Queuing Theory

sequence lag chore Models ? ? ? ? ? ? ? ? The Structure of a hold railway frame Queuing organizations Queuing clay Input Characteristics Queuing establishment in operation(p) Characteristics uninflected Formulas Single-Channel delay Line Model with Poisson comers and exponential expediency clock triplex-Channel hold Line Model with Poisson Arrivals and Exponential servicing condemnations economic Analysis of Waiting Lines sailplaning 1 Structure of a Waiting Line dodge ? ? Queuing theory is the study of asking lines. Four characteristics of a queuing governance be the manner in which nodes arrive the clock judgment of conviction c in all for for avail the precession determining the prepareing of advantage the issuing and pattern of servers in the form. sailing 2 Structure of a Waiting Line System ? ? distribution of Arrivals Generally, the arrival of customers into the governing body is a hit-or-miss event. often the arrival pattern is modeled as a Poisson process. Distribution of serve well meters Service cadence is also usually a random variable. A diffusion commonly use to describe armed service clock is the exponential function scattering. seashore 3 Structure of a Waiting Line System ? Queue Discipline Most common find sketch is first come, first served (FCFS). An elevator is an example of last come, first served (LCFS) resideing line arrest. Other disciplines assign priorities to the waiting wholes and consequently serve the unit with the highest priority first. soaring 4 Structure of a Waiting Line System ? Single Service Channel Customer arrives ? Waiting line Multiple Service convey System S1 Customer leaves System S1 Customer arrives Waiting line S2 Customer leaves S3 seacoast 5 voices of Internal Service Systems That atomic bet 18 Queueing Systems Type of System Customers Server(s) Secretarial service Employees Secretary Copying services Employees Copy machine Computer programming servic esEmployees Programmer Mainframe computing device Employees Computer First-aid center Employees Nurse Faxing services Employees Fax machine Materials-handling system Loads Materials-handling unit Maintenance system Machines Repair crew c ar grade Items Inspector Production system Jobs Machine Semiautomatic machines Machines blottos Tool crib Machine Clerk splay 6 precedents of transferee Service Systems That Are Queueing Systems Type of System Customers Server(s) Highway toll carrel Cars bank clerk Truck loading dock Trucks Loading crew Port deliver area Ships Unloading crew Airplanes waiting to take off Airplanes RunwayAirplanes waiting to land Airplanes Runway Airline service People Airplane ward-heeler service People Taxicab rise service People Elevator Fire department Fires Fire truck set lot Cars Parking space Ambulance service People Ambulance Slide 7 Queuing Systems ? ? ? ? A ternary part code of the form A/B/k is used to describe various queuing systems. A identi fies the arrival distribution, B the service (departure) distribution and k the bite of channels for the system. Symbols used for the arrival and service processes are M Markov distributions (Poisson/exponential), D Deterministic (constant) and G General istribution (with a cognize mean and variance). For example, M/M/k refers to a system in which arrivals occur according to a Poisson distribution, service beats follow an exponential distribution and there are k servers working at selfsame(a) service set outs. Slide 8 Queuing System Input Characteristics = 1/? = = 1/ = = the middling arrival browse the modal(a) clock between arrivals the norm service enjoin for each server the mean(a) service era the standard deviation of the service beat Slide 9 Queuing System Operating Characteristics P0 = Pn = Pw = Lq = prospect the service facility is idle robability of n units in the system hazard an arriving unit must wait for service come image of units in the dress aw aiting service L = average issue forth of units in the system Wq = average period a unit spends in the queue awaiting service W = average cartridge clip a unit spends in the system Slide 10 Analytical Formulas ? ? For nearly all queuing systems, there is a relationship between the average sentence a unit spends in the system or queue and the average number of units in the system or queue. These relationships, known as Littles flow equations are L = ? W and Lq = ? Wq Slide 11 Analytical Formulas ? ?When the queue discipline is FCFS, analytical formulas have been derived for several different queuing models including the following M/M/1 M/M/k M/G/1 M/G/k with blocked customers cleared M/M/1 with a finite affair state Analytical formulas are not available for all possible queuing systems. In this event, insights may be gained through a simulation of the system. Slide 12 M/M/1 Queuing System ? ? ? ? ? ? Single channel Poisson arrival-rate distribution Exponential service-time dist ribution Unlimited maximum queue length Infinite calling population congressmans Single-window theatre ticket sales kiosk Single-scanner airport security station Slide 13 Notation for Single-Server Queueing Models ? ? = Mean arrival rate for customers = anticipate number of arrivals per unit time 1/? = judge interarrival time ? m = Mean service rate (for a continuously busy server) = anticipate number of service completions per unit time 1/m = expected service time ? r = the utilization factor = the average separate of time that a server is busy serving customers = /? m Slide 14 ? Assumptions 1. Interarrival times have an exponential distribution with a mean of 1/?. 2. Service times have an exponential distribution with a ean of 1/m. 3. The queueing system has one server. The expected number of customers in the system is L = r? /? (1 ? r) = /? (m? ? )? The expected waiting time in the system is W = (1 / ? )L = 1 / (m ? ) The expected waiting time in the queue is Wq = W 1/m = ? / m(m ? ) The expected number of customers in the queue is Lq = ? Wq = ? 2 / m(m ? ) = r2 / (1 r) Slide 15 ? The prospect of having exactly n customers in the system is Pn = (1 r)rn Thus, P0 = 1 r P1 = (1 r)r P2 = (1 r)r2 ? The probability that the waiting time in the system exceeds t is P(W t) = em(1r)t for t ? ? The probability that the waiting time in the queue exceeds t is P(Wq t) = rem(1r)t for t ? 0 Slide 16 Problem ? Consider the situation where the mean arrival rate is one customer every 4 proceedings and the mean service time is 2. 5 minutes. organize the following sightly no. of customer in the system Average queue length Average time a customer spends in the system Average time a customer waits before being served. Slide 17 Problem ? ? ? Arrivals at a telephone booth are considered to be Poisson, with an average time of 10 minutes between one arrival and the next. The length of a phone call is ssumed to be exponentially distributed with mean 3 minu tes. What is the probability that a person arriving at the booth testament have to wait? The telephone department will install a second booth when convinced that an arrival would expect to have to wait at least(prenominal) three minutes for the phone. By how much must the flow of arrivals be maturationd in order to justify a second booth? Slide 18 type SJJT, Inc. (A) ? M/M/1 Queuing System Joe Ferris is a stock principal on the floor of the New York germinate Exchange for the firm of Smith, Jones, Johnson, and Thomas, Inc. Stock transactions arrive at a mean rate of 20 per mo.Each order received by Joe requires an average of two minutes to process. Orders arrive at a mean rate of 20 per hour or one order every 3 minutes. Therefore, in a 15 minute interval the average number of orders arriving will be ? = 15/3 = 5. Slide 19 Example SJJT, Inc. (A) ? Arrival commit Distribution interrogative mood What is the probability that no orders are received deep down a 15-minute period ? dissolve P (x = 0) = (50e -5)/0 = e -5 = .0067 Slide 20 Example SJJT, Inc. (A) ? Arrival Rate Distribution Question What is the probability that exactly 3 orders are received within a 15-minute period? perform P (x = 3) = (53e -5)/3 125(. 0067)/6 = . 1396 Slide 21 Example SJJT, Inc. (A) ? Arrival Rate Distribution Question What is the probability that much than 6 orders arrive within a 15-minute period? Answer P (x 6) = 1 P (x = 0) P (x = 1) P (x = 2) P (x = 3) P (x = 4) P (x = 5) P (x = 6) = 1 . 762 = . 238 Slide 22 Example SJJT, Inc. (A) ? Service Rate Distribution Question What is the mean service rate per hour? Answer Since Joe Ferris can process an order in an average time of 2 minutes (= 2/60 hr. ), then the mean service rate, , is = 1/(mean service time), or 60/2. m = 30/hr. Slide 23 Example SJJT, Inc. (A) ?Service Time Distribution Question What plowshare of the orders will take less than one minute to process? Answer Since the units are expressed in hours, P (T 1 minute) = P (T 1/60 hour). Using the exponential distribution, P (T t ) = 1 e-t. Hence, P (T 1/60) = 1 e-30(1/60) = 1 . 6065 = . 3935 = 39. 35% Slide 24 Example SJJT, Inc. (A) ? Service Time Distribution Question What percentage of the orders will be processed in exactly 3 minutes? Answer Since the exponential distribution is a continuous distribution, the probability a service time exactly equals any(prenominal) specific value is 0 . Slide 25Example SJJT, Inc. (A) ? Service Time Distribution Question What percentage of the orders will require more than 3 minutes to process? Answer The percentage of orders requiring more than 3 minutes to process is P (T 3/60) = e-30(3/60) = e -1. 5 = . 2231 = 22. 31% Slide 26 Example SJJT, Inc. (A) ? Average Time in the System Question What is the average time an order must wait from the time Joe receives the order until it is finished being processed (i. e. its turnaround time)? Answer This is an M/M/1 queue with ? = 20 per hour and m = 30 per hour. The average time an order waits in the system is W = 1/( ? ) 1/(30 20) = 1/10 hour or 6 minutes Slide 27 Example SJJT, Inc. (A) ? Average length of Queue Question What is the average number of orders Joe has waiting to be processed? Answer Average number of orders waiting in the queue is Lq = ? 2/( ? ) = (20)2/(30)(30-20) = 400/ three hundred = 4/3 Slide 28 Example SJJT, Inc. (A) ? Utilization component part Question What percentage of the time is Joe treat orders? Answer The percentage of time Joe is processing orders is equivalent to the utilization factor, ? /m. Thus, the percentage of time he is processing orders is ?/m = 20/30 = 2/3 or 66. 67% Slide 29 Example SJJT, Inc. A) Solution ? 1 2 3 4 5 6 7 8 9 A B C D E F Poisson Arrival Rate Exponential Service Rate Operating Characteristics prospect of no orders in system Average number of orders waiting Average number of orders in system Average time an order waits Average time an order is in system Probabil ity an order must wait G ? m H 20 30 Po Lg L Wq W Pw 0. 333 1. 333 2. 000 0. 067 0. 100 0. 667 Slide 30 M/M/k Queuing System ? ? ? ? ? ? Multiple channels (with one central waiting line) Poisson arrival-rate distribution Exponential service-time distribution Unlimited maximum queue length Infinite calling population Examples Four-teller transaction counter in bank devil-clerk returns counter in retail store Slide 31 1 ? P? n ? m ? P0 , for (n ? k) ? n ? ? n ? ? m ? P0 , for (n ? k) ? ? ? 1 n k ? 1 1 km ? ? ? ? n ? m ? ? k ? m ? km ? ? ? ? ? 1 ? k k n ? k P? 0 P w ? n ? k ? 1 ? n ? 0 ? n 1 ? ? P(n ? k ) ? ?m? ? k ? ? k km P0 , km ? ? k ?m ? ? m ? ? ? ? ? ? L? P0 ? 2 m (k ? 1) (km ? ? ) W? L ? , Lq ? ,r ? km Lq ? 1 ? L? , Wq ? W ? ? m m ? Slide 32 General Operating Characteristics Little s F low Equations L (or W ? ) ? Lq (or Wq ? ) ? L ? ?W L q ? ?Wq W ? Wq ? 1 m Slide 33 Problem ? ? ? ? ? ? ? ?A Tax consulting firm has four service stations (counters) in its exponent to rec eive people who have problems and complaints about their income, wealth and sales taskes. Arrivals average 80 persons in an 8 hour service day. Each tax advisor spends irregular amount of time servicing the arrivals which have been effectuate to have an exponential distribution. The average service time is 20 minutes. Calculate the average no. of customers in the system, average no. of customers waiting to be serviced, average time a customer spend in the system, average waiting time for a customer in queue. Calculate how more hours each week does a tax advisor spend erforming his business concern? What is the probability that a customer has to wait before he gets service? What is the expected no. of idle tax advisors at any undertake time? Slide 34 Example SJJT, Inc. (B) ? M/M/2 Queuing System Smith, Jones, Johnson, and Thomas, Inc. has begun a major advertizing campaign which it believes will increase its business 50%. To handle the increased volume, the company has hired an special floor trader, Fred Hanson, who works at the same speed as Joe Ferris. phone line that the new arrival rate of orders, ? , is 50% higher than that of problem (A). Thus, ? = 1. 5(20) = 30 per hour. Slide 35Example SJJT, Inc. (B) ? Sufficient Service Rate Question Why will Joe Ferris alone not be able to handle the increase in orders? Answer Since Joe Ferris processes orders at a mean rate of = 30 per hour, then ? = = 30 and the utilization factor is 1. This implies the queue of orders will grow incessantly large. Hence, Joe alone cannot handle this increase in demand. Slide 36 Example SJJT, Inc. (B) ? Probability of n Units in System Question What is the probability that neither Joe nor Fred will be working on an order at any point in time? Slide 37 Example SJJT, Inc. (B) ? Probability of n Units in System (continued)Answer Given that ? = 30, = 30, k = 2 and (? /) = 1, the probability that neither Joe nor Fred will be working is 1 P0 ? k ? 1 ( ? / m )n (? / m ) k km ? ( ) ? n k km ? ? n? 0 = 1/(1 + (1/1 )(30/30)1 + (1/2 )(1)22(30)/(2(30)-30) = 1/(1 + 1 + 1) = 1/3 = .333 Slide 38 Example SJJT, Inc. (B) ? Average Time in System Question What is the average turnaround time for an order with both Joe and Fred working? Slide 39 Example SJJT, Inc. (B) ? Average Time in System (continued) Answer The average turnaround time is the average waiting time in the system, W. Lq = ?(? /)k (k-1) (k ? )2 P0 = (30)(30)(30/30)2 (1 ((2)(30)-30))2 (1/3) = 1/3 L = Lq + (? /) = 1/3 + (30/30) = 4/3 W = L/ (4/3)/30 = 4/90 hr. = 2. 67 min. Slide 40 Example SJJT, Inc. (B) ? Average Length of Queue Question What is the average number of orders waiting to be fill with both Joe and Fred working? Answer The average number of orders waiting to be filled is Lq. This was calculated earlier as 1/3 . Slide 41 Example SJJT, Inc. (B) ? Formula Spreadsheet 1 2 3 4 5 6 7 8 9 10 A B C D E F Number of Channels Mean Arrival Rate (Poisson) Mean Service Rate (Exponential ) Operating Charact eristics Probability of no orders in system Average number of orders waitingAverage number of orders in system Average time (hrs) an order waits Average time (hrs) an order is in system Probability an order must wait G k ? m H 2 30 30 Po =Po(H1,H2,H3) Lg L =H6+H2/H3 Wq =H6/H2 W =H8+1/H3 Pw =H2/H3 Slide 42 Example SJJT, Inc. (B) ? Spreadsheet Solution 1 2 3 4 5 6 7 8 9 10 A B C D E F Number of Channels Mean Arrival Rate (Poisson) Mean Service Rate (Exponential ) Operating Characteristics Probability of no orders in system Average number of orders waiting Average number of orders in system Average time (hrs) an order waits Average time (hrs) an order is in system Probability an order must waitG k ? m H 2 30 30 Po Lg L Wq W Pw 0. 333 0. 333 1. 333 0. 011 0. 044 1. 000 Slide 43 Example SJJT, Inc. (C) ? Economic Analysis of Queuing Systems The advertising campaign of Smith, Jones, Johnson and Thomas, Inc. (see problems (A) and (B)) was so successful that business actually doubled. The m ean rate of stock orders arriving at the exchange is now 40 per hour and the company must decide how many floor traders to employ. Each floor trader hired can process an order in an average time of 2 minutes. Slide 44 Example SJJT, Inc. (C) ? Economic Analysis of Queuing Systems ground on a number of factors the brokerage firm as rigid the average waiting personify per minute for an order to be $. 50. theme traders hired will earn $20 per hour in payoff and benefits. Using this information compare the total hourly salute of hiring 2 traders with that of hiring 3 traders. Slide 45 Example SJJT, Inc. (C) ? Economic Analysis of Waiting Lines tally Hourly Cost = (Total salary cost per hour) + (Total hourly cost for orders in the system) = ($20 per trader per hour) x (Number of traders) + ($30 waiting cost per hour) x (Average number of orders in the system) = 20k + 30L. Thus, L must be headstrong for k = 2 traders and for k = 3 traders with ? = 40/hr. nd m = 30/hr. (since the av erage service time is 2 minutes (1/30 hr. ). Slide 46 Example SJJT, Inc. (C) ? Cost of Two Servers P0 ? 1 k ? 1 (? ? n? 0 / m )n ( ? / m ) k km ? ( ) n k km ? ? P0 = 1 / 1+(1/1 )(40/30)+(1/2 )(40/30)2(60/(60-40)) = 1 / 1 + (4/3) + (8/3) = 1/5 Slide 47 Example SJJT, Inc. (C) ? Cost of Two Servers (continued) Thus, Lq = ?(? /)k (k-1) (k -? )2 P0 = (40)(30)(40/30)2 1 (60-40)2 (1/5) = 16/15 L = Lq + (? /) = 16/15 + 4/3 = 12/5 Total Cost = (20)(2) + 30(12/5) = $112. 00 per hour Slide 48 Example SJJT, Inc. (C) ? Cost of Three Servers P0 ? 1 k ? 1 (? ? n? 0 / m )n ( ? / m ) k km ( ) n k km ? ? P0 = 1/1+(1/1 )(40/30)+(1/2 )(40/30)2+ (1/3 )(40/30)3(90/(90-40)) = 1 / 1 + 4/3 + 8/9 + 32/45 = 15/59 Slide 49 Example SJJT, Inc. (C) ? Cost of Three Servers (continued) (30)(40)(40/30)3 Hence, Lq = (15/59) = 128/885 = . 1446 (2 )(3(30)-40)2 Thus, L = 128/885 + 40/30 = 1308/885 (= 1. 4780) Total Cost = (20)(3) + 30(1308/885) = $104. 35 per hour Slide 50 Example SJJT, Inc. (C) ? System Cost Compariso n 2 Traders 3 Traders engage Cost/Hr $40. 00 60. 00 Waiting Cost/Hr $82. 00 44. 35 Total Cost/Hr $112. 00 104. 35 Thus, the cost of having 3 traders is less than that of 2 traders. Slide 51