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- 1. QUEUING THEORY<br />Introduction<br />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<br /><ul><li>Waiting for service in bank and at reservation counter.
- 2. Waiting for a train or bus.
- 3. Waiting at barber saloon.
- 4. Waiting at doctors’ clinic.</li></ul>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.<br /> 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.<br />Symbols and notations:<br />n = total number of customers in the system, both waiting and in service<br />µ = average number of customers being serviced per unit of time.<br />λ = average number of customers arriving per unit of time.<br />C = number of parallel service channels<br />Ls or E(n) = average number of customers in the system, both waiting in the service.<br />Lq or E(m) = average number of customers waiting in the queue<br />Ws or E(w) = average wating time of a customer in the system both waiting and in service<br />Wq or E(w) = average waiting time of a customer in the queue<br />Pn (t = probability that there are n customer in the queue<br />total cost of the system<br />cost <br />cost of service<br /> cost of waiting<br /> optical service level level of service<br />Queuing system<br />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.<br /> SERVICE SYSTEM <br />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. <br />There are two aspects of a service system—(a) the configuration of the service system and (b) the speed of the service.<br /> <br /> Configuration of the service system <br />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. <br /> Single Server – Single Queue<br /> <br /> 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.<br /> <br />Several (Parallel) Servers – Single Queue <br /> In this type of model there is more than one server and each server provides the same type<br /> of facility. The customers wait in a single queue until one of the service channels is ready to take them in for servicing<br /> <br /> <br /> <br />Several Servers – Several Queues<br /> <br /> 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.<br /> <br /> Service facilities in a series <br /> 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.<br />Characteristics of Queuing System<br />In designing a good queuing system, it is necessary to have a good <br />Information about the model. The characteristic listed below would<br />Provide sufficient information. <br /><ul><li>The Arrival pattern.
- 5. The service mechanism.
- 6. The queue discipline.
- 7. The number of service channels.
- 8. Number of Service Stages</li></ul>1. The Arrival pattern.<br />Arrivals can be measured as the arrival rate or the interarrival time<br />(time between arrivals).<br />Interarrival time =1/ arrival rate<br />These quantities may be deterministic or stochastic (given by a<br />propbability distribution).<br />Arrivals may also come in batches of multiple customers, which is<br />called batch or bulk arrivals. The batch size may be either deter-<br />ministic or stochastic.<br /> (i) Balking: The customer may decide not to enter the queue upon<br />Arrival, perhaps because it is too long.<br />(ii) Reneging: The customer may decide to leave the queue after<br />Waiting a certain time in it.<br />(iii) Jockeying: If there are multiple queues in parallel the customers<br />May switch between them.<br />(iv) Drop-o®s: Customers may be dropped from the queue for rea-<br />Sons outside of their control. (This can be viewed as a general-<br />Isation of reneging.)<br />2. Service Pattern<br />As with arrival patterns, service patterns may be deterministic or<br />stochastic. There may also be batched services.<br />The service rate may be state-dependent. (This is the analoge of<br />impatience with arrivals.)<br />Note that there is an important di®erence between arrivals and ser-<br />vices. Services do not occur when the queue is empty (i.e. in this<br />case it is a no-op).<br />3. Queue Discipline<br />This is the manner by which customers are selected for service.<br /><ul><li>First in First Out (FIFO).
- 9. Last in First Out (LIFO), also called
- 10. Service in Random Order (SIRO).</li></ul> (iv) Priority Schemes. Priority schemes are either:<br />Preemptive: A customer of higher priority immediately displaces<br />any customers of lower priority already in service. The displaced customer's service may be either resumed from where<br />it was left o®, or started a new.<br />Non-Preemptive: Customers with higher priority wait current<br />service completes, before being served.<br />4.The number of service channels<br />5. Number of Service Stages<br />Customers are served by multiple servers in series.<br />In general, a multistage queue may be a complex network with feed-<br />Back<br />Application of queuing theory:<br />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-<br /><ul><li>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 .
- 11. Waiting line theory can be used to analyze the delays at the toll booths of bridges and tunnels.
- 12. Waiting line theory can be used to improve the customers service at restaurants,cafeteria ,gasoline service station , airline counters,hospitals etc,
- 13. 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.
- 14. 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.
- 15. Queuing theory has been extended to study a wage incentive plan
- 16. Queuing theory (Limitations)
- 17. Most of the queuing models are quite complex & cannot be easily understood.
- 18. Many times form of theoretical distribution applicable to given queuing situations is not known.
- 19. If the queuing discipline is not in” first in, first out”, the study of queuing problems become more difficult.</li></ul> <br />BASIC POINTS<br />Customer:> (Arrival)<br />The arrival unit that requires some services to performed.<br />Queue:>The number of Customer waiting to be served. <br />Arrival Rate (λ):>The rate which customer arrive to the service station.<br />Service rate (µ) :> The rate at which the service unit can provide sevices to the customer<br />If Utilization Ratio Or Traffic intensity i.e λ /µ<br />λ / µ > 1 Queue is growing without end.<br />λ / µ < 1 Length of Queue is go on diminishing.<br />λ /µ = 1 Queue length remain constant.<br />When Arrival Rate (λ) is less than Service rate (µ) the system is working .<br />i.e λ< µ (system work)<br />Formulas<br />µ=Service Rate<br />λ= Arrival Rate<br />1. Traffic Intensity (P)= λ /µ<br />2. Probability Of System Is Ideal (P0) =1-P<br /> P0 = 1- λ /µ<br />3. Expected Waiting Time In The System (Ws) = 1/ (µ- λ)<br />4. Expected Waiting Time In Quie (Wq) = λ / µ(µ- λ)<br />5. Expected Number Of Customer In The System (Ls)= λ / µ(µ-λ)<br /> Ls=Length Of System<br />6. Expected Number Of Customers In The Quie (Lq)= λ 2/ µ(µ- λ)<br />7. Expected Length Of Non-Empty Quie (Lneq)= µ/ (µ- λ)<br />8. What Is The Probability Track That That K Or More Than K Customers In The System.<br /> P >=K (P Is Greater Than Equal To K)<br /> = (λ /µ)K<br />9. What Is The Probability That More Than K Customers Are In The System ( P>K)= (λ /µ)K+1<br />10. What Is The Probability That Atleast One Customer Is Standing In Quie. P=K=(λ /µ)2<br />11. What Is The Probability That Atleast Two Customer In The System<br />P=K=(λ /µ)2<br />Solved Example.<br /> 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.<br />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?<br />Solution:<br />Arrival rate λ=25/hr<br />Service rate µ= 2/min=30/hr<br />Length of Queue (Lq)= λ 2/ µ(µ- λ) <br /> = 252/(30(30-25))<br /> =4.17 persone<br /> Expected Waiting Time In Quie (Wq) = λ / µ(µ- λ)<br /> =25/(30(30-25))<br /> =1/6 hr= 10 min<br />Expected Waiting Time In The System (Ws) = 1/ (µ- λ)<br /> =1/(30-25)<br /> =1/5hr= 12 min<br />Utilization Ratio = λ /µ<br /> =25/30<br /> =0.8334 = 83.34%<br />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.<br /> 1.what is the value of utilization factor?<br /> 2.What is the expected waiting time in the system per customer?<br /> 3.what is the probability of zero customer in the system?<br />Solution:<br />Arrival rate λ=20 customer per hour<br />Service rate µ= 30 customer per hour<br /> 1.Utilization Ratio = λ /µ <br /><ul><li> = 20/30 = 2/3 </li></ul> 2. Expected Waiting Time In The System (Ws) = 1/ (µ- λ)<br /> =1/(30-20)<br /> =1/10 hour = 6 min<br />3. Probability of zero customers in the system P0 = 1 – P<br /> =1- 2/3 = 1/3<br />Question 3 : Abc company has one hob regrinding machine. The hobs needing grinding are sent from company’s 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.<br /><ul><li>Calculate the present utilization of hob regrinding machine.
- 20. What is average time for the hob to be in the regrinding section?
- 21. 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?</li></ul>Solution: : Let us calculate arrival rate and service rate per shift of 8 hours.<br />Arrival rate λ=8 shift<br />Service rate µ=8x60/30=16 /shift<br /><ul><li>Percentage of the time the machine is busy
- 22. Pb =arrival rate/service rate=8/16=0.50=50%
- 23. Average time for the hob to be in the grinding section.</li></ul> i.e., average time in the queue system=ws<br />ws = 1/( µ- λ)=1/16-8=1/8 shift=1/8x8=1 hour<br /><ul><li> Let λ’ =arrival rate for which utilization of the machine will be 80%,
- 24. Therefore, Pb’ = λ’ / µ
- 25. i.e., λ’ = Pb’ . µ=0.80x16=12.8 per shift.</li></ul>Question: 4<br />(a) calculate expected number of persons in the system if average waiting time pf a customer is 45 or more than 45 minutes .<br />b)if service rate is same.<br />c)if arrival rate is same.<br />Solution:-(a)expected no. of persons in a system(Ls )=λ/μ- λ<br /> =45/65-45<br /> =9/4 <br /> =3/4=1/65- λ<br /> λ =191/3<br />(b)Ws= 1/ μ –λ=1/65-45<br /> =1/20 x60/1=3 mins.<br />(c)ws =1/ μ- λ=1/6-4<br />= 3/4=1/ μ-45<br />=3 μ-135=4<br />=3 μ=139<br />μ =46.33<br />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.<br />Solution : Given λ = 4, n = 6, t = 2 p<br />P(n,t) = (6,4) when λ = 4<br />We know, p (n,t) = (λt)n e-λt/ n!<br />p(6,2) = (4×2)6 e-4×2/ 6!<br />=86 e-8/720<br />=0.1221<br />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.<br />Solution: <br /> Given, λ = 6 customers / hour<br /> t = 30 minutes = 0.5 hour<br /> n = 2<br /> <br /> we know, p(n,t) = (λt)n e-λt/n!<br /> p(6,2) = (6×0.5)2 e-6×0.5/2!<br /> = 0.22404<br /> <br /> <br />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<br /><ul><li>The probability that an arrival will get a chair to sit on,
- 26. The probability that an arrival will have to stand, and
- 27. Expected waiting time of a customer.</li></ul>Solution following assumption are made for solving the given queuing problem :<br /><ul><li>The arrival rate is randomly distributed according to poission distribution.
- 28. The mean value of the arrival rate is λ .
- 29. The services time distribution approximated by an exponiential distribution and a nmean rate of services is μ.
- 30. The rate of services is greather than the rate of arrival (μ>λ)
- 31. The queue discipline id FIFO.</li></ul> Arrival rate λ= 12min or 5 customer / hr<br />Services rate μ = 6 min or 10 customer/ hr<br />λ/μ = 5/10 = ½<br />there are two chairs including services one.<br /><ul><li>The probality that an arrival get a chair to seat on is given by:
- 32. Pn (n<=2) = 1- Pn(n>2)
- 33. 1-(λ/μ)3
- 34. 1-(1/2)3 = 7/8
- 35. (II) The probability that an arrival will have to stand is given by
- 36. 1-(P0+p1+P2)
- 37. = 1-(7/8)= 1/8
- 38. (III)Expected waiting time of a customer in the queue is given by
- 39. Wq =λ/μ(μ-λ)
- 40. =5/10(10-5) = 1/(2*5) hr = 6 min</li></ul> 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 repairman’s expected idle time each day? How many jobs are ahead of the average set just brought in?<br />Solution from data of problem, we have<br /> Λ= 10/8=5/4 set per hour; and μ=(1/30)60= 2set per hour<br /><ul><li>Expected idle time of repairmen each day
- 41. Number of hour for a repairman remains busy in 8 hour day( traffic intensity) is given by
- 42. (8) (λ/μ)=(8) (5/8)= 5 hour
- 43. Hence , the idle time for a repairman in an 8 hour day will be : (8-5) =3 hour
- 44. Expected (or average) number of TV set in the system
- 45.
- 46. LS = λ/μ-λ = 5/4/2-(5/4)
- 47. =5/3
- 48. =2 (APPROX) T.V sets</li></ul>Unsolved question<br /><ul><li>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 .
- 49. Calculate Arrival rate if service rate is same .
- 50. Calculate service rate if arrival rate is same.
- 51. (answer: Ws=1/5 hr,
- 52. λ =13
- 53. µ = 2)</li></ul>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.<br />Find out:-<br /><ul><li>Traffic intensity
- 54. What would be the average quieting length?
- 55. What is the expected number of customer at petrol pump?
- 56. What is the expected number time one spend at petrol pump?
- 57. What would we expected waiting time?
- 58. What would be the proportion time the petrol pump is idle?</li></ul>Answer<br /><ul><li>0.4
- 59. 0.26
- 60. 0.66
- 61. 0.02
- 62. 0.05
- 63. 0.6</li></ul>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.<br /><ul><li>Calculate:
- 64. Number of machines not working at any point of time.
- 65. Average time that a machine is waiting for the repairman.
- 66. Cost of non-productive time of the machine operator.
- 67. Expected cost of system per hour.</li></ul>Answer. a:: 2 machines<br /> b :- 2/3 hours<br /> c: Rs. 60<br /> d: Rs.110<br /> <br /> 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<br /><ul><li>Calculate the percentage of time the teller is busy.
- 68. Calculate the average time a person is expected to wait.
- 69. Answer </li></ul> a)3/4<br /><ul><li> b)6 minutes</li></ul>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.<br />Answer a)0.6<br /> b)wq=3/40hrs.<br />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.<br />Answer <br />λ=1/48 <br />μ=λ/16<br />p=0.75<br />Po=o.28<br />Lq=1.55<br />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.<br />Answer: 2.48 minutes /service<br />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.<br />Answer Rs 30 per hour<br />GROUPS MEMBER NAME <br /><ul><li>Shikha
- 70. Dayashankar Yadav
- 71. Suriender Singh Prajapati
- 72. Gaurav Gupta
- 73. Himanshu Saxena
- 74. Gauri Shankar Mishra
- 75. Pankaj Gangwar
- 76. Sanjeev kumar
- 77. Amit .kr yadav</li></ul>10.Amit sinha<br />

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