairmans expected idle time each day? How many jobs are ahead of the average set just brought in? Solution from data of problem, we have = 10/8=5/4 set per hour; and (i) =(1/30)60= 2set per hour

(ii)

Expected idle time of repairmen each day Number of hour for a repairman remains busy in 8 hour day( traffic intensity) is given by (8) ( / )=(8) (5/8)= 5 hour Hence , the idle time for a repairman in an 8 hour day will be : (8-5) =3 hour Expected (or average) number of TV set in the system

LS = / - = 5/4/2-(5/4) =5/3 =2 (APPROX) T.V sets

Unsolved question Question 1 Calculate expected number of person in the system. If average waiting time of customer is 30 min or more than 30 min , then services provider starts another windows . Calculate Arrival rate if service rate is same . Calculate service rate if arrival rate is same. (answer: Ws=1/5 hr, =13

 = 2)

Question 2 At a certain petrol pump , Customer arrive according to a passion process with a average time at 5 min between the arrivals. The service time is exponential distribution with mean 2 mins on the basic of this information. Find out:a. b. c. d. e. f. Traffic intensity What would be the average quieting length? What is the expected number of customer at petrol pump? What is the expected number time one spend at petrol pump? What would we expected waiting time? What would be the proportion time the petrol pump is idle?

Answer a. b. c. d. e. f. 0.4 0.26 0.66 0.02 0.05 0.6

Question3. The machines in production shop breakdown at an average of 2 per hour. The non productive time of any machine costs rs.30 per hour. If the cost of repairman is Rs.50 per hour. Calculate: a. Number of machines not working at any point of time. b. Average time that a machine is waiting for the repairman. c. Cost of non-productive time of the machine operator.

d. Expected cost of system per hour. Answer. a:: 2 machines b :- 2/3 hours c: Rs. 60 d: Rs.110

Question 4.In a bank cheques are cashed at a single teller counter. Customers arrived at the counter in a Poisson manner at and average rate of 30 customers /hour. The teller takes on an average, a minute and a half to cash cheque. The service time has been shown to be exponentially distributed a) Calculate the percentage of time the teller is busy. b) Calculate the average time a person is expected to wait. Answer a)3/4 b)6 minutes

Question 5 Telephone users arrive at a booth following a Poisson distribution with an average time of 5 minutes between one arrival and the next. The time taken for a telephone call is on a average 3 minutes and it follows an exponential distribution. What is the probability that the booth is busy? How many more booths should be established to reduce the waiting time less than or equal to half of the present waiting time.

Answer

a)0.6 b)wq=3/40hrs.

Question 6 Assume that goods trains are coming in a yard @ 30 trains per day and suppose that the inter arrival times follow an exponential distribution . the service time for each train is assumed to be exponential with an average of 36 minutes if the yard can admit 9 trains at a time(there being 10 lines one of which is reserved for shunting purpose).calculate the probability that the yard is empty and find the average queue length. Answer =1/48 = /16 p=0.75 Po=o.28 Lq=1.55

Question 7 At what average rate must a clerk at a supermarket in order to ensure a probability of 0.90 so that the customer will not wait longer than 12 minutes ? It is assumed that there is only one counter at which customers arrive in a Poisson fashion at an average rate of 15/hour. The length of service by the clerk has an exponential distribution.

Answer:

2.48 minutes /service

Question 8 The beta company s quality control deptt. Is managed by a single clerk, who takes an average 5 minutes in checking part of each of the machine coming for inspection. The machine arrive once in every 10 min. on the average one hour of the machine is valued at Rs 25 and cost for the clerk

is at rs 5 per hour. What are the average hourly queueing system cost associated with the quality control department. Answer Rs 30 per hour

Queueing theory is the study of queues as based on probability theory, statistics and other sub-fields of mathematics. The idea behind queueing theory is to propose models to apply to describe queues and the processes behind them. In queueing theory, queues tend to be modeled by stochastic processes, which are random functions based on probability distributions. Queueing theory has many applications, including the design of computer systems, customer service and Internet database management.

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PROS N CONS

1. Coefficient of Variation
o

Because queueing theory models are based on the exponential distribution, these models work through applying the traits of the exponential distribution. The main problem lies in that the exponential distribution has a coefficient of variation of one. This fact precludes the modeling of any process that has a coefficient of variation significantly different from one. Because of the low likelihood of a random process having a coefficient of variation of one, queueing theory has the disadvantage of low applicability.

Simplicity
o

Queuing theory offers us a method to easily and definitely describe queues in mathematical terms. This advantage of queueing theory is an advantage that plain language, economic models and pure observation lack. Through applying basic probabilistic distributions, such as the Poisson and exponential distributions, mathematicians can model the complex phenomenon of waiting in a queue as an elegantly simplistic mathematical equation. Mathematicians can later analyze these equations to understand and predict behavior.

Assumptions
o

While the assumptions for most applications of queueing models are few, the assumptions that are needed tend to be somewhat irrational. Especially in regard to human queues, queueing theory requires assumptions that cannot possibly hold true in the real world. In general, queueing theory presumes that human behavior is deterministic. These assumptions usually are a set of rules for what a person will do. For example, one assumption may be that a person will not enter a queue if there are too many people already queued up. In reality, this is not true; otherwise, there would be no lines outside stores or for

store openings, and holiday shoppers who waited too late to buy gifts would just give up.

Simulation
o

Queueing theory has flourished due to the advent of the computer age. The past difficulty of arriving at numerical solutions for queueing models is no longer a disadvantage, as mathematicians can run simulations to arrive at approximate answers. The simulation of queueing theory models also allows researchers to change the value of variables involved and analyze the results of the change, which can help in the optimization of queue design.

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EXAMPLES OF QUEUING THEORY APPLICATIONS

The most unusual recurring period is the "busy hour," that provides a pattern upon which the system should be engineered. For example, if a system receives its highest number of calls between 9 and 10 a.m., the office should be equipped with enough switchboards to handle that level of requests. The issue of how many operators to assign depends on calling patterns from one hour to the next. What is interesting about the Poisson and Erlang formulas is that the relationship between operators and congestion is not parallel. For example, assume that 10 operators are inundated by 30 percent more calls than they usually handle. A supervisor calls in an eleventh operator and, even though the rate of incoming service requests remains constant, the backlog will gradually fall. After the backlog is eliminated, the eleventh operator may actually force others to go idle for extended periods. We might assume that 10 operators handling 130 percent of their normal volume would require 13 operators. In fact, the addition of only one is more than enough to resolve the problem. This is because repeated calls are disposed of, and the aggregate wait of everyone holding (which grows geometrically) is reduced one factor at a time. The backlog simply cannot regenerate itself fast enough. Put differently, 11 operators may be able to dispose of service requests at a faster rate than they are coming in. It may take a few minutes to eliminate the backlog, but the backlog will decline eventually. Consider the situation in a grocery store where there are five lines open and 12 people in each line. The addition of only one extra cashier will quickly reduce the lines to one or two people, even though the same number of people are entering checkout lines. When the backlog is eliminated, the sixth cashier may be taken off and put on some other job. As well as a system may be engineered, unusual nonrandom disturbances can cause the system to collapse in spectacular fashion. This was demonstrated by a problem with the New York water system during the 1950s. Engineers discovered that water pressure dropped

significantlyand for firemen, perilouslyduring a period of hours every Sunday evening. A study revealed an unusual culprit: Milton Berle. The comedian hosted an immensely popular weekly television show every Sunday that was watched by nearly everyone with a set. When the show went to a commercial break, tens of thousands of people, having finished dinner, retreated to their bathrooms at the same time. With thousands of toilets being flushed within minutes of each other, sewers were inundated. More importantly, toilet tanks were refilling, each consuming two or three gallons of fresh water. The coordinated demand for water in a brief period of time virtually eliminated water pressure. In fact, some toilets took a half hour to refill, and water pressure took hours to recover. Serious consideration was given to canceling the show. The solution, however, was relatively simple. The addition of only a few more water towers was sufficient to maintain adequate water pressure. In essence, the system was reengineered to handle more demanding peaks. This situation may be repeated in a telephone system when everyone is motivated to place a call at the same time. During the 1989 San Francisco earthquake, vast numbers of people in the metropolitan area attempted to make a call at the same timeimmediately after the quake subsidedhoping to learn whether friends and relatives were safe. Although the switching systems were automated, they were completely unable to handle the volume of requests for dial tone. Only a small percentage of calls (enough to meet the capacity of the system) were allowed to go through. Indeed, radio and television reporters urged people to stay off the lines so that emergency calls could be handled. There was no need to reengineer the system because the occurrence of earthquakes, while random, are not consistently repeated. It would be uneconomic to engineer the telephone network for peak usages that occur only once every decade or so. As a result, every earthquake yields a temporary breakdown in the telephone network. Other slightly less offensive instances occur every time a radio host offers a prize to "caller number x." Telephone companies and public officials have convinced many radio stations to discontinue the practice.

GROUPS MEMBER NAME 1. 2. 3. 4. 5. 6. 7. 8. 9. Shikha Dayashankar Yadav Suriender Singh Prajapati Gaurav Gupta Himanshu Saxena Gauri Shankar Mishra Pankaj Gangwar Sanjeev kumar Amit .kr yadav

10.Amit sinha